Another problem in related rates.. hope you can help me find how to have the correct answer.

Water, at the rate 10 ft3/min, is pouring into a leaky cistern whose shape is a cone 16' deep and 8' in diameter at the top. At the time the water is 12' deep, the water level is observed to be rising 4"/min. How fast is the water leaking away?
Answer: (10 - 3pi)ft3/min

Thank you! :)

Proceed according to the following steps:
1. Find a relation between radius(r) and height(h) of the cone.
2. Substitute for 'r' in terms of 'h'in the volume equation of the cone.
3. Find dV/dh---we are finding this at a particular instant of time.So use h=12.
4. The rate of increase of water is given to you.ie, dh/dt--convert it into the units used in the above calculations.
5.Find outflow given by dV/dt using the relation dV/dt=(dV/dh)(dh/dt).
6.The final answer is inflow-outflow.

Ok, let's put all the data given in a simple way:

(rate of inflow of water, and removing the volume 'v' which is leaking)

(rate of increase in height h)





When h = 16, d = 8 and r = 8/2 = 4

Ratio h/r = 16/4 = 4/1

Therefore h = 4r

From the chain rule;

Finding dV/dh:

,







Substituting in chain rule:

, dh/dt = 4

, h = 12,





Solve for v, which is the volume leaking.

:)

Here is a quick way to do these sorts of related rates when you are dealing with a change in height at a certain time, dh/dt.

Find the area of the water surface at that particular moment.

In this case, when it is 12' deep the radius of the water is 3 feet.

Note that the change in height is equal to the change in volume divided by the area at that moment.



The area of the water when the water is 12 feet deep is simply

4 inches is 1/3 feet, so we have:



Solve for x and we get

That's it.

Ok I got! Thanks a lot! :)

Hey, sorry if I didn't take into consideration the units. I'm not, really not at all, used to those british system units of inches, feet, etc. :o