Post #11

Calculus Week #11

In this week of Calculus, we focused on reviewing for our short quiz we had, and then we all turned in ideas that we were uncomfortable with so that we could get packets on those topics in preparation for our next major test, which is a test on all of derivatives before we move on to another topic in Calculus.

As for my example, I'm going to explain how to do a related rate problem with a conical tank.

A conical tank (with vertex down) is 10 feet across the top and 12 feet deep. If water is flowing into the tank at a rate of 10 cubic feet per minute, find the rate of change of the depth of the water when the water is 8 feet deep.

Now, to start, we know that it is a related rates problem. We also know that dV/dt = 10ft/min and we want to find dh/dt (the depth) when the h=8ft. Also, we know for a conical tank, the volume is given by V = 1/3 pi r^2 h. Now, if we would take the implicit derivative with respect to time at this particular point we would have two unknowns. So, to work this out, we will solve for h in terms of r.

Setting up the proportion:

r/h = 10/12

We can solve this for r to find 5h/6.

Now we can plug this into our original and then take the implicit derivative.

dV 25 pi h^2 dh
-- = --------- --
dt 36 dt

We can pug in for dV/dt and for h=8 then solve for dh/dt to find our answer of:

dh/dt = 9/(40pi)

It's really not that hard of a problem. Just knowing the formula and the fact that you have to use a secondary equation (kind of like optimization) so that you only have one unknown after taking the derivative is all that is needed.

Anyway, good luck on this!