Question

The area of a circle of radius r is given by $A = \pi r^2$ and its circumference is given by C = 2πr. At a certain point in time, the radius of the circle is r = 8 inches and the area of the circle is changing at a rate of $\frac{dA}{dt}=\pi\sqrt{2}$ inches per second. How fast is the radius of the circle changing at this time?

Answer 1

Answer:

the radius of the circle is changing $\frac{dr}{dt}=\frac{\sqrt{2}}{8}\ in/sec$ at that time.

Step-by-step explanation:

Given:

Area of Circle $A=\pi r^2$

Radius $r=8.in$

$\frac{dA}{dt}=\pi \sqrt{2} . in/sec$

We need to find the radius of the circle changing at this time $\frac{dr}{dt}$.

Solution:

To find the rate at which the radius of the circle is changing with respect to time we will apply derivative on Area of the circle.

Now we know that;

$A=\pi r^2$

Applying derivative we get;

$\frac{dA}{dt}=\frac{d}{dt} \pi r^2\\\\\frac{dA}{dt}=\pi r\frac{dr}{dt}$

Now substituting the given value we get;

$\pi \sqrt{2} =\pi \times 8 \frac{dr}{dt}\\\\ \frac{dr}{dt}=\frac{\pi \sqrt{2}}{\pi \times 8}\\\\ \frac{dr}{dt}=\frac{\sqrt{2}}{8}\ in/sec$

Hence the radius of the circle is changing $\frac{dr}{dt}=\frac{\sqrt{2}}{8}\ in/sec$ at that time.