Helium is pumped into a spherical balloon at a rate of 3 cubic feet per second. How fast is the radius increasing after 2 minutes? Note: The volume of a sphere is given by V=43πr3V=43πr3. Rate of change of radius (in feet per second) =

The rate of change of radius is 0.0530 ft/sec.

Step-by-step explanation:

Given : Helium is pumped into a spherical balloon at a rate of 3 cubic feet per second.

To find : How fast is the radius increasing after 2 minutes?

Solution :

The volume of the sphere is

Differentiating w.r.t. r,

We have given, Helium is pumped into a spherical balloon at a rate of 3 cubic feet per second.

i.e.

Integrate w.r.t. t,

Assuming V=0 when  t=0 then C=0

So,

Now, we apply chain rule

i.e.  

The radius increasing after 2 minutes i.e. at

Therefore, the rate of change of radius is 0.0530 ft/sec.