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A $2.00 \mathrm{~kg}$ particle moves along an $x$ axis in one-dimensional motion while a conservative force along that axis acts on it. The potential energy $U(x)$ associated with the force is plotted in Fig. 8-10a. That is, if the particle were placed at any position between $x=0$ and $x=7.00 \mathrm{~m}$, it would have the plotted value of $U$. At $x=6.5 \mathrm{~m}$, the particle has velocity $\vec{v}_0=(-4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}$ From Figure, determine the particle's speed at $x_1=4.5 \mathrm{~m}$.

If a force is conservative, it is possible to assign a numerical value for the potential at any point and conversely, when an object moves from one location to another, the force changes the potential energy of the object by an amount that does not depend on the path taken, contributing to the mechanical energy and the overall conservation of energy. The work done by a conservative force is equal to the negative of change in potential energy during that process. Since the position, velocity, and acceleration are collinear (parallel, and lie on the same line) – only the magnitudes of these vectors are necessary, and because the motion is along a straight line, the problem effectively reduces from three dimensions to one. \begin{align} v & = at+v_0 & [1]\\\ r & = r_0 + v_0 t + \tfrac12 {a}t^2 & [2]\\\ r & = r_0 + \tfrac12 \left( v+v_0 \right )t & [3]\\\ v^2 & = v_0^2 + 2a\left( r - r_0 \right) & [4]\\\ r & = r_0 + vt - \tfrac12 {a}t^2 & [5]\\\ \end{align} where: * is the particle's initial position * is the particle's final position * is the particle's initial velocity * is the particle's final velocity * is the particle's acceleration * is the time interval Equations [1] and [2] are from integrating the definitions of velocity and acceleration, subject to the initial conditions and ; \begin{align} \mathbf{v} & = \int \mathbf{a} dt = \mathbf{a}t+\mathbf{v}_0 \,, & [1] \\\ \mathbf{r} & = \int (\mathbf{a}t+\mathbf{v}_0) dt = \frac{\mathbf{a}t^2}{2}+\mathbf{v}_0t +\mathbf{r}_0 \,, & [2] \\\ \end{align} in magnitudes, \begin{align} v & = at+v_0 \,, & [1] \\\ r & = \frac{{a}t^2}{2}+v_0t +r_0 \,. A conservative force depends only on the position of the object. Graph of the Lennard-Jones potential function: Intermolecular potential energy as a function of the distance of a pair of particles. (Velocity is on the y-axis and time on the x-axis. If the force is not conservative, then defining a scalar potential is not possible, because taking different paths would lead to conflicting potential differences between the start and end points. If the velocity or positions change non-linearly over time, such as in the example shown in the figure, then differentiation provides the correct solution. thumb|360px|v vs t graph for a moving particle under a non-uniform acceleration a. In physics, a conservative force is a force with the property that the total work done in moving a particle between two points is independent of the path taken.HyperPhysics - Conservative force Equivalently, if a particle travels in a closed loop, the total work done (the sum of the force acting along the path multiplied by the displacement) by a conservative force is zero. The position of the particle is \mathbf{r} =\mathbf{r}\left ( r(t),\theta(t) \right ) = r \mathbf{\hat{e}}_r where and are the polar unit vectors. For example, if a child slides down a frictionless slide, the work done by the gravitational force on the child from the start of the slide to the end is independent of the shape of the slide; it only depends on the vertical displacement of the child. ==Mathematical description== A force field F, defined everywhere in space (or within a simply-connected volume of space), is called a conservative force or conservative vector field if it meets any of these three equivalent conditions: # The curl of F is the zero vector: \vec{ abla} \times \vec{F} = \vec{0}. where in two dimensions this reduces to: \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} = 0 # There is zero net work (W) done by the force when moving a particle through a trajectory that starts and ends in the same place: W \equiv \oint_C \vec{F} \cdot \mathrm{d}\vec r = 0. The Lennard-Jones model describes the potential intermolecular energy V between two particles based on the outlined principles. The mean potential energy per particle is negative. Accordingly, some authors classify the magnetic force as conservative,For example, : "In general, a force which depends explicitly upon the velocity of the particle is not conservative. Specific potential energy is potential energy of an object per unit of mass of that object. Suppose a particle starts at point A, and there is a force F acting on it. In mechanics, the derivative of the position vs. time graph of an object is equal to the velocity of the object. The variation of energy for the particle, taking path 1 from A to B and then path 2 backwards from B to A, is 0; thus, the work is the same in path 1 and 2, i.e., the work is independent of the path followed, as long as it goes from A to B. Using equation [4] in the set above, we have: s= \frac{v^2 - u^2}{-2g}. Therefore, the slope of the curve gives the change in position divided by the change in time, which is the definition of the average velocity for that interval of time on the graph. Since the velocity of the object is the derivative of the position graph, the area under the line in the velocity vs. time graph is the displacement of the object.

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A playful astronaut releases a bowling ball, of mass $m=$ $7.20 \mathrm{~kg}$, into circular orbit about Earth at an altitude $h$ of $350 \mathrm{~km}$. What is the mechanical energy $E$ of the ball in its orbit?

Hence, mechanical energy E_\text{mechanical} of the satellite-Earth system is given by E_\text{mechanical} = U + K E_\text{mechanical} = - G \frac{M m}{r}\ + \frac{1}{2}\, m v^2 If the satellite is in circular orbit, the energy conservation equation can be further simplified into E_\text{mechanical} = - G \frac{M m}{2r} since in circular motion, Newton's 2nd Law of motion can be taken to be G \frac{M m}{r^2}\ = \frac{m v^2}{r} ==Conversion== Today, many technological devices convert mechanical energy into other forms of energy or vice versa. Thus, the total energy of the system remains unchanged though the mechanical energy of the system has reduced. ===Satellite=== thumb|plot of kinetic energy K, gravitational potential energy, U and mechanical energy E_\text{mechanical} versus distance away from centre of earth, r at R= Re, R= 2*Re, R=3*Re and lastly R = geostationary radius A satellite of mass m at a distance r from the centre of Earth possesses both kinetic energy, K, (by virtue of its motion) and gravitational potential energy, U, (by virtue of its position within the Earth's gravitational field; Earth's mass is M). Specific mechanical energy is the mechanical energy of an object per unit of mass. The relations are used. p= \frac{h^2}{\mu} = a(1-{e^2}) = r_{p}(1+e) where * p\,\\! is the conic section semi-latus rectum. * r_p\,\\! is distance at periastron of the body from the center of mass. v = \sqrt{\mu\left({2\over{r}} - {1\over{a}}\right)} where *\mu\, is the standard gravitational parameter, G(m1+m2), often expressed as GM when one body is much larger than the other. *r\, is the distance between the orbiting body and center of mass. *a\,\\! is the length of the semi-major axis. === Orbital Mechanics === When calculating the specific mechanical energy of a satellite in orbit around a celestial body, the mass of the satellite is assumed to be negligible: \mu = G(M + m) \approx GM where M is the mass of the celestial body. The elastic potential energy equation is used in calculations of positions of mechanical equilibrium. If the satellite's orbit is an ellipse the potential energy of the satellite, and its kinetic energy, both vary with time but their sum remains constant. Gravitational energy or gravitational potential energy is the potential energy a massive object has in relation to another massive object due to gravity. It is defined as: \epsilon= \epsilonk+\epsilonp where * \epsilonk is the specific kinetic energy * \epsilonp it the specific potential energy == Astrodynamics == In the gravitational two-body problem, the specific mechanical energy of one body \epsilon is given as: \begin{align} \epsilon &= \frac{v^2}{2} - \frac{\mu}{r} = -\frac{1}{2} \frac{\mu^2}{h^2} \left(1 - e^2\right) = -\frac{\mu}{2a} \end{align} where * v\,\\! is the orbital speed of the body; relative to center of mass. * r\,\\! is the orbital distance between the body and center of mass; * \mu = {G}(m_1 + m_2)\,\\! is the standard gravitational parameter of the bodies; * h\,\\! is the specific relative angular momentum of the same body referenced to the center of mass. The gravitational potential energy of an object is equal to the weight W of the object multiplied by the height h of the object's center of gravity relative to an arbitrary datum: U = W h The potential energy of an object can be defined as the object's ability to do work and is increased as the object is moved in the opposite direction of the direction of the force. Specific potential energy is potential energy of an object per unit of mass of that object. Elastic energy is the mechanical potential energy stored in the configuration of a material or physical system as it is subjected to elastic deformation by work performed upon it. The pendulum reaches greatest kinetic energy and least potential energy when in the vertical position, because it will have the greatest speed and be nearest the Earth at this point. Though energy cannot be created or destroyed, it can be converted to another form of energy. ===Swinging pendulum=== 200px|thumb|A swinging pendulum with the velocity vector (green) and acceleration vector (blue). The change in potential energy moving from the surface (a distance R from the center) to a height h above the surface is \begin{align} \Delta U &= \frac{GMm}{R}-\frac{GMm}{R+h} \\\ &= \frac{GMm}{R}\left(1-\frac{1}{1+h/R}\right). \end{align} If h/R is small, as it must be close to the surface where g is constant, then this expression can be simplified using the binomial approximation \frac{1}{1+h/r} \approx 1-\frac{h}{R} to \begin{align} \Delta U &\approx \frac{GMm}{R}\left[1-\left(1-\frac{h}{R}\right)\right] \\\ \Delta U &\approx \frac{GMmh}{R^2}\\\ \Delta U &\approx m\left(\frac{GM}{R^2}\right)h. \end{align} As the gravitational field is g = GM / R^2, this reduces to \Delta U \approx mgh. The gravitational potential energy is the potential energy an object has because it is within a gravitational field. thumb|250px|An example of a mechanical system: A satellite is orbiting the Earth influenced only by the conservative gravitational force; its mechanical energy is therefore conserved. In a mechanical system like a swinging pendulum subjected to the conservative gravitational force where frictional forces like air drag and friction at the pivot are negligible, energy passes back and forth between kinetic and potential energy but never leaves the system. The force between a point mass, M, and another point mass, m, is given by Newton's law of gravitation: Extract of page 10 F = \frac {GMm}{r^2} To get the total work done by an external force to bring point mass m from infinity to the final distance R (for example the radius of Earth) of the two mass points, the force is integrated with respect to displacement: W = \int_\infty^R \frac {GMm}{r^2}dr = -\left . \frac{G M m}{r} \right|_{\infty}^{R} Because \lim_{r\to \infty} \frac{1}{r} = 0, the total work done on the object can be written as: Extract of page 143 In the common situation where a much smaller mass m is moving near the surface of a much larger object with mass M, the gravitational field is nearly constant and so the expression for gravitational energy can be considerably simplified. The energy is potential as it will be converted into other forms of energy, such as kinetic energy and sound energy, when the object is allowed to return to its original shape (reformation) by its elasticity. In an elastic collision, mechanical energy is conserved – the sum of the mechanical energies of the colliding objects is the same before and after the collision. On the other hand, it will have its least kinetic energy and greatest potential energy at the extreme positions of its swing, because it has zero speed and is farthest from Earth at these points. In physical sciences, mechanical energy is the sum of potential energy and kinetic energy.

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Let the disk in Figure start from rest at time $t=0$ and also let the tension in the massless cord be $6.0 \mathrm{~N}$ and the angular acceleration of the disk be $-24 \mathrm{rad} / \mathrm{s}^2$. What is its rotational kinetic energy $K$ at $t=2.5 \mathrm{~s}$ ?

The rotational energy depends on the moment of inertia for the system, I . Rotational energy or angular kinetic energy is kinetic energy due to the rotation of an object and is part of its total kinetic energy. Knowledge of the Euler angles as function of time t and the initial coordinates \mathbf{r}(0) determine the kinematics of the rigid rotor. === Classical kinetic energy === The following text forms a generalization of the well- known special case of the rotational energy of an object that rotates around one axis. The rotor is modeled as an infinitely thin disc, inducing a constant velocity along the axis of rotation. The classical kinetic energy T of the rigid rotor can be expressed in different ways: * as a function of angular velocity * in Lagrangian form * as a function of angular momentum * in Hamiltonian form. The kinetic energy T of the linear rigid rotor is given by 2T = \mu R^2 \left[\dot{\theta}^2 + (\dot\varphi\,\sin\theta)^2\right] = \mu R^2 \begin{pmatrix}\dot{\theta} & \dot{\varphi}\end{pmatrix} \begin{pmatrix} 1 & 0 \\\ 0 & \sin^2\theta \\\ \end{pmatrix} \begin{pmatrix}\dot{\theta} \\\ \dot{\varphi}\end{pmatrix} = \mu \begin{pmatrix}\dot{\theta} & \dot{\varphi}\end{pmatrix} \begin{pmatrix} h_\theta^2 & 0 \\\ 0 & h_\varphi^2 \\\ \end{pmatrix} \begin{pmatrix}\dot{\theta} \\\ \dot{\varphi}\end{pmatrix}, where h_\theta = R\, and h_\varphi= R\sin\theta\, are scale (or Lamé) factors. Note the close relationship between the result for rotational energy and the energy held by linear (or translational) motion: :E_\mathrm{translational} = \tfrac{1}{2} m v^2 In the rotating system, the moment of inertia, I, takes the role of the mass, m, and the angular velocity, \omega , takes the role of the linear velocity, v. As the Earth has a sidereal rotation period of 23.93 hours, it has an angular velocity of .Launching From Florida: Life in the Fast Lane!, NASA The Earth has a moment of inertia, I = .Moment of inertia--Earth, Wolfram Therefore, it has a rotational kinetic energy of . Looking at rotational energy separately around an object's axis of rotation, the following dependence on the object's moment of inertia is observed:Resnick, R. and Halliday, D. (1966) PHYSICS, Equation 12-11 :E_\mathrm{rotational} = \tfrac{1}{2} I \omega^2 where : \omega \ is the angular velocity : I \ is the moment of inertia around the axis of rotation : E \ is the kinetic energy The mechanical work required for or applied during rotation is the torque times the rotation angle. – Page 37 of 45 (graphic) For a miniature disc with a diameter of 8 cm (radius of 4 cm), the speed ratio of outer to inner data edge is 1.6. The rotor is rigid if R is independent of time. For a stationary open rotor with no outer duct, such as a helicopter in hover, the power required to produce a given thrust is: :P = \sqrt{\frac{T^3}{2 \rho A}} where: * T is the thrust * \rho is the density of air (or other medium) * A is the area of the rotor disc * P is power A device which converts the translational energy of the fluid into rotational energy of the axis or vice versa is called a Rankine disk actuator. Note that a different rotation matrix would result from a different choice of Euler angle convention used. ==== Lagrange form ==== Backsubstitution of the expression of \boldsymbol{\omega} into T gives the kinetic energy in Lagrange form (as a function of the time derivatives of the Euler angles). The rotational energy of a rolling cylinder varies from one half of the translational energy (if it is massive) to the same as the translational energy (if it is hollow). A linear (data reading and writing) speeds of 2.4 times higher can be reached at the outer disc edge with the same angular (rotation) speed. Seiffert's spherical spiral is a curve on a sphere made by moving on the sphere with constant speed and angular velocity with respect to a fixed diameter. This is because, although the model does account for bond stretching due to rotation, it ignores any bond stretching due to vibrational energy in the bond (anharmonicity in the potential). == Arbitrarily shaped rigid rotor == An arbitrarily shaped rigid rotor is a rigid body of arbitrary shape with its center of mass fixed (or in uniform rectilinear motion) in field-free space R3, so that its energy consists only of rotational kinetic energy (and possibly constant translational energy that can be ignored). In matrix-vector notation, 2 T = \begin{pmatrix} \dot{\alpha} & \dot{\beta} & \dot{\gamma} \end{pmatrix} \; \mathbf{g} \; \begin{pmatrix} \dot{\alpha} \\\ \dot{\beta} \\\ \dot{\gamma}\\\ \end{pmatrix}, where \mathbf{g} is the metric tensor expressed in Euler angles--a non-orthogonal system of curvilinear coordinates-- \mathbf{g}= \begin{pmatrix} I_1 \sin^2\beta \cos^2\gamma+I_2\sin^2\beta\sin^2\gamma+I_3\cos^2\beta & (I_2-I_1) \sin\beta\sin\gamma\cos\gamma & I_3\cos\beta \\\ (I_2-I_1) \sin\beta\sin\gamma\cos\gamma & I_1\sin^2\gamma+I_2\cos^2\gamma & 0 \\\ I_3\cos\beta & 0 & I_3 \\\ \end{pmatrix}. ==== Angular momentum form ==== Often the kinetic energy is written as a function of the angular momentum \mathbf{L} of the rigid rotor. (Note: The corresponding eigenvalue equation gives the Schrödinger equation for the rigid rotor in the form that it was solved for the first time by Kronig and Rabi (for the special case of the symmetric rotor). thumb|An actuator disk accelerating a fluid flow from right to left In fluid dynamics, momentum theory or disk actuator theory is a theory describing a mathematical model of an ideal actuator disk, such as a propeller or helicopter rotor, by W.J.M. Rankine (1865), Alfred George Greenhill (1888) and (1889). This disc creates a flow around the rotor. The instantaneous power of an angularly accelerating body is the torque times the angular velocity.

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A food shipper pushes a wood crate of cabbage heads (total mass $m=14 \mathrm{~kg}$ ) across a concrete floor with a constant horizontal force $\vec{F}$ of magnitude $40 \mathrm{~N}$. In a straight-line displacement of magnitude $d=0.50 \mathrm{~m}$, the speed of the crate decreases from $v_0=0.60 \mathrm{~m} / \mathrm{s}$ to $v=0.20 \mathrm{~m} / \mathrm{s}$. What is the increase $\Delta E_{\text {th }}$ in the thermal energy of the crate and floor?

When heat flows into (respectively, out of) a material, its temperature increases (respectively, decreases), in proportion to the amount of heat divided by the amount (mass) of material, with a proportionality factor called the specific heat capacity of the material. The equation is much simpler and can help to understand better the physics of the materials without focusing on the dynamic of the heat transport process. The Ergun equation, derived by the Turkish chemical engineer Sabri Ergun in 1952, expresses the friction factor in a packed column as a function of the modified Reynolds number. ==Equation== f_p = \frac {150}{Gr_p}+1.75 where f_p and Gr_p are defined as f_p = \frac{\Delta p}{L} \frac{D_p}{\rho v_s^2} \left(\frac{\epsilon^3}{1-\epsilon}\right) and Gr_p = \frac{\rho v_s D_p}{(1-\epsilon)\mu} = \frac{Re}{(1-\epsilon)}; where: Gr_p is the modified Reynolds number, f_p is the packed bed friction factor \Delta p is the pressure drop across the bed, L is the length of the bed (not the column), D_p is the equivalent spherical diameter of the packing, \rho is the density of fluid, \mu is the dynamic viscosity of the fluid, v_s is the superficial velocity (i.e. the velocity that the fluid would have through the empty tube at the same volumetric flow rate) \epsilon is the void fraction (porosity) of the bed. In either case, one uses the heat equation :c_t = D \Delta c, or :P_t = D \Delta P. By the combination of these observations, the heat equation says the rate \dot u at which the material at a point will heat up (or cool down) is proportional to how much hotter (or cooler) the surrounding material is. The equation becomes :q = -k \,\frac{\partial u}{\partial x} Let Q=Q(x,t) be the internal heat energy per unit volume of the bar at each point and time. Specific mechanical energy is the mechanical energy of an object per unit of mass. The heat loss due to linear thermal bridging (H_{TB}) is a physical quantity used when calculating the energy performance of buildings. This is an efficient way of increasing the rate, since the alternative way of doing so is by increasing either the heat transfer coefficient (which depends on the nature of materials being used and the conditions of use) or the temperature gradient (which depends on the conditions of use). In the absence of heat energy generation, from external or internal sources, the rate of change in internal heat energy per unit volume in the material, \partial Q/\partial t, is proportional to the rate of change of its temperature, \partial u/\partial t. One of the experimentally obtained equations for heat transfer coefficient for the fin surface for low wind velocities is: k=2.11 v^{0.71} \theta^{0.44} a^{-0.14} where k= Fin surface heat transfer coefficient [W/m2K ] a=fin length [mm] v=wind velocity [km/hr] θ=fin pitch [mm] Another equation for high fluid velocities, obtained from experiments conducted by Gibson, is k= 241.7[0.0247-0.00148(a^{0.8}/\theta^{0.4})] v^{0.73} where k=Fin surface heat transfer coefficient[W/m2K ] a=Fin length[mm] θ=Fin pitch[mm] v=Wind velocity[km/hr] A more accurate equation for fin surface heat transfer coefficient is: k_{avg} = (2.47-2.55/\theta^{0.4}) v^{0.9} 0.0872 \theta + 4.31 where k (avg)= Fin surface heat transfer coefficient[W/m2K ] θ=Fin pitch[mm] v=Wind velocity[km/hr] All these equations can be used to evaluate average heat transfer coefficient for various fin designs. == Design == The momentum conservation equation for this case is given as follows: {\partial(\rho v)\over\partial t} + v abla . (\rho v) = - abla P + abla . \tau + F + \rho g This is used in combination with the continuity equation. thumb|upright=1.8|Animated plot of the evolution of the temperature in a square metal plate as predicted by the heat equation. Applying the law of conservation of energy to a small element of the medium centered at x, one concludes that the rate at which heat accumulates at a given point x is equal to the derivative of the heat flow at that point, negated. The energy equation is also needed, which is: {\partial (\rho E)\over\partial t} + abla.[v(\rho E + p)] = abla.[k_{eff} abla T- \Sigma_j h_j J_j +(\tau.v)]. Since heat density is proportional to temperature in a homogeneous medium, the heat equation is still obeyed in the new units. The heat equation is a consequence of Fourier's law of conduction (see heat conduction). Thus instead of being for thermodynamic temperature (Kelvin - K), units of u should be J/L. It appears in both United Kingdom and Irish methodologies. ==Calculation== The calculation of the heat loss due to linear thermal bridging is relatively simple, given by the formula below: :H_{TB} = y \sum A_{exp} In the formula, y = 0.08 if Accredited Construction details used, and y = 0.15 otherwise, and \sum A_{exp} is the sum of all the exposed areas of the building envelope, ==References== Category:Energy economics Category:Thermodynamic properties This results in velocity profiles and temperature profiles for various surfaces and this knowledge can be used to design the fin. == References == Category:Unit operations Category:Transport phenomena Category:Heat transfer Thus the rate of heat flow into V is also given by the surface integral q_t(V)= - \int_{\partial V} \mathbf{H}(x) \cdot \mathbf{n}(x) \, dS where n(x) is the outward pointing normal vector at x. When solved as a scalar equation, it can be used to calculate the temperatures at the fin and cylinder surfaces, by reducing to: abla^2 T + {\overset{.}{q}\over k} = {1 \over \alpha} {\partial T\over\partial t} Where: q = internal heat generation = 0 (in this case). * The coefficient κ(x) is the inverse of specific heat of the substance at x × density of the substance at x: \kappa = 1/(\rho c_p).

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While you are operating a Rotor (a large, vertical, rotating cylinder found in amusement parks), you spot a passenger in acute distress and decrease the angular velocity of the cylinder from $3.40 \mathrm{rad} / \mathrm{s}$ to $2.00 \mathrm{rad} / \mathrm{s}$ in $20.0 \mathrm{rev}$, at constant angular acceleration. (The passenger is obviously more of a "translation person" than a "rotation person.") What is the constant angular acceleration during this decrease in angular speed?

In physics, angular acceleration refers to the time rate of change of angular velocity. Therefore, the instantaneous angular acceleration α of the particle is given by : \alpha = \frac{d}{dt} \left(\frac{v_{\perp}}{r}\right). The angular velocity satisfies equations of motion known as Euler's equations (with zero applied torque, since by assumption the rotor is in field-free space). The angular velocity ω is the rate of change of angular position with respect to time, which can be computed from the cross-radial velocity as: : \omega = \frac{d\phi}{dt} = \frac{v_\perp}{r}. Spin angular acceleration refers to the angular acceleration of a rigid body about its centre of rotation, and orbital angular acceleration refers to the angular acceleration of a point particle about a fixed origin. The vector \boldsymbol{\omega} = (\omega_x, \omega_y, \omega_z) on the left hand side contains the components of the angular velocity of the rotor expressed with respect to the body-fixed frame. Angular acceleration is measured in units of angle per unit time squared (which in SI units is radians per second squared), and is usually represented by the symbol alpha (α). Thus the formula for Advance ratio is \mu = \frac {V}{u} = \frac {V}{\Omega\cdot R} where Omega (Ω) is the rotor's angular velocity, and R is the rotor radius (about the length of one rotor blade)Jackson, Dave. The SI unit of angular velocity is radians per second, Extract of page 27 with the radian being a dimensionless quantity, thus the SI units of angular velocity may be listed as s−1. The radian per second (symbol: rad⋅s−1 or rad/s) is the unit of angular velocity in the International System of Units (SI). The same equations for the angular speed can be obtained reasoning over a rotating rigid body. Therefore, the orbital angular acceleration is the vector \boldsymbol\alpha defined by : \boldsymbol\alpha = \frac{d}{dt} \left(\frac{\mathbf r \times \mathbf v}{r^2}\right). Angular acceleration then may be termed a pseudoscalar, a numerical quantity which changes sign under a parity inversion, such as inverting one axis or switching the two axes. === Particle in three dimensions === In three dimensions, the orbital angular acceleration is the rate at which three-dimensional orbital angular velocity vector changes with time. : Quantity correspondence Angular frequency \omega Frequency u = \omega/{2\pi} 2π rad/s 1 Hz 1 rad/s ≈ 0.159155 Hz 1 rad/s ≈ 9.5493 rpm 0.1047 rad/s ≈ 1 rpm == Coherent units == A use of the unit radian per second is in calculation of the power transmitted by a shaft. As there are two types of angular velocity, namely spin angular velocity and orbital angular velocity, there are naturally also two types of angular acceleration, called spin angular acceleration and orbital angular acceleration respectively. The radian per second is defined as the angular frequency that results in the angular displacement increasing by one radian every second. * Spin angular velocity refers to how fast a rigid body rotates with respect to its center of rotation and is independent of the choice of origin, in contrast to orbital angular velocity. 300px|right|thumb|Rotation of a rigid body P about a fixed axis O. Angular displacement of a body is the angle (in radians, degrees or turns) through which a point revolves around a centre or a specified axis in a specified sense. The instantaneous angular velocity ω at any point in time is given by : \omega = \frac{v_{\perp}}{r}, where r is the distance from the origin and v_{\perp} is the cross-radial component of the instantaneous velocity (i.e. the component perpendicular to the position vector), which by convention is positive for counter-clockwise motion and negative for clockwise motion. Angular velocity is usually represented by the symbol omega (, sometimes ). In two dimensions, angular acceleration is a number with plus or minus sign indicating orientation, but not pointing in a direction. For rigid bodies, angular acceleration must be caused by a net external torque.

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A living room has floor dimensions of $3.5 \mathrm{~m}$ and $4.2 \mathrm{~m}$ and a height of $2.4 \mathrm{~m}$. What does the air in the room weigh when the air pressure is $1.0 \mathrm{~atm}$ ?

This can be seen by using the ideal gas law as an approximation. ==Dry air== The density of dry air can be calculated using the ideal gas law, expressed as a function of temperature and pressure: \begin{align} \rho &= \frac{p}{R_\text{specific} T}\\\ R_\text{specific} &= \frac{R}{M} = \frac{k_{\rm B}}{m}\\\ \rho &= \frac{pM}{RT} = \frac{pm}{k_{\rm B}T}\\\ \end{align} where: *\rho, air density (kg/m3)In the SI unit system. Air returns from the room at ceiling level or the maximum allowable height above the occupied zone. Depending on the measuring instruments used, different sets of equations for the calculation of the density of air can be applied. The partial pressure of dry air p_\text{d} is found considering partial pressure, resulting in: p_\text{d} = p - p_\text{v} where p simply denotes the observed absolute pressure. ==Variation with altitude== thumb|upright=2.0|Standard atmosphere: , , ===Troposphere=== To calculate the density of air as a function of altitude, one requires additional parameters. At 101.325kPa (abs) and , air has a density of approximately , which is about that of water, according to the International Standard Atmosphere (ISA). * At 70°F and 14.696psi, dry air has a density of 0.074887lb/ft3. Air density is a property used in many branches of science, engineering, and industry, including aeronautics;Olson, Wayne M. (2000) AFFTC-TIH-99-01, Aircraft Performance FlightICAO, Manual of the ICAO Standard Atmosphere (extended to 80 kilometres (262 500 feet)), Doc 7488-CD, Third Edition, 1993, .Grigorie, T.L., Dinca, L., Corcau J-I. and Grigorie, O. (2010) Aircrafts' Altitude Measurement Using Pressure Information:Barometric Altitude and Density Altitude gravimetric analysis;A., Picard, R.S., Davis, M., Gläser and K., Fujii (CIPM-2007) Revised formula for the density of moist air the air-conditioningS. * At 20°C and 101.325kPa, dry air has a density of 1.2041 kg/m3. The following table illustrates the air density–temperature relationship at 1 atm or 101.325 kPa: ==Humid air== thumb|right|400px|Effect of temperature and relative humidity on air density The addition of water vapor to air (making the air humid) reduces the density of the air, which may at first appear counter-intuitive. thumb|350px|alt=Diagram of underfloor air distribution showing cool, fresh air moving through the underfloor plenum and supplied via floor diffusers and desktop vents. Therefore: * At IUPAC standard temperature and pressure (0°C and 100kPa), dry air has a density of approximately 1.2754kg/m3. *R_\text{specific}, the specific gas constant for dry air, which using the values presented above would be approximately in J⋅kg−1⋅K−1. At 101.325 kPa (abs) and 20 °C (68 °F), air has a density of approximately , according to the International Standard Atmosphere (ISA). thumb|upright=1.75|An illustration of Dalton's law using the gases of air at sea level. Both the pressure and density obey this law, so, denoting the height of the border between the troposphere and the tropopause as U: \begin{align} p &= p(U) e^{-\frac{h - U}{H_\text{TP}}} = p_0 \left(1 - \frac{L U}{T_0}\right)^\frac{g M}{R L} e^{-\frac{h - U}{H_\text{TP}}} \\\ \rho &= \rho(U) e^{-\frac{h - U}{H_\text{TP}}} = \rho_0 \left(1 - \frac{L U}{T_0}\right)^{\frac{g M}{R L} - 1} e^{-\frac{h - U}{H_\text{TP}}} \end{align} ==Composition== ==See also== *Air *Atmospheric drag *Lighter than air *Density *Atmosphere of Earth *International Standard Atmosphere *U.S. Standard Atmosphere *NRLMSISE-00 ==Notes== ==References== ==External links== *Conversions of density units ρ by Sengpielaudio *Air density and density altitude calculations and by Richard Shelquist *Air density calculations by Sengpielaudio (section under Speed of sound in humid air) *Air density calculator by Engineering design encyclopedia *Atmospheric pressure calculator by wolfdynamics *Air iTools - Air density calculator for mobile by JSyA *Revised formula for the density of moist air (CIPM-2007) by NIST Category:Atmospheric thermodynamics A The density of air or atmospheric density, denoted ρ, is the mass per unit volume of Earth's atmosphere. Also, the investigation of energy saving has shown that this amount varies for buildings located in different climates, suggesting further studies should investigate this factor prior to designing a suitable HVAC system. ==Applications== Underfloor air distribution is frequently used in office buildings, particularly highly-reconfigurable and open plan offices where raised floors are desirable for cable management. Room air distribution is characterizing how air is introduced to, flows through, and is removed from spaces.Fundamentals volume of the ASHRAE Handbook, Atlanta, GA, USA, 2005 HVAC airflow in spaces generally can be classified by two different types: mixing (or dilution) and displacement. ==Mixing systems== Mixing systems generally supply air such that the supply air mixes with the room air so that the mixed air is at the room design temperature and humidity. Displacement room airflow presents an opportunity to improve both the thermal comfort and indoor air quality (IAQ) of the occupied space. In architecture, construction, and real estate, floor area, floor space, or floorspace is the area (measured as square feet or square metres) taken up by a building or part of it. At a certain plane in the room, the airflow rate returned to the UZ is equal to the supply air. thumb|upright=1.25|Different air masses which affect North America as well as other continents, tend to be separated by frontal boundaries In meteorology, an air mass is a volume of air defined by its temperature and humidity.

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An astronaut whose height $h$ is $1.70 \mathrm{~m}$ floats "feet down" in an orbiting space shuttle at distance $r=6.77 \times 10^6 \mathrm{~m}$ away from the center of Earth. What is the difference between the gravitational acceleration at her feet and at her head?

It is calculated as the distance between the centre of gravity of a ship and its metacentre. The acceleration of a body near the surface of the Earth is due to the combined effects of gravity and centrifugal acceleration from the rotation of the Earth (but the latter is small enough to be negligible for most purposes); the total (the apparent gravity) is about 0.5% greater at the poles than at the Equator."Curious About Astronomy?", Cornell University, retrieved June 2007 Although the symbol is sometimes used for standard gravity, (without a suffix) can also mean the local acceleration due to local gravity and centrifugal acceleration, which varies depending on one's position on Earth (see Earth's gravity). Acceleration due to gravity, acceleration of gravity or gravity acceleration may refer to: *Gravitational acceleration, the acceleration caused by the gravitational attraction of massive bodies in general *Gravity of Earth, the acceleration caused by the combination of gravitational attraction and centrifugal force of the Earth *Standard gravity, or g, the standard value of gravitational acceleration at sea level on Earth ==See also== *g-force, the acceleration of a body relative to free-fall The standard acceleration due to gravity (or standard acceleration of free fall), sometimes abbreviated as standard gravity, usually denoted by or , is the nominal gravitational acceleration of an object in a vacuum near the surface of the Earth. The metacentric height (GM) is a measurement of the initial static stability of a floating body. As the ship heels over, the centre of gravity generally remains fixed with respect to the ship because it just depends on the position of the ship's weight and cargo, but the surface area increases, increasing BMφ. Although the actual acceleration of free fall on Earth varies according to location, the above standard figure is always used for metrological purposes. Hence, a sufficiently, but not excessively, high metacentric height is considered ideal for passenger ships. ==Metacentre== When a ship heels (rolls sideways), the centre of buoyancy of the ship moves laterally. Geopotential height or geopotential altitude is a vertical coordinate referenced to Earth's mean sea level (assumed zero potential) that represents the work done by lifting one unit mass one unit distance through a region in which the acceleration of gravity is uniformly 9.80665 m/s2. thumb|upright=1.8|A simplified spacecraft system. thumb|upright=1.6|Ship stability diagram showing centre of gravity (G), centre of buoyancy (B), and metacentre (M) with ship upright and heeled over to one side. * Cardiovascular events and changes occurring during spaceflight: these are due to body fluids shift and redistribution, heart rhythm disturbances and decrease in maximal exercise capacity in the micro gravity environment. Geopotential height may be obtained from normalizing geopotential by the acceleration of gravity: :{H} = \frac{\Phi}{g_{0}}\ = \frac{1}{g_{0}}\int_0^Z\ g(\phi,Z)\,dZ where g_0 = 9.80665 m/s2, the standard gravity at mean sea level. This position meant that a person's legs experienced only one sixth of their weight, which was the equivalent of being on the lunar surface. The value of defined above is a nominal midrange value on Earth, originally based on the acceleration of a body in free fall at sea level at a geodetic latitude of 45°. thumb|250px|A test subject being suited up for studies on the Reduced Gravity Walking Simulator. This illusion will alter the astronaut's perception of the orienting force of gravity and then lose spatial direction. "G", is the center of gravity. While objects are weightless in space, an astronaut has to be familiar with an object's forces of inertia and understand how the object will respond to simple motions to avoid losing it in space. Geopotential height (altitude) differs from geometric (tapeline) height but remains a historical convention in aeronautics as the altitude used for calibration of aircraft barometric altimeters. ==Definition== Geopotential is the gravitational potential energy per unit mass at elevation Z: :\Phi(Z) = \int_0^Z\ g(\phi,Z)\,dZ where g(\phi,Z) is the acceleration due to gravity, \phi is latitude, and Z is the geometric elevation. Because of this any future medical criteria for commercial spaceflight participants needs to focus specifically on the detrimental effects of rapidly changing gravitational levels, and which individuals will be capable of tolerating this. The centre of gravity of the ship is commonly denoted as point G or CG.

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If the particles in a system all move together, the com moves with them-no trouble there. But what happens when they move in different directions with different accelerations? Here is an example. The three particles in Figure are initially at rest. Each experiences an external force due to bodies outside the three-particle system. The directions are indicated, and the magnitudes are $F_1=6.0 \mathrm{~N}, F_2=12 \mathrm{~N}$, and $F_3=14 \mathrm{~N}$. What is the acceleration of the center of mass of the system?

And any system of particles that move under Newtonian gravitation as if they are a rigid body must do so in a central configuration. In Euler's three-body problem we assume that the two centres of attraction are stationary. Together with Euler's collinear solutions, these solutions form the central configurations for the three-body problem. In celestial mechanics and the mathematics of the -body problem, a central configuration is a system of point masses with the property that each mass is pulled by the combined gravitational force of the system directly towards the center of mass, with acceleration proportional to its distance from the center. The central-force problem concerns an ideal situation (a "one-body problem") in which a single particle is attracted or repelled from an immovable point O, the center of force.Goldstein, p. 71; Landau and Lifshitz, p. 30; Whittaker, p. Thus, the equation of motion for r can be written in the form \mu \ddot{\mathbf{r}} = \mathbf{F} where \mu is the reduced mass \mu = \frac{1}{\frac{1}{m_1} + \frac{1}{m_2}} = \frac{m_1 m_2}{m_1 + m_2} As a special case, the problem of two bodies interacting by a central force can be reduced to a central-force problem of one body. ==Qualitative properties== ===Planar motion=== thumb|right|alt=The image shows a yellow disc with three vectors. In an extended modern sense, a three- body problem is any problem in classical mechanics or quantum mechanics that models the motion of three particles. ==Mathematical description== The mathematical statement of the three-body problem can be given in terms of the Newtonian equations of motion for vector positions \mathbf{r_i} = (x_i, y_i, z_i) of three gravitationally interacting bodies with masses m_i: :\begin{align} \ddot\mathbf{r}_{\mathbf{1}} &= -G m_2 \frac{\mathbf{r_1} - \mathbf{r_2}}{|\mathbf{r_1} - \mathbf{r_2}|^3} - G m_3 \frac{\mathbf{r_1} - \mathbf{r_3}}{|\mathbf{r_1} - \mathbf{r_3}|^3}, \\\ \ddot\mathbf{r}_{\mathbf{2}} &= -G m_3 \frac{\mathbf{r_2} - \mathbf{r_3}}{|\mathbf{r_2} - \mathbf{r_3}|^3} - G m_1 \frac{\mathbf{r_2} - \mathbf{r_1}}{|\mathbf{r_2} - \mathbf{r_1}|^3}, \\\ \ddot\mathbf{r}_{\mathbf{3}} &= -G m_1 \frac{\mathbf{r_3} - \mathbf{r_1}}{|\mathbf{r_3} - \mathbf{r_1}|^3} - G m_2 \frac{\mathbf{r_3} - \mathbf{r_2}}{|\mathbf{r_3} - \mathbf{r_2}|^3}. \end{align} where G is the gravitational constant. In physics and classical mechanics, the three-body problem is the problem of taking the initial positions and velocities (or momenta) of three point masses and solving for their subsequent motion according to Newton's laws of motion and Newton's law of universal gravitation. This is the only central configuration for these masses that does not lie in a lower- dimensional subspace. ==Dynamics== Under Newton's law of universal gravitation, bodies placed at rest in a central configuration will maintain the configuration as they collapse to a collision at their center of mass. An additional mass (which may be zero) is placed at the center of the system. Systems of bodies in a two-dimensional central configuration can orbit stably around their center of mass, maintaining their relative positions, with circular orbits around the center of mass or in elliptical orbits with the center of mass at a focus of the ellipse. Those equations are an accurate description of a particular form of the three-body problem. Therefore, both bodies are accelerated if a force is present between them; there is no perfectly immovable center of force. If either center were absent, the particle would move on one of these ellipses, as a solution of the Kepler problem. Special cases of these generalized problems include Darboux's problemDarboux JG, Archives Néerlandaises des Sciences (ser. 2), 6, 371-376 and Velde's problem.Velde (1889) Programm der ersten Höheren Bürgerschule zu Berlin ==Overview and history== Euler's three-body problem is to describe the motion of a particle under the influence of two centers that attract the particle with central forces that decrease with distance as an inverse-square law, such as Newtonian gravity or Coulomb's law. The simplest generalization of Euler's three-body problem is to add a third center of force midway between the original two centers, that exerts only a linear Hooke force (confer Hooke's law). The simplest bodies or elements of a multibody system were treated by Newton (free particle) and Euler (rigid body). Each multibody system formulation may lead to a different mathematical appearance of the equations of motion while the physics behind is the same. In other words, a central force must act along the line joining O with the present position of the particle. Moreover, the motion of three bodies is generally non-repeating, except in special cases. Magnus, Dynamics of multibody systems, Springer Verlag, Berlin (1978). In physics and astronomy, Euler's three-body problem is to solve for the motion of a particle that is acted upon by the gravitational field of two other point masses that are fixed in space.

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An asteroid, headed directly toward Earth, has a speed of $12 \mathrm{~km} / \mathrm{s}$ relative to the planet when the asteroid is 10 Earth radii from Earth's center. Neglecting the effects of Earth's atmosphere on the asteroid, find the asteroid's speed $v_f$ when it reaches Earth's surface.

Due to its very small perihelion and comparably large aphelion, achieves the fastest speed of any known asteroid bound to the Solar System with a velocity of 157 km/s (565,000 km/h; 351,000 mi/h) at perihelionAs calculated with the vis-viva-equation : v^2 = GM \left({ 2 \over r} - {1 \over a}\right) where: * v is the relative speed of the two bodies * r is the distance between the two bodies * a is the semi-major axis (a > 0 for ellipses, a = ∞ or 1/a = 0 for parabolas, and a < 0 for hyperbolas) * G is the gravitational constant * M is the mass of the central body (there are comets, however, which obtain much higher speeds). == See also== * List of Mercury- crossing minor planets * List of Venus-crossing minor planets * Apollo asteroids * List of Mars-crossing minor planets == References == == External links == * * * # Category:Minor planet object articles (unnumbered) Category:Mercury-crossing asteroids Category:Venus-crossing asteroids 20050430 is an Aten near-Earth asteroid less than 20 meters in diameter crudely estimated to have passed roughly 6500 km above the surface of Earth on 31 March 2004. The estimated 4 to 6 meter sized body made one of the closest known approaches to Earth. == Description == On 31 March 2004, around 15:35 UTC, the asteroid is crudely estimated to have passed within approximately 1 Earth radius () or 6,400 kilometers of the surface of the Earth (or 2.02 from Earth's center). Due to this elongated orbit, the Aten asteroid and near-Earth asteroid also classifies as Earth-crosser, Venus-crosser and Mercury-grazer. 1989 VA was the first asteroid discovered with such a small semi-major axis (0.728 AU, about the same as Venus), breaking 2100 Ra-Shalom's distance record (0.832 AU), which had held for over a decade. This asteroid orbits the Sun with a short orbital period at a distance of 0.3–1.2 AU once every 227 days. is a very eccentric, stony asteroid and near-Earth object, approximately 1 kilometer in diameter. is a sub-kilometer asteroid that orbits near Mars's Lagrangian point, on average trailing 60° behind it. It passed closest approach to Earth on 3 March 2016 05:17 UT at a distance of and was quickly approaching the glare of the Sun thus preventing further optical observations. == 2021 approach == It was recovered on 17 February 2021 by Pan-STARRS when the uncertainty in the asteroid's sky position covered about 1.2° of the sky. is an asteroid and near-Earth object approximately in diameter. Another, larger near-Earth asteroid, 2004 FH passed just two weeks prior to . is a near-Earth asteroid estimated to be roughly in diameter. It remained the asteroid with the smallest known semi-major axis for five years until the discovery of (0.683 AU), which was the first asteroid discovered closer to the Sun than Venus. Due to its eccentric orbit, is also a Mars-crosser, crossing the orbit of the Red Planet at 1.66 AU. == 2016 discovery == It was first observed by the Mount Lemmon Survey on 28 February 2016, when the asteroid was about from Earth and had a solar elongation of 174°. By early February 2021 the asteroid was brighter than apparent magnitude 24, which still placed it near the limiting magnitude of even the best automated astronomical surveys. With an exceptionally high eccentricity of 0.59, it was the most eccentric Aten asteroid known at the time of discovery, more eccentric than previously discovered Aten, 3753 Cruithne. The formerly poorly known trajectory of this asteroid was further complicated by close approaches to Venus and Mercury. It was not until (0.277 AU) was discovered that an Aten asteroid with a lower perihelion was found. 's eccentric orbit takes it out past the Earth, where it has encounters of about 0.15 to 0.20 AU about every 3 to 5 years around October–November. While listed on the Sentry Risk Table, virtual clones of the asteroid that fit the uncertainty in the known trajectory showed 116 potential impacts between 2054 and 2109. On 26 March 2010, it may have come within 0.0825 AU (12.3 million km) of Earth, but with an uncertainty parameter of 9, the orbit is poorly determined. It was the eighth Aten asteroid discovered. The combination of a small semi- major axis and high eccentricity made the first Aten asteroid discovered to get closer to the Sun (0.295 AU) than Mercury ever does. 2340 Hathor (the second Aten discovered, in 1976) had the smallest perihelion (0.464 AU) earlier, which was about the same distance as Mercury's aphelion (0.467 AU). Its orbit has an eccentricity of 0.40 and an inclination of 7° with respect to the ecliptic.

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The huge advantage of using the conservation of energy instead of Newton's laws of motion is that we can jump from the initial state to the final state without considering all the intermediate motion. Here is an example. In Figure, a child of mass $m$ is released from rest at the top of a water slide, at height $h=8.5 \mathrm{~m}$ above the bottom of the slide. Assuming that the slide is frictionless because of the water on it, find the child's speed at the bottom of the slide.

The principle represents an accurate statement of the approximate conservation of kinetic energy in situations where there is no friction. Classically, conservation of energy was distinct from conservation of mass. The concept of mass conservation is widely used in many fields such as chemistry, mechanics, and fluid dynamics. The kinetic energy, as determined by the velocity, is converted to potential energy as it reaches the same height as the initial ball and the cycle repeats. thumb|An idealized Newton's cradle with five balls when there are no energy losses and there is always a small separation between the balls, except for when a pair is colliding thumb|Newton's cradle three-ball swing in a five-ball system. Academics such as John Playfair were quick to point out that kinetic energy is clearly not conserved. Many physicists at that time, such as Newton, held that the conservation of momentum, which holds even in systems with friction, as defined by the momentum: :\sum_{i} m_i v_i was the conserved vis viva. All enforce the conservation of energy and momentum. The caloric theory maintained that heat could neither be created nor destroyed, whereas conservation of energy entails the contrary principle that heat and mechanical work are interchangeable. Thus, conservation of energy (total, including material or rest energy) and conservation of mass (total, not just rest) are one (equivalent) law. The Newton's cradle is a device that demonstrates the conservation of momentum and the conservation of energy with swinging spheres. He showed that the gravitational potential energy lost by the weight in descending was equal to the internal energy gained by the water through friction with the paddle. Energy conservation has been a foundational physical principle for about two hundred years. For systems that include large gravitational fields, general relativity has to be taken into account; thus mass–energy conservation becomes a more complex concept, subject to different definitions, and neither mass nor energy is as strictly and simply conserved as is the case in special relativity. == Formulation and examples == The law of conservation of mass can only be formulated in classical mechanics, in which the energy scales associated with an isolated system are much smaller than mc^2, where m is the mass of a typical object in the system, measured in the frame of reference where the object is at rest, and c is the speed of light. Given the stationary- action principle, conservation of energy can be rigorously proven by Noether's theorem as a consequence of continuous time translation symmetry; that is, from the fact that the laws of physics do not change over time. Some say that this behavior demonstrates the conservation of momentum and kinetic energy in elastic collisions. ===Mechanical equivalent of heat=== A key stage in the development of the modern conservation principle was the demonstration of the mechanical equivalent of heat. A consequence of the law of conservation of energy is that a perpetual motion machine of the first kind cannot exist; that is to say, no system without an external energy supply can deliver an unlimited amount of energy to its surroundings.Planck, M. (1923/1927). The law of conservation of mass and the analogous law of conservation of energy were finally generalized and unified into the principle of mass–energy equivalence, described by Albert Einstein's famous formula E = mc^2. The energy conservation law is a consequence of the shift symmetry of time; energy conservation is implied by the empirical fact that the laws of physics do not change with time itself. Engineers such as John Smeaton, Peter Ewart, , Gustave-Adolphe Hirn, and Marc Seguin recognized that conservation of momentum alone was not adequate for practical calculation and made use of Leibniz's principle. In physics and chemistry, the law of conservation of mass or principle of mass conservation states that for any system closed to all transfers of matter and energy, the mass of the system must remain constant over time, as the system's mass cannot change, so the quantity can neither be added nor be removed. In reality, the conservation of mass only holds approximately and is considered part of a series of assumptions in classical mechanics.

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Find the effective annual yield of a bank account that pays interest at a rate of 7%, compounded daily; that is, divide the difference between the final and initial balances by the initial balance.

* If one has $1000 invested for 1 year at a 7-day SEC yield of 2%, then: :(0.02 × $1000 ) / 365 ~= $0.05479 per day. * If one has $1000 invested for 30 days at a 7-day SEC yield of 5%, then: :(0.05 × $1000 ) / 365 ~= $0.137 per day. The total compound interest generated is the final value minus the initial principal: I=P\left(1+\frac{r}{n}\right)^{nt}-P ====Example 1==== Suppose a principal amount of $1,500 is deposited in a bank paying an annual interest rate of 4.3%, compounded quarterly. :Multiply by 365/7 to give the 7-day SEC yield. It is also referred to as the 7-day Annualized Yield. The calculation is performed as follows: :Take the net interest income earned by the fund over the last 7 days and subtract 7 days of management fees. It is important to note that the 7-day SEC yield is only an estimate of the fund's actual yield, and may not necessarily reflect the yield that an investor would receive if they held the fund for a longer period of time. ==Examples== The examples assume interest is withdrawn as it is earned and not allowed to compound. To calculate approximately how much interest one might earn in a money fund account, take the 7-day SEC yield, multiply by the amount invested, divide by the number of days in the year, and then multiply by the number of days in question. The amount of interest received can be calculated by subtracting the principal from this amount. 1921.24-1500=421.24 The interest is less compared with the previous case, as a result of the lower compounding frequency. ===Accumulation function=== Since the principal P is simply a coefficient, it is often dropped for simplicity, and the resulting accumulation function is used instead. Multiply by 30 days to yield $4.11 in interest. *For instance, if you were to invest $100 with compounding interest at a rate of 9% per annum, the rule of 72 gives 72/9 = 8 years required for the investment to be worth $200; an exact calculation gives ln(2)/ln(1+0.09) = 8.0432 years. The Summa de arithmetica of Luca Pacioli (1494) gives the Rule of 72, stating that to find the number of years for an investment at compound interest to double, one should divide the interest rate into 72. Subtracting the original principal from this amount gives the amount of interest received: 1938.84-1500=438.84 ====Example 2==== Suppose the same amount of $1,500 is compounded biennially (every 2 years). The E-M rule could thus be written also as : t \approx \frac{70}{r} \times \frac{198}{200-r} or t \approx \frac{72}{r} \times \frac{192}{200-r} In these variants, the multiplicative correction becomes 1 respectively for r=2 and r=8, the values for which the rules of 70 and 72 are most accurate. ===Padé approximant=== The third-order Padé approximant gives a more accurate answer over an even larger range of r, but it has a slightly more complicated formula: : t \approx \frac{69.3}{r} \times \frac{600+4r}{600+r} which simplifies to: : t \approx \frac{207900+1386r}{3000r+5r^2} ==Derivation== ===Periodic compounding === For periodic compounding, future value is given by: :FV = PV \cdot (1+r)^t where PV is the present value, t is the number of time periods, and r stands for the interest rate per time period. :Divide that dollar amount by the average size of the fund's investments over the same 7 days. When interest is continuously compounded, use \delta=n\ln{\left(1+\frac{r}{n}\right)}, where \delta is the interest rate on a continuous compounding basis, and r is the stated interest rate with a compounding frequency n. ===Monthly amortized loan or mortgage payments=== The interest on loans and mortgages that are amortized—that is, have a smooth monthly payment until the loan has been paid off—is often compounded monthly. As another example, if one wants to know the number of periods it takes for the initial value to rise by 50%, replace the constant 2 with 1.5. == Using the rule to estimate compounding periods == To estimate the number of periods required to double an original investment, divide the most convenient "rule-quantity" by the expected growth rate, expressed as a percentage. Similarly, replacing the "R" in R/200 on the third line with 2.02 gives 70 on the numerator, showing the rule of 70 is most accurate for periodically compounded interests around 2%. Then the balance after 6 years is found by using the formula above, with P = 1500, r = 0.043 (4.3%), n = 1/2 (the interest is compounded every two years), and t = 6 : A=1500\times(1+(0.043\times 2))^{\frac{6}{2}}\approx 1921.24 So, the balance after 6 years is approximately $1,921.24. The effective annual rate is the total accumulated interest that would be payable up to the end of one year, divided by the principal sum. The 7-day SEC Yield is a measure of performance in the interest rates of money market mutual funds offered by US mutual fund companies. The function f(r) is shown to be accurate in the approximation of t for a small, positive interest rate when r=.08 (see derivation below). f(.08)\approx1.03949, and we therefore approximate time t as: : t=\bigg(\frac{\ln2}{r}\bigg)f(.08) \approx \frac{.72}{r} Written as a percentage: : \frac{.72}{r}=\frac{72}{100r} This approximation increases in accuracy as the compounding of interest becomes continuous (see derivation below). 100 r is r written as a percentage.

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Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains $200 \mathrm{~L}$ of a dye solution with a concentration of $1 \mathrm{~g} / \mathrm{L}$. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of $2 \mathrm{~L} / \mathrm{min}$, the well-stirred solution flowing out at the same rate. Find the time that will elapse before the concentration of dye in the tank reaches $1 \%$ of its original value.

Instead, residence time models developed for gas and fluid dynamics, chemical engineering, and bio-hydrodynamics can be adapted to generate residence times for sub-volumes of lakes. == Renewal time == One useful mathematical model the measurement of how quickly inflows are able to refill a lake. 500px|right|thumb|The lake retention time for a body of water with the volume and the exit flow of is 20 hours. If one adds 1 litre of water to this solution, the salt concentration is reduced. Renewal time simply becomes a question how quickly could the inflows of the lake fill the entire volume of the basin (this does still assume the outflows are unchanged). Time of concentration is a concept used in hydrology to measure the response of a watershed to a rain event. It roughly expresses the amount of time taken for a substance introduced into a lake to flow out of it again. This can be important for infrastructure development (design of bridges, culverts, etc.) and management, as well as to assess flood risk such as the ARkStorm-scenario. ==Example== thumb|left|400px|NOAA diagram illustrating the concept underlying time of concentration This image shows the basic principle which leads to determination of the time of concentration. thumb|Time lapse video of diffusion of a dye dissolved in water into a gel. thumb|Three-dimensional rendering of diffusion of purple dye in water. A number of methods can be used to calculate time of concentration, including the Kirpich (1940) and NRCS (1997) methods. D_t=\left [ \frac{V}{Q} \right ] \cdot \ln \left [ \frac{C_\text{initial}}{C_\text{ending}}\right ] Sometimes the equation is also written as: \ln \left [ \frac{C_\text{ending}}{C_\text{initial}}\right ] \quad = {-}\frac{Q}{V} \cdot (t_\text{ending} - t_\text{initial}) where t_\text{initial} = 0 *Dt = time required; the unit of time used is the same as is used for Q *V = air or gas volume of the closed space or room in cubic feet, cubic metres or litres *Q = ventilation rate into or out of the room in cubic feet per minute, cubic metres per hour or litres per second *Cinitial = initial concentration of a vapor inside the room measured in ppm *Cfinal = final reduced concentration of the vapor inside the room in ppm ==Dilution ventilation equation== The basic room purge equation can be used only for purge scenarios. The retention time is particularly important where downstream flooding or pollutants are concerned. ==Global retention time== The global retention time for a lake (the overall mean time that water spends in the lake) is calculated by dividing the lake volume by either the mean rate of inflow of all tributaries, or by the mean rate of outflow (ideally including evaporation and seepage). Mathematically this relationship can be shown by equation: c_1 V_1 = c_2 V_2 where *c1 = initial concentration or molarity *V1 = initial volume *c2 = final concentration or molarity *V2 = final volume .... ==Basic room purge equation== The basic room purge equation is used in industrial hygiene. The mean free time for a molecule in a fluid is the average time between collisions. Time of concentration is useful in predicting flow rates that would result from hypothetical storms, which are based on statistically derived return periods through IDF curves.Sherman, C. (1931): Frequency and intensity of excessive rainfall at Boston, Massachusetts, Transactions, American Society of Civil Engineers, 95, 951–960. (pdf) For many (often economic) reasons, it is important for engineers and hydrologists to be able to accurately predict the response of a watershed to a given rain event. The concentration of this admixture should be small and the gradient of this concentration should be also small. Lake retention time (also called the residence time of lake water, or the water age or flushing time) is a calculated quantity expressing the mean time that water (or some dissolved substance) spends in a particular lake. In the following test tubes, the blue dye is dissolved in a lower concentration (and at the same time in a smaller amount, since the volume is approximately the same). upright=1.35|thumb|Diluting a solution by adding more solvent Dilution is the process of decreasing the concentration of a solute in a solution, usually simply by mixing with more solvent like adding more water to the solution. In a scenario where a liquid continuously evaporates from a container in a ventilated room, a differential equation has to be used: \frac{dC}{dt} = \frac{G - Q' C}{V} where the ventilation rate has been adjusted by a mixing factor K: Q' = \frac{Q}{K} *C = concentration of a gas *G = generation rate *V = room volume *Q′ = adjusted ventilation rate of the volume ==Welding== The dilution in welding terms is defined as the weight of the base metal melted divided by the total weight of the weld metal. The solutions on the left are more dilute, compared to the more concentrated solutions on the right. The equation can only be applied when the purged volume of vapor or gas is replaced with "clean" air or gas. He introduced several mechanisms of diffusion and found rate constants from experimental data.

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A certain vibrating system satisfies the equation $u^{\prime \prime}+\gamma u^{\prime}+u=0$. Find the value of the damping coefficient $\gamma$ for which the quasi period of the damped motion is $50 \%$ greater than the period of the corresponding undamped motion.

For a damped harmonic oscillator with mass m, damping coefficient c, and spring constant k, it can be defined as the ratio of the damping coefficient in the system's differential equation to the critical damping coefficient: : \zeta = \frac{c}{c_c} = \frac {\text{actual damping}} {\text{critical damping}}, where the system's equation of motion is : m\frac{d^2x}{dt^2} + c\frac{dx}{dt} + kx = 0 and the corresponding critical damping coefficient is : c_c = 2 \sqrt{k m} or : c_c = 2 m \sqrt{\frac{k}{m}} = 2m \omega_n where : \omega_n = \sqrt{\frac{k}{m}} is the natural frequency of the system. The damping ratio is a system parameter, denoted by (zeta), that can vary from undamped (), underdamped () through critically damped () to overdamped (). The damping ratio is a parameter, usually denoted by ζ (Greek letter zeta), that characterizes the frequency response of a second-order ordinary differential equation. Damping is an influence within or upon an oscillatory system that has the effect of reducing or preventing its oscillation. The general equation for an exponentially damped sinusoid may be represented as: y(t) = A e^{-\lambda t} \cos(\omega t - \varphi) where: *y(t) is the instantaneous amplitude at time ; *A is the initial amplitude of the envelope; *\lambda is the decay rate, in the reciprocal of the time units of the independent variable ; *\varphi is the phase angle at ; *\omega is the angular frequency. The damping ratio is dimensionless, being the ratio of two coefficients of identical units. == Derivation == Using the natural frequency of a harmonic oscillator \omega_n = \sqrt{{k}/{m}} and the definition of the damping ratio above, we can rewrite this as: : \frac{d^2x}{dt^2} + 2\zeta\omega_n\frac{dx}{dt} + \omega_n^2 x = 0. The key difference between critical damping and overdamping is that, in critical damping, the system returns to equilibrium in the minimum amount of time. ==Damped sine wave== thumb|350px|Plot of a damped sinusoidal wave represented as the function y(t) = e^{- t} \cos(2 \pi t) A damped sine wave or damped sinusoid is a sinusoidal function whose amplitude approaches zero as time increases. * Between the overdamped and underdamped cases, there exists a certain level of damping at which the system will just fail to overshoot and will not make a single oscillation. The damping ratio is a measure describing how rapidly the oscillations decay from one bounce to the next. The damping ratio provides a mathematical means of expressing the level of damping in a system relative to critical damping. If the system has n degrees of freedom un and is under application of m damping forces. * Q factor: Q = 1 / (2 \zeta) is another non-dimensional characterization of the amount of damping; high Q indicates slow damping relative to the oscillation. == Damping ratio definition == thumb|400px|upright=1.3|The effect of varying damping ratio on a second-order system. This turns out to be a desirable outcome in many cases where engineering design of a damped oscillator is required (e.g., a door closing mechanism). == Q factor and decay rate == The Q factor, damping ratio ζ, and exponential decay rate α are related such that : \zeta = \frac{1}{2 Q} = { \alpha \over \omega_n }. Two such solutions, for the two values of s satisfying the equation, can be combined to make the general real solutions, with oscillatory and decaying properties in several regimes: thumb|Phase portrait of damped oscillator, with increasing damping strength. *Damped harmonic motion, see animation (right). The equation of motion is \ddot x + 2\gamma \dot x + \omega^2 x = 0. A lower damping ratio implies a lower decay rate, and so very underdamped systems oscillate for long times. It corresponds to the underdamped case of damped second-order systems, or underdamped second-order differential equations.Douglas C. Giancoli (2000). Critically damped systems have a damping ratio of exactly 1, or at least very close to it. In physical systems, damping is produced by processes that dissipate the energy stored in the oscillation. The damping ratio is a dimensionless measure describing how oscillations in a system decay after a disturbance. Low damping materials may be utilized in musical instruments where sustained mechanical vibration and acoustic wave propagation is desired.

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Find the value of $y_0$ for which the solution of the initial value problem $$ y^{\prime}-y=1+3 \sin t, \quad y(0)=y_0 $$ remains finite as $t \rightarrow \infty$

A solution to an initial value problem is a function y that is a solution to the differential equation and satisfies :y(t_0) = y_0. Indeed, rather than being unique, this equation has three solutions:, p. 7 :y(t) = 0, \qquad y(t) = \pm\left (\tfrac23 t\right)^{\frac{3}{2}}. ;Second example The solution of : y'+3y=6t+5,\qquad y(0)=3 can be found to be : y(t)=2e^{-3t}+2t+1. Evidently, the functions are computing the Taylor series expansion of our known solution y=\tan(t). Since the equation being studied is a first-order equation, the initial conditions are the initial x and y values. Beginning with another initial condition , the solution y(t) tends toward the stationary point, but reaches it only at the limit of infinite time, so the uniqueness of solutions (over all finite times) is guaranteed. Then, there exists some such that the initial value problem y'(t)=f(t,y(t)),\qquad y(t_0)=y_0. has a unique solution y(t) on the interval [t_0-\varepsilon, t_0+\varepsilon]., Theorem I.3.1 == Proof sketch == The proof relies on transforming the differential equation, and applying the Banach fixed-point theorem. See Newton's method of successive approximation for instruction. == Example of Picard iteration == thumb|Four Picard iteration steps and their limit Let y(t)=\tan(t), the solution to the equation y'(t)=1+y(t)^2 with initial condition y(t_0)=y_0=0,t_0=0. We are trying to find a formula for y(t) that satisfies these two equations. Rearrange the equation so that y is on the left hand side : \frac{y'(t)}{y(t)} = 0.85 Now integrate both sides with respect to t (this introduces an unknown constant B). : \int \frac{y'(t)}{y(t)}\,dt = \int 0.85\,dt : \ln |y(t)| = 0.85t + B Eliminate the logarithm with exponentiation on both sides : | y(t) | = e^Be^{0.85t} Let C be a new unknown constant, C = \pm e^B, so : y(t) = Ce^{0.85t} Now we need to find a value for C. Use y(0) = 19 as given at the start and substitute 0 for t and 19 for y : 19 = C e^{0.85 \cdot 0} : C = 19 this gives the final solution of y(t) = 19e^{0.85t}. Initial value problems are extended to higher orders by treating the derivatives in the same way as an independent function, e.g. y(t)=f(t,y(t),y'(t)). == Existence and uniqueness of solutions == The Picard–Lindelöf theorem guarantees a unique solution on some interval containing t0 if f is continuous on a region containing t0 and y0 and satisfies the Lipschitz condition on the variable y. right|thumb|390px|Solutions to the differential equation \frac{dy}{dx} = \frac{1}{2y} subject to the initial conditions y(0)=0, 1 and 2 (red, green and blue curves respectively). Another solution is given by : y_s(x) = 0 . In multivariable calculus, an initial value problem (IVP) is an ordinary differential equation together with an initial condition which specifies the value of the unknown function at a given point in the domain. \, Indeed, : \begin{align} y'+3y &= \tfrac{d}{dt} (2e^{-3t}+2t+1)+3(2e^{-3t}+2t+1) \\\ &= (-6e^{-3t}+2)+(6e^{-3t}+6t+3) \\\ &= 6t+5. \end{align} ==Notes== ==See also== * Boundary value problem * Constant of integration * Integral curve == References == * * * * * * Category:Boundary conditions el:Αρχική τιμή it:Problema ai valori iniziali sv:Begynnelsevärdesproblem In mathematics, specifically the study of differential equations, the Picard–Lindelöf theorem gives a set of conditions under which an initial value problem has a unique solution. This happens for example for the equation , which has at least two solutions corresponding to the initial condition such as: or :y(t)=\begin{cases} \left (\tfrac{at}{3} \right )^{3} & t<0\\\ \ \ \ \ 0 & t \ge 0, \end{cases} so the previous state of the system is not uniquely determined by its state after t = 0. By considering the two sets of solutions above, one can see that the solution fails to be unique when y=0. By integrating both sides, any function satisfying the differential equation must also satisfy the integral equation :y(t) - y(t_0) = \int_{t_0}^t f(s,y(s)) \, ds. Starting with \varphi_0(t)=0, we iterate :\varphi_{k+1}(t)=\int_0^t (1+(\varphi_k(s))^2)\,ds so that \varphi_n(t) \to y(t): :\varphi_1(t)=\int_0^t (1+0^2)\,ds = t :\varphi_2(t)=\int_0^t (1+s^2)\,ds = t + \frac{t^3}{3} :\varphi_3(t)=\int_0^t \left(1+\left(s + \frac{s^3}{3}\right)^2\right)\,ds = t + \frac{t^3}{3} + \frac{2t^5}{15} + \frac{t^7}{63} and so on. In that context, the differential initial value is an equation which specifies how the system evolves with time given the initial conditions of the problem. == Definition == An initial value problem is a differential equation :y'(t) = f(t, y(t)) with f\colon \Omega \subset \mathbb{R} \times \mathbb{R}^n \to \mathbb{R}^n where \Omega is an open set of \mathbb{R} \times \mathbb{R}^n, together with a point in the domain of f :(t_0, y_0) \in \Omega, called the initial condition. An older proof of the Picard–Lindelöf theorem constructs a sequence of functions which converge to the solution of the integral equation, and thus, the solution of the initial value problem.

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A certain spring-mass system satisfies the initial value problem $$ u^{\prime \prime}+\frac{1}{4} u^{\prime}+u=k g(t), \quad u(0)=0, \quad u^{\prime}(0)=0 $$ where $g(t)=u_{3 / 2}(t)-u_{5 / 2}(t)$ and $k>0$ is a parameter. Suppose $k=2$. Find the time $\tau$ after which $|u(t)|<0.1$ for all $t>\tau$.

The modified KdV–Burgers equation is a nonlinear partial differential equationAndrei D. Polyanin, Valentin F. Zaitsev, Handbook of Nonlinear Partial Differential Equations, second edition, p 1041 CRC PRESS :u_t+u_{xxx}-\alpha u^2\,u_x - \beta u_{xx}=0. ==See also== *Burgers' equation *Korteweg–de Vries equation *modified KdV equation ==References== #Graham W. Griffiths William E. Shiesser Traveling Wave Analysis of Partial Differential Equations Academy Press # Richard H. Enns George C. McCGuire, Nonlinear Physics Birkhauser,1997 #Inna Shingareva, Carlos Lizárraga-Celaya,Solving Nonlinear Partial Differential Equations with Maple Springer. The Kaup–Kupershmidt equation (named after David J. Kaup and Boris Abram Kupershmidt) is the nonlinear fifth-order partial differential equation :u_t = u_{xxxxx}+10u_{xxx}u+25u_{xx}u_x+20u^2u_x = \frac16 (6u_{xxxx}+60uu_{xx}+45u_x^2+40u^3)_x. In the second equation, the derivative at u = 2 should be taken as u approaches 2 from the right. Unnormalized KdV equation is a nonlinear partial differential equationAndrei D. Polyanin,Valentin F. Zaitsev, HANDBOOK OF NONLINEAR PARTIAL DIFFERENTIAL EQUATIONS, SECOND EDITION p540-542 CRC PRESS u_{t}+\alpha*u_{xxx}+\beta*u*u_{x}=0 ==References== #Graham W. Griffiths William E.Shiesser Traveling Wave Analysis of Partial Differential p135 Equations Academy Press # Richard H. Enns George C. McCGuire, Nonlinear Physics Birkhauser,1997 #Inna Shingareva, Carlos Lizárraga-Celaya,Solving Nonlinear Partial Differential Equations with Maple Springer. The case f(u) = 3u2 is the original Korteweg–De Vries equation. ==References== * Category:Partial differential equations The term for 0 < k < u, k even, may be simplified using the properties of the gamma function to :\operatorname E(T^k)= u^{\frac{k}{2}} \, \prod_{i=1}^{k/2} \frac{2i-1}{ u - 2i} \qquad k\text{ even},\quad 0 For a t-distribution with u degrees of freedom, the expected value is 0 if u>1, and its variance is \frac{ u}{ u-2} if u>2. Also, \omega(u)-e^{-\gamma} oscillates in a regular way, alternating between extrema and zeroes; the extrema alternate between positive maxima and negative minima. \\! | cdf =\begin{matrix} \frac{1}{2} + x \Gamma \left( \frac{ u+1}{2} \right) \times\\\\[0.5em] \frac{\,_2F_1 \left ( \frac{1}{2},\frac{ u+1}{2};\frac{3}{2}; -\frac{x^2}{ u} \right)} {\sqrt{\pi u}\,\Gamma \left(\frac{ u}{2}\right)} \end{matrix} where 2F1 is the hypergeometric function | mean =0 for u > 1, otherwise undefined | median =0 | mode =0 | variance =\textstyle\frac{ u}{ u-2} for u > 2, ∞ for 1 < u \le 2, otherwise undefined | skewness =0 for u > 3, otherwise undefined | kurtosis =\textstyle\frac{6}{ u-4} for u > 4, ∞ for 2 < u \le 4, otherwise undefined | entropy =\begin{matrix} \frac{ u+1}{2}\left[ \psi \left(\frac{1+ u}{2} \right) \- \psi \left(\frac{ u}{2} \right) \right] \\\\[0.5em] \+ \ln{\left[\sqrt{ u}B \left(\frac{ u}{2},\frac{1}{2} \right)\right]}\,{\scriptstyle\text{(nats)}} \end{matrix} * ψ: digamma function, * B: beta function | mgf = undefined | char =\textstyle\frac{K_{ u/2} \left(\sqrt{ u}|t|\right) \cdot \left(\sqrt{ u}|t| \right)^{ u/2}} {\Gamma( u/2)2^{ u/2-1}} for u > 0 * K_ u(x): modified Bessel function of the second kind | ES =\mu + s \left( \frac{ u + T^{-1}(1-p)^2 \tau(T^{-1}(1-p)^2 )} {( u-1)(1-p)} \right) Where T^{-1} is the inverse standardized student-t CDF, and \tau is the standardized student-t PDF. In mathematics, a generalized Korteweg–De Vries equation is the nonlinear partial differential equation :\partial_t u + \partial_x^3 u + \partial_x f(u) = 0.\, The function f is sometimes taken to be f(u) = uk+1/(k+1) + u for some positive integer k (where the extra u is a "drift term" that makes the analysis a little easier). This may also be written as :f(t) = \frac{1}{\sqrt{ u}\,\mathrm{B} (\frac{1}{2}, \frac{ u}{2})} \left(1+\frac{t^2} u \right)^{-( u+1)/2}, where B is the Beta function. For t > 0, :F(t) = \int_{-\infty}^t f(u)\,du = 1 - \tfrac{1}{2} I_{x(t)}\left(\tfrac{ u}{2}, \tfrac{1}{2}\right), where :x(t) = \frac{ u}. It can be easily calculated from the cumulative distribution function Fν(t) of the t-distribution: :A(t\mid u) = F_ u(t) - F_ u(-t) = 1 - I_{\frac{ u}{ u +t^2}}\left(\frac{ u}{2},\frac{1}{2}\right), where Ix is the regularized incomplete beta function (a, b). Here a, b, and k are parameters. An alternative formula, valid for t^2 < u, is :\int_{-\infty}^t f(u)\,du = \tfrac{1}{2} + t\frac{\Gamma \left( \tfrac{1}{2}( u+1) \right)} {\sqrt{\pi u}\,\Gamma \left(\tfrac{ u}{2}\right)} \, {}_2F_1 \left( \tfrac{1}{2}, \tfrac{1}{2}( u+1); \tfrac{3}{2}; -\tfrac{t^2}{ u} \right), where 2F1 is a particular case of the hypergeometric function. For information on its inverse cumulative distribution function, see . ===Special cases=== Certain values of u give a simple form for Student's t-distribution. u PDF CDF notes 1 \frac{1}{\pi (1+t^2)} \frac{1}{2} + \frac{1}{\pi}\arctan(t) See Cauchy distribution 2 \frac{1}{2\sqrt{2}\left(1+\frac{t^2}{2}\right)^{3/2}} \frac{1}{2}+\frac{t}{2\sqrt{2}\sqrt{1+\frac{t^2}{2}}} 3 \frac{2}{\pi\sqrt{3}\left(1+\frac{t^2}{3}\right)^2} \frac{1}{2}+\frac{1}{\pi}{\left[\frac{1}{\sqrt{3}}{\frac{t}{1+\frac{t^2}{3}}}+\arctan\left(\frac{t}{\sqrt{3}}\right)\right]} 4 \frac{3}{8\left(1+\frac{t^2}{4}\right)^{5/2}} \frac{1}{2}+\frac{3}{8}{\frac{t}{\sqrt{1+\frac{t^2}{4}}}}{\left[1-\frac{1}{12}{\frac{t^2}{1+\frac{t^2}{4}}}\right]} 5 \frac{8}{3\pi\sqrt{5}\left(1+\frac{t^2}{5}\right)^3} \frac{1}{2}+\frac{1}{\pi}{ \left[\frac{t}{\sqrt{5}\left(1+\frac{t^2}{5}\right)} \left(1+\frac{2}{3\left(1+\frac{t^2}{5}\right)}\right) +\arctan\left(\frac{t}{\sqrt{5}}\right)\right]} \infty \frac{1}{\sqrt{2\pi}} e^{-t^2/2} \frac{1}{2}{\left[1+\operatorname{erf}\left(\frac{t}{\sqrt{2}}\right)\right]} See Normal distribution, Error function ===Moments=== For u > 1, the raw moments of the t-distribution are :\operatorname E(T^k)=\begin{cases} 0 & k \text{ odd},\quad 0 Moments of order u or higher do not exist. 300px|thumbnail|Graph of the Buchstab function ω(u) from u = 1 to u = 4\. The Buchstab function (or Buchstab's function) is the unique continuous function \omega: \R_{\ge 1}\rightarrow \R_{>0} defined by the delay differential equation :\omega(u)=\frac 1 u, \qquad\qquad\qquad 1\le u\le 2, :{\frac{d}{du}} (u\omega(u))=\omega(u-1), \qquad u\ge 2. With :T \sim t_ u and location-scale family transformation :X = \mu + \tau T we get :X \sim lst(\mu, \tau^2, u) The resulting distribution is also called the non- standardized Student's t-distribution. ===Density and first two moments=== The location-scale t distribution has a density defined by: :p(x\mid u,\mu,\tau) = \frac{\Gamma(\frac{ u + 1}{2})}{\Gamma(\frac{ u}{2})\sqrt{\pi u}\tau\,} \left(1+\frac{1}{ u} \left( \frac{ x-\mu } {\tau } \right)^2\right)^{-( u+1)/2} Equivalently, the density can be written in terms of \tau^2: :p(x\mid u, \mu, \tau^2) = \frac{\Gamma(\frac{ u + 1}{2})}{\Gamma(\frac{ u}{2})\sqrt{\pi u\tau^2}} \left(1+\frac{1}{ u}\frac{(x-\mu)^2}{{\tau}^2}\right)^{-( u+1)/2} Other properties of this version of the distribution are: :\begin{align} \operatorname{E}(X) &= \mu & \text{ for } u > 1 \\\ \operatorname{var}(X) &= \tau^2\frac{ u}{ u-2} & \text{ for } u > 2 \\\ \operatorname{mode}(X) &= \mu \end{align} ===Special cases=== * If X follows a location-scale t-distribution X \sim \mathrm{lst}\left(\mu, \tau^2, u\right) then for u \rightarrow \infty X is normally distributed X \sim \mathrm{N}\left(\mu, \tau^2\right) with mean \mu and variance \tau^2. #Eryk Infeld and George Rowlands, Nonlinear Waves, Solitons and Chaos,Cambridge 2000 #Saber Elaydi,An Introduction to Difference Equationns, Springer 2000 #Dongming Wang, Elimination Practice, Imperial College Press 2004 # David Betounes, Partial Differential Equations for Computational Science: With Maple and Vector Analysis Springer, 1998 # George Articolo Partial Differential Equations and Boundary Value Problems with Maple V Academic Press 1998 Category:Nonlinear partial differential equations #Eryk Infeld and George Rowlands,Nonlinear Waves,Solitons and Chaos,Cambridge 2000 #Saber Elaydi,An Introduction to Difference Equationns, Springer 2000 #Dongming Wang, Elimination Practice,Imperial College Press 2004 # David Betounes, Partial Differential Equations for Computational Science: With Maple and Vector Analysis Springer, 1998 # George Articolo Partial Differential Equations & Boundary Value Problems with Maple V Academic Press 1998 Category:Nonlinear partial differential equations It is the first equation in a hierarchy of integrable equations with the Lax operator : \partial_x^3 + 2u\partial_x + u_x, . The interval between consecutive extrema approaches 1 as u approaches infinity, as does the interval between consecutive zeroes.p. 131, Cheer and Goldston 1990. ==Applications== The Buchstab function is used to count rough numbers.

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Suppose that a sum $S_0$ is invested at an annual rate of return $r$ compounded continuously. Determine $T$ if $r=7 \%$.

The function f(r) is shown to be accurate in the approximation of t for a small, positive interest rate when r=.08 (see derivation below). f(.08)\approx1.03949, and we therefore approximate time t as: : t=\bigg(\frac{\ln2}{r}\bigg)f(.08) \approx \frac{.72}{r} Written as a percentage: : \frac{.72}{r}=\frac{72}{100r} This approximation increases in accuracy as the compounding of interest becomes continuous (see derivation below). 100 r is r written as a percentage. *For instance, if you were to invest $100 with compounding interest at a rate of 9% per annum, the rule of 72 gives 72/9 = 8 years required for the investment to be worth $200; an exact calculation gives ln(2)/ln(1+0.09) = 8.0432 years. The total compound interest generated is the final value minus the initial principal: I=P\left(1+\frac{r}{n}\right)^{nt}-P ====Example 1==== Suppose a principal amount of $1,500 is deposited in a bank paying an annual interest rate of 4.3%, compounded quarterly. The E-M rule could thus be written also as : t \approx \frac{70}{r} \times \frac{198}{200-r} or t \approx \frac{72}{r} \times \frac{192}{200-r} In these variants, the multiplicative correction becomes 1 respectively for r=2 and r=8, the values for which the rules of 70 and 72 are most accurate. ===Padé approximant=== The third-order Padé approximant gives a more accurate answer over an even larger range of r, but it has a slightly more complicated formula: : t \approx \frac{69.3}{r} \times \frac{600+4r}{600+r} which simplifies to: : t \approx \frac{207900+1386r}{3000r+5r^2} ==Derivation== ===Periodic compounding === For periodic compounding, future value is given by: :FV = PV \cdot (1+r)^t where PV is the present value, t is the number of time periods, and r stands for the interest rate per time period. Similarly, replacing the "R" in R/200 on the third line with 2.02 gives 70 on the numerator, showing the rule of 70 is most accurate for periodically compounded interests around 2%. * P = principal deposit * r = rate of return (monthly) * M = monthly deposit, and * t = time, in months The compound interest for each deposit is: M'=M(1+r)^{t} and adding all recurring deposits over the total period t (i starts at 0 if deposits begin with the investment of principal; i starts at 1 if deposits begin the next month) : M'=\sum^{t-1}_{i=0}{M(1+r)^{t-i}} recognizing the geometric series: M'=M\sum^{t-1}_{i=0}(1+r)^{t}\frac{1}{(1+r)^{i}} and applying the closed-form formula (common ratio :1/(1+r)) we obtain: P' = M\frac{(1+r)^{t}-1}{r}+P(1+r)^t If two or more types of deposits occur (either recurring or non-recurring), the compound value earned can be represented as \text{Value}=M\frac{(1+r)^{t}-1}{r}+P(1+r)^t+k\frac{(1+r)^{t-x}-1}{r}+C(1+r)^{t-y} where C is each lump sum and k are non-monthly recurring deposits, respectively, and x and y are the differences in time between a new deposit and the total period t is modeling. Then the balance after 6 years is found by using the formula above, with P = 1500, r = 0.043 (4.3%), n = 1/2 (the interest is compounded every two years), and t = 6 : A=1500\times(1+(0.043\times 2))^{\frac{6}{2}}\approx 1921.24 So, the balance after 6 years is approximately $1,921.24. The amount after t periods of continuous compounding can be expressed in terms of the initial amount P0 as P(t)=P_0 e ^ {rt}. ===Force of interest=== As the number of compounding periods n tends to infinity in continuous compounding, the continuous compound interest rate is referred to as the force of interest \delta. Let RSt be the simple rate of return on the security from t − 1 to t. Accumulation functions for simple and compound interest are a(t)=1 + r t a(t) = \left(1 + \frac {r} {n}\right) ^ {nt} If n t = 1, then these two functions are the same. ===Continuous compounding=== As n, the number of compounding periods per year, increases without limit, the case is known as continuous compounding, in which case the effective annual rate approaches an upper limit of , where is a mathematical constant that is the base of the natural logarithm. Then the balance after 6 years is found by using the formula above, with P = 1500, r = 0.043 (4.3%), n = 4, and t = 6: A=1500\times\left(1+\frac{0.043}{4}\right)^{4\times 6}\approx 1938.84 So the amount A after 6 years is approximately $1,938.84. Subtracting the original principal from this amount gives the amount of interest received: 1938.84-1500=438.84 ====Example 2==== Suppose the same amount of $1,500 is compounded biennially (every 2 years). The continuously compounded rate of return or instantaneous rate of return RCt obtained during that period is : RC_{t}=\ln\left (\frac{P_{t}}{P_{t-1}}\right ). For continuous compounding, 69 gives accurate results for any rate, since ln(2) is about 69.3%; see derivation below. A practical estimate for reverse calculation of the rate of return when the exact date and amount of each recurring deposit is not known, a formula that assumes a uniform recurring monthly deposit over the period, is:http://moneychimp.com/features/portfolio_performance_calculator.htm "recommended by The Four Pillars of Investing and The Motley Fool" r=\left(\frac{P'-P-\sum{M}}{P+\sum{M}/2}\right)^{1/t} or r=\left(\frac{P'-\sum{M}/2}{P+\sum{M}/2}\right)^{1/t}-1 ==See also== * Credit card interest * Exponential growth * Fisher equation * Interest * Interest rate * Rate of return * Rate of return on investment * Real versus nominal value (economics) * Yield curve ==References== Category:Interest Category:Exponentials Category:Mathematical finance Category:Actuarial science it:Anatocismo When interest is continuously compounded, use \delta=n\ln{\left(1+\frac{r}{n}\right)}, where \delta is the interest rate on a continuous compounding basis, and r is the stated interest rate with a compounding frequency n. ===Monthly amortized loan or mortgage payments=== The interest on loans and mortgages that are amortized—that is, have a smooth monthly payment until the loan has been paid off—is often compounded monthly. * A rate of 1% per month is equivalent to a simple annual interest rate (nominal rate) of 12%, but allowing for the effect of compounding, the annual equivalent compound rate is 12.68% per annum (1.0112 − 1). As another example, if one wants to know the number of periods it takes for the initial value to rise by 50%, replace the constant 2 with 1.5. == Using the rule to estimate compounding periods == To estimate the number of periods required to double an original investment, divide the most convenient "rule-quantity" by the expected growth rate, expressed as a percentage. Compound interest is standard in finance and economics. For every three percentage points away from 8%, the value of 72 could be adjusted by 1: : t \approx \frac{72 + (r - 8)/3}{r} or, for the same result: : t \approx \frac{70 + (r - 2)/3}{r} Both of these equations simplify to: : t \approx \frac{208}{3r} + \frac{1}{3} Note that \frac{208}{3} is quite close to 69.3. ===E-M rule=== The Eckart–McHale second- order rule (the E-M rule) provides a multiplicative correction for the rule of 69.3 that is very accurate for rates from 0% to 20%, whereas the rule is normally only accurate at the lowest end of interest rates, from 0% to about 5%. Thus, continuing the above nominal example, the final value of the investment expressed in real terms is :P_t^{real} = P_t \cdot \frac{PL_{t-1}}{PL_t}. For periodic compounding, the exact doubling time for an interest rate of r percent per period is :t = \frac{\ln(2)}{\ln(1+r/100)}\approx \frac{72}{r}, where t is the number of periods required.

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A mass weighing $2 \mathrm{lb}$ stretches a spring 6 in. If the mass is pulled down an additional 3 in. and then released, and if there is no damping, determine the position $u$ of the mass at any time $t$. Find the frequency of the motion.

The general differential equation of motion is: :I\frac{d^2\theta}{dt^2} + C\frac{d\theta}{dt} + \kappa\theta = \tau(t) If the damping is small, C \ll \sqrt{\kappa I}\,, as is the case with torsion pendulums and balance wheels, the frequency of vibration is very near the natural resonant frequency of the system: :f_n = \frac{\omega_n}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{\kappa}{I}}\, Therefore, the period is represented by: :T_n = \frac{1}{f_n} = \frac{2\pi}{\omega_n} = 2\pi \sqrt{\frac{I}{\kappa}}\, The general solution in the case of no drive force (\tau = 0\,), called the transient solution, is: :\theta = Ae^{-\alpha t} \cos{(\omega t + \phi)}\, where: ::\alpha = C/2I\, ::\omega = \sqrt{\omega_n^2 - \alpha^2} = \sqrt{\kappa/I - (C/2I)^2}\, ===Applications=== thumb|Animation of a torsion spring oscillating The balance wheel of a mechanical watch is a harmonic oscillator whose resonant frequency f_n\, sets the rate of the watch. Since the inertia of the beam can be found from its mass, the spring constant can be calculated. thumb In physics and mathematics, in the area of dynamical systems, an elastic pendulum (also called spring pendulum or swinging spring) is a physical system where a piece of mass is connected to a spring so that the resulting motion contains elements of both a simple pendulum and a one-dimensional spring-mass system. The weight rotates about the axis of the spring, twisting it, instead of swinging like an ordinary pendulum. In the case of =1, the considered problem has a closed solution: y(\tau )=\left[\frac{4}{3}\tau (1-\tau) -\frac{4}{3}\tau \left( 1+2 \tau\ln (1-\tau )+2\ln (1-\tau )\right)\right]\ . ==References== Category:Mechanical vibrations thumb|The second pendulum, with a period of two seconds so each swing takes one second A seconds pendulum is a pendulum whose period is precisely two seconds; one second for a swing in one direction and one second for the return swing, a frequency of 0.5 Hz.Seconds pendulum == Pendulum == A pendulum is a weight suspended from a pivot so that it can swing freely. When the oscillatory motion of the balance dies out, the deflection will be proportional to the force: :\theta = FL/\kappa\, To determine F\, it is necessary to find the torsion spring constant \kappa\,. We assume dimensionless displacements of the string and dimensionless time : thumb|240px|Massless string and a moving mass - mass trajectory. : y(\tau)=\frac{w(vt,t)}{w_{st}}\ ,\ \ \ \ \tau\ =\ \frac{vt}{l}\ , where st is the static deflection in the middle of the string. More complex problems can be solved by the finite element method or space-time finite element method. massless load inertial load thumb|321px|Vibrations of a string under a moving massless force (v=0.1c); c is the wave speed. thumb|321px|Vibrations of a string under a moving massless force (v=0.5c); c is the wave speed. thumb|321px|Vibrations of a string under a moving inertial force (v=0.1c); c is the wave speed. thumb|321px|Vibrations of a string under a moving inertial force (v=0.5c); c is the wave speed. It has been the basis of all the most significant > experiments on gravitation ever since. ==Torsional harmonic oscillators== Definition of terms Term Unit Definition \theta\, rad Angle of deflection from rest position I\, kg m2 Moment of inertia C\, joule s rad−1 Angular damping constant \kappa\, N m rad−1 Torsion spring constant \tau\, \mathrm{N\,m}\, Drive torque f_n\, Hz Undamped (or natural) resonant frequency T_n\, s Undamped (or natural) period of oscillation \omega_n\, \mathrm{rad\,s^{-1}}\, Undamped resonant frequency in radians f\, Hz Damped resonant frequency \omega\, \mathrm{rad\,s^{-1}}\, Damped resonant frequency in radians \alpha\, \mathrm{s^{-1}}\, Reciprocal of damping time constant \phi\, rad Phase angle of oscillation L\, m Distance from axis to where force is applied Torsion balances, torsion pendulums and balance wheels are examples of torsional harmonic oscillators that can oscillate with a rotational motion about the axis of the torsion spring, clockwise and counterclockwise, in harmonic motion. The value of damping that causes the oscillatory motion to settle quickest is called the critical dampingC_c\,: :C_c = 2 \sqrt{\kappa I}\, ==See also== * Beam (structure) * Slinky, helical toy spring ==References== ==Bibliography== * . thumb|Video of a model torsion pendulum oscillating A torsion spring is a spring that works by twisting its end along its axis; that is, a flexible elastic object that stores mechanical energy when it is twisted. The motion of an elastic pendulum is governed by a set of coupled ordinary differential equations. ==Analysis and interpretation== thumb|300px|2 DOF elastic pendulum with polar coordinate plots. If the damping is low, this can be obtained by measuring the natural resonant frequency of the balance, since the moment of inertia of the balance can usually be calculated from its geometry, so: :\kappa = (2\pi f_n)^2 I\, In measuring instruments, such as the D'Arsonval ammeter movement, it is often desired that the oscillatory motion die out quickly so the steady state result can be read off. *The torsion pendulum used in torsion pendulum clocks is a wheel-shaped weight suspended from its center by a wire torsion spring. Inertial load in numerical models is described in Unexpected property of differential equations that govern the motion of the mass particle travelling on the string, Timoshenko beam, and Mindlin plate is described in. It is also possible that the spring has a range that is overtaken by the motion of the pendulum, making it practically neutral to the motion of the pendulum. ===Lagrangian=== The spring has the rest length l_0 and can be stretched by a length x. In the case of inertial moving load, the analytical solutions are unknown. Made in the USA of cherry wood and power coated steel, it simulates a walking motion, using see-saw oscillations from 0-10mm, with a frequency of 0 to 15.5 Hz. The equation of motion is increased by the term related to the inertia of the moving load. We will use the word "torsion" in the following for a torsion spring according to the definition given above, whether the material it is made of actually works by torsion or by bending. ==Torsion coefficient== As long as they are not twisted beyond their elastic limit, torsion springs obey an angular form of Hooke's law: : \tau = -\kappa\theta\, where \tau\, is the torque exerted by the spring in newton- meters, and \theta\, is the angle of twist from its equilibrium position in radians. \kappa\, is a constant with units of newton-meters / radian, variously called the spring's torsion coefficient, torsion elastic modulus, rate, or just spring constant, equal to the change in torque required to twist the spring through an angle of 1 radian. The kinetic energy is given by: :T=\frac{1}{2}mv^2 where v is the velocity of the mass.

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If $\mathbf{x}=\left(\begin{array}{c}2 \\ 3 i \\ 1-i\end{array}\right)$ and $\mathbf{y}=\left(\begin{array}{c}-1+i \\ 2 \\ 3-i\end{array}\right)$, find $(\mathbf{y}, \mathbf{y})$.

More precisely, given two sets of variables represented as coordinate vectors and y, then each equation of the system can be written y^TA_ix=g_i, where, is an integer whose value ranges from 1 to the number of equations, each A_i is a matrix, and each g_i is a real number. There are several possible ways to compute these quantities for a given implicit curve. XHJTA-FM is a radio station on 94.3 FM in Irapuato, Guanajuato. 300px|thumb|Cassini ovals: (1) a=1.1, c=1 (above), (2) a=c=1 (middle), (3) a=1, c=1.05 (below) 300px|thumb|Implicit curve: \sin(x+y)-\cos(xy)+1=0 In mathematics, an implicit curve is a plane curve defined by an implicit equation relating two coordinate variables, commonly x and y. Implicit means that the equation is not expressed as a solution for either x in terms of y or vice versa. is a passenger railway station located in the city of Himeji, Hyōgo Prefecture, Japan, operated by West Japan Railway Company (JR West). ==Lines== Yobe Station is served by the Kishin Line, and is located 6.1 kilometers from the terminus of the line at . ==Station layout== The station consists of two ground-level opposed side platforms connected by a level crossing. Finally, we calculate the value c via linear interpolation of c_{0} and c_{1} In practice, a trilinear interpolation is identical to two bilinear interpolation combined with a linear interpolation: :c \approx l\left( b(c_{000}, c_{010}, c_{100}, c_{110}),\, b(c_{001}, c_{011}, c_{101}, c_{111})\right) ===Alternative algorithm=== An alternative way to write the solution to the interpolation problem is :f(x, y, z) \approx a_0 + a_1 x + a_2 y + a_3 z + a_4 x y + a_5 x z + a_6 y z + a_7 x y z where the coefficients are found by solving the linear system :\begin{align} \begin{bmatrix} 1 & x_0 & y_0 & z_0 & x_0 y_0 & x_0 z_0 & y_0 z_0 & x_0 y_0 z_0 \\\ 1 & x_1 & y_0 & z_0 & x_1 y_0 & x_1 z_0 & y_0 z_0 & x_1 y_0 z_0 \\\ 1 & x_0 & y_1 & z_0 & x_0 y_1 & x_0 z_0 & y_1 z_0 & x_0 y_1 z_0 \\\ 1 & x_1 & y_1 & z_0 & x_1 y_1 & x_1 z_0 & y_1 z_0 & x_1 y_1 z_0 \\\ 1 & x_0 & y_0 & z_1 & x_0 y_0 & x_0 z_1 & y_0 z_1 & x_0 y_0 z_1 \\\ 1 & x_1 & y_0 & z_1 & x_1 y_0 & x_1 z_1 & y_0 z_1 & x_1 y_0 z_1 \\\ 1 & x_0 & y_1 & z_1 & x_0 y_1 & x_0 z_1 & y_1 z_1 & x_0 y_1 z_1 \\\ 1 & x_1 & y_1 & z_1 & x_1 y_1 & x_1 z_1 & y_1 z_1 & x_1 y_1 z_1 \end{bmatrix}\begin{bmatrix} a_0 \\\ a_1 \\\ a_2 \\\ a_3 \\\ a_4 \\\ a_5 \\\ a_6 \\\ a_7 \end{bmatrix} = \begin{bmatrix} c_{000} \\\ c_{100} \\\ c_{010} \\\ c_{110} \\\ c_{001} \\\ c_{101} \\\ c_{011} \\\ c_{111} \end{bmatrix}, \end{align} yielding the result :\begin{align} a_0 ={} &\frac{-c_{000} x_1 y_1 z_1 + c_{001} x_1 y_1 z_0 + c_{010} x_1 y_0 z_1 - c_{011} x_1 y_0 z_0}{(x_0 - x_1) (y_0 - y_1) (z_0 - z_1)} +{} \\\ &\frac{ c_{100} x_0 y_1 z_1 - c_{101} x_0 y_1 z_0 - c_{110} x_0 y_0 z_1 + c_{111} x_0 y_0 z_0}{(x_0 - x_1) (y_0 - y_1) (z_0 - z_1)}, \\\\[4pt] a_1 ={} &\frac{ c_{000} y_1 z_1 - c_{001} y_1 z_0 - c_{010} y_0 z_1 + c_{011} y_0 z_0}{(x_0 - x_1) (y_0 - y_1) (z_0 - z_1)} +{} \\\ &\frac{-c_{100} y_1 z_1 + c_{101} y_1 z_0 + c_{110} y_0 z_1 - c_{111} y_0 z_0}{(x_0 - x_1) (y_0 - y_1) (z_0 - z_1)}, \\\\[4pt] a_2 ={} &\frac{ c_{000} x_1 z_1 - c_{001} x_1 z_0 - c_{010} x_1 z_1 + c_{011} x_1 z_0}{(x_0 - x_1) (y_0 - y_1) (z_0 - z_1)} +{} \\\ &\frac{-c_{100} x_0 z_1 + c_{101} x_0 z_0 + c_{110} x_0 z_1 - c_{111} x_0 z_0}{(x_0 - x_1) (y_0 - y_1) (z_0 - z_1)}, \\\\[4pt] a_3 ={} &\frac{ c_{000} x_1 y_1 - c_{001} x_1 y_1 - c_{010} x_1 y_0 + c_{011} x_1 y_0}{(x_0 - x_1) (y_0 - y_1) (z_0 - z_1)} +{} \\\ &\frac{-c_{100} x_0 y_1 + c_{101} x_0 y_1 + c_{110} x_0 y_0 - c_{111} x_0 y_0}{(x_0 - x_1) (y_0 - y_1) (z_0 - z_1)}, \\\\[4pt] a_4 ={} &\frac{-c_{000} z_1 + c_{001} z_0 + c_{010} z_1 - c_{011} z_0 + c_{100} z_1 - c_{101} z_0 - c_{110} z_1 + c_{111} z_0}{(x_0 - x_1) (y_0 - y_1) (z_0 - z_1)}, \\\\[4pt] a_5 = &\frac{-c_{000} y_1 + c_{001} y_1 + c_{010} y_0 - c_{011} y_0 + c_{100} y_1 - c_{101} y_1 - c_{110} y_0 + c_{111} y_0}{(x_0 - x_1) (y_0 - y_1) (z_0 - z_1)}, \\\\[4pt] a_6 ={} &\frac{-c_{000} x_1 + c_{001} x_1 + c_{010} x_1 - c_{011} x_1 + c_{100} x_0 - c_{101} x_0 - c_{110} x_0 + c_{111} x_0}{(x_0 - x_1) (y_0 - y_1) (z_0 - z_1)}, \\\\[4pt] a_7 ={} &\frac{ c_{000} - c_{001} - c_{010} + c_{011} - c_{100} + c_{101} + c_{110} - c_{111}}{(x_0 - x_1) (y_0 - y_1) (z_0 - z_1)}. \end{align} ==See also== * Linear interpolation * Bilinear interpolation * Tricubic interpolation * Radial interpolation * Tetrahedral interpolation * Spherical Linear Interpolation ==External links== *pseudo-code from NASA, describes an iterative inverse trilinear interpolation (given the vertices and the value of C find Xd, Yd and Zd). Trilinear interpolation is a method of multivariate interpolation on a 3-dimensional regular grid. Plane curves can be represented in Cartesian coordinates (x, y coordinates) by any of three methods, one of which is the implicit equation given above. 'Systems of bilinear equations'. 250px|thumb|right|Scalar multiplication of a vector by a factor of 3 stretches the vector out. 250px|thumb|right|The scalar multiplications −a and 2a of a vector a In mathematics, scalar multiplication is one of the basic operations defining a vector space in linear algebra (or more generally, a module in abstract algebra). Vector \mathbf t(x_0,y_0,z_0) is a tangent vector of the curve at point (x_0,y_0,z_0). 300px|thumb|Intersection curve between a sphere and a cylinder Examples: (1)\quad x+y+z-1=0 \ ,\ x-y+z-2=0 ::is a line. (2)\quad x^2+y^2+z^2-4=0 \ , \ x+y+z-1=0 ::is a plane section of a sphere, hence a circle. (3)\quad x^2+y^2-1=0 \ , \ x+y+z-1=0 ::is an ellipse (plane section of a cylinder). (4)\quad x^2+y^2+z^2-16=0 \ , \ (y-y_0)^2+z^2-9=0 ::is the intersection curve between a sphere and a cylinder. One method is to use implicit differentiation to compute the derivatives of y with respect to x. For a real scalar and matrix: : \lambda = 2, \quad \mathbf{A} =\begin{pmatrix} a & b \\\ c & d \\\ \end{pmatrix} : 2 \mathbf{A} = 2 \begin{pmatrix} a & b \\\ c & d \\\ \end{pmatrix} = \begin{pmatrix} 2 \\!\cdot\\! a & 2 \\!\cdot\\! b \\\ 2 \\!\cdot\\! c & 2 \\!\cdot\\! d \\\ \end{pmatrix} = \begin{pmatrix} a \\!\cdot\\! 2 & b \\!\cdot\\! 2 \\\ c \\!\cdot\\! 2 & d \\!\cdot\\! 2 \\\ \end{pmatrix} = \begin{pmatrix} a & b \\\ c & d \\\ \end{pmatrix}2= \mathbf{A}2. For quaternion scalars and matrices: : \lambda = i, \quad \mathbf{A} = \begin{pmatrix} i & 0 \\\ 0 & j \\\ \end{pmatrix} : i\begin{pmatrix} i & 0 \\\ 0 & j \\\ \end{pmatrix} = \begin{pmatrix} i^2 & 0 \\\ 0 & ij \\\ \end{pmatrix} = \begin{pmatrix} -1 & 0 \\\ 0 & k \\\ \end{pmatrix} e \begin{pmatrix} -1 & 0 \\\ 0 & -k \\\ \end{pmatrix} = \begin{pmatrix} i^2 & 0 \\\ 0 & ji \\\ \end{pmatrix} = \begin{pmatrix} i & 0 \\\ 0 & j \\\ \end{pmatrix}i\,, where are the quaternion units. The result of trilinear interpolation is independent of the order of the interpolation steps along the three axes: any other order, for instance along x, then along y, and finally along z, produces the same value. ) :(P3) until the distance between the points (x_{j+1},y_{j+1}),\, (x_j,y_j) is small enough. Next, we perform linear interpolation between c_{000} and c_{100} to find c_{00}, c_{001} and c_{101} to find c_{01}, c_{011} and c_{111} to find c_{11}, c_{010} and c_{110} to find c_{10}. First we interpolate along x (imagine we are "pushing" the face of the cube defined by C_{0jk} to the opposing face, defined by C_{1jk}), giving: : \begin{align} c_{00} &= c_{000} (1 - x_\text{d}) + c_{100} x_\text{d} \\\ c_{01} &= c_{001} (1 - x_\text{d}) + c_{101} x_\text{d} \\\ c_{10} &= c_{010} (1 - x_\text{d}) + c_{110} x_\text{d} \\\ c_{11} &= c_{011} (1 - x_\text{d}) + c_{111} x_\text{d} \end{align} Where c_{000} means the function value of (x_0, y_0, z_0). There are several ways to arrive at trilinear interpolation, which is equivalent to 3-dimensional tensor B-spline interpolation of order 1, and the trilinear interpolation operator is also a tensor product of 3 linear interpolation operators. ==Method== right|thumb|Eight corner points on a cube surrounding the interpolation point C right|thumb|Depiction of 3D interpolation thumb|A geometric visualisation of trilinear interpolation. Contains a very clever and simple method to find trilinear interpolation that is based on binary logic and can be extended to any dimension (Tetralinear, Pentalinear, ...). A normal vector to the curve at the point is given by : \mathbf{n}(x_0,y_0) = (F_x(x_0,y_0), F_y(x_0,y_0)) (here written as a row vector). === Curvature === For readability of the formulas, the arguments (x_0,y_0) are omitted.

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15. Consider the initial value problem $$ 4 y^{\prime \prime}+12 y^{\prime}+9 y=0, \quad y(0)=1, \quad y^{\prime}(0)=-4 . $$ Determine where the solution has the value zero.

Since the equation being studied is a first-order equation, the initial conditions are the initial x and y values. right|thumb|390px|Solutions to the differential equation \frac{dy}{dx} = \frac{1}{2y} subject to the initial conditions y(0)=0, 1 and 2 (red, green and blue curves respectively). A solution to an initial value problem is a function y that is a solution to the differential equation and satisfies :y(t_0) = y_0. By considering the two sets of solutions above, one can see that the solution fails to be unique when y=0. Indeed, rather than being unique, this equation has three solutions:, p. 7 :y(t) = 0, \qquad y(t) = \pm\left (\tfrac23 t\right)^{\frac{3}{2}}. Furthermore, for a given x ot=0, this is the unique solution going through (x,y(x)). ==Failure of uniqueness== Consider the differential equation : y'(x)^2 = 4y(x) . In multivariable calculus, an initial value problem (IVP) is an ordinary differential equation together with an initial condition which specifies the value of the unknown function at a given point in the domain. If p' = 0 it means that y' = p = c = constant, and the general solution of this new equation is: : y_c(x) = c \cdot x + c^2 where c is determined by the initial value. Hence, : y_s(x) = -\tfrac{1}{4} \cdot x^2 \,\\! is tangent to every member of the one- parameter family of solutions : y_c(x) = c \cdot x + c^2 \,\\! of this Clairaut equation: : y(x) = x \cdot y' + (y')^2. \,\\! ==See also== * Chandrasekhar equation * Chrystal's equation * Caustic (mathematics) * Envelope (mathematics) * Initial value problem * Picard–Lindelöf theorem ==Bibliography== * Category:Differential equations Initial value problems are extended to higher orders by treating the derivatives in the same way as an independent function, e.g. y(t)=f(t,y(t),y'(t)). == Existence and uniqueness of solutions == The Picard–Lindelöf theorem guarantees a unique solution on some interval containing t0 if f is continuous on a region containing t0 and y0 and satisfies the Lipschitz condition on the variable y. ;Second example The solution of : y'+3y=6t+5,\qquad y(0)=3 can be found to be : y(t)=2e^{-3t}+2t+1. A singular solution ys(x) of an ordinary differential equation is a solution that is singular or one for which the initial value problem (also called the Cauchy problem by some authors) fails to have a unique solution at some point on the solution. Another solution is given by : y_s(x) = 0 . Solutions which are singular in the sense that the initial value problem fails to have a unique solution need not be singular functions. In mathematics, specifically the study of differential equations, the Picard–Lindelöf theorem gives a set of conditions under which an initial value problem has a unique solution. We solve : c \cdot x + c^2 = y_c(x) = y_s(x) = -\tfrac{1}{4} x^2 to find the intersection point, which is (-2c , -c^2). Rearrange the equation so that y is on the left hand side : \frac{y'(t)}{y(t)} = 0.85 Now integrate both sides with respect to t (this introduces an unknown constant B). : \int \frac{y'(t)}{y(t)}\,dt = \int 0.85\,dt : \ln |y(t)| = 0.85t + B Eliminate the logarithm with exponentiation on both sides : | y(t) | = e^Be^{0.85t} Let C be a new unknown constant, C = \pm e^B, so : y(t) = Ce^{0.85t} Now we need to find a value for C. Use y(0) = 19 as given at the start and substitute 0 for t and 19 for y : 19 = C e^{0.85 \cdot 0} : C = 19 this gives the final solution of y(t) = 19e^{0.85t}. Now we shall check when these solutions are singular solutions. Uniqueness fails for these solutions on the interval c_1\leq x\leq c_2, and the solutions are singular, in the sense that the second derivative fails to exist, at x=c_1 and x=c_2. ==Further example of failure of uniqueness== The previous example might give the erroneous impression that failure of uniqueness is directly related to y(x)=0. The general solution to this equation is : y(x)= C x^{-2} . If x + 2p = 0 then we get that p = −½x and substituting in the ODE gives : y_s(x) = -\tfrac{1}{2}x^2 + (-\tfrac{1}{2}x)^2 = -\tfrac{1}{4} x^2. # Check that the solution is consistent with step 2.

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A certain college graduate borrows $8000 to buy a car. The lender charges interest at an annual rate of 10%. What monthly payment rate is required to pay off the loan in 3 years?

If, in the second case, equal monthly payments are made of $946.01 against 9.569% compounded monthly then it takes 240 months to pay the loan back. Over the life of a 30-year loan, this amounts to $23,070.86, which is over 11% of the original loan amount. ===Certain fees are not considered=== Some classes of fees are deliberately not included in the calculation of APR. So the present value of the drawdowns is equal to the present value of the repayments, given the APR as the interest rate. For example, consider a 30-year loan of $200,000 with a stated APR of 10.00%, i.e., 10.0049% APR or the EAR equivalent of 10.4767%. There are two primary methods of borrowing money to buy a car: direct and indirect. thumb|279px|Parts of total cost and effective APR for a 12-month, 5% monthly interest, $100 loan paid off in equally sized monthly payments. If the $1000 one-time fees are taken into account then the yearly interest rate paid is effectively equal to 10.31%. The monthly payments, using APR, would be $1755.87. If the fee is not considered, this loan has an effective APR of approximately 80% (1.0512 = 1.7959, which is approximately an 80% increase). If the $10 fee were considered, the monthly interest increases by 10% ($10/$100), and the effective APR becomes approximately 435% (1.1512 = 5.3503, which equals a 435% increase). * APR is also an abbreviation for "Annual Principal Rate" which is sometimes used in the auto sales in some countries where the interest is calculated based on the "Original Principal" not the "Current Principal Due", so as the Current Principal Due decreases, the interest due does not. ==Rate format== An effective annual interest rate of 10% can also be expressed in several ways: * 0.7974% effective monthly interest rate, because 1.00797412=1.1 * 9.569% annual interest rate compounded monthly, because 12×0.7974=9.569 * 9.091% annual rate in advance, because (1.1-1)÷1.1=0.09091 These rates are all equivalent, but to a consumer who is not trained in the mathematics of finance, this can be confusing. In that case the formula becomes: :: S -A = R (1 + \mathrm{APR}/100)^{-t_N} + \sum_{k=1}^N A_k (1 + \mathrm{APR}/100)^{-t_k} :where: :: S is the borrowed amount or principal amount. Over 85% of new cars and half of used cars are financed (as opposed to being paid for in a lump sum with cash). If the consumer pays the loan off early, the effective interest rate achieved will be significantly higher than the APR initially calculated. Typically, the indirect auto lender will set an interest rate, known as the "buy rate". Using the improved notation of directive 2008/48/EC, the basic equation for calculation of APR in the EU is: :: \sum_{i=1}^M C_i (1 + \mathrm{APR}/100)^{-t_i} = \sum_{j=1}^N D_j (1 + \mathrm{APR}/100)^{-s_j} :where: :: M is the total number of drawdowns paid by the lender :: N is the total number of repayments paid by the borrower :: i is the sequence number of a drawdown paid by the lender :: j is the sequence number of a repayment paid by the borrower :: Ci is the cash flow amount for drawdown number i :: Dj is the cash flow amount for repayment number j :: ti is the interval, expressed in years and fractions of a year, between the date of the first drawdown* and the date of drawdown i :: sj is the interval, expressed in years and fractions of a year, between the date of the first drawdown* and the date of repayment j. In this equation the left side is the present value of the drawdowns made by the lender and the right side is the present value of the repayments made by the borrower. Suppose that the complete amount including the interest is withdrawn after exactly one year. Buy Here Pay Here financing accounts for 6% of the total financing market. Consumers can, of course, use the nominal interest rate and any costs on the loan (or savings account) and compute the APR themselves, for instance using one of the calculators on the internet. An interest rate is the amount of interest due per period, as a proportion of the amount lent, deposited, or borrowed (called the principal sum). This can be expressed mathematically by : p = \frac{P_0\cdot r\cdot (1+r)^n}{(1+r)^n-1} :where: ::p is the payment made each period :: P0 is the initial principal :: r is the percentage rate used each payment :: n is the number of payments This also explains why a 15-year mortgage and a 30-year mortgage with the same APR would have different monthly payments and a different total amount of interest paid.

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Consider the initial value problem $$ y^{\prime \prime}+\gamma y^{\prime}+y=k \delta(t-1), \quad y(0)=0, \quad y^{\prime}(0)=0 $$ where $k$ is the magnitude of an impulse at $t=1$ and $\gamma$ is the damping coefficient (or resistance). Let $\gamma=\frac{1}{2}$. Find the value of $k$ for which the response has a peak value of 2 ; call this value $k_1$.

* Determine the system steady-state gain k=A_0with k=\lim_{t\to\infty} \dfrac{y(t)}{x(t)} * Calculate r=\dfrac{t_{25}}{t_{75}} P=-18.56075\,r+\dfrac{0.57311}{r-0.20747}+4.16423 X=14.2797\,r^3-9.3891\,r^2+0.25437\,r+1.32148 * Determine the two time constants \tau_2=T_2=\dfrac{t_{75}-t_{25}}{X\,(1+1/P)} \tau_1=T_1=\dfrac{T_2}{P} * Calculate the transfer function of the identified system within the Laplace-domain G(s) = \dfrac{k}{(1+s\,T_1)\cdot(1+s\,T_2)} ====Phase margin==== thumbnail|280px|Figure 5: Bode gain plot to find phase margin; scales are logarithmic, so labeled separations are multiplicative factors. Here damping ratio is greater than one. ==Properties== thumb|Typical second order transient system properties Transient response can be quantified with the following properties. In particular, the unit step response of the system is: :S(t) = \left(\frac {A_0} {1+ \beta A_0}\right)\left(1 - e^{- \rho t} \ \frac { \sin \left( \mu t + \phi \right)}{ \sin \phi}\right)\ , which simplifies to :S(t) = 1 - e^{- \rho t} \ \frac { \sin \left( \mu t + \phi \right)}{ \sin \phi} when A0 tends to infinity and the feedback factor β is one. * Determine the time-spans t_{25}and t_{75}where the step response reaches 25% and 75% of the steady state output value. : \Delta = e^{- \rho t_S }\text{ or }t_S = \frac { \ln \frac{1}{\Delta} } { \rho } = \tau_2 \frac {2 \ln \frac{1} { \Delta} } { 1 + \frac { \tau_2 } { \tau_1} } \approx 2 \tau_2 \ln \frac{1} { \Delta}, where the τ1 ≫ τ2 is applicable because of the overshoot control condition, which makes τ1 = αβAOL τ2. The final value of the step response is 1, so the exponential is the actual overshoot itself. Using the equations above, the amount of overshoot can be found by differentiating the step response and finding its maximum value. Its step response is of the same form: an exponential decay toward the new equilibrium value. Notice that the damping of the response is set by ρ, that is, by the time constants of the open-loop amplifier. The equation reads as :\frac{d^2y}{d\zeta^2} =(y^2-\zeta^2)e^{-\delta^{-1/3}(y+\gamma \zeta)} subjected to the boundary conditions : \begin{align} \zeta\rightarrow -\infty : &\quad \frac{dy}{d\zeta}=-1,\\\ \zeta\rightarrow \infty : &\quad \frac{dy}{d\zeta}=1 \end{align} where \delta is the reduced or rescaled Damköhler number and \gamma is the ratio of excess heat conducted to one side of the reaction sheet to the total heat generated in the reaction zone. In electrical engineering and mechanical engineering, a transient response is the response of a system to a change from an equilibrium or a steady state. For \delta>\delta_E with |\gamma|<1, the equation possess two solutions, of which one is an unstable solution. The impulse response and step response are transient responses to a specific input (an impulse and a step, respectively). This forward amplifier has unit step response :S_{OL}(t) = A_0(1 - e^{-t / \tau}), an exponential approach from 0 toward the new equilibrium value of A0. The time dependence of the amplifier is easy to discover by switching variables to s = jω, whereupon the gain becomes: : A_{FB} = \frac {A_0} { \tau_1 \tau_2 } \; \cdot \; \frac {1} {s^2 +s \left( \frac {1} {\tau_1} + \frac {1} {\tau_2} \right) + \frac {1+ \beta A_0} {\tau_1 \tau_2}} The poles of this expression (that is, the zeros of the denominator) occur at: :2s = - \left( \frac {1} {\tau_1} + \frac {1} {\tau_2} \right) \pm \sqrt { \left( \frac {1} {\tau_1} - \frac {1} {\tau_2} \right) ^2 -\frac {4 \beta A_0 } {\tau_1 \tau_2 } }, which shows for large enough values of βA0 the square root becomes the square root of a negative number, that is the square root becomes imaginary, and the pole positions are complex conjugate numbers, either s+ or s−; see Figure 2: : s_{\pm} = -\rho \pm j \mu, with : \rho = \frac {1}{2} \left( \frac {1} {\tau_1} + \frac {1} {\tau_2} \right ), and : \mu = \frac {1} {2} \sqrt { \frac {4 \beta A_0} { \tau_1 \tau_2} - \left( \frac {1} {\tau_1} - \frac {1} {\tau_2} \right)^2 }. The result for maximum step response Smax is: :S_\max= 1 + \exp \left( - \pi \frac { \rho }{ \mu } \right). It is clear the overshoot is zero if μ = 0, which is the condition: : \frac {4 \beta A_0} { \tau_1 \tau_2} = \left( \frac {1} {\tau_1} - \frac {1} {\tau_2} \right)^2. Comparing the top panel (α = 4) with the lower panel (α = 0.5) shows lower values for α increase the rate of response, but increase overshoot. As an example of this formula, if the settling time condition is tS = 8 τ2. The step response can be described by the following quantities related to its time behavior, *overshoot *rise time *settling time *ringing In the case of linear dynamic systems, much can be inferred about the system from these characteristics. The transient response is not necessarily tied to abrupt events but to any event that affects the equilibrium of the system. If there is a second pole, then as the closed-loop time constant approaches the time constant of the second pole, a two-pole analysis is needed. ===Two-pole amplifiers=== In the case that the open-loop gain has two poles (two time constants, τ1, τ2), the step response is a bit more complicated.

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If a series circuit has a capacitor of $C=0.8 \times 10^{-6} \mathrm{~F}$ and an inductor of $L=0.2 \mathrm{H}$, find the resistance $R$ so that the circuit is critically damped.

In the case of the series RLC circuit, the damping factor is given by :\zeta = \frac{\, R \,}{2} \sqrt{ \frac{C}{\, L \,} \,} = \frac{1}{\ 2 Q\ } ~. For the parallel circuit, the attenuation is given byNilsson and Riedel, p. 286. : \alpha = \frac{1}{\,2\,R\,C\,} and the damping factor is consequently :\zeta = \frac{1}{\,2\,R\,} \sqrt{\frac{L}{C}~}\,~. The value of the damping factor determines the type of transient that the circuit will exhibit.Irwin, pp. 217–220. ===Transient response=== thumb|350px|Plot showing underdamped and overdamped responses of a series RLC circuit to a voltage input step of 1 V. Rearranging for the case where is known – capacitance: : C = \frac{~\alpha + \beta~}{R\,\alpha\,\beta} \,, inductance (total of circuit and load): : L = \frac{R}{\,\alpha + \beta~} \,, initial terminal voltage of capacitor: : V_0 = \frac{\,-I_0 R\,\alpha\,\beta~}{\alpha + \beta} \left(\frac{1}{\beta} - \frac{1}{\alpha}\right) \,. ==See also== *RC circuit *RL circuit *Linear circuit == Footnotes == ==References== ==Bibliography== * * * * * Category:Analog circuits Category:Electronic filter topology For the case of the series RLC circuit these two parameters are given by:Agarwal and Lang, p. 641. :\begin{align} \alpha &= \frac{R}{\, 2L \,} \\\ \omega_0 &= \frac{1}{\, \sqrt{L\,C\,} \,} \;. \end{align} A useful parameter is the damping factor, , which is defined as the ratio of these two; although, sometimes is not used, and is referred to as damping factor instead; hence requiring careful specification of one's use of that term.Agarwal and Lang, p. 646. : \zeta \equiv \frac{\alpha}{\, \omega_0 \,} \;. These equations show that a series RC circuit has a time constant, usually denoted being the time it takes the voltage across the component to either rise (across the capacitor) or fall (across the resistor) to within of its final value. If the inductance is known, then the remaining parameters are given by the following – capacitance: : C = \frac{1}{~L\,\alpha\,\beta\,~} \,, resistance (total of circuit and load): : R = L\,(\,\alpha + \beta\,) \,, initial terminal voltage of capacitor: : V_0 = -I_0 L\,\alpha\,\beta\,\left(\frac{1}{\beta} - \frac{1}{\alpha}\right) \,. Parallel RC, series L circuit with resistance in parallel with the capacitor In the same vein, a resistor in parallel with the capacitor in a series LC circuit can be used to represent a capacitor with a lossy dielectric. This results in the linear differential equation :C\frac{dV}{dt} + \frac{V}{R}=0 \,, where is the capacitance of the capacitor. Series RL, parallel C circuit with resistance in series with the inductor is the standard model for a self-resonant inductor A series resistor with the inductor in a parallel LC circuit as shown in Figure 4 is a topology commonly encountered where there is a need to take into account the resistance of the coil winding and its self- capacitance. thumb|350px|A series RLC network (in order): a resistor, an inductor, and a capacitor An RLC circuit is an electrical circuit consisting of a resistor (R), an inductor (L), and a capacitor (C), connected in series or in parallel. With complex impedances: :\begin{align} I_R &= \frac{V_\mathrm{in}}{R} \\\ I_C &= j\omega C V_\mathrm{in}\,. \end{align} This shows that the capacitor current is 90° out of phase with the resistor (and source) current. The critically damped response represents the circuit response that decays in the fastest possible time without going into oscillation. As a result, : \omega_\mathrm{d} \approx \omega_0 \,. === Voltage multiplier === In a series RLC circuit at resonance, the current is limited only by the resistance of the circuit : I = \frac{V}{R}\,. As a result, \sigma = 0 and the impedance becomes :Z_C = \frac{1}{j\omega C} = - \frac{j}{\omega C} \,. ==Series circuit== By viewing the circuit as a voltage divider, the voltage across the capacitor is: :V_C(s) = \frac{\frac{1}{Cs}}{R + \frac{1}{Cs}}V_\mathrm{in}(s) = \frac{1}{1 + RCs}V_\mathrm{in}(s) and the voltage across the resistor is: :V_R(s) = \frac{R}{R + \frac{1}{Cs}}V_\mathrm{in}(s) = \frac{RCs}{1 + RCs}V_\mathrm{in}(s)\,. ===Transfer functions=== The transfer function from the input voltage to the voltage across the capacitor is :H_C(s) = \frac{ V_C(s) }{ V_\mathrm{in}(s) } = \frac{ 1 }{ 1 + RCs } \,. The current through the resistor must be equal in magnitude (but opposite in sign) to the time derivative of the accumulated charge on the capacitor. Solving this equation for yields the formula for exponential decay: :V(t)=V_0 e^{-\frac{t}{RC}} \,, where is the capacitor voltage at time . A resistor–capacitor circuit (RC circuit), or RC filter or RC network, is an electric circuit composed of resistors and capacitors. The article next gives the analysis for the series RLC circuit in detail. Considering the expression for again, when :R \ll \frac{1}{\omega C}\,, so :\begin{align} I &\approx \frac{V_\mathrm{in}}\frac{1}{j\omega C} \\\ V_\mathrm{in} &\approx \frac{I}{j\omega C} = V_C \,.\end{align} Now, :\begin{align} V_R &= IR = C\frac{dV_C}{dt}R \\\ V_R &\approx RC\frac{dV_{in}}{dt}\,, \end{align} which is a differentiator across the resistor. RLC parallel circuit – the voltage source powering the circuit – the current admitted through the circuit – the equivalent resistance of the combined source, load, and components – the inductance of the inductor component – the capacitance of the capacitor component The properties of the parallel RLC circuit can be obtained from the duality relationship of electrical circuits and considering that the parallel RLC is the dual impedance of a series RLC. This article considers the RC circuit, in both series and parallel forms, as shown in the diagrams below. == Natural response == 200px|thumb|right| RC circuit The simplest RC circuit consists of a resistor and a charged capacitor connected to one another in a single loop, without an external voltage source.

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If $y_1$ and $y_2$ are a fundamental set of solutions of $t y^{\prime \prime}+2 y^{\prime}+t e^t y=0$ and if $W\left(y_1, y_2\right)(1)=2$, find the value of $W\left(y_1, y_2\right)(5)$.

In classical mechanics, a Liouville dynamical system is an exactly solvable dynamical system in which the kinetic energy T and potential energy V can be expressed in terms of the s generalized coordinates q as follows: : T = \frac{1}{2} \left\\{ u_{1}(q_{1}) + u_{2}(q_{2}) + \cdots + u_{s}(q_{s}) \right\\} \left\\{ v_{1}(q_{1}) \dot{q}_{1}^{2} + v_{2}(q_{2}) \dot{q}_{2}^{2} + \cdots + v_{s}(q_{s}) \dot{q}_{s}^{2} \right\\} : V = \frac{w_{1}(q_{1}) + w_{2}(q_{2}) + \cdots + w_{s}(q_{s}) }{u_{1}(q_{1}) + u_{2}(q_{2}) + \cdots + u_{s}(q_{s}) } The solution of this system consists of a set of separably integrable equations : \frac{\sqrt{2}}{Y}\, dt = \frac{d\varphi_{1}}{\sqrt{E \chi_{1} - \omega_{1} + \gamma_{1}}} = \frac{d\varphi_{2}}{\sqrt{E \chi_{2} - \omega_{2} + \gamma_{2}}} = \cdots = \frac{d\varphi_{s}}{\sqrt{E \chi_{s} - \omega_{s} + \gamma_{s}}} where E = T + V is the conserved energy and the \gamma_{s} are constants. This is a Liouville dynamical system if ξ and η are taken as φ1 and φ2, respectively; thus, the function Y equals : Y = \cosh^{2} \xi - \cos^{2} \eta and the function W equals : W = -\mu_{1} \left( \cosh \xi + \cos \eta \right) - \mu_{2} \left( \cosh \xi - \cos \eta \right) Using the general solution for a Liouville dynamical system below, one obtains : \frac{ma^{2}}{2} \left( \cosh^{2} \xi - \cos^{2} \eta \right)^{2} \dot{\xi}^{2} = E \cosh^{2} \xi + \left( \frac{\mu_{1} + \mu_{2}}{a} \right) \cosh \xi - \gamma : \frac{ma^{2}}{2} \left( \cosh^{2} \xi - \cos^{2} \eta \right)^{2} \dot{\eta}^{2} = -E \cos^{2} \eta + \left( \frac{\mu_{1} - \mu_{2}}{a} \right) \cos \eta + \gamma Introducing a parameter u by the formula : du = \frac{d\xi}{\sqrt{E \cosh^{2} \xi + \left( \frac{\mu_{1} + \mu_{2}}{a} \right) \cosh \xi - \gamma}} = \frac{d\eta}{\sqrt{-E \cos^{2} \eta + \left( \frac{\mu_{1} - \mu_{2}}{a} \right) \cos \eta + \gamma}}, gives the parametric solution : u = \int \frac{d\xi}{\sqrt{E \cosh^{2} \xi + \left( \frac{\mu_{1} + \mu_{2}}{a} \right) \cosh \xi - \gamma}} = \int \frac{d\eta}{\sqrt{-E \cos^{2} \eta + \left( \frac{\mu_{1} - \mu_{2}}{a} \right) \cos \eta + \gamma}}. Multiplying both sides by 2 Y \dot{\varphi}_{r}, re-arranging, and exploiting the relation 2T = YF yields the equation : 2 Y \dot{\varphi}_{r} \frac{d}{dt} \left(Y \dot{\varphi}_{r}\right) = 2T\dot{\varphi}_{r} \frac{\partial Y}{\partial \varphi_{r}} - 2 Y \dot{\varphi}_{r} \frac{\partial V}{\partial \varphi_{r}} = 2 \dot{\varphi}_{r} \frac{\partial}{\partial \varphi_{r}} \left[ (E-V) Y \right], which may be written as : \frac{d}{dt} \left(Y^{2} \dot{\varphi}_{r}^{2} \right) = 2 E \dot{\varphi}_{r} \frac{\partial Y}{\partial \varphi_{r}} - 2 \dot{\varphi}_{r} \frac{\partial W}{\partial \varphi_{r}} = 2E \dot{\varphi}_{r} \frac{d\chi_{r} }{d\varphi_{r}} - 2 \dot{\varphi}_{r} \frac{d\omega_{r}}{d\varphi_{r}}, where E = T + V is the (conserved) total energy. The Battle of the Ypres–Comines Canal was a battle of the Second World War fought between the British Expeditionary Force (BEF) and German Army Group B during the BEF's retreat to Dunkirk in 1940. In scientific computation and simulation, the method of fundamental solutions (MFS) is a technique for solving partial differential equations based on using the fundamental solution as a basis function. Inverting, taking the square root and separating the variables yields a set of separably integrable equations: : \frac{\sqrt{2}}{Y} dt = \frac{d\varphi_{1}}{\sqrt{E \chi_{1} - \omega_{1} + \gamma_{1}}} = \frac{d\varphi_{2}}{\sqrt{E \chi_{2} - \omega_{2} + \gamma_{2}}} = \cdots = \frac{d\varphi_{s}}{\sqrt{E \chi_{s} - \omega_{s} + \gamma_{s}}}. ==References== ==Further reading== * Category:Classical mechanics Since these are elliptic integrals, the coordinates ξ and η can be expressed as elliptic functions of u. ===Constant of motion=== The bicentric problem has a constant of motion, namely, : r_{1}^{2} r_{2}^{2} \left( \frac{d\theta_{1}}{dt} \right) \left( \frac{d\theta_{2}}{dt} \right) - 2c \left[ \mu_{1} \cos \theta_{1} + \mu_{2} \cos \theta_{2} \right], from which the problem can be solved using the method of the last multiplier. ==Derivation== ===New variables=== To eliminate the v functions, the variables are changed to an equivalent set : \varphi_{r} = \int dq_{r} \sqrt{v_{r}(q_{r})}, giving the relation : v_{1}(q_{1}) \dot{q}_{1}^{2} + v_{2}(q_{2}) \dot{q}_{2}^{2} + \cdots + v_{s}(q_{s}) \dot{q}_{s}^{2} = \dot{\varphi}_{1}^{2} + \dot{\varphi}_{2}^{2} + \cdots + \dot{\varphi}_{s}^{2} = F, which defines a new variable F. Using the new variables, the u and w functions can be expressed by equivalent functions χ and ω. The Second Anglo-Sikh War was a military conflict between the Sikh Empire and the East India Company that took place in 1848 and 1849. Similarly, denoting the sum of the ω functions by W : W = \omega_{1}(\varphi_{1}) + \omega_{2}(\varphi_{2}) + \cdots + \omega_{s}(\varphi_{s}), the potential energy V can be written as : V = \frac{W}{Y}. ===Lagrange equation=== The Lagrange equation for the rth variable \varphi_{r} is : \frac{d}{dt} \left( \frac{\partial T}{\partial \dot{\varphi}_{r}} \right) = \frac{d}{dt} \left( Y \dot{\varphi}_{r} \right) = \frac{1}{2} F \frac{\partial Y}{\partial \varphi_{r}} -\frac{\partial V}{\partial \varphi_{r}}. It follows that : \frac{d}{dt} \left(Y^{2} \dot{\varphi}_{r}^{2} \right) = 2\frac{d}{dt} \left( E \chi_{r} - \omega_{r} \right), which may be integrated once to yield : \frac{1}{2} Y^{2} \dot{\varphi}_{r}^{2} = E \chi_{r} - \omega_{r} + \gamma_{r}, where the \gamma_{r} are constants of integration subject to the energy conservation : \sum_{r=1}^{s} \gamma_{r} = 0. Later developments indicated that the MFS can be used to solve partial differential equations with variable coefficients.C.M. Fan, C.S. Chen, J. Monroe, The method of fundamental solutions for solving convection- diffusion equations with variable coefficients, Advances in Applied Mathematics and Mechanics. 1 (2009) 215–230 The MFS has proved particularly effective for certain classes of problems such as inverse,Y.C. Hon, T. Wei, The method of fundamental solution for solving multidimensional inverse heat conduction problems, CMES Comput. A differential equation can be homogeneous in either of two respects. Introducing elliptic coordinates, : x = a \cosh \xi \cos \eta, : y = a \sinh \xi \sin \eta, the potential energy can be written as : V(\xi, \eta) = \frac{-\mu_{1}}{a\left( \cosh \xi - \cos \eta \right)} - \frac{\mu_{2}}{a\left( \cosh \xi + \cos \eta \right)} = \frac{-\mu_{1} \left( \cosh \xi + \cos \eta \right) - \mu_{2} \left( \cosh \xi - \cos \eta \right)}{a\left( \cosh^{2} \xi - \cos^{2} \eta \right)}, and the kinetic energy as : T = \frac{ma^{2}}{2} \left( \cosh^{2} \xi - \cos^{2} \eta \right) \left( \dot{\xi}^{2} + \dot{\eta}^{2} \right). Terso Solutions, Inc., located in Madison, Wisconsin, USA, is the developer and distributor of an automated system for storage and distribution of high value research reagents and medical supplies. It follows that, if is a solution, so is , for any (non-zero) constant . Denoting the sum of the χ functions by Y, : Y = \chi_{1}(\varphi_{1}) + \chi_{2}(\varphi_{2}) + \cdots + \chi_{s}(\varphi_{s}), the kinetic energy can be written as : T = \frac{1}{2} Y F. Sci. 7 (2005) 119–132 unbounded domain, and free-boundary problems.A.K. G. Fairweather, The method of fundamental solutions for elliptic boundary value problems, Advances in Computational Mathematics. 9 (1998) 69–95. Retrieved on July 7, 2008. thumb|right|300px|Terso Solutions, Madison, WI, USA. == History == Developed initially as an on-site inventory supplier for Promega products, the privately held Terso Solutions, Inc. was spun off from Promega Corporation in 2005. As described below, the variables have been changed from qs to φs, and the functions us and ws substituted by their counterparts χs and ωs. However, the method was first proposed as a computational technique much later by R. Mathon and R. L. Johnston in the late 1970s,R. Mathon, R.L. Johnston, The approximate solution of elliptic boundary-value problems by fundamental solutions, SIAM Journal on Numerical Analysis. (1977) 638–650. followed by a number of papers by Mathon, Johnston and Graeme Fairweather with applications. The MFS then gradually became a useful tool for the solution of a large variety of physical and engineering problems.Z. Fu, W. Chen, W. Yang, Winkler plate bending problems by a truly boundary-only boundary particle method, Computational Mechanics. 44 (2009) 757–763.W. Chen, J. Lin, F. Wang, Regularized meshless method for nonhomogeneous problems , Engineering Analysis with Boundary Elements. 35 (2011) 253–257.W. Chen, F.Z. Wang, A method of fundamental solutions without fictitious boundary , Engineering Analysis with Boundary Elements. 34 (2010) 530–532.JIANG Xin-rong, CHEN Wen, Method of fundamental solution and boundary knot method for helmholtz equations: a comparative study, Chinese Journal of Computational Mechanics, 28:3(2011) 338–344 (in Chinese) In the 1990s, M. A. Golberg and C. S. Chen extended the MFS to deal with inhomogeneous equations and time-dependent problems, greatly expanding its applicability.M.A. Golberg, C.S. Chen, The theory of radial basis functions applied to the BEM for inhomogeneous partial differential equations, Boundary Elements Communications. 5 (1994) 57–61.M. a. Golberg, C.S. Chen, H. Bowman, H. Power, Some comments on the use of Radial Basis Functions in the Dual Reciprocity Method, Computational Mechanics. 21 (1998) 141–148.

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Consider the initial value problem $$ 5 u^{\prime \prime}+2 u^{\prime}+7 u=0, \quad u(0)=2, \quad u^{\prime}(0)=1 $$ Find the smallest $T$ such that $|u(t)| \leq 0.1$ for all $t>T$.

A solution to an initial value problem is a function y that is a solution to the differential equation and satisfies :y(t_0) = y_0. ;Second example The solution of : y'+3y=6t+5,\qquad y(0)=3 can be found to be : y(t)=2e^{-3t}+2t+1. In continuous time, the problem of finding a closed form solution for the state variables as a function of time and of the initial conditions is called the initial value problem. \, Indeed, : \begin{align} y'+3y &= \tfrac{d}{dt} (2e^{-3t}+2t+1)+3(2e^{-3t}+2t+1) \\\ &= (-6e^{-3t}+2)+(6e^{-3t}+6t+3) \\\ &= 6t+5. \end{align} ==Notes== ==See also== * Boundary value problem * Constant of integration * Integral curve == References == * * * * * * Category:Boundary conditions el:Αρχική τιμή it:Problema ai valori iniziali sv:Begynnelsevärdesproblem Assuming W_c is nonsingular (if and only if the system is controllable), the minimum energy control is then : u(t) = -B^*e^{A^*(t_1-t)}W_c^{-1}(t_1)[e^{A(t_1-t_0)}x_0-x_1]. Rearrange the equation so that y is on the left hand side : \frac{y'(t)}{y(t)} = 0.85 Now integrate both sides with respect to t (this introduces an unknown constant B). : \int \frac{y'(t)}{y(t)}\,dt = \int 0.85\,dt : \ln |y(t)| = 0.85t + B Eliminate the logarithm with exponentiation on both sides : | y(t) | = e^Be^{0.85t} Let C be a new unknown constant, C = \pm e^B, so : y(t) = Ce^{0.85t} Now we need to find a value for C. Use y(0) = 19 as given at the start and substitute 0 for t and 19 for y : 19 = C e^{0.85 \cdot 0} : C = 19 this gives the final solution of y(t) = 19e^{0.85t}. Modeling a system in physics or other sciences frequently amounts to solving an initial value problem. Initial value problems are extended to higher orders by treating the derivatives in the same way as an independent function, e.g. y(t)=f(t,y(t),y'(t)). == Existence and uniqueness of solutions == The Picard–Lindelöf theorem guarantees a unique solution on some interval containing t0 if f is continuous on a region containing t0 and y0 and satisfies the Lipschitz condition on the variable y. In the second equation, the derivative at u = 2 should be taken as u approaches 2 from the right. In mathematics and particularly in dynamic systems, an initial condition, in some contexts called a seed value, is a value of an evolving variable at some point in time designated as the initial time (typically denoted t = 0). Also, \omega(u)-e^{-\gamma} oscillates in a regular way, alternating between extrema and zeroes; the extrema alternate between positive maxima and negative minima. Here the number of initial conditions necessary for obtaining a closed form solution is the dimension n = 1 times the order k, or simply k. In multivariable calculus, an initial value problem (IVP) is an ordinary differential equation together with an initial condition which specifies the value of the unknown function at a given point in the domain. In control theory, the minimum energy control is the control u(t) that will bring a linear time invariant system to a desired state with a minimum expenditure of energy. Hiroshi Okamura obtained a necessary and sufficient condition for the solution of an initial value problem to be unique. The interval between consecutive extrema approaches 1 as u approaches infinity, as does the interval between consecutive zeroes.p. 131, Cheer and Goldston 1990. ==Applications== The Buchstab function is used to count rough numbers. Here the dimension is n = 1 and the order is k, so the necessary number of initial conditions to trace the system through time, either iteratively or via closed form solution, is nk = k. 300px|thumbnail|Graph of the Buchstab function ω(u) from u = 1 to u = 4\. We are trying to find a formula for y(t) that satisfies these two equations. The Banach fixed point theorem is then invoked to show that there exists a unique fixed point, which is the solution of the initial value problem. One seeks an input u(t) so that the system will be in the state x_1 at time t_1, and for any other input \bar{u}(t), which also drives the system from x_0 to x_1 at time t_1, the energy expenditure would be larger, i.e., : \int_{t_0}^{t_1} \bar{u}^*(t) \bar{u}(t) dt \ \geq \ \int_{t_0}^{t_1} u^*(t) u(t) dt. An older proof of the Picard–Lindelöf theorem constructs a sequence of functions which converge to the solution of the integral equation, and thus, the solution of the initial value problem.

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Consider the initial value problem $$ y^{\prime}=t y(4-y) / 3, \quad y(0)=y_0 $$ Suppose that $y_0=0.5$. Find the time $T$ at which the solution first reaches the value 3.98.

A solution to an initial value problem is a function y that is a solution to the differential equation and satisfies :y(t_0) = y_0. Rearrange the equation so that y is on the left hand side : \frac{y'(t)}{y(t)} = 0.85 Now integrate both sides with respect to t (this introduces an unknown constant B). : \int \frac{y'(t)}{y(t)}\,dt = \int 0.85\,dt : \ln |y(t)| = 0.85t + B Eliminate the logarithm with exponentiation on both sides : | y(t) | = e^Be^{0.85t} Let C be a new unknown constant, C = \pm e^B, so : y(t) = Ce^{0.85t} Now we need to find a value for C. Use y(0) = 19 as given at the start and substitute 0 for t and 19 for y : 19 = C e^{0.85 \cdot 0} : C = 19 this gives the final solution of y(t) = 19e^{0.85t}. Indeed, rather than being unique, this equation has three solutions:, p. 7 :y(t) = 0, \qquad y(t) = \pm\left (\tfrac23 t\right)^{\frac{3}{2}}. Since the equation being studied is a first-order equation, the initial conditions are the initial x and y values. In continuous time, the problem of finding a closed form solution for the state variables as a function of time and of the initial conditions is called the initial value problem. We are trying to find a formula for y(t) that satisfies these two equations. ;Second example The solution of : y'+3y=6t+5,\qquad y(0)=3 can be found to be : y(t)=2e^{-3t}+2t+1. Then, there exists some such that the initial value problem y'(t)=f(t,y(t)),\qquad y(t_0)=y_0. has a unique solution y(t) on the interval [t_0-\varepsilon, t_0+\varepsilon]., Theorem I.3.1 == Proof sketch == The proof relies on transforming the differential equation, and applying the Banach fixed-point theorem. This happens for example for the equation , which has at least two solutions corresponding to the initial condition such as: or :y(t)=\begin{cases} \left (\tfrac{at}{3} \right )^{3} & t<0\\\ \ \ \ \ 0 & t \ge 0, \end{cases} so the previous state of the system is not uniquely determined by its state after t = 0. In that context, the differential initial value is an equation which specifies how the system evolves with time given the initial conditions of the problem. == Definition == An initial value problem is a differential equation :y'(t) = f(t, y(t)) with f\colon \Omega \subset \mathbb{R} \times \mathbb{R}^n \to \mathbb{R}^n where \Omega is an open set of \mathbb{R} \times \mathbb{R}^n, together with a point in the domain of f :(t_0, y_0) \in \Omega, called the initial condition. Beginning with another initial condition , the solution y(t) tends toward the stationary point, but reaches it only at the limit of infinite time, so the uniqueness of solutions (over all finite times) is guaranteed. Initial value problems are extended to higher orders by treating the derivatives in the same way as an independent function, e.g. y(t)=f(t,y(t),y'(t)). == Existence and uniqueness of solutions == The Picard–Lindelöf theorem guarantees a unique solution on some interval containing t0 if f is continuous on a region containing t0 and y0 and satisfies the Lipschitz condition on the variable y. In multivariable calculus, an initial value problem (IVP) is an ordinary differential equation together with an initial condition which specifies the value of the unknown function at a given point in the domain. By integrating both sides, any function satisfying the differential equation must also satisfy the integral equation :y(t) - y(t_0) = \int_{t_0}^t f(s,y(s)) \, ds. In both differential equations in continuous time and difference equations in discrete time, initial conditions affect the value of the dynamic variables (state variables) at any future time. In mathematics, specifically the study of differential equations, the Picard–Lindelöf theorem gives a set of conditions under which an initial value problem has a unique solution. Modeling a system in physics or other sciences frequently amounts to solving an initial value problem. A singular solution ys(x) of an ordinary differential equation is a solution that is singular or one for which the initial value problem (also called the Cauchy problem by some authors) fails to have a unique solution at some point on the solution. Its behavior through time can be traced with a closed form solution conditional on an initial condition vector X_0. See Newton's method of successive approximation for instruction. == Example of Picard iteration == thumb|Four Picard iteration steps and their limit Let y(t)=\tan(t), the solution to the equation y'(t)=1+y(t)^2 with initial condition y(t_0)=y_0=0,t_0=0. In mathematics and particularly in dynamic systems, an initial condition, in some contexts called a seed value, is a value of an evolving variable at some point in time designated as the initial time (typically denoted t = 0). This large class of methods from numerical analysis is based on the exact integration of the linear part of the initial value problem.

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25. Consider the initial value problem $$ 2 y^{\prime \prime}+3 y^{\prime}-2 y=0, \quad y(0)=1, \quad y^{\prime}(0)=-\beta, $$ where $\beta>0$. Find the smallest value of $\beta$ for which the solution has no minimum point.

Although the first derivative (3x2) is 0 at x = 0, this is an inflection point. (2nd derivative is 0 at that point.) \sqrt[x]{x} Unique global maximum at x = e. (See figure at top of page.) x3 \+ 3x2 − 2x + 1 defined over the closed interval (segment) [−4,2] Local maximum at x = −1−/3, local minimum at x = −1+/3, global maximum at x = 2 and global minimum at x = −4. This is illustrated by the function :f(x,y)= x^2+y^2(1-x)^3,\qquad x,y \in \R, whose only critical point is at (0,0), which is a local minimum with f(0,0) = 0\. A solution to an initial value problem is a function y that is a solution to the differential equation and satisfies :y(t_0) = y_0. For example, if a bounded differentiable function f defined on a closed interval in the real line has a single critical point, which is a local minimum, then it is also a global minimum (use the intermediate value theorem and Rolle's theorem to prove this by contradiction). ;Second example The solution of : y'+3y=6t+5,\qquad y(0)=3 can be found to be : y(t)=2e^{-3t}+2t+1. \, Indeed, : \begin{align} y'+3y &= \tfrac{d}{dt} (2e^{-3t}+2t+1)+3(2e^{-3t}+2t+1) \\\ &= (-6e^{-3t}+2)+(6e^{-3t}+6t+3) \\\ &= 6t+5. \end{align} ==Notes== ==See also== * Boundary value problem * Constant of integration * Integral curve == References == * * * * * * Category:Boundary conditions el:Αρχική τιμή it:Problema ai valori iniziali sv:Begynnelsevärdesproblem A function need not have a least fixed point, but if it does then the least fixed point is unique. ==Examples== With the usual order on the real numbers, the least fixed point of the real function f(x) = x2 is x = 0 (since the only other fixed point is 1 and 0 < 1). From the sign of the second derivative, we can see that −1 is a local maximum and +1 is a local minimum. Then the second derivative test provides a sufficient condition for the point to be a local maximum or local minimum. ==Search techniques== Local search or hill climbing methods for solving optimization problems start from an initial configuration and repeatedly move to an improving neighboring configuration. In multivariable calculus, an initial value problem (IVP) is an ordinary differential equation together with an initial condition which specifies the value of the unknown function at a given point in the domain. If the first derivative exists everywhere, it can be equated to zero; if the function has an unbounded domain, for a point to be a local optimum it is necessary that it satisfy this equation. The definition of global minimum point also proceeds similarly. This function has no global maximum or minimum. x| Global minimum at x = 0 that cannot be found by taking derivatives, because the derivative does not exist at x = 0. cos(x) Infinitely many global maxima at 0, ±2, ±4, ..., and infinitely many global minima at ±, ±3, ±5, .... 2 cos(x) − x Infinitely many local maxima and minima, but no global maximum or minimum. cos(3x)/x with Global maximum at x = 0.1 (a boundary), a global minimum near x = 0.3, a local maximum near x = 0.6, and a local minimum near x = 1.0. Initial value problems are extended to higher orders by treating the derivatives in the same way as an independent function, e.g. y(t)=f(t,y(t),y'(t)). == Existence and uniqueness of solutions == The Picard–Lindelöf theorem guarantees a unique solution on some interval containing t0 if f is continuous on a region containing t0 and y0 and satisfies the Lipschitz condition on the variable y. Setting the first derivative to 0 and solving for x gives stationary points at −1 and +1. Rearrange the equation so that y is on the left hand side : \frac{y'(t)}{y(t)} = 0.85 Now integrate both sides with respect to t (this introduces an unknown constant B). : \int \frac{y'(t)}{y(t)}\,dt = \int 0.85\,dt : \ln |y(t)| = 0.85t + B Eliminate the logarithm with exponentiation on both sides : | y(t) | = e^Be^{0.85t} Let C be a new unknown constant, C = \pm e^B, so : y(t) = Ce^{0.85t} Now we need to find a value for C. Use y(0) = 19 as given at the start and substitute 0 for t and 19 for y : 19 = C e^{0.85 \cdot 0} : C = 19 this gives the final solution of y(t) = 19e^{0.85t}. thumb|right|220px|Attraction basins around locally optimal points thumb|right|233px|Polynomial of degree 4: the trough on the right is a local minimum and the one on the left is the global minimum. (See figure at right) x−x Unique global maximum over the positive real numbers at x = 1/e. x3/3 − x First derivative x2 − 1 and second derivative 2x. One can distinguish whether a critical point is a local maximum or local minimum by using the first derivative test, second derivative test, or higher-order derivative test, given sufficient differentiability. thumb|150px|The function f(x) = x2 − 4 has two fixed points, shown as the intersection with the blue line; its least one is at 1/2 − /2. However, not all critical points are extrema.

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Consider the initial value problem $$ y^{\prime}=t y(4-y) /(1+t), \quad y(0)=y_0>0 . $$ If $y_0=2$, find the time $T$ at which the solution first reaches the value 3.99.

A solution to an initial value problem is a function y that is a solution to the differential equation and satisfies :y(t_0) = y_0. Rearrange the equation so that y is on the left hand side : \frac{y'(t)}{y(t)} = 0.85 Now integrate both sides with respect to t (this introduces an unknown constant B). : \int \frac{y'(t)}{y(t)}\,dt = \int 0.85\,dt : \ln |y(t)| = 0.85t + B Eliminate the logarithm with exponentiation on both sides : | y(t) | = e^Be^{0.85t} Let C be a new unknown constant, C = \pm e^B, so : y(t) = Ce^{0.85t} Now we need to find a value for C. Use y(0) = 19 as given at the start and substitute 0 for t and 19 for y : 19 = C e^{0.85 \cdot 0} : C = 19 this gives the final solution of y(t) = 19e^{0.85t}. In continuous time, the problem of finding a closed form solution for the state variables as a function of time and of the initial conditions is called the initial value problem. We are trying to find a formula for y(t) that satisfies these two equations. Initial value problems are extended to higher orders by treating the derivatives in the same way as an independent function, e.g. y(t)=f(t,y(t),y'(t)). == Existence and uniqueness of solutions == The Picard–Lindelöf theorem guarantees a unique solution on some interval containing t0 if f is continuous on a region containing t0 and y0 and satisfies the Lipschitz condition on the variable y. right|thumb|390px|Solutions to the differential equation \frac{dy}{dx} = \frac{1}{2y} subject to the initial conditions y(0)=0, 1 and 2 (red, green and blue curves respectively). In multivariable calculus, an initial value problem (IVP) is an ordinary differential equation together with an initial condition which specifies the value of the unknown function at a given point in the domain. In that context, the differential initial value is an equation which specifies how the system evolves with time given the initial conditions of the problem. == Definition == An initial value problem is a differential equation :y'(t) = f(t, y(t)) with f\colon \Omega \subset \mathbb{R} \times \mathbb{R}^n \to \mathbb{R}^n where \Omega is an open set of \mathbb{R} \times \mathbb{R}^n, together with a point in the domain of f :(t_0, y_0) \in \Omega, called the initial condition. ;Second example The solution of : y'+3y=6t+5,\qquad y(0)=3 can be found to be : y(t)=2e^{-3t}+2t+1. In both differential equations in continuous time and difference equations in discrete time, initial conditions affect the value of the dynamic variables (state variables) at any future time. Modeling a system in physics or other sciences frequently amounts to solving an initial value problem. In mathematics and particularly in dynamic systems, an initial condition, in some contexts called a seed value, is a value of an evolving variable at some point in time designated as the initial time (typically denoted t = 0). Its behavior through time can be traced with a closed form solution conditional on an initial condition vector X_0. In the second equation, the derivative at u = 2 should be taken as u approaches 2 from the right. This large class of methods from numerical analysis is based on the exact integration of the linear part of the initial value problem. \, Indeed, : \begin{align} y'+3y &= \tfrac{d}{dt} (2e^{-3t}+2t+1)+3(2e^{-3t}+2t+1) \\\ &= (-6e^{-3t}+2)+(6e^{-3t}+6t+3) \\\ &= 6t+5. \end{align} ==Notes== ==See also== * Boundary value problem * Constant of integration * Integral curve == References == * * * * * * Category:Boundary conditions el:Αρχική τιμή it:Problema ai valori iniziali sv:Begynnelsevärdesproblem While a closed form solution is not always possible to obtain, future values of a discrete time system can be found by iterating forward one time period per iteration, though rounding error may make this impractical over long horizons. ==Linear system== ===Discrete time=== A linear matrix difference equation of the homogeneous (having no constant term) form X_{t+1}=AX_t has closed form solution X_t=A^tX_0 predicated on the vector X_0 of initial conditions on the individual variables that are stacked into the vector; X_0 is called the vector of initial conditions or simply the initial condition, and contains nk pieces of information, n being the dimension of the vector X and k = 1 being the number of time lags in the system. We use their notation, and assume that the unknown function is u, and that we have a known solution u_n at time t_n. Exponential integrators are a class of numerical methods for the solution of ordinary differential equations, specifically initial value problems. Here the constants c_1, \dots , c_k are found by solving a system of k different equations based on this equation, each using one of k different values of t for which the specific initial condition x_t Is known. ===Continuous time=== A differential equation system of the first order with n variables stacked in a vector X is :\frac{dX}{dt}=AX. These problems can come from a more typical initial value problem :u'(t) = f(u(t)), \qquad u(t_0)=u_0, after linearizing locally about a fixed or local state u^*: : L = \frac{\partial f}{\partial u}(u^*); \qquad N = f(u) - L u. The characteristic equation of this dynamic equation is \lambda^k+a_{k-1}\lambda^{k-1}+\cdots +a_1\lambda +a_0=0, whose solutions are the characteristic values \lambda_1,\dots , \lambda_k; these are used in the solution equation :x(t)=c_1e^{\lambda_1t}+\cdots + c_ke^{\lambda_kt}.

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28. A mass of $0.25 \mathrm{~kg}$ is dropped from rest in a medium offering a resistance of $0.2|v|$, where $v$ is measured in $\mathrm{m} / \mathrm{s}$. If the mass is to attain a velocity of no more than $10 \mathrm{~m} / \mathrm{s}$, find the maximum height from which it can be dropped.

If the falling object is spherical in shape, the expression for the three forces are given below: where *d is the diameter of the spherical object, *g is the gravitational acceleration, *\rho is the density of the fluid, *\rho_s is the density of the object, *A = \frac{1}{4} \pi d^2 is the projected area of the sphere, *C_d is the drag coefficient, and *V is the characteristic velocity (taken as terminal velocity, V_t ). Some heights are difficult to verify due to lack of documentation and are approximated. ==List of the highest falls survived without a parachute== Name Image Distance of fall Date Notes and References Feet Meters Vesna Vulović 33,330 10160 1972 Flight attendant from Serbia who was the sole survivor of an airplane bombing mid-air. The mass m0 used in the fall is 80 kg. The same terminal speed is reached for a typical .30-06 bullet dropping downwards—when it is returning to the ground having been fired upwards, or dropped from a tower—according to a 1920 U.S. Army Ordnance study. the current record is held by Felix Baumgartner who jumped from an altitude of and reached , though he achieved this speed at high altitude where the density of the air is much lower than at the Earth's surface, producing a correspondingly lower drag force. thumb|250px|The climber will fall about the same height h in both cases, but they will be subjected to a greater force at position 1, due to the greater fall factor. The biologist J. B. S. Haldane wrote, ==Physics== Using mathematical terms, terminal speed—without considering buoyancy effects—is given by V_t= \sqrt\frac{2 m g}{\rho A C_d} where *V_t represents terminal velocity, *m is the mass of the falling object, *g is the acceleration due to gravity, *C_d is the drag coefficient, *\rho is the density of the fluid through which the object is falling, and *A is the projected area of the object. This article attempts to list of the highest falls survived without a parachute. We first state an equation for this quantity and describe its interpretation, and then show its derivation and how it can be put into a more convenient form. ===Equation for the impact force and its interpretation=== When modeling the rope as an undamped harmonic oscillator (HO) the impact force Fmax in the rope is given by: :F_{max} = mg + \sqrt{(mg)^2 + 2mghk}, where mg is the climber's weight, h is the fall height and k is the spring constant of the portion of the rope that is in play. thumb|The aftermath of a hypervelocity impact, with a projectile the same size as the one that impacted for scale Hypervelocity is very high velocity, approximately over 3,000 meters per second (6,700 mph, 11,000 km/h, 10,000 ft/s, or Mach 8.8). When climbing from the ground up, the maximum possible fall factor is 1, since any greater fall would mean that the climber hit the ground. Solving for Vt yields Derivation of the solution for the velocity v as a function of time t The drag equation is—assuming ρ, g and Cd to be constants: m a = m \frac{\mathrm{d}v}{\mathrm{d}t} = m g - \frac{1}{2} \rho v^2 A C_d. After reaching the local terminal velocity, while continuing the fall, speed decreases to change with the local terminal speed. ===Derivation for terminal velocity=== Using mathematical terms, defining down to be positive, the net force acting on an object falling near the surface of Earth is (according to the drag equation): F_\text{net} = m a = m g - \frac{1}{2} \rho v^2 A C_d, with v(t) the velocity of the object as a function of time t. For objects falling through the atmosphere, for every of fall, the terminal speed decreases 1%. The amount by which the liquid rises in the cylinder (∆V) is equal to the volume of the object. The terminal velocity in such cases will have a negative value, corresponding to the rate of rising up. ==See also== * Stokes's law * Terminal ballistics ==References== ==External links== *Terminal Velocity - NASA site *Onboard video of Space Shuttle Solid Rocket Boosters rapidly decelerating to terminal velocity on entry to the thicker atmosphere, from at 5:15 in the video, to 220 mph at 6:45 when the parachutes are deployed 90 seconds later—NASA video and sound, @ io9.com. *Terminal settling velocity of a sphere at all realistic Reynolds Numbers, by Heywood Tables approach. The mass of the displaced fluid can be expressed in terms of the density and its volume, . We will see below that when varying the height of the fall while keeping the fall factor fixed, the quantity hk stays constant. At some speed, the drag or force of resistance will equal the gravitational pull on the object (buoyancy is considered below). Terminal velocity is the maximum velocity (speed) attainable by an object as it falls through a fluid (air is the most common example). As a numerical example, consider a fall of 20 feet that occurs with 10 feet of rope out (i.e., the climber has placed no protection and falls from 10 feet above the belayer to 10 feet below—a factor 2 fall). This fall produces far more force on the climber and the gear than if a similar 20 foot fall had occurred 100 feet above the belayer. Therefore, the weight of the displaced fluid can be expressed as .

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A home buyer can afford to spend no more than $\$ 800$ /month on mortgage payments. Suppose that the interest rate is $9 \%$ and that the term of the mortgage is 20 years. Assume that interest is compounded continuously and that payments are also made continuously. Determine the maximum amount that this buyer can afford to borrow.

Mortgage calculators can be used to answer such questions as: If one borrows $250,000 at a 7% annual interest rate and pays the loan back over thirty years, with $3,000 annual property tax payment, $1,500 annual property insurance cost and 0.5% annual private mortgage insurance payment, what will the monthly payment be? "A capital budgeting model of the supply and demand of loanable funds", Journal of Macroeconomics 12, Summer 1990, pp. 427-436 (specifically p. 430). = \frac {rP(1+r)^N}{(1+r)^N-1}, & r e 0; \\\ \frac{P}{N}, & r = 0. \end{cases} }} For example, for a home loan of $200,000 with a fixed yearly interest rate of 6.5% for 30 years, the principal is P=200000, the monthly interest rate is r=0.065/12, the number of monthly payments is N=30\cdot 12=360, the fixed monthly payment equals $1,264.14. If in practice the loan is being repaid via monthly instalments, the decimal portion can be converted to months and rounded so this answer would equate to 172 months. ==Calculation of interest rate== In the discrete time interval model, calculation of a mortgage based interest rate given the remaining parameters has not been possible using analytic methods. For a 24-month loan, the denominator is 300. Finally the model demonstrates that it is to the modest advantage of the mortgage holder to increase frequency of payment where practically possible. ==Summary of formulae and online calculators== Annual payment rate (mortgage loan): M_a=\lim_{N\to\infty}N\cdot x(N)=\frac{P_0\cdot r}{1 - e^{-rT}} Annual payment rate (sinking fund): M_a=\lim_{N\to\infty}N\cdot x(N)=\frac{P_T\cdot r}{e^{rT}-1} Future value: F_v(t) = \frac{M_a}{r}(e^{rt}-1) Present value: P_v(t) = \frac{M_a}{r}(1 - e^{-rt}) Loan balance: P(t) = \frac{M_a}{r}(1 - e^{-r(T-t)}) Loan period: T=-\frac{1}{r}\ln\left(1-\frac{P_0 r}{M_a}\right) Half- life of loan: t_{\frac{1}{2}}=\frac{1}{r}\ln\left(\frac{1+e^{rT}}{2}\right) Interest rate: r\approx\frac{2}{T}\ln{\frac{M_aT}{P_0}} r=\frac{1}{T}\left (W(-se^{-s})+s\right )\text{ with }s=\frac{M_at}{P_0} Universal mortgage calculator. Each payment accumulates compound interest from time of deposit to the end of the mortgage timespan at which point the sum of the payments with their accumulated interest equals the value of the loan with interest compounded over the entire timespan. Figure 1 Month Numerator Denominator Percentage of total interest Monthly interest 1 12 78 15.4% $77.00 2 11 78 14.1% $70.50 3 10 78 12.8% $64.00 4 9 78 11.5% $57.50 5 8 78 10.3% $51.50 6 7 78 9.0% $45.00 7 6 78 7.7% $38.50 8 5 78 6.4% $32.00 9 4 78 5.1% $25.50 10 3 78 3.8% $19.00 11 2 78 2.6% $13.00 12 1 78 1.3% $6.50 ==History== Prior to 1935, a borrower might have entered a contract with the lender to repay off a principal plus the pre-calculated total interest divided equally into the monthly repayments. Although this article focuses primarily on mortgages, the methods employed are relevant to any situation in which payment or saving is effected by a regular stream of fixed interval payments (annuity). ==Derivation of time-continuous equation== The classical formula for the present value of a series of n fixed monthly payments amount x invested at a monthly interest rate i% is: :P_v(n) = \frac{x(1 - (1 + i)^{-n})}{i} The formula may be re-arranged to determine the monthly payment x on a loan of amount P0 taken out for a period of n months at a monthly interest rate of i%: :x = \frac{P_0\cdot i}{1 - (1 + i)^{-n}} We begin with a small adjustment of the formula: replace i with r/N where r is the annual interest rate and N is the annual frequency of compounding periods (N = 12 for monthly payments). thumb|385px|30 year mortgage on a $250,000 loan thumb|30 year mortgage of $250,000 at different interest rates Mortgage calculators are automated tools that enable users to determine the financial implications of changes in one or more variables in a mortgage financing arrangement. The expansion P\approx P_0 \left(1 + X + \frac{X^2}{3}\right) is valid to better than 1% provided X\le 1 . ====Example of mortgage payment==== For a $10,000 mortgage with a term of 30 years and a note rate of 4.5%, payable yearly, we find: T=30 I=0.045 which gives X=\frac{1}{2}IT=.675 so that P\approx P_0 \left(1 + X + \frac{1}{3}X^2 \right)=\$333.33 (1+.675+.675^2/3)=\$608.96 The exact payment amount is P=\$608.02 so the approximation is an overestimate of about a sixth of a percent. ===Investing: monthly deposits=== Given a principal (initial) deposit and a recurring deposit, the total return of an investment can be calculated via the compound interest gained per unit of time. When interest is continuously compounded, use \delta=n\ln{\left(1+\frac{r}{n}\right)}, where \delta is the interest rate on a continuous compounding basis, and r is the stated interest rate with a compounding frequency n. ===Monthly amortized loan or mortgage payments=== The interest on loans and mortgages that are amortized—that is, have a smooth monthly payment until the loan has been paid off—is often compounded monthly. Loan (P) Period (T) Annual payment rate (Ma) Initial estimate: 2 ln(MaT/P)/T 10000 3 6000 39.185778% Newton–Raphson iterations n r(n) f[r(n)] f'[r(n)] 0 39.185778% −229.57 4444.44 1 44.351111% 21.13 5241.95 2 43.948044% 0.12 5184.06 3 43.945798% 0 5183.74 ==Present value and future value formulae== Corresponding to the standard formula for the present value of a series of fixed monthly payments, we have already established a time continuous analogue: :P_v(t)=\frac{M_a}{r}(1-e^{-rt}). If n were 24, the sum of the numbers from 1 to 24 is 24 * (24+1) / 2 = (24 * 25) / 2 = 300, which is the loan's denominator, D. thumb|300px|Effective interest rates thumb|right|310px|The effect of earning 20% annual interest on an initial $1,000 investment at various compounding frequencies Compound interest is the addition of interest to the principal sum of a loan or deposit, or in other words, interest on principal plus interest. thumb|right|350px|The effect of earning 20% annual interest on an initial $1,000 investment at various compounding frequencies Analogous to continuous compounding, a continuous annuity \- Entry on continuous annuity, p. 86 is an ordinary annuity in which the payment interval is narrowed indefinitely. Given an annual interest rate r and a borrower with an annual payment capability MN (divided into N equal payments made at time intervals Δt where Δt = 1/N years), we may write: :\begin{align} P_{t+\Delta t} & = P_t+(rP_t- M_N)\Delta t\\\\[12pt] \dfrac{P_{t+\Delta t}-P_t}{\Delta t} & = rP_t-M_N \end{align} If N is increased indefinitely so that Δt → 0, we obtain the continuous time differential equation: :{\operatorname{d}P(t)\over\operatorname{d}t}=rP(t)-M_a Beckwith: Equation (25) p. 123Hackman: Equation (2) p.1 Note that for there to be a continually diminishing mortgage balance, the following inequality must hold: :P_0 \leqslant \frac{M_a}{r} Where equality holds, the mortgage becomes a perpetuity. This can be expressed mathematically by : p = \frac{P_0\cdot r\cdot (1+r)^n}{(1+r)^n-1} :where: ::p is the payment made each period :: P0 is the initial principal :: r is the percentage rate used each payment :: n is the number of payments This also explains why a 15-year mortgage and a 30-year mortgage with the same APR would have different monthly payments and a different total amount of interest paid. For example, for interest rate of 6% (0.06/12), 25 years * 12 p.a., PV of $150,000, FV of 0, type of 0 gives: = PMT(0.06/12, 25 * 12, -150000, 0, 0) = $966.45 ====Approximate formula for monthly payment==== A formula that is accurate to within a few percent can be found by noting that for typical U.S. note rates (I<8\% and terms T=10–30 years), the monthly note rate is small compared to 1: r << 1 so that the \ln(1+r)\approx r which yields a simplification so that c\approx \frac{Pr}{1-e^{-nr}}= \frac{P}{n}\frac{nr}{1-e^{-nr}} which suggests defining auxiliary variables Y\equiv n r = IT c_0\equiv \frac{P}{n} . Starting in one month's time he decides to make equal monthly payments into an account that pays interest at 12% per annum compounded monthly. The amount after t periods of continuous compounding can be expressed in terms of the initial amount P0 as P(t)=P_0 e ^ {rt}. ===Force of interest=== As the number of compounding periods n tends to infinity in continuous compounding, the continuous compound interest rate is referred to as the force of interest \delta. If, in the second case, equal monthly payments are made of $946.01 against 9.569% compounded monthly then it takes 240 months to pay the loan back. During the second month the borrower has use of two $1000 (2/3) amounts and so the payment should be $1000 plus two $10 interest fees.

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A spring is stretched 6 in by a mass that weighs $8 \mathrm{lb}$. The mass is attached to a dashpot mechanism that has a damping constant of $0.25 \mathrm{lb} \cdot \mathrm{s} / \mathrm{ft}$ and is acted on by an external force of $4 \cos 2 t \mathrm{lb}$. If the given mass is replaced by a mass $m$, determine the value of $m$ for which the amplitude of the steady state response is maximum.

As such, m cannot be simply added to M to determine the frequency of oscillation, and the effective mass of the spring is defined as the mass that needs to be added to M to correctly predict the behavior of the system. ==Ideal uniform spring== right|frame|vertical spring-mass system The effective mass of the spring in a spring-mass system when using an ideal spring of uniform linear density is 1/3 of the mass of the spring and is independent of the direction of the spring- mass system (i.e., horizontal, vertical, and oblique systems all have the same effective mass). *Reduced mass ==References== ==External links== *http://tw.knowledge.yahoo.com/question/question?qid=1405121418180 *http://tw.knowledge.yahoo.com/question/question?qid=1509031308350 *https://web.archive.org/web/20110929231207/http://hk.knowledge.yahoo.com/question/article?qid=6908120700201 *https://web.archive.org/web/20080201235717/http://www.goiit.com/posts/list/mechanics- effective-mass-of-spring-40942.htm *http://www.juen.ac.jp/scien/sadamoto_base/spring.html *"The Effective Mass of an Oscillating Spring" Am. J. Phys., 38, 98 (1970) *"Effective Mass of an Oscillating Spring" The Physics Teacher, 45, 100 (2007) Category:Mechanical vibrations Category:Mass By differentiation of the equation with respect to time, the equation of motion is: :\left( \frac{-m}{3}-M \right) \ a = kx -\tfrac{1}{2} mg - Mg The equilibrium point x_{\mathrm{eq}} can be found by letting the acceleration be zero: :x_{\mathrm{eq}} = \frac{1}{k}\left(\tfrac{1}{2}mg + Mg \right) Defining \bar{x} = x - x_{\mathrm{eq}}, the equation of motion becomes: :\left( \frac{m}{3}+M \right) \ a = -k\bar{x} This is the equation for a simple harmonic oscillator with period: :\tau = 2 \pi \left( \frac{M + m/3}{k} \right)^{1/2} So the effective mass of the spring added to the mass of the load gives us the "effective total mass" of the system that must be used in the standard formula 2 \pi\sqrt{\frac{m}{k}} to determine the period of oscillation. ==General case == As seen above, the effective mass of a spring does not depend upon "external" factors such as the acceleration of gravity along it. This unexpected behavior of the effective mass can be explained in terms of the elastic after- effect (which is the spring's not returning to its original length after the load is removed). ==See also== *Simple harmonic motion (SHM) examples. The effective mass of the spring can be determined by finding its kinetic energy. The homogeneous equation for the mass spring system is: :\ddot x + 2 \zeta \omega_n \dot x + \omega_n^2 x = 0 This has the solution: : x = A e^{-\omega_n t \left(\zeta + \sqrt{\zeta^2-1}\right)} + B e^{-\omega_n t \left(\zeta - \sqrt{\zeta^2-1}\right)} If \zeta < 1 then \zeta^2-1 is negative, meaning the square root will be negative the solution will have an oscillatory component. ==See also== * Numerical methods * Soft body dynamics#Spring/mass models * Finite element analysis ==References== Category:Classical mechanics Category:Mechanical vibrations Using this result, the total energy of system can be written in terms of the displacement x from the spring's unstretched position (ignoring constant potential terms and taking the upwards direction as positive): : T(Total energy of system) ::= \tfrac{1}{2}(\frac{m}{3})\ v^2 + \tfrac{1}{2}M v^2 + \tfrac{1}{2} k x^2 - \tfrac{1}{2}m g x - M g x Note that g here is the acceleration of gravity along the spring. Jun-ichi Ueda and Yoshiro Sadamoto have found that as M/m increases beyond 7, the effective mass of a spring in a vertical spring-mass system becomes smaller than Rayleigh's value m/3 and eventually reaches negative values. Since not all of the spring's length moves at the same velocity v as the suspended mass M, its kinetic energy is not equal to \tfrac{1}{2} m v^2. This requires adding all the mass elements' kinetic energy, and requires the following integral, where u is the velocity of mass element: :K =\int_m\tfrac{1}{2}u^2\,dm Since the spring is uniform, dm=\left(\frac{dy}{L}\right)m, where L is the length of the spring at the time of measuring the speed. In fact, for a non-uniform spring, the effective mass solely depends on its linear density \rho(x) along its length: ::K = \int_m\tfrac{1}{2}u^2\,dm ::: = \int_0^L\tfrac{1}{2}u^2 \rho(x) \,dx ::: = \int_0^L\tfrac{1}{2}\left(\frac{v x}{L} \right)^2 \rho(x) \,dx ::: = \tfrac{1}{2} \left[ \int_0^L \frac{x^2}{L^2} \rho(x) \,dx \right] v^2 So the effective mass of a spring is: :m_{\mathrm{eff}} = \int_0^L \frac{x^2}{L^2} \rho(x) \,dx This result also shows that m_{\mathrm{eff}} \le m, with m_{\mathrm{eff}} = m occurring in the case of an unphysical spring whose mass is located purely at the end farthest from the support. ==Real spring== The above calculations assume that the stiffness coefficient of the spring does not depend on its length. The frequency response of this oscillator describes the amplitude z of steady state response of the equation (i.e. x(t)) at a given frequency of excitation \omega. Hence, :K = \int_0^L\tfrac{1}{2}u^2\left(\frac{dy}{L}\right)m\\! ::=\tfrac{1}{2}\frac{m}{L}\int_0^L u^2\,dy The velocity of each mass element of the spring is directly proportional to length from the position where it is attached (if near to the block then more velocity and if near to the ceiling then less velocity), i.e. u=\frac{vy}{L}, from which it follows: :K =\tfrac{1}{2}\frac{m}{L}\int_0^L\left(\frac{vy}{L}\right)^2\,dy :=\tfrac{1}{2}\frac{m}{L^3}v^2\int_0^L y^2\,dy :=\tfrac{1}{2}\frac{m}{L^3}v^2\left[\frac{y^3}{3}\right]_0^L :=\tfrac{1}{2}\frac{m}{3}v^2 Comparing to the expected original kinetic energy formula \tfrac{1}{2}mv^2, the effective mass of spring in this case is m/3. thumb|Phase portrait of damped oscillator, with increasing damping strength. The forcing amplitude increases from \gamma=0.20 to \gamma=0.65. The Duffing equation can be seen as describing the oscillations of a mass attached to a nonlinear spring and a linear damper. For the case with linearly distributed load of maximum intensity q_0, :M_{\mathrm{right}}^{\mathrm{fixed}} = \int_{0}^{L} q_0 \frac{x}{L} dx \frac{ x^2 (L-x)}{L^2} = \frac{q_0 L^2}{20} :M_{\mathrm{left}}^{\mathrm{fixed}} = \int_{0}^{L} \left \\{ - q_0 \frac{x}{L} dx \frac{x (L-x)^2}{L^2} \right \\} = - \frac{q_0 L^2}{30} == See also == * Moment distribution method * Statically Indeterminate * Slope deflection method * Matrix method == References == * Category:Structural analysis *Damped harmonic motion, see animation (right). The lower overhanging side is unstable – i.e. the dashed-line parts in the figures of the frequency response – and cannot be realized for a sustained time. The restoring force provided by the nonlinear spring is then \alpha x + \beta x^3. The dashed parts of the frequency response are unstable. In a real spring–mass system, the spring has a non-negligible mass m.

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A recent college graduate borrows $\$ 100,000$ at an interest rate of $9 \%$ to purchase a condominium. Anticipating steady salary increases, the buyer expects to make payments at a monthly rate of $800(1+t / 120)$, where $t$ is the number of months since the loan was made. Assuming that this payment schedule can be maintained, when will the loan be fully paid?

If, in the second case, equal monthly payments are made of $946.01 against 9.569% compounded monthly then it takes 240 months to pay the loan back. If in practice the loan is being repaid via monthly instalments, the decimal portion can be converted to months and rounded so this answer would equate to 172 months. ==Calculation of interest rate== In the discrete time interval model, calculation of a mortgage based interest rate given the remaining parameters has not been possible using analytic methods. Loan (P) Period (T) Annual payment rate (Ma) Initial estimate: 2 ln(MaT/P)/T 10000 3 6000 39.185778% Newton–Raphson iterations n r(n) f[r(n)] f'[r(n)] 0 39.185778% −229.57 4444.44 1 44.351111% 21.13 5241.95 2 43.948044% 0.12 5184.06 3 43.945798% 0 5183.74 ==Present value and future value formulae== Corresponding to the standard formula for the present value of a series of fixed monthly payments, we have already established a time continuous analogue: :P_v(t)=\frac{M_a}{r}(1-e^{-rt}). Although this article focuses primarily on mortgages, the methods employed are relevant to any situation in which payment or saving is effected by a regular stream of fixed interval payments (annuity). ==Derivation of time-continuous equation== The classical formula for the present value of a series of n fixed monthly payments amount x invested at a monthly interest rate i% is: :P_v(n) = \frac{x(1 - (1 + i)^{-n})}{i} The formula may be re-arranged to determine the monthly payment x on a loan of amount P0 taken out for a period of n months at a monthly interest rate of i%: :x = \frac{P_0\cdot i}{1 - (1 + i)^{-n}} We begin with a small adjustment of the formula: replace i with r/N where r is the annual interest rate and N is the annual frequency of compounding periods (N = 12 for monthly payments). Dividing by loan time period t will then give the equivalent simple interest rate. Starting in one month's time he decides to make equal monthly payments into an account that pays interest at 12% per annum compounded monthly. A graduated payment loan typically involves negative amortization, and is intended for students in the case of student loans, and homebuyers in the case of real estate, who currently have moderate incomes and anticipate their income will increase over the next 5–10 years. However this would contradict the primary assumption upon which the "continuous payment" model is based: namely that the annual payment rate is defined as: :M_a=\lim_{N\to\infty}N\cdot x(N) \, Since it is of course impossible for an investor to make an infinitely small payment infinite times per annum, a bank or other lending institution wishing to offer "continuous payment" annuities or mortgages would in practice have to choose a large but finite value of N (annual frequency of payments) such that the continuous time formula will always be correct to within some minimal pre-specified error margin. For a theoretical continuous payment savings annuity we can only calculate an annual rate of payment: :M_a=\frac{500000 \times 12\%}{e^{0.12\cdot 10}-1}=25860.77 At this point there is a temptation to simply divide by 12 to obtain a monthly payment. Given an annual interest rate r and a borrower with an annual payment capability MN (divided into N equal payments made at time intervals Δt where Δt = 1/N years), we may write: :\begin{align} P_{t+\Delta t} & = P_t+(rP_t- M_N)\Delta t\\\\[12pt] \dfrac{P_{t+\Delta t}-P_t}{\Delta t} & = rP_t-M_N \end{align} If N is increased indefinitely so that Δt → 0, we obtain the continuous time differential equation: :{\operatorname{d}P(t)\over\operatorname{d}t}=rP(t)-M_a Beckwith: Equation (25) p. 123Hackman: Equation (2) p.1 Note that for there to be a continually diminishing mortgage balance, the following inequality must hold: :P_0 \leqslant \frac{M_a}{r} Where equality holds, the mortgage becomes a perpetuity. In a (theoretical) continuous-repayment mortgage the payment interval is narrowed indefinitely until the discrete interval process becomes continuous and the fixed interval payments become—in effect—a literal cash "flow" at a fixed annual rate. As with many similar examples the discrete interval problem and its solution is closely approximated by calculations based on the continuous repayment model - Dr Hahn's solution for interest rate is 40.8% as compared to the 41.6% calculated above. ==Period of a loan== If a borrower can afford an annual repayment rate Ma, then we can re-arrange the formula for calculating Ma to obtain an expression for the time period T of a given loan P0: : \begin{align} & M_a = \frac{P_0 r}{1-e^{-rT}} \\\\[8pt] \Rightarrow & T = \frac{1}{r}\ln\frac{M_a}{M_a-P_0 r} = -\frac{1}{r}\ln\left(1 - \frac{P_0 r}{M_a} \right) \end{align} ==Minimum payment ratio== The minimum payment ratio of a loan is the ratio of minimum possible payment rate to actual payment rate. For example, consider a $100 loan which must be repaid after one month, plus 5%, plus a $10 fee. \, Another way to calculate balance due P(t) on a continuous-repayment loan is to subtract the future value (at time t) of the payment stream from the future value of the loan (also at time t): :P(t)=P_0 e^{rt}-\frac{M_a}{r}(e^{rt}-1). Finally the model demonstrates that it is to the modest advantage of the mortgage holder to increase frequency of payment where practically possible. ==Summary of formulae and online calculators== Annual payment rate (mortgage loan): M_a=\lim_{N\to\infty}N\cdot x(N)=\frac{P_0\cdot r}{1 - e^{-rT}} Annual payment rate (sinking fund): M_a=\lim_{N\to\infty}N\cdot x(N)=\frac{P_T\cdot r}{e^{rT}-1} Future value: F_v(t) = \frac{M_a}{r}(e^{rt}-1) Present value: P_v(t) = \frac{M_a}{r}(1 - e^{-rt}) Loan balance: P(t) = \frac{M_a}{r}(1 - e^{-r(T-t)}) Loan period: T=-\frac{1}{r}\ln\left(1-\frac{P_0 r}{M_a}\right) Half- life of loan: t_{\frac{1}{2}}=\frac{1}{r}\ln\left(\frac{1+e^{rT}}{2}\right) Interest rate: r\approx\frac{2}{T}\ln{\frac{M_aT}{P_0}} r=\frac{1}{T}\left (W(-se^{-s})+s\right )\text{ with }s=\frac{M_at}{P_0} Universal mortgage calculator. Beware of extremely long repayment periods, as generally speaking, the longer the term, the more you will owe because the interest accrues over a long period of time. The minimum possible payment rate is that which just covers the loan interest – a borrower would in theory pay this amount forever because there is never any decrease in loan capital. This can be expressed mathematically by : p = \frac{P_0\cdot r\cdot (1+r)^n}{(1+r)^n-1} :where: ::p is the payment made each period :: P0 is the initial principal :: r is the percentage rate used each payment :: n is the number of payments This also explains why a 15-year mortgage and a 30-year mortgage with the same APR would have different monthly payments and a different total amount of interest paid. If the consumer pays the loan off early, the effective interest rate achieved will be significantly higher than the APR initially calculated. The sum of these interest and principal payments must equal the cumulative fixed payments at time t i.e. Mat. Evaluating the first integral on the right we obtain an expression for I(t), the interest paid: :I(t)=M_at-\frac{M_a(e^{rt}-1)}{re^{rT}} Unsurprisingly the second integral evaluates to P0 − P(t) and therefore: :I(t)=M_at-P_0+P(t) \, The reader may easily verify that this expression is algebraically identical to the one above. ==Loan cost factor== The cost of a loan is simply the annual rate multiplied by loan period: : C = M_aT= \frac{P_0 rT}{1-e^{-rT}} Let s = rT. Each payment accumulates compound interest from time of deposit to the end of the mortgage timespan at which point the sum of the payments with their accumulated interest equals the value of the loan with interest compounded over the entire timespan. A (theoretical) continuous repayment mortgage is a mortgage loan paid by means of a continuous annuity.

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