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Differentiation-dont really get the question

  1. Jul 20, 2010 #1
    1. The problem statement, all variables and given/known data

    see attachment please

    2. Relevant equations



    3. The attempt at a solution

    y= sin(2t)+t
    x=e2t
    ln(x)=2t
    t=ln(x)/2
    y=sin(2(lnx/2) )+ln(x)/2
    y=sin(ln(x))+ln(x)/2

    the question says leave in terms of t but when i differentiate i get
    y'=cos(ln(x))/x+1/2x

    this may not even be what the question is asking??? but if it is how do i put that back in terms of t???
     

    Attached Files:

  2. jcsd
  3. Jul 20, 2010 #2

    hunt_mat

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    What you have done is perfectly correct, but you haven't answered the question. Use the chain rule:
    [tex]
    \frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}
    [/tex]
    and the fact that [tex]dt/dx=1/(dx/dt)[/tex] and leave your answer in terms of t, just like it said. You have answered the harder question, well done.
     
  4. Jul 20, 2010 #3
    sorry I dont understand that, how do I leave it in terms of t??????
     
  5. Jul 20, 2010 #4

    hunt_mat

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    calculate dy/dt and dx/dt, these will be functions of t divide dy/dt by dx/dt, this will also be a function of t, is is this function they're asking for. You have done 99% of the problem already.
     
  6. Jul 20, 2010 #5
    what would dx/dt be????
     
  7. Jul 20, 2010 #6

    hunt_mat

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    you have the following:
    y= sin(2t)+t
    x=e^(2t)
    so what is dy/dt and dx/dt? It's just simple differentiation.
     
  8. Jul 20, 2010 #7
    are ok thats simple
    dx/dt=2e^(2t)
    dy/dt=2cos(2t)+1
    then I multiply them??
     
  9. Jul 20, 2010 #8

    hunt_mat

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    No, you divide them in a previous post I wrote:
    [tex]
    \frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}
    [/tex]
    How do you get from dx/dt to dt/dx?
     
  10. Jul 20, 2010 #9

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Equivalently
    [tex]\frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}}[/tex]
    You get hunt mat's formula by "invert and multiply".
     
  11. Jul 20, 2010 #10
    so i get 2cos(2t)+1 * (2e^(2t))^-1
    is that the right answer??? my 89 wont simplify it at all.
     
  12. Jul 20, 2010 #11

    hunt_mat

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    probably because you can't. It is correct, you can check it with your previous answer, find t in terms of x(you already did this), put it into the answer above and check that they're the same.
     
  13. Jul 20, 2010 #12
    ok thanks for your help
     
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