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

1 Answer
May 30, 2018

Answer:

#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)}#