Chapter 11, Problem 43RE

To determine

To calculate: The position vector.

Answer to Problem 43RE

The required position vector is $\overrightarrow{r}\left(t\right)=\left(\cos t-1,\sin t+1\right)$ .

Explanation of Solution

Given information:

The values of vectors: $\overrightarrow{v}\left(t\right)=\langle -\sin t,\cos t\rangle$

  $r\left(0\right)=\langle 0,1\rangle$

Calculation:

The given vectors are $\overrightarrow{v}\left(t\right)=\langle -\sin t,\cos t\rangle$ and $r\left(0\right)=\langle 0,1\rangle$ .

It is known that the first derivative of the position vector $\overrightarrow{r}\left(t\right)=\langle \overrightarrow{x}\left(t\right),\overrightarrow{y}\left(t\right)\rangle$ is the velocity vector $\overrightarrow{v}\left(t\right)=\langle {\overrightarrow{v}}_1\left(t\right),{\overrightarrow{v}}_2\left(t\right)\rangle$ .

So, $\frac{d\overrightarrow{x}\left(t\right)}{dt}={\overrightarrow{v}}_1\left(t\right)$ , $\frac{d\overrightarrow{y}\left(t\right)}{dt}={\overrightarrow{v}}_2\left(t\right)$

Now, $\begin{array}{c}\frac{d\overrightarrow{x}\left(t\right)}{dt}=-\sin t\\ \overrightarrow{x}\left(t\right)={\displaystyle \int \left(-\sin t\right)dt}\\ \overrightarrow{x}\left(t\right)=\cos t+C\end{array}$

Substitute 0 for x and 0 for t in the above expression.

  $\begin{array}{c}\cos \left(0\right)+C=0\\ C+1=0\\ C=-1\end{array}$

Substitute $-1$ for C in the expression $\overrightarrow{x}\left(t\right)=\cos t+C$ .

  $\overrightarrow{x}\left(t\right)=\cos t-1$

Thus, the position of the particle is $\overrightarrow{x}\left(t\right)=\cos t-1$ .

Also, $\begin{array}{c}\frac{d\left(\overrightarrow{y}\left(t\right)\right)}{dt}=\cos t\\ \overrightarrow{y}\left(t\right)={\displaystyle \int \left(\cos t\right)dt}\\ \overrightarrow{y}\left(t\right)=\sin t+C\end{array}$

Substitute 1 for y and 0 for t in the above expression.

  $\begin{array}{c}\sin \left(0\right)+C=1\\ 0+C=1\\ C=1\end{array}$

Substitute 1 for C in the expression $\overrightarrow{y}\left(t\right)=\sin t+C$ .

  $\overrightarrow{y}\left(t\right)=\sin t+1$

Thus, the position of the particle is $\overrightarrow{y}\left(t\right)=\sin t+1$ .

Hence, the required position vector is $\overrightarrow{r}\left(t\right)=\left(\cos t-1,\sin t+1\right)$ .