Chapter 4, Problem 7RWDT

To determine

To find: If the particle is moving clockwise or counterclockwise.

Answer to Problem 7RWDT

The particle is moving counterclockwise.

Explanation of Solution

Given information:

The partial moves along the circle of radius $r$ with center $\left(0,0\right)$ .

Calculation:

The general equation of the circle is given as:

  ${\left(x-a\right)}^2+{\left(y-b\right)}^2=r^2$

Here $\left(a,b\right)$ are center of the circle and $r$ is radius.

The equation for the circle with radius $r$ and center $\left(0,0\right)$ is:

  $\begin{array}{c}{\left(x-0\right)}^2+{\left(y-0\right)}^2=r^2\\ x^2+y^2=r^2\end{array}$

In order to find whether the particle is moving clockwise or anticlockwise differentiate $x$ and $y$ with respect to $t$ gives:

  $\begin{array}{l}2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\\ x\frac{dx}{dt}+y\frac{dy}{dt}=0\\ x\frac{dx}{dt}=-y\frac{dy}{dt}\\ \frac{dx}{dt}=\frac{-y}{x}\frac{dy}{dt}\end{array}$

If $\frac{dx}{dt}$ and $y$ has opposite sign say $\frac{dx}{dt}$ is positive and $y$ is negative this will lie in third quadrant the equation $\frac{dx}{dt}=\frac{-y}{x}\frac{dy}{dt}$ becomes:

  $\begin{array}{l}\frac{dx}{dt}=\frac{-\left(-y\right)}{x}\frac{dy}{dt}\\ \frac{dx}{dt}=\frac{y}{x}\frac{dy}{dt}\end{array}$

The rate of change in horizontal axis and vertical axis will be positive which lies in first quadrant therefore particle moves from third to first quadrant hence its moving counterclockwise.