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Related rates, simple question

  1. Jun 1, 2012 #1
    1. The problem statement, all variables and given/known data
    A snowball melts such that the volume decreases at a rate of 1cm3/min. At what rate is the diameter decreasing when diameter=10?

    I know the answer is [itex]\frac{-1cm}{50∏ per ?}[/itex]. My problem is with the units on the bottom. It was given as seconds, but shouldn't it be per minute? There was no conversion done anywhere in the problem. The answer is small, and it seems like it would fit better as per seconds rather than per minute, but I can't find out how we went from minutes to seconds.

    d=derivative
    D=diameter

    volume = [itex]\frac{4∏r^3}{3} = \frac{4∏(\frac{D}{2})^3}{3} = \frac{∏}{6}D^3[/itex]

    [itex]\frac{dv}{dt}=\frac{-1cm^3}{min}=-1[/itex]

    [itex]\frac{dv}{dt} = \frac{∏}{6}(3D^2)\frac{dD}{dt}[/itex]

    [itex]-1 = \frac{∏}{6}(3D^2)\frac{dD}{dt}[/itex]

    [itex]-1 = (\frac{D^2∏}{2})(\frac{dD}{dt})[/itex]

    [itex]\frac{dD}{dt} = \frac{-1}{(D^2∏)/2}[/itex]

    [itex]\frac{dD}{dt} = \frac{-2}{D^2∏}[/itex]

    [itex]\left. \frac{dD}{dt}\right|_{D = 10} = \frac{-2}{(10)^2∏} = \frac{-2}{100∏} = \frac{-1}{50∏}[/itex]
     
    Last edited: Jun 1, 2012
  2. jcsd
  3. Jun 1, 2012 #2

    Doc Al

    User Avatar

    Staff: Mentor

    I'd say you are correct. The units should be cm/min, not cm/s.
     
  4. Jun 1, 2012 #3

    Mark44

    Staff: Mentor

    Your work would be easier to follow if you used upper case D for diameter instead of lower case d. dd/dt is somewhat confusing, while dD/dt is probably less so.

    In your last line you have d(10)/dt = some nonzero number. I know what you mean, but this isn't the way to say it, since the derivative of every constant is zero.

    I think this is what you meant to say:
    $$\left. \frac{dD}{dt}\right|_{D = 10} = \text{whatever}$$
     
  5. Jun 1, 2012 #4
    Clarified, thanks
     
  6. Jun 2, 2012 #5

    Mark44

    Staff: Mentor

    Much better!
     
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