Chapter 11, Problem 42RE

To determine

To prove: That the angle between $\overrightarrow{r}$ and the acceleration vector $\overrightarrow{a}$ never changes.

Answer to Problem 42RE

The required value of the angle is $90°$ .

Explanation of Solution

Given information:

The position of the particle: $\overrightarrow{r}\left(t\right)=\langle e^t\cos t,e^t\sin t\rangle$

Calculation:

The given position of the particle is $\overrightarrow{r}\left(t\right)=\langle e^t\cos t,e^t\sin t\rangle$ .

It is known that if $\overrightarrow{r}\left(t\right)$ is the position vector of a particle, then the first derivative of $\overrightarrow{r}\left(t\right)$ is the velocity vector $\overrightarrow{v}\left(t\right)$ and the second derivative of $\overrightarrow{r}\left(t\right)$ is the acceleration $\overrightarrow{a}\left(t\right)$ of the particle.

So, $\overrightarrow{v}\left(t\right)=\frac{d\left(\overrightarrow{r}\left(t\right)\right)}{dt}$ and $\overrightarrow{a}\left(t\right)=\frac{d^2\left(\overrightarrow{r}\left(t\right)\right)}{d{t}^2}=\frac{d\left(\overrightarrow{v}\left(t\right)\right)}{dt}$

Substitute $\langle e^t\cos t,e^t\sin t\rangle$ for $\overrightarrow{r}\left(t\right)$ in the expression $\overrightarrow{v}\left(t\right)=\frac{d\left(\overrightarrow{r}\left(t\right)\right)}{dt}$ .

  $\begin{array}{c}\overrightarrow{v}\left(t\right)=\frac{d\left(\overrightarrow{r}\left(t\right)\right)}{dt}\\ =\frac{d}{dt}\langle e^t\cos t,e^t\sin t\rangle \\ =\langle \frac{d}{dt}\left(e^t\cos t\right),\frac{d}{dt}\left(e^t\sin t\right)\rangle \\ =\langle e^t\cos t-e^t\sin t,e^t\sin t+e^t\cos t\rangle \end{array}$

Substitute $\langle e^t\cos t-e^t\sin t,e^t\sin t+e^t\cos t\rangle$ for $\overrightarrow{v}\left(t\right)$ in the expression $\overrightarrow{a}\left(t\right)=\frac{d\left(\overrightarrow{v}\left(t\right)\right)}{dt}$ .

  $\begin{array}{c}\overrightarrow{a}\left(t\right)=\frac{d}{dt}\langle e^t\cos t-e^t\sin t,e^t\sin t+e^t\cos t\rangle \\ =\langle \frac{d}{dt}\left(e^t\cos t-e^t\sin t\right),\frac{d}{dt}\left(e^t\sin t+e^t\cos t\right)\rangle \\ =\langle -2e^t\sin t,2e^t\cos t\rangle \end{array}$

Now, the slope of the position vector is:

  $\frac{e^t\sin t}{e^t\cos t}=\tan t$ :

And, the slope of the acceleration vector is

  $\frac{2e^t\sin t}{-2e^t\cos t}=-\frac{1}{\tan t}$

It is seen that the slope of the acceleration vector is the negative reciprocal of the slope of the position vector.

Thus, the angle between them is $90°$ .

Hence, it is proved that the angle between $\overrightarrow{r}$ and the acceleration vector $\overrightarrow{a}$ never changes and the required value of the angle is $90°$ .