Question

The top half of a storage tank is a circular cylinder that is 5 meters tall and has a diameter 2 meters. bottom half of the tank is shaped like an 8-meter inverted cone (pointed down). Let h represent the depth of the tank's contents. At t = 0 minutes, a release valve at the bottom of the tank is opened and its contents flow out at a rate of 0.5 cubic meters per minute. Assuming the tank is completely full when the release valve is opened, what is the value of dh/dt when t = 30 minutes?

Do I need to derive the volume of cylinder (V=pi*h*r^3) with respect to t first?

Answer 1

Well, the volume is not only the cylinder, its the cone as well. So you need to find the volume of the cone and add it to that of the cylinder, first.

Answer 2

Okay, I got 23/3pi. What do I do next? I really don't get how to do this problem...

Answer 3

Hmm.. I think it's writing h in terms of V. Once you complete the implicit differentiation you're left with a differential equation that you can plug in values for. But honestly i'm not sure I haven't done this for a while..

Answer 4

do you mean dV/dh?

Answer 5

for the volume of both the cylinder and the cone?

Answer 6

I mean like

\[V=\pi*h*r^3\] =>
\[h = \frac{ V }{ \pi * r^3 }\]

Answer 7

oh...

Answer 8

then
\[\frac{ dh }{ dt } = \frac{ dV }{ dt }* \frac{ 1 }{ \pi*r^3 }\]

Answer 9

okay i wil try

Answer 10

But I don't remember what to do with that or do after. Sorry :/ But good luck with it

Answer 11

you can try to search google there's lots of problems like that.

Answer 12

alright, thanks for helping me up to this point :) i appreciate it