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. See
motion graphs and derivatives. 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 •
electric 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.
Example: circular motion (
r,
θ). For example, consider a particle moving in a circular path. Its position is given by the displacement vector r=x\hat{\imath}+y\hat{\jmath}, related to the angle,
θ, and radial distance,
r, as defined in the figure: :\begin{align} x &= r \cos(\theta) \\ y &= r \sin(\theta) \end{align} For this example, we assume that . Hence, the displacement (position) at any time
t is given by :\mathbf{r}(t) = r\cos(t)\hat{\imath}+r\sin(t)\hat{\jmath} This form shows the motion described by
r(
t) is in a circle of radius
r because the
magnitude of
r(
t) is given by :|\mathbf{r}(t)| = \sqrt{\mathbf{r}(t) \cdot \mathbf{r}(t)}=\sqrt {x(t)^2 + y(t)^2 } = r\, \sqrt{\cos^2(t) + \sin^2(t)} = r using the
trigonometric identity and where \cdot is the usual Euclidean dot product. 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: : \begin{align} \mathbf{v}(t) = \frac {d\,\mathbf{r}(t) }{dt} &= r \left[\frac{d\, \cos(t)}{dt}, \frac{d\, \sin(t)}{dt} \right] \\ &= r\ [ -\sin(t),\ \cos(t)] \\ &= [-y (t), x(t)]. \end{align} Thus the velocity of the particle is nonzero even though the magnitude of the position (that is, the radius of the path) is constant. The velocity is directed perpendicular to the displacement, as can be established using the
dot product: :\mathbf{v} \cdot \mathbf{r} = [-y, x] \cdot [x, y] = -yx + xy = 0\, . Acceleration is then the time-derivative of velocity: :\mathbf{a}(t) = \frac {d\, \mathbf{v}(t)}{dt} = [-x(t), -y(t)] = -\mathbf{r}(t)\, . The acceleration is directed inward, toward the axis of rotation. It points opposite to the position vector and perpendicular to the velocity vector. This inward-directed acceleration is called
centripetal acceleration. ==In differential geometry==