Chapter 10.2, Problem 60E

To determine

To show: The position vector is orthogonal to acceleration vector and position vector.

Explanation of Solution

Given information:

The position vector of a particles is $〈\cos t,\sin t〉$ .

Calculation:

The position vector of the particle is,

  $\begin{array}{ccc}x\left(t\right)=\cos t& \text{and}& y\left(t\right)=\sin t\\ x\text{'}\left(t\right)=-\sin t& \text{and}& y\text{'}\left(t\right)=\cos t\\ x\text{'}\text{'}\left(t\right)=-\cos t& \text{and}& y\text{'}\text{'}\left(t\right)=-\sin t\end{array}$

The position vector is $〈\cos t,\sin t〉$ .

The velocity vector is $〈-\sin t,\cos t〉$ .

The acceleration vector is $〈-\cos t,-\sin t〉$ .

If two vector are orthogonal then their dot product is 0.

The dot product of velocity vector and position vector,

  $\begin{array}{c}〈-\sin t,\cos t〉\cdot 〈\cos t,\sin t〉=-\sin t\cos t+\cos t\sin t\\ =0\end{array}$

Since the dot product is 0, the velocity vector and position vector is orthogonal.

The dot product of velocity vector and acceleration vector,

  $\begin{array}{c}〈-\sin t,\cos t〉\cdot 〈-\cos t,-\sin t〉=\sin t\cos t-\cos t\sin t\\ =0\end{array}$

Since the dot product is 0, the velocity vector and acceleration vector is orthogonal.

Hence Proved.