Answer:
Step-by-step explanation:
If I'm understanding this correctly, your problem is as follows:
The area of a circle is given by the formula
[tex]A=\pi r^2[/tex]
The area of the circle is changing at a rate of [tex]\frac{dA}{dt}=\sqrt{2}\pi[/tex]. Find the rate of change of the radius, [tex]\frac{dr}{dt}[/tex] , 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.
[tex]\frac{dA}{dt}=\pi2r\frac{dr}{dt}[/tex]
Filling in what we have:
[tex]\sqrt{2}\pi=\pi(2)(8)\frac{dr}{dt}[/tex] which simplifies a bit to
[tex]\sqrt{2}\pi=16\pi\frac{dr}{dt}[/tex]
Divide both sides by 16π to get:
[tex]\frac{\sqrt{2}\pi }{16\pi}=\frac{dr}{dt}[/tex]
The π's cancel leaving the rate of change of the radius as
[tex]\frac{dr}{dt}=.0883883476[/tex] inches per second
