# Another related rates problem.

• Sep 18th 2012, 07:24 PM
larrydevlin1770
Another related rates problem.
I have tried to solve the following problem. When I perform the differentiation before supplying the given variables, I am left with two unknowns: dr/dt and dh/dt. When I try to remove a variable by supplying the radius as a constant before differentiation, I receive the wrong answer (it is exactly 3x too big). Several other problems later on give me the same trouble, so I am hoping that an understanding of this problem will enlighten me to the proper method to solve the others.

"A water tank is in the shape of a cone with vertical axis and vertex downward. The tank's radius is 3 ft. and the tank is 5 ft. high. At first the tank is full of water, but at time t = 0 (in seconds), a small hole at the vertex is opened, and the water begins to drain. When the height of water in the tank has dropped to 3 ft, the water is flowing out at 0.02 ft^3/s. At what rate, in feet per second, is the water level dropping then?"

I am working with a radius of 9/5 ft of the surface of the water.

• Sep 18th 2012, 07:36 PM
MarkFL
Re: Another related rates problem.
You have the correct radius of the water's surface at the time the depth of the water is 3 ft.

Can you show your working for the problem, so we may identify where you are erring?
• Sep 19th 2012, 07:36 PM
larrydevlin1770
Re: Another related rates problem.
Ok. The knowns: h0=5 ft. r0=3 ft, h1=3 ft, dV/dt = -0.02 ft3/s. Formula for the volume of a cone: V = 1/3 pi r^2 h. Because the h1 is 3/5 of h0, r1 = 9/5.

Sought variable: dh/dt at t1

Now, one of the cardinal rules for doing this type of problem is that we don't insert known quantities until after differentiation. So my first attempt went like this:

V = 1/3 pi r^2 h.

dV/dt = (((2pi rh)/3) dr/dt) + (((pi r^2)/3)dh/dt)

-0.02 = (18pi/5)dr/dt + 81pi/75 dh/dt.

This leaves me with two unknowns: dr/dt and dh/dt.

My second approach was to supply the radius as a constant before differentiation, to leave just one unknown: dh/dt.

So:

V = 1/3 pi r^2 h.

V = 1/3 pi (81/25) h

dV/Dt = 1/3 pi (81/25) dh/dt

-0.02 = 1/3 pi (81/25) dh/dt

dh/dt = ((3)(-0.02)(25))/81pi

= -1.5/81 pi

= -1/54pi ft./s.

But this answer is exactly three times two big according to the answer key. The answer key has -1/162 pi. What did I do wrong?
• Sep 19th 2012, 09:25 PM
MarkFL
Re: Another related rates problem.
Your mistake with the second approach is to make r a constant. I would work it as follows:

$\displaystyle V=\frac{1}{3}\pi r^2h$

$\displaystyle r=\frac{3}{5}h$

Thus:

$\displaystyle V(h)=\frac{1}{3}\pi\left(\frac{3}{5}h \right)^2h=\frac{3\pi}{25}h^3$

Now, the chain rule gives us:

$\displaystyle \frac{dV}{dt}=\frac{dV}{dh}\cdot\frac{dh}{dt}\,\to \,\frac{dh}{dt}=\frac{\frac{dV}{dt}}{\frac{dV}{dh} }$

$\displaystyle \frac{dh}{dt}=\frac{-0.02}{\frac{9\pi}{25}h^2}$

Hence, at $\displaystyle h=3$, we find in ft/s:

$\displaystyle \frac{dh}{dt}=-\frac{1}{162\pi}$
• Sep 21st 2012, 06:44 AM
larrydevlin1770
Re: Another related rates problem.
This makes sense. Thank you so much for your help. I think this solution will help me with my other problems, too.