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:

\frac{dV}{dt} = 10 - v (rate of inflow of water, and removing the volume 'v' which is leaking)

\frac{dh}{dt} = 4 (rate of increase in height h)

h=12

V = \frac13 \pi r^2h

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

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

Therefore h = 4r

From the chain rule; \frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}

Finding dV/dh:

V = \frac13 \pi r^2h, h = 4r

V = \frac13 \pi (\frac{h}{4})^2h

V =\frac{\pi h^3}{48}

\frac{dV}{dh} = 3\pi h^2

Substituting in chain rule:

\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}, dh/dt = 4

(10-v) = 3\pi h^2 \times 4, h = 12,

(10-v) = 3\pi (12)^2 \times 4

(10-v) = 3\pi (12)^2 \times 4

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.

\frac{dh}{dt}=\frac{\frac{dV}{dt}}{A(t)}

The area of the water when the water is 12 feet deep is simply {\pi}(3)^{2}=9{\pi}

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

\frac{1}{3}=\frac{10-x}{9\pi}

Solve for x and we get x=10-3{\pi}

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