# Homework Help: 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

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

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