Chapter 11.2, Problem 59E

To determine

To calculate: The acceleration vector in terms of x . Also, show that it its orthogonal to the corresponding velocity vector.

Answer to Problem 59E

The required acceleration vector is $\langle x,\frac{x^2}{\sqrt{1-x^2}}\rangle$ .

Explanation of Solution

Given information:

The value of $\frac{dx}{dt}=-x$

The value of $\frac{dy}{dt}=\sqrt{1-x^2}$

Calculation:

The given value of $\frac{dx}{dt}$ is $-x$ and the value of $\frac{dy}{dt}$ is $\sqrt{1-x^2}$ .

It is known that the acceleration vector is $\overrightarrow{a}\left(t\right)=\frac{d\left(\overrightarrow{v}\left(t\right)\right)}{dt}$ .

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

The slope of the velocity vector is $-\frac{\sqrt{1-x^2}}{x}$ .

It is known that the slope of a velocity vector is a negative reciprocal of the slope of the acceleration vector.

Thus, the slope of the acceleration vector is $\frac{x^2}{\sqrt{1-x^2}}$ .

This means that the vector are orthogonal.

Hence, the required acceleration vector is $\langle x,\frac{x^2}{\sqrt{1-x^2}}\rangle$ .