Question

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

Answer 1

Explanation:

If I'm understanding this correctly, your problem is as follows:

The area of a circle is given by the formula

`A=\pi r^2`

The area of the circle is changing at a rate of
`(dA)/(dt)=√(2)\pi`. Find the rate of change of the radius,
`(dr)/(dt)` , when r = 8.

Assuming that is what you are asking, we will begin by finding the derivative of the area of a circle using implicit differentiation.

`(dA)/(dt)=\pi2r(dr)/(dt)`

Filling in what we have:

`√(2)\pi=\pi(2)(8)(dr)/(dt)` which simplifies a bit to

`√(2)\pi=16\pi(dr)/(dt)`

Divide both sides by 16π to get:

`(√(2)\pi )/(16\pi)=(dr)/(dt)`

The π's cancel leaving the rate of change of the radius as

`(dr)/(dt)=.0883883476` inches per second