related rates balloon problem

Question

A spherical balloon is inflated so that its volume is increasing at the rate of 3.6 ft^3/min. How rapidly is the diameter of the balloon increasing when the diameter is 1.8 feet?

Answer 1

v=(4/3)pi(d/2)^3

v= (1/6)pid^3

v'= (1/2)pi(d^2)d'

3.6=(.5)pi(1.8)^2(d')

d'= 3.6/.5(1.8^2)pi

= 7.2/pi(1.8^2)

=about .707 feet per minute = rate of change of the diameter

Answer 2

V = (4/3)πr³ = (4/3)π(1/2 D)³ = (1/6)πD³

dV/dt = (1/2)πD²(dD/dt)

3.6 = (1/2)π(1.8)²(dD/dt)

Solve for dD/dt.