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Integration Word Problem

  1. Jan 30, 2007 #1
    1. The problem statement, all variables and given/known data

    A spherical raindrops evaporates at a rate proportional to its surface area. If its radius is 3mm, and 1 hour later has been reduced to 2mm, find an expresssion for the raduis of the raindrops at anytime.

    2. Relevant equations

    [tex]Volume = \frac{4}{3}\pi R^3[/tex]

    [tex]Area = 4\pi R^2[/tex]

    3. The attempt at a solution

    [tex]\frac{d}{dt}(\frac{4}{3}\pi R^3) = -k (4\pi R^2)[/tex]

    [tex]4\pi R^3 \frac{d}{dt} = -k 4\pi R^2[/tex]

    [tex]\frac{R dR}{dt} = -k[/tex]

    then... how do I plug in 2 and 3 mm??
     
  2. jcsd
  3. Jan 30, 2007 #2

    Dick

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    There's already a problem with your DE. d(R(t)^3)/dt=?
     
  4. Jan 30, 2007 #3
    this is precalculus math?
     
  5. Jan 30, 2007 #4
    I thought it was just basic integration? Sorry, I'll try and see if this can get moved then or something
     
  6. Jan 30, 2007 #5

    Dick

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    It reduces to the world's easiest differential equation. It's pretty basic, if not precisely 'pre-calc'.
     
  7. Jan 31, 2007 #6

    HallsofIvy

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    Well, it is basic differentiation- but that is "calculus", not "pre-calculus"!

    In any case, go back and look at your work again. What is the derivative of R3 with respect to R? Using the chain rule, what is the derivative of R3 with respect to t?
     
  8. Jan 31, 2007 #7
    [tex]= 4\pi\int R^3\frac{d}{dt}[/tex]

    [tex]= \frac{4\pi R^4}{t}[/tex]

    ??
     
  9. Feb 1, 2007 #8

    HallsofIvy

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    I agree: ?? That first line makes no sense. If "d/dt" is the derivative operator, you have to have something to differentiate! Also an integral has to have a "dt" or "dx" so you will know what the variable of integration is.

    If this is in response to my previous question "what is the derivative of R3 with respect to t", that was prompted by your
    [tex]\frac{d}{dt} \frac{4}{3}\pi R^3= 4\pi R^3 \frac{d}{dt}[/tex]
    which, again, makes no sense- what is that final d/dt applied to?

    Use the chain rule:
    [tex]\frac{d R^3}{dt}= \frac{d R^3}{dR}\frac{dR}{dt}[/tex]
     
  10. Feb 1, 2007 #9
    https://www.physicsforums.com/showthread.php?t=154042

    I'm not typing all that out again:smile: :tongue2:

    This is essentially how it is derived and this question asks about how to integrate the volume of half a sphere, but really it's asking pretty much the same thing, one is calculus, the other is calculus :smile: I never learned how it was derived until about two months ago. :smile: Talking about circles and spheres at the time and started playing around with the figures and what do you know :smile:

    do all that with t, and then integrate it and there you have it.

    [tex] \int_3^2 \int 4 \pi r^2/t =\int_3^2 \frac{4}{3}(\frac{\pi r^3}{t}) dt=[/tex]

    r between the value of 2 & 3; then make the equation =r; that's how I'd do it, something like that?
     
    Last edited: Feb 1, 2007
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