8.3 Vector valued Functions
Vector Valued Function, Velocity Vector, Acceleration Vector, and Speed
If a particle moves in the $x y$-plane so that at time $t>0$ its position vector is given by
$
\mathbf{r}(t)=<x(t), y(t)>
$
then the velocity vector, acceleration vector, and speed at time $t$ are as follows.
Velocity $=\mathrm{v}(t)=\mathrm{r}^{\prime}(t)=\left\langle x^{\prime}(t), y^{\prime}(t)\right\rangle$
Accleration $=\mathrm{a}(t)=\mathrm{v}^{\prime}(t)=\left\langle x^{\prime \prime}(t), y^{\prime \prime}(t)>\right.$
Speed $=|\mathrm{v}(t)|=\sqrt{\left[x^{\prime}(t)\right]^2+\left[y^{\prime}(t)\right]^2}$
If $x(t)$ is increasing $\frac{d x}{d t}$ is positive and if $x(t)$ is decreasing $\frac{d x}{d t}$ is negative.
If $y(t)$ is increasing $\frac{d x}{d t}$ is positive and if $y(t)$ is decreasing $\frac{d x}{d t}$ is negative.
Example 1
- A particle moving in the $x y$-plane is defined by the vector-valued function $f(t)=<t-\sin t, 1-\cos t>$, for $0 \leq t \leq \pi$.
(a) Find the velocity vector for the particle at any time $t$.
(b) Find the speed of the particle when $t=\frac{\pi}{3}$.
(c) Find the acceleration vector for the particle at any time $t$.
(d) Find the average speed of the particle from time $t=0$ to time $t=\pi$.
▶️Answer/Explanation
Solution
$\begin{aligned} (a) & x(t)=t-\sin t, y(t)=1-\cos t \\ & x^{\prime}(t)=1-\cos t, y^{\prime}(t)=\sin t \\ & \mathrm{v}(t)=<x^{\prime}(t), y^{\prime}(t)>=<1-\cos t, \sin t>\end{aligned}$
$
\begin{aligned}(b)
& x^{\prime}\left(\frac{\pi}{3}\right)=1-\cos \frac{\pi}{3}=1-\frac{1}{2}=\frac{1}{2} \\
& y^{\prime}\left(\frac{\pi}{3}\right)=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2} \\
& \text { Speed }=\left|\mathrm{v}\left(\frac{\pi}{3}\right)\right|=\sqrt{\left[x^{\prime}\left(\frac{\pi}{3}\right)\right]^2+\left[y^{\prime}\left(\frac{\pi}{3}\right)\right]^2}=\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}=1
\end{aligned}
$
$\begin{aligned}(c) x^{\prime \prime}(t) & =\sin t, y^{\prime \prime}(t)=\cos t \\ \mathrm{a}(t) & =<x^{\prime \prime}(t), y^{\prime \prime}(t)>=<\sin t, \cos t>\end{aligned}$
$\begin{aligned} & \text { (d) Average speed }=\frac{1}{\pi-0} \int_0^\pi \sqrt{\left[x^{\prime}(t)\right]^2+\left[y^{\prime}(t)\right]^2} d t \\ & =\frac{1}{\pi} \int_0^\pi \sqrt{(1-\cos t)^2+(\sin t)^2} d t=\frac{4}{\pi}\end{aligned}$
Example 2
- If a particle moves in the $x y$-plane so that at time $t>0$ its position vector is $\left(t^3-1, \ln \sqrt{t^2+1}\right)$, then at time $t=1$, its velocity vector is
(A) $\left(0, \frac{1}{2}\right)$
(B) $\left(1, \frac{1}{2}\right)$
(C) $\left(3, \frac{1}{2}\right)$
(D) $\left(3, \frac{1}{4}\right)$
▶️Answer/Explanation
Ans:C
Example 3
- A particle moves in the $x y$-plane so that at any time $t$ its coordinates are $x=t^3-t^2$ and $y=t+\ln t$. At time $t=2$, its acceleration vector is
(A) $\left(4, \frac{1}{2}\right)$
(B) $\left(6, \frac{1}{4}\right)$
(C) $\left(8, \frac{3}{4}\right)$
(D) $\left(10,-\frac{1}{4}\right)$
▶️Answer/Explanation
Ans:D
Example 4
- A particle moves in the $x y$-plane so that its position at time $t>0$ is given by $x(t)=e^t \cos t$ and $y(t)=e^t \sin t$. What is the speed of the particle when $t=2$ ?
(A) $\sqrt{2} e$
(B) $\sqrt{2} e^2$
(C) $2 e$
(D) $2 e^2$
▶️Answer/Explanation
Ans:B
Example 5
- If $f$ is a vector-valued function defined by $f(t)=\left(\ln (\sin t), t^2+e^{-t}\right)$, then the acceleration vector is
(A) $\left(-\csc ^2 t, 2+e^{-t}\right)$
(B) $\left(\sec ^2 t, 2+e^{-t}\right)$
(C) $\left(\csc ^2 t, 2-e^{-t}\right)$
(D) $\left(-\csc ^2 t \cdot \cot t, 2+e^{-t}\right)$
▶️Answer/Explanation
Ans:A
Example 6
- A particle moves on the curve $y=x+\sqrt{x}$ so that the $x$-component has velocity $x^{\prime}(t)=\cos t$ for $t \geq 0$. At time $t=0$, the particle is at the point $(1,0)$. At time $t=\frac{\pi}{2}$, the particle is at the point
(A) $(0,0)$
(B) $(1,2)$
(C) $\left(\frac{\pi}{2}, \frac{\pi}{2}+\sqrt{\frac{\pi}{2}}\right)$
(D) $(2,2+\sqrt{2})$
▶️Answer/Explanation
Ans:D