Question

What is the arclength of `r=-3cos(theta/16+(pi)/16)` on `theta in [(-5pi)/16,(9pi)/16]`?

Answer 1

`L=(32pi)/128+3sum_(n=1)^oo((1/2),(n))(-255/1024)^n{((2n),(n))(7pi)/128+sum_(k=0)^(n-1)(-1)^(n-k)((2n),(k))(sin((n-k)(25pi)/128)-sin((n-k)(11pi)/128))/(n-k)}` units.

Explanation:

`r=-3cos(theta/16+pi/16)`
`r^2=9cos^2(theta/16+pi/16)`

`r'=3/16sin(theta/16+pi/16)`
`(r')^2=9/256sin^2(theta/16+pi/16)`

Arclength is given by:

`L=int_((-5pi)/16)^((9pi)/16)sqrt(9cos^2(theta/16+pi/16)+9/256sin^2(theta/16+pi/16))d theta`

Apply the substitution `theta/16+pi/16=phi`:

`L=3int_((11pi)/256)^((25pi)/256)sqrt(cos^2phi+1/256sin^2phi)dphi`

Rearrange:

`L=3int_((11pi)/256)^((25pi)/256)sqrt(1-255/256sin^2phi)dphi`

From here we can use the small-angle approximation for an easy answer.

Alternatively take the series expansion of the square root:

`L=3int_((11pi)/256)^((25pi)/256)sum_(n=0)^oo((1/2),(n))(-255/256sin^2phi)^ndphi`

Isolate the `n=0` term and simplify:

`L=3int_((11pi)/256)^((25pi)/256)dphi+3sum_(n=1)^oo((1/2),(n))(-255/256)^nint_((11pi)/256)^((25pi)/256)sin^(2n)phidphi`

Apply the trigonometric power reduction formula:

`L=(32pi)/128+3sum_(n=1)^oo((1/2),(n))(-255/256)^nint_((11pi)/256)^((25pi)/256){1/4^n((2n),(n))+2/4^nsum_(k=0)^(n-1)(-1)^(n-k)((2n),(k))cos((2n-2k)phi)}dphi`

Integrate directly:

`L=(32pi)/128+3sum_(n=1)^oo((1/2),(n))(-255/1024)^n[((2n),(n))phi+sum_(k=0)^(n-1)(-1)^(n-k)((2n),(k))sin((2n-2k)phi)/(n-k)]_((11pi)/256)^((25pi)/256)`

Hence:

`L=(32pi)/128+3sum_(n=1)^oo((1/2),(n))(-255/1024)^n{((2n),(n))(7pi)/128+sum_(k=0)^(n-1)(-1)^(n-k)((2n),(k))(sin((n-k)(25pi)/128)-sin((n-k)(11pi)/128))/(n-k)}`