The radius r and height h of a circular cone are increasing at a rate of 2 cm/min and 3 cm/min, respectively. How fast is the volume of this cone changing when r = 4cm and h = 3cm?

So far I have pi/3 [(4^2 dh/dt + 3(2pi(4)(2))]
I don't know how to find the dh/dt

Question

The radius r and height h of a circular cone are increasing at a rate of 2 cm/min and 3 cm/min, respectively. How fast is the volume of this cone changing when r = 4cm and h = 3cm?

So far I have pi/3 [(4^2 dh/dt + 3(2pi(4)(2))] 
I don't know how to find the dh/dt

anonymous · Answer

Could you retype your equation part?

anonymous · Answer

$$(\pi/3)[4^{2}*(dh/dt)+3(2\pi(4)(2))] $$

dumbcow · Answer

it is given in the problem....dh/dt = 3

dumbcow · Answer

$$\frac{dV}{dt} = \frac{dV}{dh}*\frac{dh}{dt}+ \frac{dV}{dr}*\frac{dr}{dt}$$
$$\frac{dV}{dt} = \frac{\pi r^{2}}{3}*3 + \frac{2 \pi r h}{3}*2$$
plug in ...r = 4, h=3

anonymous · Answer

Thanks definitely figured that out, but forgot to look on here to see if anyone replied! lol. but thanks again!