Question

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

What I have so far:
`ds=\sqrt(1+(dy/dx)^2)dx` with `a=0` and `b=2`

`dy/dx=1/2(e^x-e^-x)^2`
`(dy/dx)^2=1/4(e^x-e^-x)`

`ds=\sqrt(1+[1/4(e^x-e^-x)^2])dx...`

So how do I proceed from here...?

Answer 1

`L=1/2(e^2-1/e^2)`

Explanation:

So, we have `y=1/2(e^x+e^-x)`

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

`L=int_0^2sqrt(1+(dy/dx)^2)dx`

Let's find the derivative:

`dy/dx=1/2(e^x-e^-x)`

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

`y=1/2(e^x+e^-x)=coshx`

`dy/dx=1/2(e^x-e^-x)=sinhx`

Then, squaring gives us

`(dy/dx)^2=sinh^2x`

So, applying the hyperbolic identity, we get

`L=int_0^2sqrt(1+sinh^2x)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^2x-sinh^2x=1`

This tells us that

`1+sinh^2x=cosh^2x`

Thus,

`L=int_0^2sqrt(cosh^2x)dx=int_0^2coshxdx`

In general,

`intcoshxdx=sinhx+C`.

Since the derivatives hyperbolic sine and hyperbolic sine are simply one another, IE, `d/dxsinhx=coshx, d/dxcoshx=sinhx,` so their integrals also hold this property.

Thus,

`L=sinhx|_0^2`

`L=1/2(e^x-e^-x)|_0^2`

`L=1/2(e^2-e^-2-(e^0-e^0))`

`L=1/2(e^2-1/e^2-(1-1))`

`L=1/2(e^2-1/e^2)`

Answer 2

note: this answer is for the question: Find the arc length of the function `y=ln(secx)` with parameters `0 ≤ x ≤ pi/4`?

`ln(sqrt(2)+1)=0.881374...`

Explanation:

arclength `=int_0^(pi/4)sqrt(1+(y'(x))^2)dx`

[side note:
`y'(x)=1/secx*(secxtanx)=tanx`]

arclength `=int_0^(pi/4)sqrt(1+tan^2x)dx`

`=int_0^(pi/4)sqrt(sec^2x)dx`

`=int_0^(pi/4)secxdx`
[side note:
you can do this because `secx` is always positive on `0<=x<=pi/4`]

[another side note: `intsecxdx=ln|secx+tanx|+C` see this]

`=ln|sec(pi/4)+tan(pi/4)|-ln|sec0+tan0|`

`=ln|sqrt(2)+1|-ln|1+0|`

`=ln|sqrt(2)+1|`

`=0.881374...`

Answer 3

As you stated,

`y=1/2(e^x+e^-x)`

`dy/dx=1/2(e^x-e^-x)`

`(dy/dx)^2=1/4(e^x-e^-x)^2`

But now let's expand `(e^x-e^-x)^2` as `(e^x-e^-x)(e^x-e^-x)=e^(2x)+e^(-2x)-2`. Recall that `e^xe^-x=e^0=1`.

Then, we see that

`ds=sqrt(1+1/4(e^(2x)+e^(-2x)-2))dx=sqrt((e^(2x)+e^(-2x)+2)/4)dx`

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

`ds=sqrt((e^(2x)(e^(2x)+e^(-2x)+2))/(4e^(2x)))dx=sqrt((e^(4x)+2e^(2x)+1)/(4e^(2x)))dx`

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

`ds=sqrt((e^(2x)+1)^2/((2e^x)^2))dx=(e^(2x)+1)/(2e^x)dx=1/2(e^x+e^-x)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_0^2ds=1/2int_0^2(e^x+e^-x)dx=1/2(e^x-e^-x)|_0^2`

Evaluating:

`L=1/2[(e^2-1)-(e^-2-1)]=1/2(e^2-1/e^2)=(e^4-1)/(2e^2)`