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$y = 2e^x + (1/8)e^{-x}$ ... 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.

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2 Answers 2

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

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  • $\begingroup$ what happened to the +1? Can you show how this is true? My algebra clearly needs some work. $\endgroup$
    – aSilveira
    Mar 9, 2015 at 22:23
  • $\begingroup$ 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. $\endgroup$
    – nukeguy
    Mar 9, 2015 at 22:58
  • $\begingroup$ @nukeguy Now I see it, man, I'm never going to spot that by myself in a billion years... $\endgroup$
    – aSilveira
    Mar 9, 2015 at 23:42
  • $\begingroup$ A billion years? Now that you have seen this one, also consult math.stackexchange.com/questions/1177227 and maybe you can remember this trick. $\endgroup$
    – GEdgar
    Mar 10, 2015 at 1:06
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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.

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  • $\begingroup$ 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? $\endgroup$
    – aSilveira
    Mar 9, 2015 at 22:36
  • $\begingroup$ @aSilveira Yes, i think so! $\endgroup$
    – Frieder
    Mar 9, 2015 at 22:42
  • $\begingroup$ @nukeguy Thank's, now I see it. $\endgroup$
    – Frieder
    Mar 9, 2015 at 23:38

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