The formula for the volume of a cone is V = (1/3) * (pi * r²) * h.

The volume of water in the tank is a function of time:
V(t) = 10 (ft³ / min) * t (min).

As the water flows into the tank, it fills a smaller cone in the tip of the larger one, the volume of which is the volume of water that has entered the tank so far. The radius also varies with time, and this will give us trouble unless we do something about it now.

You can think of a cone as a right triangle that has been spun about its height. This triangle will let us find r as a function of h, so we can eliminate r altogether in our original formula.

For the cone given in this problem, r = h * tan[sin-¹(5/12)]. You should probably draw the triangle to convince yourself that this is true.

Ok, now plug that in for radius in our original formula:
V = (1/3) pi * h * {h * tan[sin-¹(5/12)] }²

We need h as a function of time, so start by solving for h:
h = (3V / (pi * tan[sin-¹(5/12)]² ))^(1/3)
Oof.

Remember that V is a function of time, so we can substitute that to get h as a function of time:
h(t) = (30t / (pi * tan[sin-¹(5/12)]² ))^(1/3)

Differentiate this function you must. I don't want to .

Too bad they didn't just give you a time. Instead, you have to find the rate of change of h when h = 8. You can find the time at which this happens by setting h = 8 and solving for t. Plug that t into h'(t), and you should be set.

Last edited by ryos (2005-10-15 15:46:35)

Hey ganesh, I was trying to learn from you , and I found a calculation error:

V = 1/3*pi*(5/12 *h)²*h
V = 1/3*pi*5/12*h³

You forgot to square the 5/12.

Also, dh/dt should have units of ft/min. I have no idea what, if anything, went wrong.

You'll never catch up with my record of mistakes!

Well...insomnia said he wanted to be a 'mod'. He never said he wanted to mod.