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Math Help - tanget line parallel to curve positino

  1. #1
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    tanget line parallel to curve positino

    Hello, I am trying to solve the following problem:
    At what point(s) on the curve y=2(x-cosx) is the tangent parallel to the line 3x-y=5.

    My attempt is:
    rewrite 3x-y=5 as y=3x-5
    differentiated to 3

    Differentiate y=2(x-cosx) = 2(1+sinx) = 2 + 2sinx
    so
    3 = 2+2sinx
    =
    1 = 2sinx2
    =
    1/2 = sinx
    =
    sin^-1(1/2) = x
    = 0.524
    plugging back into original curve formula gives a y value of -0.684
    so my final answer is "the tangent is parallel to the line 3x-y=5 at the point .524(x), -0.684(y)."

    I am unsure if I am correct. Could someone please cast their eyes over this and let me know if I am on the right track ?
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  2. #2
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    Re: tanget line parallel to curve positino

    Looks correct to me.
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  3. #3
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    Re: tanget line parallel to curve positino

    actually I think I got this one wrong.

    To find the points on the curve (2(x-cosx) where the tangent is parallel to the line y=3x-5.

    differentiating 2(x-cosx) = 3x-5:
    2+2sinx = 3

    Then solving for x to find where the tangent is parallel to the curve:
    2sinx=1
    sin^-1 (0.5) = x

    At this stage I am not sure what to do next. This trig equation now gives me 30 degrees and 150 degrees ?, but I am trying to find an x and a y position.
    Can someone help please.
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  4. #4
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    Re: tanget line parallel to curve positino

    You must work in radians in this question. You were correct with your first solution but you only gave one answer. Did they not ask for answers in a given range. If sinx=1/2 another answer would be x=2.618 (that is 150 degrees in radians)
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  5. #5
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    Re: tanget line parallel to curve positino

    yes, I understand now. thank you !
    Last edited by fran1942; March 29th 2012 at 12:08 AM.
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