Vector-Valued Functions - Velocity, Acceleration, and Speed

Sometimes we can use a vector-valued function $\vec{r}(t) = (x(t), y(t), z(t))$ to represent an object that traverses the curve $C$ traced by $\vec{r}(t)$ at time $t$. We can then define the velocity, acceleration, and speed of that object which we define below.

Definition: If $\vec{r}(t) = (x(t), y(t), z(t))$ is a vector-valued function that represents the position of an object at time $t$, then the Velocity of this object at time $t$ is $\vec{v}(t) = \vec{r'}(t)$. The Acceleration of this object at time $t$ is $\vec{a}(t) = \vec{v'}(t) = \vec{r''}(t)$. The Speed of this object at time $t$ is $v(t) = \| \vec{v}(t) \| = \| \vec{r'}(t) \|$.

Note that the speed of an object is represented by the single-variable function $v(t)$ which is defined to be the norm/magnitude of the velocity $\vec{v}(t) = \vec{r'}(t)$. Do not confuse the speed $v(t)$ with the velocity $\vec{v}(t)$!

Let's look at an example.

Example 1

Let $\vec{r}(t) = (t^2, \cos t, e^{2t})$ represent the position of a particle at time $t > 0$. Find the velocity, acceleration, and speed of this particle at time $t = 2$.

We first differentiate $\vec{r}(t)$ to get $\vec{v}(t) = \vec{r'}(t) = (2t, -\sin t, 2e^{2t})$. At time $t = 2$ we thus have the the velocity of this particle is $\vec{v}(2) = (4, -\sin 2, 2e^{4})$.

If we differentiate $\vec{v}(t)$ we get $\vec{a}(t) = \vec{v'}(t) = \vec{r''}(t) = (2, -\cos t, 4e^{2t})$. At time $t = 2$ we have that the acceleration of this particle is $\vec{a}(2) = (2, -\cos 2, 4e^4)$.

Lastly, we obtain the speed of the particle by taking the norm/magnitude of velocity, that is $v(t) = \| \vec{v}(t) \| = \sqrt{4t^2 + \sin^2 t + 4 e^{4t}}$. So $v(2) = \sqrt{16 + \sin ^2 2 + 4e^8}$.