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

**Twig** · July 4th 2009, 04:57 AM

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

**Moo** · July 4th 2009, 05:45 AM

Hello,

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

**Twig** · July 4th 2009, 12:10 PM

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 $g'(\frac{1}{a}) = a = f'(a)$

**mr fantastic** · July 4th 2009, 02:37 PM

Originally Posted by Twig

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?

**Twig** · July 5th 2009, 05:11 AM

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..

**earboth** · July 5th 2009, 06:47 AM

Originally Posted by Twig

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:

$y = 0.3982390482·x - 0.07929716961$

I've attached the graphs of the three functions.