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Product Rule on a triangle

  1. Sep 24, 2009 #1
    1. The problem statement, all variables and given/known data
    Question: the base b and height h change with a time t in such a way that b= [tex](t+1)^{2}[/tex] and h= [tex]t^{2}+1[/tex]
    Determine the rate of change of the area of the triangle when t=3


    2. Relevant equations

    [tex]A=\frac {1}{2}(b)(h)[/tex]

    3. The attempt at a solution

    I am not sure what to do here.

    Do you just multiply it out using the area of a triangle? Or should I take the derivative of both then use the area formula?
     
  2. jcsd
  3. Sep 24, 2009 #2

    lanedance

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    find the area in terms of t, then differentiate w.r.t t
     
  4. Sep 24, 2009 #3

    LCKurtz

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    You want to calculate dA/dt. Can you do that somehow?
     
  5. Sep 24, 2009 #4
    I was thinking of doing something like this:
    [tex]A=\frac {1}{2}(b)(h)[/tex]
    [tex]A=\frac {1}{2}((t+1)^{2})((t^{2}+1))[/tex]

    and continue from here.

    thus then I can take dA/dt?
     
  6. Sep 24, 2009 #5

    LCKurtz

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    Are you asking for permission? If nobody's looking, go ahead and do it.
     
  7. Sep 24, 2009 #6
    I was just wondering if it is correct :D
     
  8. Sep 25, 2009 #7

    lanedance

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    sounds good to me
     
  9. Sep 25, 2009 #8
    I've been wondering, is it possible to use the product rule on 3 functions? Because I could split the (t+1)^2 into (t+1)(t+1) then just 0.5(t+1)(t+1)(t^2+1)

    I'm not sure if this is allowed though since I've only see the product rule for two functions.
     
  10. Sep 25, 2009 #9

    LCKurtz

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    Yes there is a three factor product rule:

    (fgh)' = f'gh + fg'h + fgh'

    But in your example the easiest way is to leave it as (1/2)(t+1)2(t2+1) and just use the ordinary product rule, which involves using the chain rule on one of the factors.
     
  11. Sep 25, 2009 #10

    lanedance

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    or just multiply the expression out before differentiating
     
  12. Sep 25, 2009 #11
    [tex]\frac {d}{dx} A =\frac {1}{2} [/tex][tex](t+1)[/tex][tex](t+1)[/tex][tex](t^{2}+1)[/tex]

    [tex]\frac {d}{dx} A = f'(t)g(t)h(t) + f(t)g'(t)h(t) + f(t)g(t)h'(t)[/tex]

    [tex]\frac {d}{dx} A = (t+1)[/tex][tex](t^{2}+1)[/tex] +[tex](t+1)[/tex][tex](t^{2}+1)[/tex] +[tex](t+1)[/tex][tex](t+1)[/tex][tex](2t)[/tex]

    the d/dx for t+1 is just 1.

    [tex]\frac {d}{dx} A = t^{3}+t+t^{2}+1+t^{3}+t+t^{2}+1+2t^{2}+4t+2t[/tex]

    [tex]\frac {d}{dx} A = 2t^{3}+4t^{2}+8t+2[/tex]

    Now, to multiply the [tex]\frac {1}{2}[/tex] in.

    [tex]\frac {d}{dx} A = t^{3}+2t^{2}+4t+1[/tex]

    Is this correct? Something doesn't seem right.
     
  13. Sep 26, 2009 #12
    If I multiply it out, I get this as the derivative:

    [tex] 2t^{3}+3t-2t[/tex]

    which seems to be a more reasonable answer.
     
  14. Sep 26, 2009 #13

    LCKurtz

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    You both have it wrong. If you multiply it out you should get:

    2t3 + 3t2 + 2t + 1
     
  15. Sep 26, 2009 #14
    But why is that answer different if I used the product rule?
     
  16. Sep 26, 2009 #15
    Two problems.
    1) You are not taking the derivative with respect to x; there is no x in the problem which means your derivative would be 0. You want [tex]\frac{d}{dt}[/tex].
    2) The last product in the sum was not distributed properly. You should have gotten 2t3 + 4t2 + 2t, not 2t2 + 4t + 2t.
     
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