Check out some similar questions!

akotoh · Sep 22, 2009, 07:57 AM

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! :)

radiation · Sep 22, 2009, 08:29 AM

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.

Unknown008 · Sep 22, 2009, 08:37 AM

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.

:)

galactus · Sep 22, 2009, 10:09 AM

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.

akotoh · Sep 22, 2009, 03:49 PM

Ok I got! Thanks a lot! :)

Unknown008 · Sep 23, 2009, 07:49 AM

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