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Homework Help: Related Rate Problem

  1. Oct 9, 2005 #1
    Here's my problem:

    A machine is rolling a metal cylinder under pressure. The radius, r, of the cylinder is decreasing at a constant rate of .05 inches per second and the volume, V, remains constant at 128(pi) cubic inches. At what rate is the length, h, changing when the radius is 2.5 inches?

    So dr/dt= .05 v=128(pi) r=2.5 and I should be able to solve for h using the equation:

    v=(pi)(r^2)(h) right?

    So where do I go after this?
  2. jcsd
  3. Oct 9, 2005 #2


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    Go ahead and differentiate with respect to time. You will have the terms dv/dt, dr/dt, and dh/dt. Don't forget to use the product rule on the right side.

    You know dv/dt = 0, dr/dt = 0.05, r = 2.5, h = (solve from V = pi(r^2)h) which turns it into a plug in the numbers question.
  4. Oct 9, 2005 #3
    I'm looking for dh/dt, right?
  5. Oct 9, 2005 #4
    and when I took the derivative the equation, I got

    (pi)(r^2) + (2pi(r))+h

    is this correct?
  6. Oct 9, 2005 #5


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    dh/dt reads, the change of h with respect to time. Sounds like it.

    Your equation there is very incomplete. Where is the dv/dt, dh/dt, dr/dt? You are not using the product rule. r is a function of time. h is a function of time.

    Suppose we have a function
    f(x,y) = xy
    df/dt = ydx/dt + xdy/dt
    Look familiar? It's the product rule.
  7. Oct 9, 2005 #6

    so is it:

    dv/dt= dr/dt(2pi*r) + dh/dt

    i don't understand why i would have a dv/dt because the volume isnt changing though
  8. Oct 9, 2005 #7


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    You are having difficulty understanding the physical meaning of the mathematics you are doing I can see. You should be consistent following the rules of mathematics rather than let instinct simplify things. And so... from your response you aren't familiar with the product rule it seems. Look through your notes. It would say.

    f(x) = g(x)h(x)
    f'(x) = g(x)h'(x) + h(x)g'(x)

    Do you see a similarity, this time in your problem, instead of f, g, and h, you have:

    v(t) = A[r(t)]^2h(t), if you let u = A(r(t))^2. You can do this through the chain rule, then:
    v(t) = u(t)h(t)

    Don't let the changing of variables distract you. The volume of the cylinder along with the radius and height are all functions of time. Thus, so you may check your answer:

    [tex]V = \pi r^2 h [/tex]

    [tex]\frac{dV}{dt} = \pi[r^2\frac{dh}{dt} + 2rh\frac{dr}{dt}][/tex]

    If there's something you understand about the product or chain rule, let me know again.

    Edit: A bit more to explain this equation. dV/dt indicates the rate in which the volume is changing over time. Apparently the case is that dV/dt is zero. The other d/dt's mean similar things. Now you wouldn't just arbitrarily "leave them out" or omit them from an equation just because you think they are not doing something meritable. You may substitute them with proper values later on. Actually, that's why the subject is called related rates!. The rates of change of several quantities of interest are related through an equation.
    Last edited: Oct 9, 2005
  9. Oct 9, 2005 #8
    Ok, I see what I was doing wrong now... I understand the chain and product rule but instead of seeing the equation as *h I was seeing it as plus h which really messed me up.

    I also now see dV/dt thing... so basically I should be able to set the derivative equal to zero and then plug in the values to obtain the rate, correct?
  10. Oct 9, 2005 #9


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    Yups, that's right ^_^.
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