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Sketch the graph of y = sinx(1-cosx)

  1. Jul 11, 2011 #1
    1. The problem statement, all variables and given/known data
    Sketch the graph of y = sinx(1-cosx) over [-2pi, 2pi]

    2. Relevant equations
    first derivative and second derivative

    3. The attempt at a solution

    y' = cosxsinx
    0 = cosxsinx
    x = pi/2, 3pi/2, 0, pi, 2pi

    I know I have the sin values right (the answers say the same) but the cos values in the answers are actually 2pi/3 for MAX and 4pi/3 for MIN. For that cosx would have to equal -1/2. I'm not sure how to get that );

    Thank you!
  2. jcsd
  3. Jul 11, 2011 #2
    Hi Glissando! :smile:

    Your derivative is wrong. If you want to differentiate a product, then you can't just differentiate both factors and be done with it. You need to apply the product rule:

    [tex](fg)^\prime=f^\prime g+fg^\prime[/tex]
  4. Jul 11, 2011 #3


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    Once you get the first derivative right, don't forget to find the second derivative.
  5. Jul 11, 2011 #4
    Good point (wow that was embarassing...)

    K well I tried doing it and I still get cosx = 0:

    y' = cosx(1-cosx) + sin2x
    y' = cosx - cos2x + sin2x
    0 = -cosx + cos2x - sin2x
    0 = -cosx + cos2x
    0 = cosx, x = pi/2, 3pi/2

    UGH ): Thanks guys for putting up with me!
  6. Jul 11, 2011 #5


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    Well, you have y' right, but ...

    -cos(x) + cos(2x) ≠ cos(x)

    (Plug x = π/2 into your equation for y' to check your answer.)

    You could use sin2(x) =1 - cos2(x), then y' = 0 gives a quadratic in cos(x) . → Factor, or use the quadratic formula to solve for cos(x), and then solve for x.
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