Chapter 4, Problem 6RWDT

To determine

To find: The sign of $\frac{dx}{dt}$ and $\frac{dy}{dt}$ in I,II and IV quadrant.

Answer to Problem 6RWDT

In I quadrant either the sign of $\frac{dx}{dt}$ will be negative and sign of $\frac{dy}{dt}$ will be negative or vice versa.

In II quadrant sign of $\frac{dx}{dt}$ and sign of $\frac{dy}{dt}$ will be positive, In IV quadrant either the sign of $\frac{dx}{dt}$ will be positive and sign of $\frac{dy}{dt}$ will be negative or vice-versa.

Explanation of Solution

Given information:

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

Calculation:

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

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

The equation for the circle with radius $r$ and center $\left(0,0\right)$ can be obtained by substituting $a=0,b=0$ in the general equation as given below:

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

In order to find sign of $\frac{dx}{dt}$ and $\frac{dy}{dt}$ in I,II and IV quadrant differentiate $x$ and $y$ with respect to $t$ .

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

In I quadrant $x$ and $y$ both are positive then equation $\frac{dx}{dt}=\frac{-y}{x}\frac{dy}{dt}$ will remain same therefore either sign of $\frac{dx}{dt}$ will be negative or sign of $\frac{dy}{dt}$ will be negative and vice-versa.

This gives the idea that one of the terms will be positive and other will be negative.

In II quadrant $x$ is negative and $y$ is positive the equation becomes:

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

Hence sign of $\frac{dx}{dt}$ will be positive and sign of $\frac{dy}{dt}$ will also be positive.

In IV quadrant $x$ and $y$ both are negative the equation becomes:

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

Hence either the sign of $\frac{dx}{dt}$ will be positive and sign of $\frac{dy}{dt}$ will be negative or vice-versa.