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Normal lines

  1. Dec 25, 2011 #1
    1. The problem statement, all variables and given/known data

    The curve C has equation y = ln3x and crosses the x-axis at point (p,0)

    The normal to the curve at the point Q with x-co-ordinate of q passes through the origin
    Show that x=q is a solution of the equation x2 + ln3x = 0

    2. Relevant equations
    3. The attempt at a solution
    So, the gradient of ln(3x) at q is
    1/q
    therefore the gradient of the normal is -q

    How do I go one from here?
     
  2. jcsd
  3. Dec 25, 2011 #2
    oh so as x=q
    y = -qx
    ln(3x)=-x2
    so
    xx2 + ln(3x) = 0
    right?
     
  4. Dec 25, 2011 #3

    Curious3141

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    Homework Helper

    You've found the gradient of the normal line at that point to be -q.

    Since you're told the normal line passes through the origin (meaning the y-intercept is zero), its equation is:

    y = -qx. ---equation 1

    You need to determine the co-ordinates of the intersection point between the normal line and the curve. The x-coordinate of this point is q. Hence the y-coordinate is ln(3q)because the point lies on the curve as well. So the point (q,ln(3q)) lies on the normal line (in addition to the curve).

    Hence that point will also satisfy equation 1. Put those values of x and y into eqn1:

    ln(3q) = -q*q

    q^2 + ln(3q) = 0

    Hence q is a solution of x^2 + ln(3x) = 0, as required (this is just a roundabout way of saying that you can substitute q for x and the equation would hold true).
     
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