Hi

Show that has at least one common tangent.

How do one proceed smartest here?

Thx

Printable View

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

Show that has at least one common tangent.

How do one proceed smartest here?

Thx - July 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

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

Not really sure what you mean.

Let

If we take , then becomes the tangent equation in that point for the function.

Now, for , if we here take , then - July 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 .

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

These lines are not the same... :/

I must be thinking wrong.. - July 5th 2009, 06:47 AMearboth
Now, for , if we here take , then

and the equation of the tangent becomes

The equation of the tangent to the parabola becomes:

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

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

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

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