Calculus related rates question.

Question

A Balloon in the shape of a sphere is being inflated so that its surface area is increasing at a rate of 5 square centimeters per second. At the instant when the radius of the balloon is 10cm, what is the rate of change of the volume of the balloon?

Answer 1

Surface Area SA = 4 * pi * radius^2 = 4 * pi * R^2

 

Volume = 4/3*pi*radius^3 = 4/3 * pi * R^3

 

 

Implicit differentiation:

 

dSA/dt = 8*pi*r * dR/dt  

 

(1/(8*pi))(dSA/dt)(1/r) = dR/dt <--- solves first equation for dR/dt

 

dV/dt = 4*pi*R^2 * dR/dt <----- second equation

 

dV/dt = 4 * pi * R^2 * (1/(8*pi))(dSA/dt)(1/R) <--- substitutes first equation into second equation

 

dV/dt  = (1/2)* R * dSA/dt <---- pi cancels; 4/8=1/2; R cancels

 

dV/dt = (1/2)*10*5 = 25 cm^3/sec