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Thread: Tangent to a parametric curve

  1. #1
    Member Em Yeu Anh's Avatar
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    Angry Tangent to a parametric curve

    I thought I solved this correctly, but my answer was marked wrong:

    Here it is: Find the equations of the tangents to this curve that pass through the point (15,15).

    x=9t^2+6
    y=6t^3+9

    Solution:
    \frac{dy}{dx}=\frac{18t^2}{18t} = t

    y-(6t^3+9)=t(x-(9t^2+6))

    Substituting in the point (15,15):

    15-6t^3-9=t(15-9t^2-6)

    After simplifying, this turns into: (t-1)^2(t+2) = 0

    Setting t=1:
    y-15=1(x-15)
    y=x is my first tangent line, which was wrong. When I graphed it, it did appear to be tangent to the curve at that point..

    Setting t=-2:
    y-15=-2(x-15)
     y=-2x+45 also incorrect.

    If someone could fix my work, I'd be very very thankful.
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  2. #2
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    Opalg's Avatar
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    Both those answers look correct to me. Why do you say that they are wrong?
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  3. #3
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    t= -2 is incorrect because y(-2)= 6(-2)^3+ 9= -48+ 9= -39, not 15.

    However, at t= 1, (x, y)= (15, 15) and then tangent line is indeed y= x.
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  4. #4
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    t= -2 is incorrect because y(-2)= 6(-2)^3+ 9= -48+ 9= -39, not 15.
    I thought the same at first. But when t=2, x=42 and y=39, and the tangent at (42,39) is y+39 = 2(x42) which does indeed simplify to y = -2x+45.
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