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Math Help - Tangents - Parametric Equations

  1. #1
    Del
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    Tangents - Parametric Equations

    Suppose a curve is traced by the parametric equations:

    x = 4sin(t)

    y = 47 - 16cos^2(t) - 32sin(t)


    At what point on this curve is the tangent line horizontal?

    I need to find an x and a y value, please help!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Del View Post
    Suppose a curve is traced by the parametric equations:

    x = 4sin(t)

    y = 47 - 16cos^2(t) - 32sin(t)


    At what point on this curve is the tangent line horizontal?

    I need to find an x and a y value, please help!
    The slope of a curve y(x), when expressed parametrically, may be found by
    \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}

    So
    \frac{dy}{dx} = \frac{4~cos(t)}{32~sin(t)~cos(t) - 32~cos(t)} = 0

    = \frac{4}{32~sin(t) - 32} = 0

    Notice that there is no value of t such that the derivative is 0. Thus there is no solution.

    -Dan
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  3. #3
    Del
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    I see, but I am looking for the point on the graph where the tangent would be horizontal, the x-value will be 4, but what will the y-value be?

    Edit: I found it to be 15, thanks!
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  4. #4
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    errors

    topsquark, you got dx/dt and dy/dt reversed in your reply:

    \frac{dy}{dx}=\frac{32\sin(t)\cos(t)-32\cos(t)}{4\cos(t)}=0

    --Kevin C.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TwistedOne151 View Post
    topsquark, you got dx/dt and dy/dt reversed in your reply:

    \frac{dy}{dx}=\frac{32\sin(t)\cos(t)-32\cos(t)}{4\cos(t)}=0

    --Kevin C.
    Whoops! Thanks for the catch.

    However, I still say there is no such point where the tangent is horizontal. The derivative is discontinuous at t = \pi / 2 isn't it? (I always get screwed up when a function "ends" suddenly.)

    -Dan
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