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Math Help - Slope of a tangent line on a curve. Too difficult for me.

  1. #1
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    Slope of a tangent line on a curve. Too difficult for me.

    So, I have this curve C: y=2x^3 and a point P(p, 2p^3), p being a positive number. With the following graphic:




    L1 and L2 are tangent lines of C passing through P.

    I am asked to express the slope of L2 in terms of p.
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  2. #2
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    Seems like the image should have been

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  3. #3
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    hmm... I don't really understand the question since at P the tangent line is L1
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by rualin View Post
    hmm... I don't really understand the question since at P the tangent line is L1
    L2 is tangent to the part of the curve on the left of the y-axis. it intersects with the tangent line L1 at P, where L1 is tangent to the curve at P (it seems, that diagram is really small)
    Last edited by Jhevon; July 8th 2007 at 02:44 PM.
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  5. #5
    MHF Contributor red_dog's Avatar
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    Let m be the slope of L_2.
    Then the equation of L_2 is: y-2p^3=m(x-p).
    L_2 must intersect the graph in two distinct points. One point is P and the other is the point where L_2 is tangent to the graph.
    So, the system \left\{\begin{array}{ll}y-2p^3=m(x-p)\\y=2x^3\end{array}\right. must have two solutions.
    Replacing y from the second equation in the first equation we have
    2(x^3-p^3)=m(x-p)\Leftrightarrow (x-p)(2x^2+2px+2p^2-m)=0\Rightarrow x_1=p and 2x^2+2px+2p^2-m=0.
    The last equation must have equal roots. Then
    4p^2-16p^2+8m=0\Rightarrow m=\frac{3p^2}{2}
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  6. #6
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    Interesting... I had never seen this done. Thanks, red_dog!
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  7. #7
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    And here was silly me thinking of derivatives and more complex stuff.

    Seriously, I would have never thought of that.

    Thank you very much, Red Dog.
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  8. #8
    Eater of Worlds
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    Here's a calc way if you'd still like to see it.

    We can find where L2 is tangent to C by:

    y'=6x^{2}=m

    2x^{2}-2p^{3}=6x^{2}(x-p)

    Solve this for x and we get x=\frac{-p}{2}

    Sub this back into y=2x^{3} and we see y=\frac{-p^{3}}{4}

    Now, we have two end points from which to find the slope of L2. Use the slope of a line formula.

    m=\frac{2p^{3}+\frac{p^{3}}{4}}{p+\frac{p}{2}}=\fr  ac{3p^{2}}{2}

    Which agrees with red dogs clever method.
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  9. #9
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    Quote Originally Posted by galactus View Post
    Here's a calc way if you'd still like to see it.

    We can find where L2 is tangent to C by:

    y'=6x^{2}=m

    2x^{2}-2p^{3}=6x^{2}(x-p)

    Solve this for x and we get x=\frac{-p}{2}
    Could we have substituted this x into m=6x^2 to get m=6\left(\frac{-p}{2}\right)^2=\frac{3}{2}\,p^2?
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  10. #10
    Eater of Worlds
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    Why yes, even shorter. I wonder why they have theta in there when it's not really needed to find the slope of L2. L1 is easy. Now that we have L2, theta can be found by:

    tan{\theta}=\frac{m_{2}-m_{1}}{1+m_{1}m_{2}}, if need be.
    Last edited by galactus; July 9th 2007 at 09:05 AM.
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