Answer:
[tex] \frac{dA}{dr} =2\pi (6in) (0.02 \frac{in}{min})=0.754 \frac{in^2}{min}[/tex]
Step-by-step explanation:
Data given
[tex] \frac{dr}{dt}= 0.02 \frac{in}{min}[/tex]
r = 6 in
Solution to the problem
We know that the area for a circle is given by:
[tex] A= \pi r^2[/tex]
We want to find at what rate is the area of the plate increasing at the instant when the radius is 6 inches. So we need to take partial derivates in both sides of the equation for the area, and if we do this we got:
[tex] \frac{dA}{dr} = 2\pi r \frac{dr}{dt}[/tex]
Since we have all the values provided we just need to replace and we got:
[tex] \frac{dA}{dr} =2\pi (6in) (0.02 \frac{in}{min})=0.754 \frac{in^2}{min}[/tex]