Question

Find the arc length of the function below?

`y=\ln(\secx)`, with parameters `0\lex\le\pi/4`?

Answer 1

The arc has a length of `ln(sqrt2+1).`

Explanation:

For a function in the form `y=f(x),` the arc length from `[a,b]` is given by

`L=int_a^bsqrt(1+(dy/dx)^2)dx`.

The given function `y=lnsecx` is in the form `y=f(x)`, so we'll use the above formula. Furthermore, `[a,b]=[0,pi/4]`, so these are our integral's bounds.

Differentiate using the Chain Rule:

`y=lnsecx`
`(dy)/dx=1/secx*d/dxsecx`

`dy/dx=(cancelsecxtanx)/(cancelsecx)`

`dy/dx=tanx`

Then,

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

Recall the identity `1+tan^2x=sec^2x`. Then we get

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

`L=int_0^(pi/4)secxdx`

This is a common integral and worth memorizing. In general,

`intsecxdx=ln|secx+tanx|+C`

Then,

`int_0^(pi/4)secxdx=ln(secx+tanx)|_0^(pi/4)`

Absolute value bars are dropped because secant and tangent are positive on `[0, pi/4]`.

`=ln(sec(pi/4)+tan(pi/4))-ln(sec(0)+tan(0))`

`=ln(sqrt2+1)-ln(1)`

`=ln(sqrt2+1)`

The arc has a length of `ln(sqrt2+1).`