Question

What is the arc length of `f(x)=sqrt(1+64x^2)` on `x in [1,5]`?

Answer 1

A first-order approximation gives `L=4sqrt65-4/sqrt65tan^(-1)(32/321)` units.

Explanation:

`f(x)=sqrt(1+64x^2)`

`f'(x)=(64x)/sqrt(1+64x^2)`

Arc length is given by:

`L=int_1^5sqrt(1+(4096x^2)/(1+64x^2))dx`

Apply the substitution `8x=u` and rearrange:

`L=sqrt65/8int_8^40sqrt(1-64/65 1/(1+u^2))du`

For `u in [8,40]`, `64/65 1/(1+u^2)<1`. Take the series expansion of the square root:

`L=sqrt65/8int_8^40sum_(n=0)^oo((1/2),(n))(-64/65 1/(1+u^2))^ndu`

Isolate the `n=0` term and simplify:

`L=sqrt65/8int_8^40du+sqrt65/8sum_(n=1)^oo((1/2),(n))(-64/65)^nint_8^40(1/(1+u^2))^ndu`

Apply the substitution `u=tantheta`:

`L=4sqrt65+sqrt65/8sum_(n=1)^oo((1/2),(n))(-64/65)^nint_(tan^(-1)(8))^(tan^(-1)(40))cos^(2n-2)thetad theta`

Isolate the `n=1` term:

`L=4sqrt65-4/sqrt65int_(tan^(-1)(8))^(tan^(-1)(40))d theta+sqrt65/8sum_(n=2)^oo((1/2),(n))(-64/65)^nint_(tan^(-1)(8))^(tan^(-1)(40))cos^(2n-2)thetad theta`

Rescale `n`:

`L=4sqrt65-4/sqrt65(tan^(-1)(40)-tan^(-1)(8))+sqrt65/8sum_(n=1)^oo((1/2),(n+1))(-64/65)^(n+1)int_(tan^(-1)(8))^(tan^(-1)(40))cos^(2n)thetad theta`

Apply the Trigonometric power-reduction formula:

`L=4sqrt65-4/sqrt65tan^(-1)(32/321)-8/sqrt65sum_(n=1)^oo((1/2),(n+1))(-64/65)^nint_(tan^(-1)(8))^(tan^(-1)(40)){1/4^n((2n),(n))+2/4^nsum_(k=0)^(n-1)((2n),(k))cos((2n-2k)theta)}d theta`

Integrate term by term:

`L=4sqrt65-4/sqrt65tan^(-1)(32/321)-8/sqrt65sum_(n=1)^oo((1/2),(n+1))(-16/65)^n[((2n),(n))theta+sum_(k=0)^(n-1)((2n),(k))sin((2n-2k)theta)/(n-k)]_(tan^(-1)(8))^(tan^(-1)(40))`

Hence:

`L=4sqrt65-4/sqrt65tan^(-1)(32/321)-8/sqrt65sum_(n=1)^oo((1/2),(n+1))(-16/65)^n[((2n),(n))tan^(-1)(32/321)+sum_(k=0)^(n-1)((2n),(k))(sin((2n-2k)tan^(-1)40)-sin((2n-2k)tan^(-1)8))/(n-k)]`

Apply the sum-to-product formula for the sine function:

`L=4sqrt65-4/sqrt65tan^(-1)(32/321)-8/sqrt65sum_(n=1)^oo((1/2),(n+1))(-16/65)^n[((2n),(n))tan^(-1)(32/321)+2sum_(k=0)^(n-1)((2n),(k))(sin{(n-k)(tan^(-1)40-tan^(-1)8)}cos{(n-k)(tan^(-1)40+tan^(-1)8)})/(n-k)]`

Apply the angle-sum identities for the tangent function:

`L=4sqrt65-4/sqrt65tan^(-1)(32/321)-8/sqrt65sum_(n=1)^oo((1/2),(n+1))(-16/65)^n[((2n),(n))tan^(-1)(32/321)-2sum_(k=0)^(n-1)((2n),(k))(sin{(n-k)tan^(-1)(32/321)}cos{(n-k)tan^(-1)(48/319)})/(n-k)]`