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

1 Answer
Jun 7, 2018

Answer:

#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.