Question

The volume of a spherical balloon is increasing at a constant rate of 8 cubic feet per minute. How fast is the radius of the sphere increasing when the radius is exactly 10 feet?

Answer 1

Step-by-step explanation:

If the radius is r, then the rate of change of r with respect to time t,

ddt(r)=2 cm/minute

Volume as a function of radius r for a spherical object is

V(r)=43⋅π⋅r3

We need to find ddt(V) at r = 14cm

Now, ddt(V)=ddt(43⋅π⋅r3)=4π3⋅3⋅r2⋅ddt(r)=4π⋅r2⋅ddt(r)

But ddt(r) = 2cm/minute. Thus, ddt(V) at r = 14 cm is:

4π⋅142⋅2 cubic cm / minute =1568⋅π cc/minute