Apostol 4.12 #28 - Related Rates Problem

[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?

 

[micromass]

Hi 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{dr}{dt}=3\frac{dr}{dt}=3c.

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.

 

[dynamicsolo]

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

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.

Doesn't this imply ac = 3c ?

 

[process91]

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



Doesn't this imply ac = 3c ?

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

 

[dynamicsolo]

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

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

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

 

[process91]

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 .

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.

 

[micromass]

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.

Yes, but doesn't ac=3c imply that a=3??

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.

 

[process91]

Yes, but doesn't ac=3c imply that a=3??

*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.

 

[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 .