A spherical balloon is being filled with air at a constant rate of 2 . How fast is the radius increasing when the radius is 4 ?

Question

A spherical balloon is being filled with air at a constant rate of 2   .  How fast is the radius increasing when the radius is 4   ? 

Please round your answer to 3 decimal places, and do not include units.)

Answer 1

Given: Volume increasing at rate of 2

Find: How fast is radius increasing when radius is 4?

Solution:

V = (4/3)*pi*r^3

d/dt [V] = d/dt [ (4/3)*pi*r^3 ]

V' = (4/3)*pi*[ 3*r^2*( r' ) ] Note: r' = dr/dt

V' = 4*pi*r^2*r'

2 = 4*pi*(4)^2*r'

2 = 64*pi*r'

r' = 2/(64*pi)

r' = 1/(32*pi)

Radius is increasing at rate of r'

r' = 1/(32*pi)

r' = 0.00995

r' = 0.010 Rounded to 3 decimal places