Find the arc length of the function y=1/2(e^x+e^-x) with parameters 0\lex\le2?

What I have so far: $\mathrm{ds} = \setminus \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$ with $a = 0$ and $b = 2$ $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} {\left({e}^{x} - {e}^{-} x\right)}^{2}$ ${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} = \frac{1}{4} \left({e}^{x} - {e}^{-} x\right)$ $\mathrm{ds} = \setminus \sqrt{1 + \left[\frac{1}{4} {\left({e}^{x} - {e}^{-} x\right)}^{2}\right]} \mathrm{dx} \ldots$ So how do I proceed from here...?

Apr 25, 2018

$L = \frac{1}{2} \left({e}^{2} - \frac{1}{e} ^ 2\right)$

Explanation:

So, we have $y = \frac{1}{2} \left({e}^{x} + {e}^{-} x\right)$

This is in the form $y = f \left(x\right)$, so we know arc length $L$ on $\left[0 , 2\right]$ is given by

$L = {\int}_{0}^{2} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$

Let's find the derivative:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \left({e}^{x} - {e}^{-} x\right)$

Before jumping into the integration, we should realize the following hyperbolic identities:

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \left({e}^{x} - {e}^{-} x\right) = \sinh x$

Then, squaring gives us

${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} = {\sinh}^{2} x$

So, applying the hyperbolic identity, we get

$L = {\int}_{0}^{2} \sqrt{1 + {\sinh}^{2} x} \mathrm{dx}$

There is a hyperbolic Pythagorean identity we apply here, however, it looks slightly different from the normal trigonometric Pythagorean identity.

Deriving it is a pretty algebraically messy process, so I'll just give the identity:

${\cosh}^{2} x - {\sinh}^{2} x = 1$

This tells us that

$1 + {\sinh}^{2} x = {\cosh}^{2} x$

Thus,

$L = {\int}_{0}^{2} \sqrt{{\cosh}^{2} x} \mathrm{dx} = {\int}_{0}^{2} \cosh x \mathrm{dx}$

In general,

$\int \cosh x \mathrm{dx} = \sinh x + C$.

Since the derivatives hyperbolic sine and hyperbolic sine are simply one another, IE, $\frac{d}{\mathrm{dx}} \sinh x = \cosh x , \frac{d}{\mathrm{dx}} \cosh x = \sinh x ,$ so their integrals also hold this property.

Thus,

$L = \sinh x {|}_{0}^{2}$

$L = \frac{1}{2} \left({e}^{x} - {e}^{-} x\right) {|}_{0}^{2}$

$L = \frac{1}{2} \left({e}^{2} - {e}^{-} 2 - \left({e}^{0} - {e}^{0}\right)\right)$

$L = \frac{1}{2} \left({e}^{2} - \frac{1}{e} ^ 2 - \left(1 - 1\right)\right)$

$L = \frac{1}{2} \left({e}^{2} - \frac{1}{e} ^ 2\right)$

Apr 25, 2018

note: this answer is for the question: Find the arc length of the function $y = \ln \left(\sec x\right)$ with parameters 0 ≤ x ≤ pi/4?

$\ln \left(\sqrt{2} + 1\right) = 0.881374 \ldots$

Explanation:

arclength $= {\int}_{0}^{\frac{\pi}{4}} \sqrt{1 + {\left(y ' \left(x\right)\right)}^{2}} \mathrm{dx}$

[side note:
$y ' \left(x\right) = \frac{1}{\sec} x \cdot \left(\sec x \tan x\right) = \tan x$]

arclength $= {\int}_{0}^{\frac{\pi}{4}} \sqrt{1 + {\tan}^{2} x} \mathrm{dx}$

$= {\int}_{0}^{\frac{\pi}{4}} \sqrt{{\sec}^{2} x} \mathrm{dx}$

$= {\int}_{0}^{\frac{\pi}{4}} \sec x \mathrm{dx}$
[side note:
you can do this because $\sec x$ is always positive on $0 \le x \le \frac{\pi}{4}$]

[another side note: $\int \sec x \mathrm{dx} = \ln | \sec x + \tan x | + C$ see this]

$= \ln | \sec \left(\frac{\pi}{4}\right) + \tan \left(\frac{\pi}{4}\right) | - \ln | \sec 0 + \tan 0 |$

$= \ln | \sqrt{2} + 1 | - \ln | 1 + 0 |$

$= \ln | \sqrt{2} + 1 |$

$= 0.881374 \ldots$

Apr 25, 2018

As you stated,

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \left({e}^{x} - {e}^{-} x\right)$

${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} = \frac{1}{4} {\left({e}^{x} - {e}^{-} x\right)}^{2}$

But now let's expand ${\left({e}^{x} - {e}^{-} x\right)}^{2}$ as $\left({e}^{x} - {e}^{-} x\right) \left({e}^{x} - {e}^{-} x\right) = {e}^{2 x} + {e}^{- 2 x} - 2$. Recall that ${e}^{x} {e}^{-} x = {e}^{0} = 1$.

Then, we see that

$\mathrm{ds} = \sqrt{1 + \frac{1}{4} \left({e}^{2 x} + {e}^{- 2 x} - 2\right)} \mathrm{dx} = \sqrt{\frac{{e}^{2 x} + {e}^{- 2 x} + 2}{4}} \mathrm{dx}$

Let's try to clear up this numerator (get rid of negative exponents) by multiplying the fraction through by ${e}^{2 x} / {e}^{2 x}$:

$\mathrm{ds} = \sqrt{\frac{{e}^{2 x} \left({e}^{2 x} + {e}^{- 2 x} + 2\right)}{4 {e}^{2 x}}} \mathrm{dx} = \sqrt{\frac{{e}^{4 x} + 2 {e}^{2 x} + 1}{4 {e}^{2 x}}} \mathrm{dx}$

Now we recognize that our numerator is actually factorable, and in a very favorable way!

$\mathrm{ds} = \sqrt{{\left({e}^{2 x} + 1\right)}^{2} / \left({\left(2 {e}^{x}\right)}^{2}\right)} \mathrm{dx} = \frac{{e}^{2 x} + 1}{2 {e}^{x}} \mathrm{dx} = \frac{1}{2} \left({e}^{x} + {e}^{-} x\right) \mathrm{dx}$

What a remarkable amount of work to end up... exactly where we started?

Amazing and interesting coincidence aside, we can calculate the arc length now by:

$L = {\int}_{0}^{2} \mathrm{ds} = \frac{1}{2} {\int}_{0}^{2} \left({e}^{x} + {e}^{-} x\right) \mathrm{dx} = \frac{1}{2} \left({e}^{x} - {e}^{-} x\right) {|}_{0}^{2}$

Evaluating:

$L = \frac{1}{2} \left[\left({e}^{2} - 1\right) - \left({e}^{-} 2 - 1\right)\right] = \frac{1}{2} \left({e}^{2} - \frac{1}{e} ^ 2\right) = \frac{{e}^{4} - 1}{2 {e}^{2}}$