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Math Help - Finding two points...

  1. #1
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    Finding two points...

    Ive been trying to figure out this problem for a while now... I can solve it graphically but I am trying to solve it algebraically using the derivative but I am still stuck:
    On the curve y=x^3, find two points, P and Q, for which the tangent line at P crosses Q. Furthermore, the slope of the tangent line at Q is 4 times the slope of the tangent line at P.
    I got this far then I hit a roadblock:

    q = 2p.... the two tangent lines:

    3p^2(x-p) + p^3 = 12p^2(x-2p) + 8p^3
    Thanks in advance!
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  2. #2
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    Quote Originally Posted by jllinas View Post
    Ive been trying to figure out this problem for a while now... I can solve it graphically but I am trying to solve it algebraically using the derivative but I am still stuck:
    On the curve y=x^3, find two points, P and Q, for which the tangent line at P crosses Q. Furthermore, the slope of the tangent line at Q is 4 times the slope of the tangent line at P.
    I got this far then I hit a roadblock:

    q = 2p.... the two tangent lines:

    3p^2(x-p) + p^3 = 12p^2(x-2p) + 8p^3
    Thanks in advance!
    If P is the point (p,p^3) then the tangent at P has slope 3p^2 and equation y=3p^2x-2p^3. It meets the curve again at Q = (q,q^3), where q=2p (note the minus sign). The slope at Q is 3q^2 = 3(-2p)^2 = 12p^2. So the slope of the tangent line at Q is always 4 times the slope of the tangent line at P, no matter where P may be on the curve.
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