The right procedure on difficult related rates problems I'm pretty sure the sample problems my teacher gives to us violate some article of the Geneva convention. I'm in talks with my embassy about that, but in the mean time maybe you guys could look over my steps and tell me how badly I screwed up. The first part might be right, I'm afraid sure the second one isn't. I don't expect anyone to double check all the math on this, but if they could tell me whether or not I'm following the right procedure, that would be great.
The question is as follows: A tank of water in the shape of a cone is leaking water at a constant rate of 2 ft3/hr. The base radius of the tank is 5 ft and the height of the tank is 14 ft. 
(a) At what rate is the depth of the water in the tank changing when the depth of the water is 6 ft? 
(b) At what rate is the radius of the top of the water in the tank changing when the depth of the water is 6 ft?

Part A:
So, I begin by identifying what I've been told and what I need to discover, as well as writing down the equation for volume in a cylinder.
$$\frac{dv}{dt} = -2ft^3/hr$$
$$\frac{dh}{dt} = ?$$
$$V = \frac{1}{3}\pi r^2h$$
I would like to isolate the volume formula in regards to h (that's not the right terminology is it?) so I used related triangles to discover:
$$\frac{r}{h} = \frac{5}{14}$$
$$r = \frac{5h}{14}$$
I plug that back into the volume equation and get:
$$V = \frac{25\pi h^2}{588}$$
I use implicit differentiation to render:
$$\frac{dv}{dt} = \frac{25\pi h^2}{196}*\frac{dh}{dt}$$
I plug in the values for $\frac{dv}{dt}$ and height to render:
$$\frac{dh}{dt} = \frac{-98}{225\pi}$$
Part B
Now I'm trying to find $\frac{dr}{dt}$ I use related triangles again but this time I render: $$h = \frac{14r}{5}$$
I plug that into my volume formula to get:
$$V = \frac{1/3}{\pi r^3*\frac{14r}{5}}\Rightarrow V = \frac{14}{15}\pi r^3$$
I take my implicit differentiation:
$$\frac{dv}{dt} = \frac{42}{15}\pi r^2*\frac{dr}{dt}$$
Now I plug in my values and simplify a lot to get:
$$-2 = \frac{102}{7}*\frac{dr}{dt}\Rightarrow\frac{dr}{dt} = \frac{-14}{102}$$
So what do you guys think? I want to thank anyone ahead of time who took the time to read all that, it took forever to type!
 A: Good shot.
The basics are right:  $V=(1/3)\pi r^2 h$ and $h=14r/5.$
For part (a) I agree with you:
$$V(h) = \frac{25 \pi h^3}{588} $$
So,
$$\frac{dV(h)}{dt} = \frac{dV(h)}{dh} \frac{dh}{dt} = \frac{25 \pi h^2}{196}\frac{dh}{dt} = -2,$$
and
$$\frac{dh}{dt} = -\frac{392}{25 \pi h^2},$$
which at $h=6$ (feet) is
$$\frac{dh}{dt} = -\frac{98}{225 \pi}$$ feet per second.
For part (b) it seems like you're doing more work than necessary.  We are looking for the rate of change in radius at the same height we found the rate of change in height for.  But we know that $h = 14r/5$ so
$$\frac{dh}{dt} = \frac{14}{5} \frac{dr}{dt} \to \frac{dr}{dt} = \frac{5}{14} \frac{dh}{dt}.$$
So multiply your answer in (a) by $5/14$ and you're done.
A: Given:
$$r/h = 5/14$$
$$V = \frac{1}{3}\pi r^2h = \frac{1}{3}\pi(\frac{5}{14}h)^2h = \frac{1}{3}\pi r^2(\frac{14}{5}r)$$
$$\frac{dV}{dt}=-2ft^3/hr$$
To find $dh/dt$
$$dh/dt = \frac{dV}{dt}\cdot\frac{dh}{dV}=\frac{\frac{dV}{dt}}{\frac{DV}{dh}}$$
Part B is really 2 problems
What is the radius if the height is 6ft?
Find $dr/dt$
$$dr/dt = \frac{dV}{dt}\cdot\frac{dr}{dV}=\frac{\frac{dV}{dt}}{\frac{dV}{dr}}$$
or alternatively
$$dr/dt = \frac{dV}{dt}\cdot\frac{dh}{dV}\cdot\frac{dr}{dh}=\frac{\frac{dV}{dt}\cdot\frac{dr}{dh}}{\frac{dV}{dh}}$$
