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For what values y is y(t) increasing/decreasing?

  1. Feb 6, 2013 #1
    1. The problem statement, all variables and given/known data

    y'=y[itex]^{3}[/itex]-y[itex]^{2}[/itex]-12y

    For what values of y is y(t) increasing and for what values is it decreasing?

    2. Relevant equations



    3. The attempt at a solution

    I think you take the second derivative and equal it to zero to figure out the inflection pints right and then Im not so sure from there.

    y''=3y[itex]^{2}[/itex]-2y-12

    0=3y[itex]^{2}[/itex]-2y-12

    Quadratic formula

    y=[itex]\frac{2\pm2\sqrt{37}}{6}[/itex]

    Am I correct in this approach? I dont think I am.
     
  2. jcsd
  3. Feb 6, 2013 #2
    For y to be either increasing or decreasing, you just need to look at whether y' is positive or negative, so the original equation gets you half-way there already - no need to differentiate again.

    Try to factorize the RHS and it should become clear.
     
  4. Feb 7, 2013 #3

    rude man

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    You also need to take y''. Solving for y' = 0 just gets you the values of y for which the slope is zero. To also learn whether those points are (relative) minima or maxima requires taking y'' and substituting the values of y for which y' = 0.

    Yes, factor by all means.
     
  5. Feb 7, 2013 #4
    Rubbish.
     
  6. Feb 7, 2013 #5

    rude man

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    I say! I think you're right!
     
    Last edited: Feb 7, 2013
  7. Feb 8, 2013 #6

    HallsofIvy

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    Odd that oay should be ruder than rude man and then rude man agrees with him! But oay is right- the second derivative is irrelevant. You are NOT asked to find "inflection points".

    y'= y3- y2- 12y= y(y2- y- 12)= y(y- 4)(y+ 3), then if y< -3, all three factors are negative so y' is negative. If -3< y< 0 then y+3 is positive while y and y= 4 are both negative so y' is positive, etc.

    But you would still need to find what values of t give those values of y.
     
    Last edited: Feb 8, 2013
  8. Feb 8, 2013 #7

    rude man

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    I agree with hallsofivy. When you're wrong... admit it! And I do.
     
  9. Feb 8, 2013 #8
    Finding values of t wasn't in the question, though.
    And I apologize for my sharp reply of "Rubbish". I think I'd just had a bottle of Chardonnay. :redface:
     
  10. Feb 8, 2013 #9

    rude man

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    No apology required! It was just that.
     
  11. Feb 8, 2013 #10

    Dick

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    Apt comment gracefully accepted and, shall I say, "wittily" replied to. I worry a lot less about rudeness when replying to somebody with 1600+ posts under their belt. I assume that's made them at least a little bit thick skinned. Nice exchange. I don't think there was any real rudeness intended and better yet, none recieved.
     
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