Chapter 11.2, Problem 60E

To determine

To prove: That the particle’s velocity is always orthogonal to both its position and its acceleration.

Explanation of Solution

Given information:

The position vector $=\langle \cos t,\sin t\rangle$

Calculation:

The given position vector is $\langle \cos t,\sin t\rangle$ .

Consider the position vector be $\overrightarrow{r}\left(t\right)=\langle \cos t,\sin t\rangle$ .

It is known that 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 vector $\overrightarrow{a}\left(t\right)$ .

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

Write the acceleration vector as:

  $\begin{array}{c}\overrightarrow{a}\left(t\right)=\frac{d\left(v\left(t\right)\right)}{dt}\\ =\frac{d}{dt}\langle -x,\sqrt{1-x^2}\rangle \\ =\langle \frac{d}{dt}\left(-x\right),\frac{d}{dt}\left(\sqrt{1-x^2}\right)\rangle \\ =\langle -\cos t,-\sin t\rangle \end{array}$

The slope of the velocity vector is the negative reciprocal of the slopes of the position vector and the acceleration vector.

Here, the slope of the position vector is $\tan t$ .

The slope of the velocity vector is $-\frac{1}{\tan t}$ .

The slope of the acceleration vector is $\tan t$ .

This means that the velocity is orthogonal to the position and the acceleration.

Hence, the statement is proved.