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Apostol 4.12 #28 - Related Rates Problem

  1. Aug 19, 2011 #1
    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 [itex]\frac{dr}{dt}=c[/itex] 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 [itex]\frac{dh}{dt}=3\frac{dr}{dt}=3c[/itex]. Also, we have [itex]\frac{dh}{dr}=a[/itex] and so by the chain rule, [itex]\frac{dh}{dr}\frac{dr}{dt}=\frac{dh}{dt} \implies a=2c[/itex] as long as [itex]c\ne0[/itex], which we will have shortly.

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

    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 [itex]\frac{dv}{dt}=1[/itex]. This is how we know that [itex]c\ne0[/itex], since [itex]\frac{dv}{dt}=\frac{dv}{dr}\frac{dr}{dt}[/itex].

    Now we work to find n. We have [itex]6=2c+b \implies b=6-2c \implies h=2cr+6-2c[/itex]. The volume of a right-circular cylinder is given by [itex]v = \pi r^2 h =\pi r^2 (2cr+6-2c) = 2c\pi r^3+(6-2c)\pi r^2[/itex] and so [itex]\frac{dv}{dr}=6c\pi r^2 + 2(6-2c)\pi r[/itex]. Using the chain rule, we have [itex]\frac{dv}{dt}=(6c\pi r^2 + 2(6-2c)\pi r)c[/itex] and when r=6, [itex]1=(6c\pi 6^2 + 2(6-2c)\pi 6)c[/itex] which, simplified, is [itex]1=192 \pi c^2+72 \pi c[/itex]. 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 [itex]\frac{dv}{dt}[/itex] 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. jcsd
  3. Aug 19, 2011 #2

    micromass

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    Hi process91! :smile:

    I don't see that. Doesn't the information give you that [itex]\frac{dh}{dt}=3c[/itex]?? You seem to say that [itex]\frac{dr}{dt}=3c[/itex] and I don't see why you would say that.
     
  4. Aug 19, 2011 #3

    dynamicsolo

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

    Doesn't this imply ac = 3c ?
     
  5. Aug 19, 2011 #4
    Yes, that was a typo. I've corrected it in the original problem.
     
  6. Aug 19, 2011 #5

    dynamicsolo

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    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 .
     
  7. Aug 19, 2011 #6
    What part of the implication is incorrect? Just to make sure we're talking about the same spot, I have that
    [itex]\frac{dh}{dr}=a[/itex] and [itex]\frac{dr}{dt}=c[/itex] and [itex]\frac{dh}{dt}=3c[/itex]. Using the chain rule, that implies that [itex]ac=3c[/itex] and, as long as [itex]c\ne0[/itex], [itex]a=2c[/itex].

    For the part where I write [itex]6 = 2c + b[/itex], I am using the given statement that the height is 6 when the radius is 1 and the fact that [itex]h=ar+b=2cr+b[/itex] based on my above implication. Taking r=1 and h=6 yields [itex]6=2c+b[/itex].
     
  8. Aug 19, 2011 #7

    micromass

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

     
  9. Aug 19, 2011 #8
    *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.
     
  10. Aug 19, 2011 #9

    dynamicsolo

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    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
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