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Finding normal to ln curve

  1. Jan 12, 2007 #1
    I was wondering if anyone could help me with a question I was trying to work through today in our A-level math class.

    1. The problem statement, all variables and given/known data

    a. Sketch the graph of y= ln(3x)
    b. The normal to the curve at Point Q, with x coordinate q, passes through the origin. Find the equation of the normal and the value of q


    2. Relevant equations

    I don’t know which to use

    3. The attempt at a solution

    Part a was not too difficult and I was able to do that

    However I have no idea how to tackle part b. I have only been able to express the gradient of the normal as -q, but cannot solve it. I think [but am not sure] that I can get to the equation

    ln3q + q^2 = 0

    but I cannot solve this.

    I would be very grateful for some help.

    Thanks
     
  2. jcsd
  3. Jan 12, 2007 #2

    Hootenanny

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    How do you get that equation?
     
  4. Jan 12, 2007 #3
    Since when the normal and the curve y=ln(3x) intersect, the x coordinate is q, the the y coordinate is thus ln3q.

    And since the normal passes through (0,0), the gradient of the normal can also be expressed as (ln3q)/q

    I believe also that the gradient of the normal can be expressed as -q, thus

    (ln3q) / q= -q

    Rearranging gives

    ln3q + q^2 = 0

    However I don't know any method for solving such an equation
     
  5. Jan 13, 2007 #4

    Hootenanny

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    Good
    Would it not be better to express the gradient as;

    [tex]m=\frac{dy}{dx}\ln\left(3x\right) = \frac{1}{x}[/tex]

    [tex]\left.\frac{dy}{dx}\right|_{\left(q,\ln(q)\right)} = \frac{1}{q}[/tex]
     
    Last edited: Jan 13, 2007
  6. Jan 13, 2007 #5

    HallsofIvy

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    Almost. That is the slope of the tangent line. The slope of the normal line is -3q
     
  7. Jan 13, 2007 #6
    Thanks for the replies.

    Okay we haven't really covered much differentiation of ln graphs yet, but using what hootenanny/HallsofIvy posted, should the equation therefore be

    ln(3q)/q = -3q

    giving:

    ln(3q) + 3(q^2) = 0

    is this right? and how could i solve that?

    thanks again
     
  8. Jan 13, 2007 #7

    Dick

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    It might be if it were correct. But it's NOT. 1/ x not 1/(3x). Think chain rule. I think Doc G's original equation is actually correct.
     
  9. Jan 13, 2007 #8

    Hootenanny

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    My mistake, stupid slips, duly corrected.
     
  10. Jan 13, 2007 #9

    Dick

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    His problem doesn't seem to be finding the normal, etc. It's finding a solution to ln(3*x)+x^2=0 (I think this what he actually means by q). And I do think you can only solve it numerically. By inspection I would say you have a root just below 1/3.
     
  11. Jan 13, 2007 #10
    Thanks, should I now attempt to find a close solution by iteration [we've only just covered the basics of which in class]?

    And would the correct formula for which be:

    [tex]x_{n+1} = x_{n} - \frac{\ln3x_{n} + x^2_{n}}{1/x_{n} + 2x}[/tex] ?


    i tried putting in the equation ln3q + q^2 = 0 into Newton's Method. Is this okay? - i dont really have a good understanding of this area of maths yet..
     
    Last edited: Jan 13, 2007
  12. Jan 13, 2007 #11

    Dick

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    Unless this is a numerical analysis course I wouldn't open a second front on root finding algorithms. Either use some canned software or just do bisection to get a few decimal points if you want. This is easy enough to do with a calculator and a little patience...
     
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