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Math Help - AP Calc HW problem help (take 2)

  1. #1
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    AP Calc HW problem help (take 2)

    I scanned the problem in, and the work that is underneath it, is what i tried to do.

    Please help, Thanks!

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  2. #2
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    Quote Originally Posted by autounion View Post
    I scanned the problem in, and the work that is underneath it, is what i tried to do.

    Please help, Thanks!


    Where did you take this question from? F(x) is a function of one variable so if THE tangent line at some point of the graph of F(x) exists then it is unique since its slope is given by the value of F'(x) at the x-coordinate of the tangency point. How can they ask something about "tangent lineS" at some point on the graph??
    This must be a mistake or else this book's teaching stuff which I believe is not standard.

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    Where did you take this question from? F(x) is a function of one variable so if THE tangent line at some point of the graph of F(x) exists then it is unique since its slope is given by the value of F'(x) at the x-coordinate of the tangency point. How can they ask something about "tangent lineS" at some point on the graph??
    This must be a mistake or else this book's teaching stuff which I believe is not standard.

    Tonio
    It's my problem of the week for my AP Calc Class. He handed out the problem to us. After looking at it carefully, I think that they only mean 1 tangent line, still it's confusing and I have no idea on where to start?
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  4. #4
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    maybe the question wants two distinct lines tangent to the curve that are perpendicular ... ?

    if so, then given the symmetry of f(x), one point of tangency would be (t,f(t)) and the other would be (-t,f(-t)). note also that since f(x) is even, f(t) = f(-t)

    f(x) = kx^2+3

    f'(x) = 2kx

    f'(t) = 2kt

    f'(-t) = -2kt

    f'(t) and f'(-t) are opposite reciprocals ...

    2kt = -\frac{1}{-2kt}

    4k^2t^2 = 1

    t = \frac{1}{|2k|}

    slopes ...

    at (t, f(t)) , m = 2k \cdot \frac{1}{|2k|} = 1 if k > 0 and -1 if k < 0

    at (-t,f(t)) , m = -2k \cdot \frac{1}{|2k|} = -1 if k > 0 and 1 if k < 0


    it should be obvious that the two tangent lines intersect on the y-axis

    one of the tangent line equations is ...

    y - (kt^2+3) = x - t

    for x = 0 ...

    y - kt^2 - 3 = -t

    y = kt^2 - t + 3

    y = k \cdot \frac{1}{4k^2} - \frac{1}{|2k|} + 3

    y = \frac{1}{4k} - \frac{1}{|2k|} + 3

    the intersection point is \left(0,\frac{1}{4k} - \frac{1}{|2k|} + 3\right)


    really ... a very poorly worded question.
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  5. #5
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    Quote Originally Posted by skeeter View Post
    maybe the question wants two distinct lines tangent to the curve that are perpendicular ... ?

    if so, then given the symmetry of f(x), one point of tangency would be (t,f(t)) and the other would be (-t,f(-t)). note also that since f(x) is even, f(t) = f(-t)

    f(x) = kx^2+3

    f'(x) = 2kx

    f'(t) = 2kt

    f'(-t) = -2kt

    f'(t) and f'(-t) are opposite reciprocals ...

    2kt = -\frac{1}{-2kt}

    4k^2t^2 = 1

    t = \frac{1}{|2k|}

    slopes ...

    at (t, f(t)) , m = 2k \cdot \frac{1}{|2k|} = 1 if k > 0 and -1 if k < 0

    at (-t,f(t)) , m = -2k \cdot \frac{1}{|2k|} = -1 if k > 0 and 1 if k < 0


    it should be obvious that the two tangent lines intersect on the y-axis

    one of the tangent line equations is ...

    y - (kt^2+3) = x - t

    for x = 0 ...

    y - kt^2 - 3 = -t

    y = kt^2 - t + 3

    y = k \cdot \frac{1}{4k^2} - \frac{1}{|2k|} + 3

    y = \frac{1}{4k} - \frac{1}{|2k|} + 3

    the intersection point is \left(0,\frac{1}{4k} - \frac{1}{|2k|} + 3\right)


    really ... a very poorly worded question.
    Thank you soo much
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