[tex]\frac{d(Area)}{dt} =94.2ft^2/min[/tex] rate is its area changing when the radius is 5 ft.
Step-by-step explanation:
Here we have , The radius of a circle is increasing at a rate of 3 ft/min. We need to find At what rate is its area changing when the radius is 5 ft. Let's find out:
We know that area of circle = [tex]\pi r^2[/tex] .
⇒ [tex]Area = \pi r^2[/tex]
Differentiating both sides w.r.t to time we get :
⇒ [tex]\frac{d(Area)}{dt} = \pi\frac{ d(r^2)}{dt}[/tex]
⇒ [tex]\frac{d(Area)}{dt} = \pi(2r)\frac{ dr}{dt}[/tex] ............(1) { [tex]\frac{d(r^n)}{dr}= nr^{n-1}[/tex] }
But , According to Question
[tex]\frac{dr}{dt} = 3ft/min\\r=5ft[/tex]
Putting this value in equation (1):
⇒ [tex]\frac{d(Area)}{dt} = \pi(2(5))(3)[/tex]
⇒ [tex]\frac{d(Area)}{dt} =30 \pi[/tex]
⇒ [tex]\frac{d(Area)}{dt} =94.2ft^2/min[/tex]
Therefore , [tex]\frac{d(Area)}{dt} =94.2ft^2/min[/tex] rate is its area changing when the radius is 5 ft.
