What is the arclength of #r=-2sin(theta/4+(7pi)/8) # on #theta in [(pi)/4,(7pi)/4]#?
1 Answer
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
#L=2int_((15pi)/16)^((21pi)/16)sqrt(sin^2phi+1/16cos^2phi)dphi#
Apply the identity
#L=2int_((15pi)/16)^((21pi)/16)sqrt(1-15/16cos^2phi)dphi#
Since
#L=2int_((15pi)/16)^((21pi)/16)sum_(n=0)^oo((1/2),(n))(-15/16cos^2phi)^ndphi#
Isolate the
#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
#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.