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Two Calculus Problem Solving Questions.

  1. Oct 13, 2005 #1
    #1: The tangent at a point P on the curve y = x^3 intersects the curve again at a point Q. Show that the slope of the tangent at Q is four times the slope of the tangent at P.

    So far, I found the derivative of y = x^3, which is y' = 3x^2. Now I set these two equal to each other:

    X^3 = 3x^2
    x^3 - 3x^2 = 0
    x^2 (x-3) = 0
    so x = 0 or x = 3.

    I don't know where to go from here, or if I did this right at all.

    #2: Show that the x- and y- intercepts for any tangent to the curve y = 16 - 8sqrt(x) + x have a sum of 16.

    First I expanded the equation:

    y = 16 - 8sqrt(x) + x
    = 16 - 8^(1/2) + x

    Then I determined the derivative:

    y' = -4x^(-1/2) + 1
    = -4/sqrt(x) + 1

    Once again, I have no idea what do from here. :(
    Any help would be greatly appreciated.
     
  2. jcsd
  3. Oct 13, 2005 #2

    Fermat

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    Homework Helper

    For #1, assume that it's true. That slope at Q is 4 times slope at P.

    P(xp, yp)
    Q(xp, yp)

    mp = 3x²p
    mq = 3x²q

    and

    mq = 4*mp

    hence

    q = 4x²p

    Now write the slope of the line PQ in terms of the coords of P and Q. If you can show that this slope is the same as the slope of the tangent at P, then you will have proved your point. Q.E.D.
     
  4. Oct 13, 2005 #3

    Fermat

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    Homework Helper

    For #2,

    Let P(xp, yp) be a point on the curve.

    Get the slope of the tangent at P.

    Get the eqn of the tangent at P, in terms of the coords at P.

    To get the y-intercept, yi, set x = 0, in the eqn of the tangent, and solve for yi.

    To get the x-intercept, xi, set y = 0, in the eqn of the tangent, and solve for xi.

    Now add xi and yi and manipulate the resulting expression until it becomes 16.
     
  5. Oct 14, 2005 #4

    VietDao29

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    Homework Helper

    Why are you setting the two equal each other????
    That's not correct.
    The tangent line at P will be:
    g(x) = f'(xP)(x - xP) + yP
    Now you have f(x) = x3
    To find point Q, xQ is the solution to the equation:
    xQ3 = f'(xP)(xQ - xP) + yP = (3xP2) (xQ - xP) + xP3 = 3xP2xQ - 3xP3 + xP3 = 3xP2xQ - 2xP3.
    <=> xQ3 - 3xP2xQ + 2xP3 = 0
    Now you have to relate xQ and xP.
    Can you go from here?
    Note that the slope of the tangent at P is 3xP2, and the slope of the tangent at Q is 3xQ2.
     
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