Question

What is the arclength of `r=-10sin(theta/4+(9pi)/8)` on `theta in [(pi)/4,(7pi)/4]`?

Answer 1

`L=15pi+40sum_(n=1)^oo((1/2),(n))(-15/64)^n{((2n),(n))(3pi)/8+2sum_(k=0)^(n-1)((2n),(k))(sin((n-k)(3pi)/4)cos((n-k)(11pi)/2))/(n-k)}` units.

Explanation:

`r=-10sin(theta/4+(9pi)/8)`
`r^2=100sin^2(theta/4+(9pi)/8)`

`r'=-10/4cos(theta/4+(9pi)/8)`
`(r')^2=100/16cos^2(theta/4+(9pi)/8)`

Arclength is given by:

`L=int_(pi/4)^((7pi)/4)sqrt(100sin^2(theta/4+(9pi)/8)+100/16cos^2(theta/4+(9pi)/8))d theta`

Apply the substitution `theta/4+(9pi)/8=phi`

`L=40int_((19pi)/16)^((25pi)/16)sqrt(sin^2phi+1/16cos^2phi)dphi`

Apply the Trigonometric identity `sin^2x+cos^2x=1`:

`L=40int_((19pi)/16)^((25pi)/16)sqrt(1-15/16cos^2phi)dphi`

Since `15/16cos^2phi<1`, take the series expansion of the square root:

`L=40int_((19pi)/16)^((25pi)/16)sum_(n=0)^oo((1/2),(n))(-15/16cos^2phi)^ndphi`

Isolate the `n=0` term and simplify:

`L=40int_((19pi)/16)^((25pi)/16)dphi+40sum_(n=1)^oo((1/2),(n))(-15/16)^nint_((19pi)/16)^((25pi)/16)cos^(2n)phidphi`

Apply the Trigonometric power-reduction formula:

`L=40((25pi)/16-(19pi)/16)+40sum_(n=1)^oo((1/2),(n))(-15/16)^nint_((19pi)/16)^((25pi)/16){1/4^n((2n),(n))+2/4^nsum_(k=0)^(n-1)((2n),(k))cos((2n-2k)phi)}dphi`

Integrate directly:

`L=15pi+40sum_(n=1)^oo((1/2),(n))(-15/64)^n[((2n),(n))phi+sum_(k=0)^(n-1)((2n),(k))sin((2n-2k)phi)/(n-k)]_((19pi)/16)^((25pi)/16)`

Insert the limits of integration:

`L=15pi+40sum_(n=1)^oo((1/2),(n))(-15/64)^n{((2n),(n))(3pi)/8+sum_(k=0)^(n-1)((2n),(k))(sin((n-k)(25pi)/8)-sin((n-k)(19pi)/8))/(n-k)}`

Apply the Trigonometric sum-to-product identity:

`L=15pi+40sum_(n=1)^oo((1/2),(n))(-15/64)^n{((2n),(n))(3pi)/8+2sum_(k=0)^(n-1)((2n),(k))(sin((n-k)(3pi)/4)cos((n-k)(11pi)/2))/(n-k)}`