Time derivative

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A time derivative is a derivative of a function with respect to time, usually interpreted as the rate of change of the value of the function. The variable denoting time is usually written as $t\,$ .

1 Notation
2 Use in physics
3 Example: circular motion
4 See also

[edit] Notation

A variety of notations are used to denote the time derivative. In addition to the normal notation,

$\frac {dx} {dt}$

two very common shorthand notations are also used: adding a dot over the variable, $\dot{x}$ , and adding a prime to the variable, $x'\,$ . These two shorthands are generally not mixed in the same set of equations. Other shorthands include a subscripted t as in $x t$ .

Higher time derivatives are also used: the second derivative with respect to time is written as

$\frac {d^2x} {dt^2}$

with the corresponding shorthands of $\ddot{x}$ and $x''\,$ .

As a generalization, the time derivative of a vector, say:

$\vec V = \left[ v_1,\ v_2,\ v_3, \cdots \right] \ ,$

is defined as the vector whose components are the derivatives of the original vector. That is,

$\frac {d \vec V } {dt} = \left[ \frac{ d v_1 }{dt},\frac {d v_2 }{dt},\frac {d v_3 }{dt}, \cdots \right] \ .$

[edit] Use in physics

Time derivatives are a key concept in physics. For example, for a changing position $x\,$ , its time derivative $\dot{x}$ is its velocity, and its second derivative with respect to time, $\ddot{x}$ , is its acceleration. Even higher derivatives are sometimes also used: the third derivative of position with respect to time is known as the jerk.

A large number of fundamental equations in physics involve first or second time derivatives of quantities. Many other fundamental quantities in science are time derivatives of one another:

force is the time derivative of momentum
power is the time derivative of energy
electrical current is the time derivative of electric charge

and so on.

A common occurrence in physics is the time derivative of a vector, such as velocity or displacement. In dealing with such a derivative, both magnitude and orientation may depend upon time.

[edit] Example: circular motion

See also: Uniform circular motion and Centripetal force

Figure 1: Relation between x-y coordinates and polar coordinates.

For example, consider a particle moving in a circular path. Its position is given by the displacement vector $\vec r = [x, y]$ , related to the angle, θ, and radial distance, ρ, as defined in Figure 1:

$x = \rho \ \cos \ (\theta) \ ,$ $y = \rho \ \sin\ (\theta)\ .$

For purposes of this example, time dependence is introduced by setting θ = t. The displacement (position) at any time t is then:

$\vec r \ (t) = [ x(t),\ y(t) ] = \rho \ [\cos (t), \ \sin (t)] \ .$

The second form shows the motion described by $\stackrel{ \vec r (t) }{}$ is in a circle of radius ρ because the magnitude of $\vec r (t)$ is given by

$| \vec r | = \sqrt {x (t )^2\ +\ y (t )^2 } = \rho \left[ \mathrm{cos}^2 (t) + \mathrm{sin}^2 (t) \right]^{1/2} = \rho \ ,$

using the result ( valid for any choice of t ) sin² ( t ) + cos² ( t ) = 1.

With this form for the displacement, the velocity now is found. The time derivative of the displacement vector is the velocity vector. In general, the derivative of a vector is a vector made up of components each of which is the derivative of the corresponding component of the original vector. Thus, in this case, the velocity vector is:

$\vec v \ (t) = \frac {d \vec r \ (t) }{dt} = \rho \ \left[ \frac {d\ \mathrm{cos} (t)}{dt},\ \frac {d \ \mathrm {sin} (t)} {dt} \right]$

$= \rho \ \left[ - \mathrm{sin} (t),\ \mathrm {cos} (t) \right] \$

$=\left[ - y (t),\ x(t) \right] \ ,$

which shows (i) that the derivative of the displacement vector is nonzero, even though the magnitude of the displacement (the radius of the path) is constant, and (ii) that the velocity is perpendicular to the displacement, as can be established using the dot product:

$\vec v \ \cdot \ \vec r = [-y,\ x] \cdot [x, \ y ] = -yx\ +\ xy = 0 \ .$

Repeating this process for the acceleration, which is the time derivative of the velocity:

$\vec a \ (t) = \frac {d \vec v \ (t) }{dt} = [-x(t),\ -y(t)] = - \vec r \ (t) \ ,$

showing that the acceleration is inward directed, exactly opposite in direction to the displacement vector, and orthogonal to the velocity vector. This inward directed acceleration can be provided by gravitational attraction, for example, as in the case of the earth and moon, and is called centripetal acceleration.