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Rate of change of volume of a rectangular box

  1. Dec 9, 2012 #1
    1. The problem statement, all variables and given/known data

    How fast is the volume of a rectangular box changing when the length = 6 cm, width = 5 cm and depth = 4 cm and the length and height are increasing at a rate of 1 cm/s and width is decreasing at a rate of 2 cm/s.

    2. Relevant equations

    Volume of a rectangle = l*w*h

    3. The attempt at a solution

    I know how to solve these kinds of the problems, done about 20 so far. The only trouble I am having is that I don't know which variable to take the derivative with since I have 3. I just can't figure out where to start this problem.
     
    Last edited: Dec 9, 2012
  2. jcsd
  3. Dec 9, 2012 #2
    Then you should take the derivative of all three variables! Just take the derivative of the exression of volume (lwh) and use Leibnitz's rule to compute it: you should have three terms and each of them has only one derivative. Then replace the given instantaneous values of lenght, width and height and also the derivatives with the rates of change of them
     
  4. Dec 9, 2012 #3
    what's Leibniz's rule? I don't think I have covered that in class
     
  5. Dec 9, 2012 #4
    When you want to take the derivative of a function that is the product of two other functions, then you take the derivative of one function multiplied by the other plus the derivative on the other multiplied by the first function. That is:

    (fg)' = f' g + f g'

    When you have a product of three functions then you use the above realation twice to get:

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

    That is Leibniz's rule for differentiation.
     
  6. Dec 9, 2012 #5
    All three dimensions (width, length, height) are dependent on time, so I'd use that as the only real variable:

    Volume of ractangle = V(t) = l(t) * w(t) * h(t).
     
  7. Dec 9, 2012 #6
    Take the derivative with respect to time. Whatever your given rates are wit respect to, that's what you'll differentiate wrt.
     
  8. Dec 9, 2012 #7
    Apparently there is some confusion when using the term "variable" to refer to lenght, widht or height and when using it to refer to time. In the first case, these are "depentent variables" because they depent on time, while in the second case, time is an "indepentant variable" .
     
  9. Dec 9, 2012 #8
    That's called the product rule in my book and this still doesn't make sense to me. If I go lwht3 I end up with t in my derivative and I don't know how to get rid of it
     
  10. Dec 9, 2012 #9
    lwht^3 ???? What is that? The volume is lwh, so just take it's derivative w.r.t. time:

    V = lwh → V' = l'wh + lw'h+lwh'

    replace in that expression l' , w' and h' with the given rates and replace l, w and h with the given instantaneous values.
     
  11. Dec 9, 2012 #10
    ahh that makes more sense I was going off of Michael Redei's equation. thanks for all the help
     
  12. Dec 9, 2012 #11
    Btw, you made a mistake typing out your question. You said, "the length and width are increasing at a rate of 1 cm/s and width is decreasing at a rate of 2 cm/s."

    Please specify the correct rates.
     
  13. Dec 9, 2012 #12
    length and height* but the question is solved so I guess it doesn't matter anymore.
     
  14. Dec 9, 2012 #13

    Ray Vickson

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    Well, his equation did not give you lwht^3, so I can't figure out where the t^3 comes from. Iit is not there, when you write things carefully.
     
  15. Dec 9, 2012 #14
    I think I can see what may have caused the misunderstanding:
    V(t) = l(t) * w(t) * h(t)​
    could be read as
    V*t = l*t * w*t * h*t = (l*w*h) * (t*t*t) = lwh * t³​
    But if you confuse w(t) and w*t, you're in real trouble...
     
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