Apostol 4.12 #28 - Related Rates Problem

1. Aug 19, 2011

process91

Here is the question verbatim:

The radius of a right-circular cylinder increases at a constant rate. Its altitude is a linear function of the radius and increases three times as fast as the radius. When the radius is 1 foot, the altitude is 6 feet. When the radius is 6 feet, the volume is increasing at a rate of 1 cubic feet per second. When the radius is 36 feet, the volume is increasing at a rate of n cubic feet per second, where n is an integer. Compute n.

I set the problem up as follows:
The radius of a right-circular cylinder increases at a constant rate. Then let r be the radius and t be the number of seconds. This implies that $\frac{dr}{dt}=c$ for some constant c.

Its altitude is a linear function of the radius Letting h be the altitude, we have h=ar+b for some constants a and b.
and increases three times as fast as the radius. Then $\frac{dh}{dt}=3\frac{dr}{dt}=3c$. Also, we have $\frac{dh}{dr}=a$ and so by the chain rule, $\frac{dh}{dr}\frac{dr}{dt}=\frac{dh}{dt} \implies a=2c$ as long as $c\ne0$, which we will have shortly.

When the radius is 1 foot, the altitude is 6 feet. So $6=2c+b$.

When the radius is 6 feet, the volume is increasing at a rate of 1 cubic feet per second. Letting v be the volume, we have $\frac{dv}{dt}=1$. This is how we know that $c\ne0$, since $\frac{dv}{dt}=\frac{dv}{dr}\frac{dr}{dt}$.

Now we work to find n. We have $6=2c+b \implies b=6-2c \implies h=2cr+6-2c$. The volume of a right-circular cylinder is given by $v = \pi r^2 h =\pi r^2 (2cr+6-2c) = 2c\pi r^3+(6-2c)\pi r^2$ and so $\frac{dv}{dr}=6c\pi r^2 + 2(6-2c)\pi r$. Using the chain rule, we have $\frac{dv}{dt}=(6c\pi r^2 + 2(6-2c)\pi r)c$ and when r=6, $1=(6c\pi 6^2 + 2(6-2c)\pi 6)c$ which, simplified, is $1=192 \pi c^2+72 \pi c$. From this alone, we can see that whatever c is, it is a very complicated number because it must remove the values of pi in the equation. Solving this using the quadratic equation will give us two possible values:
Solved with WolframAlpha

Using the positive one, since the radius is increasing, we should be able to compute $\frac{dv}{dt}$ directly, however for r=36 we don't get an integer.
Final Solution from WolframAlpha

The book's answer is 33. Where did I go wrong?

Last edited: Aug 19, 2011
2. Aug 19, 2011

micromass

Staff Emeritus
Hi process91!

I don't see that. Doesn't the information give you that $\frac{dh}{dt}=3c$?? You seem to say that $\frac{dr}{dt}=3c$ and I don't see why you would say that.

3. Aug 19, 2011

dynamicsolo

Have a look at this line again (I corrected a typo):

Doesn't this imply ac = 3c ?

4. Aug 19, 2011

process91

Yes, that was a typo. I've corrected it in the original problem.

5. Aug 19, 2011

dynamicsolo

OK, that takes care of the typo, but your implication is still incorrect. Also,

I think you want to write 6 = 3r + b here. This equation for the height involves r , not dr/dt .

6. Aug 19, 2011

process91

What part of the implication is incorrect? Just to make sure we're talking about the same spot, I have that
$\frac{dh}{dr}=a$ and $\frac{dr}{dt}=c$ and $\frac{dh}{dt}=3c$. Using the chain rule, that implies that $ac=3c$ and, as long as $c\ne0$, $a=2c$.

For the part where I write $6 = 2c + b$, I am using the given statement that the height is 6 when the radius is 1 and the fact that $h=ar+b=2cr+b$ based on my above implication. Taking r=1 and h=6 yields $6=2c+b$.

7. Aug 19, 2011

micromass

Staff Emeritus
Yes, but doesnt ac=3c imply that a=3??

8. Aug 19, 2011

process91

*blink, blink* Yes, it does. I must be tired or something. Thank you so much - I was tearing my hair out. Sometimes it's the little things.

Thanks to dynamicsolo as well, I see that you were also pointing to this flaw.

9. Aug 19, 2011

dynamicsolo

Sleep deprivation (or fatigue) and mathematics don't mix...

By the way, you will never need to use dr/dt for anything -- it can be divided out when you compare dV/dt at r = 6 with dV/dt at r = 36 .

Last edited: Aug 19, 2011