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Math Help - 3cosx and Perpendiculars/Parallel

  1. #1
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    3cosx and Perpendiculars/Parallel

    Let f be the function given by f(x)=3cosx. The graph of f crosses the y-axis at point P (0,3) and x-axis at point Q (pi/2,0).

    Find the x coordinate of the point of the graph f, between points P and Q, at which the line tangent to the graph of f is parallel to the line PQ.


    I found the equation for line PQ:
    y-3=(-6/pi)x

    And the equation for the line tangent to f at point Q:
    y=-3[x-(pi/2)]

    But I can't find the x-coordinate (the bolded part), i know that the slope is (-6/pi) and you set that equal to -3sinx, but i got an x-coordinate of sin^(-1)(2/pi). But I am supposed to do it all by hand and I don't think that is a valid answer.
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  2. #2
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    Quote Originally Posted by warriors837 View Post
    Let f be the function given by f(x)=3cosx. The graph of f crosses the y-axis at point P (0,3) and x-axis at point Q (pi/2,0).

    Find the x coordinate of the point of the graph f, between points P and Q, at which the line tangent to the graph of f is parallel to the line PQ.

    I found the equation for line PQ:
    y-3=(-6/pi)x

    And the equation for the line tangent to f at point Q:
    y=-3[x-(pi/2)]

    But I can't find the x-coordinate (the bolded part), i know that the slope is (-6/pi) and you set that equal to -3sinx, but i got an x-coordinate of sin^(-1)(2/pi). But I am supposed to do it all by hand and I don't think that is a valid answer.

    I think you did it alright and, imo, you can write x=Arcsin\frac{2}{\pi} =0.69\,\,\,aprox., which indeed is between P and Q on the x-axis

    tonio
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  3. #3
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    Quote Originally Posted by warriors837 View Post
    Let f be the function given by f(x)=3cosx. The graph of f crosses the y-axis at point P (0,3) and x-axis at point Q (pi/2,0).

    Find the x coordinate of the point of the graph f, between points P and Q, at which the line tangent to the graph of f is parallel to the line PQ.


    I found the equation for line PQ:
    y-3=(-6/pi)x

    And the equation for the line tangent to f at point Q:
    y=-3[x-(pi/2)]

    But I can't find the x-coordinate (the bolded part), i know that the slope is (-6/pi) and you set that equal to -3sinx, but i got an x-coordinate of sin^(-1)(2/pi). But I am supposed to do it all by hand and I don't think that is a valid answer.
    it's a valid solution ...

    x = \arcsin\left(\frac{2}{\pi}\right)

    just doesn't come out "pretty" , that's all.
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