Question

What is the arclength of `f(x)=x/(x-5) in [0,3]`?

Answer 1

`L=9/2-sqrt5cot^(-1)(sqrt5)-2sqrt5sum_(n=1)^oo((1/2),(n+1))(-1/8)^n{((2n),(n))cot^(-1)(sqrt5)+sum_(m=0)^(n-1)(-1)^(n-m)/(n-m)((2n),(m))sin((2n-2m)cot^(-1)(sqrt5))cos((2n-2m)tan^(-1)(4/sqrt5))}`

Explanation:

`f(x)=x/(x-5)`, `x in [0,3]`

Let `y=f(x)`. We can invert:

`x=5/(y-1)+5`, `y in [-3/2,0]`

Take the derivative with respect to `y`:

`x'=-5/(y-1)^2`

Arclength is given by:

`L=int_(-3/2)^0sqrt(1+25/(y-1)^4)dy`

Apply the substitution `1-y=u/2`:

`L=1/2int_2^5sqrt(1+400/u^4)du`

Complete the square:

`L=1/2int_2^5sqrt((1+20/u^2)^2-40/u^2)du`

Factor out the larger piece:

`L=1/2int_2^5(1+20/u^2)sqrt(1-(40u^2)/(u^2+20)^2)du`

For `u in [2,5]`, `(40u^2)/(u^2+20)^2<1`. Take the series expansion of the square root:

`L=1/2int_2^5(1+20/u^2){sum_(n=0)^oo((1/2),(n))(-(40u^2)/(u^2+20)^2)^n}du`

Isolate the `n=0` term and simplify:

`L=1/2int_2^5(1+20/u^2)du+1/2sum_(n=1)^oo((1/2),(n))(-40)^nint_2^5 1/u(u/(u^2+20))^(2n-1)du`

Apply the substitution `u=sqrt20tantheta`:

`L=1/2[u-20/u]_ 2^5+sqrt5sum_(n=1)^oo((1/2),(n))(-2)^nint((tantheta)/(sec^2theta))^(2n-2)d theta`

Isolate the `n=1` term and simplify:

`L=9/2-sqrt5intd theta+sqrt5sum_(n=2)^oo((1/2),(n))(-2)^nint((tantheta)/(sec^2theta))^(2n-2)d theta`

Apply the appropriate half-angle Trigonometric identities:

`L=9/2-sqrt5[tan^(-1)(u/sqrt20)]_ 2^5+4sqrt5sum_(n=2)^oo((1/2),(n))((-1)/2)^nintsin^(2n-2)(2theta)d theta`

Rescale `n`:

`L=9/2-sqrt5(tan^(-1)(sqrt5/2)-tan^(-1)(1/sqrt5))-2sqrt5sum_(n=1)^oo((1/2),(n+1))(-1/2)^nintsin^(2n)(2theta)d theta`

Apply the appropriate Trigonometric power-reduction formula:

`L=9/2-sqrt5cot^(-1)(sqrt5)-2sqrt5sum_(n=1)^oo((1/2),(n+1))(-1/2)^nint{1/4^n((2n),(n))+2/4^nsum_(m=0)^(n-1)(-1)^(n-m)((2n),(m))cos((4n-4m)theta)}d theta`

Integrate term by term:

`L=9/2-sqrt5cot^(-1)(sqrt5)-2sqrt5sum_(n=1)^oo((1/2),(n+1))(-1/8)^n[((2n),(n))theta+1/2sum_(m=0)^(n-1)(-1)^(n-m)/(n-m)((2n),(m))sin((4n-4m)theta)]`

Reverse the last substitution:

`L=9/2-sqrt5cot^(-1)(sqrt5)-2sqrt5sum_(n=1)^oo((1/2),(n+1))(-1/8)^n[((2n),(n))tan^(-1)(u/sqrt20)+1/2sum_(m=0)^(n-1)(-1)^(n-m)/(n-m)((2n),(m))sin((4n-4m)tan^(-1)(u/sqrt20))]_2^5`

Insert the limits of integration:

`L=9/2-sqrt5cot^(-1)(sqrt5)-2sqrt5sum_(n=1)^oo((1/2),(n+1))(-1/8)^n{((2n),(n))cot^(-1)(sqrt5)+1/2sum_(m=0)^(n-1)(-1)^(n-m)/(n-m)((2n),(m))(sin((4n-4m)tan^(-1)(sqrt5/2))-sin((4n-4m)tan^(-1)(1/sqrt5)))}`

Apply the appropriate Trigonometric angle-addition identities:

`L=9/2-sqrt5cot^(-1)(sqrt5)-2sqrt5sum_(n=1)^oo((1/2),(n+1))(-1/8)^n{((2n),(n))cot^(-1)(sqrt5)+sum_(m=0)^(n-1)(-1)^(n-m)/(n-m)((2n),(m))sin((2n-2m)cot^(-1)(sqrt5))cos((2n-2m)tan^(-1)(4/sqrt5))}`