# Find the exact length of the arc of this curve

$y = 2e^x + (1/8)e^{-x}$ ... on the interval $[0, \ln(2)]$

I know am supposed to user the Arc Length formula, but I'm not sure if I have the derivative of this function correct.

I came up with:

$$f'^2 (x)= 4e^{2x} - \frac{1}{2} + \frac{1}{64} e^{-2x}$$

I'm really rusty on this stuff though and am probably wrong.

And even if this is right, I'm not sure what to do next with all that + 1 under the square root.

hint:$$1+(f'(x))^2 = \left(2e^x + \frac{1}{8}e^{-x}\right)^2$$

• what happened to the +1? Can you show how this is true? My algebra clearly needs some work. Mar 9, 2015 at 22:23
• The $+1$ combines with the $-1/2$ in $(f')^2$ to give $+1/2$. Then $1 + (f')^2$ is a perfect square of the form shown above. Mar 9, 2015 at 22:58
• @nukeguy Now I see it, man, I'm never going to spot that by myself in a billion years... Mar 9, 2015 at 23:42
• A billion years? Now that you have seen this one, also consult math.stackexchange.com/questions/1177227 and maybe you can remember this trick. Mar 10, 2015 at 1:06

You can use the formulae:

$$\int_a^b \sqrt{y'(x)^2+1} \, dx$$

This is arclength for cartesian coordinates. Evaluate derivative $y'$, square it and add one, should be:

$1+\left(2 e^x-\frac{e^{-x}}{8}\right)^2=\left(2 e^x+\frac{e^{-x}}{8}\right)^2$

Now use formulae from above.

• Ok, this is what I had then, but I wasn't sure what to do next so I multiplied it out, and it's messy. Is that what Im supposed to do? Mar 9, 2015 at 22:36
• @aSilveira Yes, i think so! Mar 9, 2015 at 22:42
• @nukeguy Thank's, now I see it. Mar 9, 2015 at 23:38