# Math Help - Related Rates Question 2

1. ## Related Rates Question 2

Water is draining from a conical tank with a height of 12 feet and diameter of 8 feet into a cylindrical tank that has a base with area 400pi square feet. The depth h, in feet, of the water in the conical tank is changing at a rate of (h-12) feet per minute.

(a) Write an expression for the volume of water in the conival tank as a function of h.
(b) At what rate is the volume of water in the conical tank changing when h=3?
Thanks for your help.

2. The motivation for related rates is expressing one rate in terms of another. This is useful because some object's propriets are easier to measure than others. A famous textbook example is the spherical baloon one. It's way easier to measure the baloon volume than the radius of the baloon. If you need to know the rate at which the radius is growing you can derivate a formula for this from the volume formula.

So, for (a):

$V(h) = \frac{1}{3} 4^2\pi h$

In (b), you want to know how these quantities vary over TIME. But first you need to know what's the relation between the variation of V and h:

$\frac{dV}{dh} = \frac{16}{3} (I)$

Now derivating with respect to time (remember that V is a function of h, so you use the chain rule):

$\frac{dV}{dT} = \frac{dV}{dh}\frac{dh}{dt}$

(I) gives you dV/dh, and the problem statement gives you $\frac{dh}{dt} = (h-12) ft/min$:

$\frac{dV}{dt} = \frac{16(h-12)}{3} ft^3/min$

It should be clear from now on. I hope I didn't make any major mistake in my explanations or calculations, since I don't see this subject for a while now.

Regards,

3. Another way to look at some related rates, especially when dealing with a

volume change and depth change is to note that the rate of change of the

volume is equal to the cross-sectional area at that moment times the rate of

change of the depth.

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

You have H=12, r=4, dh/dt=-9 when h=3.

By similar triangles, the radius of the surface of the water at that moment is $\frac{4}{12}=\frac{r}{3}, \;\ r=1$

Thererfore, the area of the surface of the water when h=3 is ${\pi}(1)^{2}={\pi}$

Now, all we need is to plug them into the formula:

$\frac{dV}{dt}=({\pi})\cdot (-9)= \boxed{-9{\pi} \;\ \frac{ft^{3}}{min}}$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Also, by similar triangles before, $\frac{4}{12}=\frac{r}{h}$

$r=\frac{h}{3}$

Sub into the volume of cone formula:

$V=\frac{\pi}{3}(\frac{h}{3})^{2}h$

$=\frac{\pi}{27}h^{3}$

$\frac{dV}{dt}=\frac{\pi}{9}h^{2}\cdot \frac{dh}{dt}$

Plug in the knowns and we see:

$\frac{dV}{dt}=\frac{\pi}{9}(3)^{2}(-9)=\boxed{-9{\pi} \;\ \frac{ft^{3}}{min}}$

Same as previous method.

I like the first method better because it is easier. As long as it is easy to find the surface area of the water at that moment.