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Homework Help: Derivatives Problem (Calculus I)

  1. Nov 7, 2012 #1
    1. The problem statement, all variables and given/known data
    find d/dt for a rectangle.

    2. Relevant equations
    product rule for derivatives (the first times the derivative of the second plus the second times the derivative of the first)
    Chain rule for derivatives

    3. The attempt at a solution
    If b is a constant, then I know that dA/dh = b (this was the previous problem in which I could solve)
    my issue with the current problem is that this variable t that I'm supposed to take the derivative with respect to is not in the problem- so how would I go about doing this?

    My attempt:
    d/dt (A=bh)
    dA/dt = bh
    dA/dt = (b(dA/dh)) + (h(dA/db))
    My only problem with this solution is that I think there should be a d(b or h)/dt on the right side of the equation because of the chain rule

    could someone please explain where/why the variable t comes in and what the correct answer is? Thank you!!
  2. jcsd
  3. Nov 7, 2012 #2
    I have no clue what this means. Can you quote the full problem as it appears in your book?
  4. Nov 7, 2012 #3
    that's my problem too... that IS the quote!

    "Find d/dt for a rectangle" and then the formula given is A=bh
    so find d/dt for A=bh
  5. Nov 7, 2012 #4
    Well, that makes no sense. I advice asking your teacher for more explanations.
  6. Nov 7, 2012 #5
    On a previous problem:
    Find d/dt for a square

    and the answer was:
    d/dt (A=s2)

    dA/dt = 2s (ds/dt)

    perhaps you understand this one? (I don't)
    well i know the derivative of s2 = 2s. so we've taken the derivative of A with respect to t... so perhaps multiplying 2s by ds/dt is part of the chain rule or something... I don't really know where ds/dt came from
  7. Nov 7, 2012 #6


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    Assume that b and h are both functions of time.

    There should be no derivatives of A on the right hand side of the following:

    dA/dt = (b(dA/dh)) + (h(dA/db))

    Use the product rule to find [itex]\displaystyle \frac{d}{dt}(b\,h)\ .[/itex
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