Finding point where slope of line equals curve

  1. 1. The problem statement, all variables and given/known data

    At what point on the curve y=2(x-cosx) is the tangent parallel to the line 3x-y=5.

    3. The attempt at a solution

    1. rewrite 3x-y=5 as y-3x-5
    2. equate 2(x-cosx) = y-3x-5
    3. differentiate: 2+2sinx = 3
    4. solve for x: sin^-1(,5) = 0.524
    5. plug into y=2(x-cosx) to get y value: y = -0.684

    I am not sure I have this correct though. At stage 4 above, I have a trig equation. So doesn't that give me an infinite number of solutions. (I was not given a range in this question). Or do I just look at the graph of 2(x-cosx) and y-3x-5 to see logically where the point of equal slope is ?

    Thanks for any help.

    Attached Files:

  2. jcsd
  3. Mark44

    Staff: Mentor

    No. The problem isn't asking where the two curves intersect, which is what you might be thinking you're doing in the step above.

    What you want is to find any point on the graph of y=2(x-cosx) whose slope is 3, the slope ope of the line.

    This looks OK, but I suspect there are a whole lot of points that are solutions. The inverse sine gives you only one.
  4. You are right, there is an infinite number of solutions, solving for x gives you your answer or pi/6, and so the tangent of y=2(x-cosx) is parallel to 3x-y=5 whenever x=pi/6 +2pi*n where n is an integer. As said above, instead of considering the intersection of the two functions, you are looking for points where the tangent line (whose slope is the derivative of y=2(x-cosx)) is parallel to 3x-y=5.
  5. thanks guys.
    I understand what you are saying.

    So to clarify, when the question asks: at what point(s) on the curve is the tangent parallel to the line, then the answer would just be:
    "the tangent of y=2(x-cosx) is parallel to 3x-y=5 whenever x=pi/6 +2pi*n where n is an integer" ?
    i.e. I would not need to include y value reference in the answer ?
    Last edited: Mar 29, 2012
  6. Mark44

    Staff: Mentor

    To identify the points they're asking for, you need to supply a y value as well.
  7. sorry to persist with this, but I just don't know how to express the recurring y values (due to the infinite amount of possible answers).

    Can someone please show me to express the y values ?

    Perhaps I would just express the points as:
    x=pi/6 +2pi*n, y = f(pi/6 +2pi*n)
    x=3pi/6 +2pi*n, y = f(3pi/6 +2pi*n)
    Last edited: Mar 29, 2012
  8. Mark44

    Staff: Mentor

    Use the equation you're given: y = 2(x - cos(x))

    If x = ##\pi/6## + 2n*##\pi## , then y = 2(##\pi/6## + 2n*##\pi## -√3/2)
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