okay last problem, I solved the others myself, having trouble with the pieces of this... A conical tank has height 10 m and radius 7 m at the top. Water flows in at a rate of 4 m3/min. How fast is the water level rising when the water level is 3 m? okay... so 1/3*pi*(r^2)*h is volume... the volume of the filled portion would be a ratio to the overall volume? but how do you fit all that into a differentiable formula so that it can be implicitly differentiated for rates and then solved?

Question

anonymous · Answer

Here is what I attempted, but the answer was rejected by the computer (computer based assignment - I have several tries before I get it wrong).
$$Volume=1/3\pi(\pi*R^2)*H $$ for the volume of the container.
but to figure out the volume of the filled portion I used a radius in respect to height of the filled portion from the over all dimension of 10m high and 7m radius. So the radius of the filled portion is: $$r=7*h/10$$ using this in the over all formula as the radius gives me now this formula to implicitly differentiate:  $$V=1/3\pi*(7/10*h)^2*h$$ if I did that right it should just be a function of using chain rule and product rule to differentiate. I pulled the constant out first by squaring it to be 49/100 (maybe I can't do that here?) Which I got:
$$dv/dt = 1/3 * 49/100 * \pi * 2h*dh/dt*h + 1/3*49/100* \pi h^2*dh/dt$$ and when simplified and solved for dh/dt I got $$4*dV/dt * 100/49 * 1/(\pi*h^2) = dh/dt$$
and when I enter the values for height of 3m and rate of 4m^3/sec filling I get the following (wrong) answer... $$400/441 * 1/\pi =dh/dt$$
if someone sees where I went wrong, or can point me in the right direction to continue it would be helpful. Thank you.