Thevolumer V of a cone (V=1/3pir^2h) is increasing at the rate of 28pi cubic units per second. At the instant when the radius r of the cone is 3 units, its volume is 12pi cubic units and the radius is increasing at 1/2 unit per second….
1) at the instant when the radius of the cone is 3 units, what is the rate of change of the area of the base?

PLEASE HELP….

Question

Thevolumer V of a cone (V=1/3pir^2h) is increasing at the rate of 28pi cubic units per second. At the instant when the radius r of the cone is 3 units, its volume is 12pi cubic units and the radius is increasing at 1/2 unit per second….
1) at the instant when the radius of the cone is 3 units, what is the rate of change of the area of the base?

PLEASE HELP….

anonymous · Answer

$$V(t)=\frac{1}{3}r^2(t)h(t)$$
$$V'(t)=\frac{1}{3}\left(2r(t)r'(t)h(t)+r^2(t)h'(t)\right)$$ is a start

anonymous · Answer

that being the product rule (plus chain rule)
you are told $V'=28,r=3,r'=\frac{1}{2}$

anonymous · Answer

you have to solve for $h'$ and you are done

anonymous · Answer

well i guess you also have to solve for $h$ but you are given enough information for that too

anonymous · Answer

how did you get the first equation?@satellite73