1. ## related rates question?

Assume the volume V of a cube is increasing at a constant rate of 3cm^3 per second. Let t0 be the instant (t>0) when the rate of change of the volume (cm^3/sec) is numerically equal to the rate of change of the surface area (cm^2/sec) for the cube. Assume V=0 when t=0.

a. Find the rate of change of the length of a side when t=t0
b. Find the rate of change of the surface area when t=t0
c. Find the value of t0.

Thanks a lot everyone.

2. Originally Posted by GlobalCooling
Assume the volume V of a cube is increasing at a constant rate of 3cm^3 per second. Let t0 be the instant (t>0) when the rate of change of the volume (cm^3/sec) is numerically equal to the rate of change of the surface area (cm^2/sec) for the cube. Assume V=0 when t=0.

a. Find the rate of change of the length of a side when t=t0
b. Find the rate of change of the surface area when t=t0
c. Find the value of t0.

Thanks a lot everyone.
V = r^3 and A= 6r^2 where r depends on t.

dV/dt = 3 r^2 dr/dt (by the chain rule)

dA/dt = 12 r dr/dt

part c) first:

Setting the two equal to each other, we get 4r = r^2, with solutions r=0 and r = 4. So when the side is 4 units the two rates (dV/dt and dA/dt) are equal.

a) Use the fact that dV/dt is constant and r = 2, to find dr/dt:
3 r^2 dr/dt = 3
3 4^2 dr/dt = 3
dr/dt = 1/16 cm/s

part b) should be straight-forward now.

Good luck!!