$y = 2e^x + (1/8)e^{-x}$ ... 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.
$\endgroup$ 02 Answers
$\begingroup$hint:$$1+(f'(x))^2 = \left(2e^x + \frac{1}{8}e^{-x}\right)^2$$
$\endgroup$ 4 $\begingroup$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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