Question about finding a tangent

[kwal0203]

Homework Statement

Show that, for each real number t $\in$ the interval (0, 1], the curve given by:

$y=ln(\frac{x+\sqrt{1+x^{2}}}{1+\sqrt{2}})$

has a tangent line with slope t. Find the points on the curve at which the tangent line has slope 2/3.

The Attempt at a Solution

I found the first derivative of this curve to be:

$dy/dx=1/\sqrt{(1+x^{2})}$

but now not sure how to proceed.

What do they mean when they ask me to show that for each t has a tangent line with slope t?

any help appreciated!

Also to find the points with dy/dx=3 I did this:

$dy/dx=1/\sqrt{(1+x^{2})} =2/3$ >>>> $x=\pm \sqrt{5}/2$

 

[LCKurtz]

Homework Statement

Show that, for each real number t $\in$ the interval (0, 1], the curve given by:

$y=ln(\frac{x+\sqrt{1+x^{2}}}{1+\sqrt{2}})$

has a tangent line with slope t. Find the points on the curve at which the tangent line has slope 2/3.

The Attempt at a Solution

I found the first derivative of this curve to be:

$dy/dx=1/\sqrt{(1+x^{2})}$

but now not sure how to proceed.

What do they mean when they ask me to show that for each t has a tangent line with slope t?

any help appreciated!

Also to find the points with dy/dx=2/3 I did this:

$dy/dx=1/\sqrt{(1+x^{2})} =2/3$ >>>> $x=\pm \sqrt{5}/2$

Do what you just did for $t$ instead of 2/3. For the points on the curve you probably want the y values too.

 

[kwal0203]

Do what you just did for $t$ instead of 2/3. For the points on the curve you probably want the y values too.

Oh of course, lol! thanks