Related Rates Anyone?

Question

Ok here goes:

Gas is being pumped into a spherical ballon at the rate of 3cm^3/s.
a) How fast is the radius increasing when the radius is 15cm?
b) Without using the result from part 'a', find the rate at which the surface area of the balloon is increasing when the radius is 15 cm.

I get the first part but how would you calculate the second without the result from the first!?!!

Answer 1

b) The answer is 2/5 = 0.4 cm^2 /sec. Do it as follows:

You are given the Volume of the sphere as a function of time:

V(t) = 3t cm^3 (t in sec.)

and so dV(t)/dt = 3 cm^3 /sec

The surface area (SA) is a function of t, and:

SA(t) = 4 pi r(t)^2

(1) d/dt SA(t) = 8 pi r(t) dr(t)/dt

And V(t) = 4/3 pi r(t)^3

(2) d/dt V(t) = 4 pi r(t)^2 dr(t)/dt

From (1) and (2) we see that:

d/dt SA(t) = (2/r(t)) d/dt V(t)

Since we are given r(t) = 15cm and we know dV(t)/dt = 3cm^3/sec we have:

d/dt SA(t) = (2/15cm)x 3cm^3 /sec = 2/5 cm^2 /sec

Answer 2

Normally, you would relate the surface area to the radius. Instead, relate it to the volume, and use the given fact that dV/dt = 3 cm^3/s.

V = 4/3pi*r^3, and S = 4pi*r^2, so r^3 = 3V/(4pi), meaning

S = 4pi*(r^3)^(2/3) = 4pi*(3V/(4pi))^(2/3).