Hi

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

How do one proceed smartest here?

Thx

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- Jul 4th 2009, 04:57 AMTwigShow(1/2)x^2 and ln(x) has common tangent
Hi

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

How do one proceed smartest here?

Thx - Jul 4th 2009, 05:45 AMMoo
Hello,

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

:) - Jul 4th 2009, 12:10 PMTwig
Hi

Not really sure what you mean.

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

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

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

Now, for $\displaystyle g(x)$ , if we here take $\displaystyle x=\frac{1}{a} $ , then $\displaystyle g'(\frac{1}{a}) = a = f'(a) $ - Jul 4th 2009, 02:37 PMmr fantastic
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 $\displaystyle x = \frac{1}{x}$.

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? - Jul 5th 2009, 05:11 AMTwig
hi

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

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

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

These lines are not the same... :/

I must be thinking wrong.. - Jul 5th 2009, 06:47 AMearboth
Now, for $\displaystyle g(x)$ , if we here take $\displaystyle x=b $ , then $\displaystyle g'(b) = \dfrac1b $

and the equation of the tangent becomes

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

The equation of the tangent to the parabola becomes:

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

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

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

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.

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Here I am again:

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

$\displaystyle y = 0.3982390482·x - 0.07929716961$

I've attached the graphs of the three functions. - Jul 5th 2009, 10:52 AMearboth
... and here comes the second tangent:

$\displaystyle y = 1.773751172·x - 1.573096610$

I've attached the complete graph. - Jul 5th 2009, 11:51 AMTwig
Thank you Earboth! I knew I was thinking about it totally wrong.