Math Help!!!!?

Question

The radius of a right circular cylinder is given by sqrt(t+2) and its height is (1/6)sqrt(t), where t is time in seconds and the dimensions are in inches. Find the rate of change of the volume with respect to time.

Answer 1

V=Pi r^2 h
V=Pi (sqrt(t+2))^2 (1/6)sqrt(t)
V=Pi(t+2)(1/6)sqrt(t)

Distribute and then take the derivative!

Answer 2

V = pi*(t+2)* 1/6 sqrt (t)
V´ = pi/6[ sqrt(t) +(t+2)/(2sqrt(t)]= pi/(12sqrt(t) *(3t+2)

Answer 3

V= pi r^2h = pi (t+2)*sqrt(t)/6
V=[pi*t sqrt(t) +2pi sqrt(t)]/6
dV/dt =[pi t^.5 +pi t *1/2t^.5 + 2pi *1/2t^.5]/6
= pi/6[t^.5 + t/2t^.5 +t^.5]
= pi/6(2t^.5 +t/2t^.5)
=pi/6(5t/2t^.5)
= 5pisqrt(t)/12