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
`(dA)/(dt)=\pi√(2)` inches per second. How fast is the radius of the circle changing at this time?

Answer 1

the radius of the circle is changing
`(dr)/(dt)=(√(2))/(8)\ in/sec` at that time.

Explanation:

Given:

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

Radius
`r=8.in`

`(dA)/(dt)=\pi √(2) . in/sec`

We need to find the radius of the circle changing at this time
`(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;

`(dA)/(dt)=(d)/(dt) \pi r^2\\\\(dA)/(dt)=\pi r(dr)/(dt)`

Now substituting the given value we get;

`\pi √(2) =\pi * 8 (dr)/(dt)\\\\ (dr)/(dt)=(\pi √(2))/(\pi * 8)\\\\ (dr)/(dt)=(√(2))/(8)\ in/sec`

Hence the radius of the circle is changing
`(dr)/(dt)=(√(2))/(8)\ in/sec` at that time.