The Surface Area of a Spherical Balloon is Increasing at the Rate of 2 cm^2 / Sec. at What Rate is The Volume of the Balloon is Increasing When the Radius of the Balloon is 6 Cm? - Mathematics and Statistics

The surface area of a spherical balloon is increasing at the rate of 2cm2 / sec. At what rate is the volume of the balloon is increasing when the radius of the balloon is 6 cm?

Solution

Let r be the radius, A be the surface area and V be the volume of the spherical ballon at time t seconds.
Then,A=4pir2

dA/dt=4pi(2r((dr)/dt))

Given

2=8pirdr/dt

r dr/dt=2/8pi=1/4pi .............(1)

Now , V=4/3 pir^3

(dV)/dt=4pi/3xx3r^2 (dr)/dt=(4pir)(r (dr)/dt)

Given :  r=6, (d)V/dt=?

therefore (dV)/dt=4pi(6)(1/(4pi))=6...........from(1)

Therefore, the volume of the balloon is increasing at the rate of 6cm3/sec.

Is there an error in this question or solution?