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Math Help - Show(1/2)x^2 and ln(x) has common tangent

  1. #1
    Senior Member Twig's Avatar
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    Show(1/2)x^2 and ln(x) has common tangent

    Hi

    Show that y=\frac{x^{2}}{2} \mbox{ and } y = ln(x) \; \; , x > 0 has at least one common tangent.

    How do one proceed smartest here?

    Thx
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  2. #2
    Moo
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    Hello,

    They'll have the same (at least one) tangent if there are more than one solution a to the equation f'(a)=g'(a)

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  3. #3
    Senior Member Twig's Avatar
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    Hi

    Not really sure what you mean.

    Let f(x)=\frac{1}{2}x^{2} \mbox{ and } g(x)=ln(x)

    f'(x)=x \mbox{ and } g'(x) = \frac{1}{x}

    If we take  x = a , then  y-\frac{1}{2}a^{2}=a(x-a) becomes the tangent equation in that point for the f(x) function.

    Now, for g(x) , if we here take x=\frac{1}{a} , then  g'(\frac{1}{a}) = a = f'(a)
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  4. #4
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    Quote Originally Posted by Twig View Post
    Hi

    Not really sure what you mean.

    Let f(x)=\frac{1}{2}x^{2} \mbox{ and } g(x)=ln(x)

    f'(x)=x \mbox{ and } g'(x) = \frac{1}{x}

    If we take  x = a , then  y-\frac{1}{2}a^{2}=a(x-a) becomes the tangent equation in that point for the f(x) function.

    Now, for g(x) , if we here take x=\frac{1}{a} , then  g'(\frac{1}{a}) = a = f'(a)
    Obviously if there is a common tangent that means that there has to be a point on each curve where the gradient is the same.

    So do as moo suggested and solve x = \frac{1}{x}.

    Now determine the point on each curve that has this x-coordinate. Now calculate the tangent at this point for each curve. Do you get the same line?
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  5. #5
    Senior Member Twig's Avatar
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    hi

    x=\frac{1}{x} \Rightarrow x = 1 \mbox{ since } \; x>0

    y-\frac{1}{2}=1\cdot(x-1)

    y-0=1\cdot (x-1)

    These lines are not the same... :/

    I must be thinking wrong..
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  6. #6
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    Quote Originally Posted by Twig View Post
    Hi

    Not really sure what you mean.

    Let f(x)=\frac{1}{2}x^{2} \mbox{ and } g(x)=ln(x)

    f'(x)=x \mbox{ and } g'(x) = \frac{1}{x}

    If we take  x = a , then  y-\frac{1}{2}a^{2}=a(x-a) becomes the tangent equation in that point for the f(x) function.
    Now, for g(x) , if we here take x=b , then  g'(b) = \dfrac1b
    and the equation of the tangent becomes

    y-\ln(b)=\dfrac1b ( x-b)~\implies~y=\dfrac1b \cdot x -1+\ln(b)

    The equation of the tangent to the parabola becomes:

     y=ax-\frac{1}{2}a^{2}

    Compare these two equations. You'll get a system of equations:

    \left|\begin{array}{rcl}a&=&\dfrac1b \\ -\frac{1}{2}a^{2} &=& -1+\ln(b) \end{array}\right.

    Solve for a and b.

    Remark I: If you have drawn a rough sketch of both graphs you probably would have noticed that there must be at least one common tangent.

    Remark II: I haven't found a solution of this system - but if I succeed I'll be back.
    - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

    Here I am again:

    I substituted b=\dfrac1a in the second equation. I wasn't able to solve this equation algebraically so I used Newtons method and got an approximate result:

    y = 0.3982390482x - 0.07929716961

    I've attached the graphs of the three functions.
    Attached Thumbnails Attached Thumbnails Show(1/2)x^2 and ln(x) has common tangent-par_logtang.png  
    Last edited by earboth; July 5th 2009 at 08:11 AM. Reason: found one possible solution
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  7. #7
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    ... and here comes the second tangent:

    y = 1.773751172x - 1.573096610

    I've attached the complete graph.
    Attached Thumbnails Attached Thumbnails Show(1/2)x^2 and ln(x) has common tangent-par_tang_log.png  
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  8. #8
    Senior Member Twig's Avatar
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    Thank you Earboth! I knew I was thinking about it totally wrong.
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