Question

What is the arclength of `r=-2sin(theta/4+(7pi)/8)` on `theta in [(pi)/4,(7pi)/4]`?

Answer 1

`L=(3pi)/4+sum_(n=1)^oo((1/2),(n))(-15/64)^n{((2n),(n))(3pi)/4+4sum_(k=0)^(n-1)sum_(m=0)^n((n),(k))(-1)^m/(n-k)delta_((n-k)(4m+2))}`

Explanation:

`r=-2sin(theta/4+(7pi)/8)`
`r^2=4sin^2(theta/4+(7pi)/8)`

`r'=-1/2costheta(theta/4+(7pi)/8)`
`(r')^2=1/4cos^2(theta/4+(7pi)/8)`

Arclength is given by:

`L=int_(pi/4)^((7pi)/4)sqrt(4sin^2(theta/4+(7pi)/8)+1/4cos^2(theta/4+(7pi)/8))d theta`

Apply the substitution `theta/4+(7pi)/8=phi`:

`L=2int_((15pi)/16)^((21pi)/16)sqrt(sin^2phi+1/16cos^2phi)dphi`

Apply the identity `sin^2phi=1-cos^2phi`:

`L=2int_((15pi)/16)^((21pi)/16)sqrt(1-15/16cos^2phi)dphi`

Since `15/16cos^2phi<1`, apply a series expansion for the square root:

`L=2int_((15pi)/16)^((21pi)/16)sum_(n=0)^oo((1/2),(n))(-15/16cos^2phi)^ndphi`

Isolate the `n=0` term and simplify:

`L=2int_((15pi)/16)^((21pi)/16)dphi+2sum_(n=1)^oo((1/2),(n))(-15/16)^nint_((15pi)/16)^((21pi)/16)cos^(2n)phidphi`

Apply the Trigonometric power-reduction formula:

`L=(3pi)/4+2sum_(n=1)^oo((1/2),(n))(-15/16)^nint_((15pi)/16)^((21pi)/16){1/4^n((2n),(n))+2/4^nsum_(k=0)^(n-1)((n),(k))cos((2n-2k)phi)}dphi`

Integrate directly:

`L=(3pi)/4+2sum_(n=1)^oo((1/2),(n))(-15/64)^n[((2n),(n))phi+sum_(k=0)^(n-1)((n),(k))sin((2n-2k)phi)/(n-k)]int_((15pi)/16)^((21pi)/16)`

Insert limits:

`L=(3pi)/4+2sum_(n=1)^oo((1/2),(n))(-15/64)^n{((2n),(n))(3pi)/8+sum_(k=0)^(n-1)((n),(k))(sin((n-k)(21pi)/8)-sin((n-k)(15pi)/8))/(n-k)}`

Apply the Trigonometric sum-to-product formula:

`L=(3pi)/4+sum_(n=1)^oo((1/2),(n))(-15/64)^n{((2n),(n))(3pi)/4+4sum_(k=0)^(n-1)((n),(k))(sin((n-k)(3pi)/4)cos((n-k)(9pi)/2))/(n-k)}`

Since `sin((n-k)(3pi)/4)cos((n-k)(9pi)/2)` gives either `0` or `+-1`:

`L=(3pi)/4+sum_(n=1)^oo((1/2),(n))(-15/64)^n{((2n),(n))(3pi)/4+4sum_(k=0)^(n-1)sum_(m=0)^n((n),(k))(-1)^m/(n-k)delta_((n-k)(4m+2))}`

This can probably be further simplified.