Expert: Paul Klarreich - 11/11/2008

Question
QUESTION: My question refers to this problem from Michael that you answered on February 28th:

The radius of a cylinder increases at a constant rate.  Its height is linear function of its radius, and increases three times as fast as the radius.  When the radius is 1 foot, the height it 6 feet.  When the radius is 6 feet, the rate of change in volume is 1 foot^3/second.  Find the rate at which the volume increases when the radius is 36 feet.

You concluded that
dV/dt = 3 pi(3r^2 + 2r)(c)

and then to find for c, you substituted 1 for dv/dt
dv/dt(at r = 6) = 1

1 = 3 pi(3 + 2)(c)

what I can't understand is why you didn't also substitute 6 for r, since dv/dt = 1 when r=6. It looks like you assigned r=1 but why?

Thanks,

Adam

ANSWER: Questioner:   Adam
Category:  Calculus
Private:  No
 
Subject:  Question regarding your related rates calculus answer
Question:  My question refers to this problem from Michael that you answered on February 28th:

The radius of a cylinder increases at a constant rate.  Its height is linear function of its radius, and increases three times as fast as the radius.  When the radius is 1 foot, the height it 6 feet.  When the radius is 6 feet, the rate of change in volume is 1 foot^3/second.  Find the rate at which the volume increases when the radius is 36 feet.

You concluded that
dV/dt = 3 pi(3r^2 + 2r)(c)

and then to find for c, you substituted 1 for dv/dt
dv/dt(at r = 6) = 1

1 = 3 pi(3 + 2)(c)

what I can't understand is why you didn't also substitute 6 for r, since dv/dt = 1 when r=6. It looks like you assigned r=1 but why?

Thanks,

Adam
.........................
Hi, Adam,

It looks like a boo-boo on my part. (They get more frequent as I get older.)

I think it should go like this:

The radius of a cylinder increases at a constant rate, so

dr/dt = c

Its height is linear function of its radius, and increases three times as fast as the radius, so

h = 3r + h0

When the radius is 1 foot, the height is 6 feet, so

6 = 3(1) + h0.
h0 = 3, so

h = 3r + 3  is the relation between r and h.

Now

V = pi r^2 h

V = pi r^2 (3r + 3)
V = 3 pi (r^3 + r^2)

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

That = 1 when  r = 6:

1 = 3 pi(3(36) + 6) c

1 = 3 pi(108 + 6) c

1 = 3 pi(114) c

1 = 342 pi c

c = 1/(342 pi)

etc.

I could not find the problem.  If you tell me where it is, I will post a correction.


---------- FOLLOW-UP ----------

QUESTION: Thank you. Here is the link to original question:

http://en.allexperts.com/q/Calculus-2063/2008/2/Related-Rates-30.htm

Answer
OK. I found it. You made a slight booboo, too. It was on Feb 26, not Feb 28.

I corrected it, and also fixed a small error in what I had sent you.

Thank you.  It is always encouraging (and sometimes embarrassing) to find people reading my answers.  Although I note from the record that Micheal, the original questioner, never read the answer.  The nerve of him!