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Related rate expanding rectangle

  1. Oct 25, 2009 #1
    1. The problem statement, all variables and given/known data
    The length of a rectangle is increasing at a rate of 8cm/s and its width is increasing at a rate of 3cm/s when the length is 20 cm and the width is 10cm, how fast is the area of the rectangle increasing?

    2. Relevant equations

    V=LW


    3. The attempt at a solution

    dL/dt=8 dW/dt=3
    l=20 w= 10
    dA/dt=?

    (V=LW)'

    dA/dt= dL/dt * dW/dt

    before I go any further is this correct both the L and the W become one when derived right?
     
  2. jcsd
  3. Oct 25, 2009 #2

    Dick

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    A(t)=L(t)W(t), right? Take a look at the 'product rule'. And why would the derivative of L and W be 1???
     
  4. Oct 25, 2009 #3
    I have really just not been using my brain lately. They would be one if they were being added which they're not. (x + a)'= 2 does it not? if they are both variables.

    And you have t in parenthesis to make clear that L and W and A change with time and are therefore not constants right?

    so according to the product rule->

    dA/dt=L(dW/dt)+W(dW/dt)
     
  5. Oct 25, 2009 #4

    Dick

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    Ok, you've got the product rule. But (x+a)'=x'+a' if by ' you mean d/dt. Whether that's 2 or not depends what x and a are. dx/dx=1 but dx/dt doesn't necessarily equal 1.
     
  6. Oct 25, 2009 #5
    Ok I was just using the power rule (x^1 + a^1)'= 1*x^0 + 1*a^0= 1+1
    am I assuming something here I shouldn't be?
     
  7. Oct 25, 2009 #6

    Dick

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    d(x(t)^1)/dt=1*x(t)^0*dx/dt, chain rule again.
     
  8. Oct 25, 2009 #7
    Could you explain quickly why you need to use the chain rule here?
     
  9. Oct 25, 2009 #8

    Dick

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    Because x(t) is a function of t. d(x^1)/dx is 1*x^0 (though you could still throw the chain rule in and write it as 1*x^0*dx/dx, but dx/dx=1). d(x^1)/dt=1*x^0*dx/dt. dx/dx is always 1. dx/dt is not.
     
    Last edited: Oct 25, 2009
  10. Oct 25, 2009 #9
    Ok thank you its starting to come together.
     
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