Question

The radius of a circle is increasing at a rate of 3 ft/min. At what rate is its area changing when the radius is 5 ft.

Answer 1

 $\frac{d(Area)}{dt} =94.2ft^2/min$ 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  = $\pi r^2$ .

⇒ $Area = \pi r^2$

Differentiating both sides w.r.t to time we get :

⇒ $\frac{d(Area)}{dt} = \pi\frac{ d(r^2)}{dt}$

⇒ $\frac{d(Area)}{dt} = \pi(2r)\frac{ dr}{dt}$    ............(1)              {  $\frac{d(r^n)}{dr}= nr^{n-1}$  }

But , According to Question

$\frac{dr}{dt} = 3ft/min\\r=5ft$

Putting this value in equation (1):

⇒ $\frac{d(Area)}{dt} = \pi(2(5))(3)$

⇒ $\frac{d(Area)}{dt} =30 \pi$

⇒ $\frac{d(Area)}{dt} =94.2ft^2/min$

Therefore ,  $\frac{d(Area)}{dt} =94.2ft^2/min$ rate is its area changing when the radius is 5 ft.