Chapter 4.1, Problem 8E

To determine

To find: The value of $\frac{dx}{dt}$ when $y=4$ .

Answer to Problem 8E

The value of $\frac{dx}{dt}$ when $y=4$ is $-8$ .

Explanation of Solution

Given information: The equation is $x^2+y^2=25$ and $\frac{dy}{dt}=6$ .

Calculation:

Consider the equation.

  $x^2+y^2=25$

Substitute $y=4$ in above equation.

  $\begin{array}{l}{x}^2+4^2=25\\ x^2+16=25\\ x=3\end{array}$

Differentiate above equation.

  $\begin{array}{c}2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\\ 2x\frac{dx}{dt}=0-2y\frac{dy}{dt}\\ \frac{dx}{dt}=-\frac{y}{x}\frac{dy}{dt}\\ \frac{dx}{dt}=-\frac{4}{3}\left(6\right)\\ =-8\end{array}$