What is the arclength of #r=-10sin(theta/4+(5pi)/16) # on #theta in [(-5pi)/16,(9pi)/16]#?

1 Answer
May 27, 2018

Answer:

#L=(133pi)/20-48/5sin(7/16pi)cos(11/8pi)+40sum_(n=2)^oo((1/2),(n))(-6/25)^n{((2n),(n))(7pi)/32+2sum_(k=0)^(n-1)((2n),(k))(sin((n-k)(7pi)/16)cos((n-k)(11pi)/8))/(n-k)}# units.

Explanation:

#r=−10sin(θ/4+(5pi)/16)#
#r^2=100sin^2(θ/4+(5pi)/16)#

#r'=−10/4cos(θ/4+(5pi)/16)#
#(r')^2=100/25cos^2(θ/4+(5pi)/16)#

Arclength is given by:

#L=int_((-5pi)/16)^((9pi)/16)sqrt(100sin^2(θ/4+(5pi)/16)+100/25cos^2(θ/4+(5pi)/16))d theta#

Apply the substitution #θ/4+(5pi)/16=phi# and rearrange:

#L=40int_((15pi)/64)^((29pi)/64)sqrt(1-24/25cos^2phi)dphi#

For #phi in [(15pi)/64,(29pi)/64]#, #24/25cos^2phi<1#. Take the series expansion of the square root:

#L=40int_((15pi)/64)^((29pi)/64)sum_(n=0)^oo((1/2),(n))(-24/25cos^2phi)^ndphi#

Isolate the #n=0# term and rearrange:

#L=40int_((15pi)/64)^((29pi)/64)dphi+40sum_(n=1)^oo((1/2),(n))(-24/25)^nint_((15pi)/64)^((29pi)/64)cos^(2n)phidphi#

Apply the trigonometric power-reduction formula:

#L=(35pi)/4+40sum_(n=1)^oo((1/2),(n))(-24/25)^nint_((15pi)/64)^((29pi)/64){1/4^n((2n),(n))+2/4^nsum_(k=0)^(n-1)((2n),(k))cos((2n-2k)phi)}dphi#

Integrate term by term:

#L=(35pi)/4+40sum_(n=1)^oo((1/2),(n))(-6/25)^n[((2n),(n))phi+sum_(k=0)^(n-1)((2n),(k))sin((2n-2k)phi)/(n-k)]_((15pi)/64)^((29pi)/64)#

Insert the limits of integration:

#L=(35pi)/4+40sum_(n=1)^oo((1/2),(n))(-6/25)^n{((2n),(n))(7pi)/32+sum_(k=0)^(n-1)((2n),(k))(sin((n-k)(29pi)/32)-sin((n-k)(15pi)/32))/(n-k)}#

Apply the trigonometric sum-to-product formula:

#L=(35pi)/4+40sum_(n=1)^oo((1/2),(n))(-6/25)^n{((2n),(n))(7pi)/32+2sum_(k=0)^(n-1)((2n),(k))(sin((n-k)(7pi)/16)cos((n-k)(11pi)/8))/(n-k)}#

Isolate the #n=1# terms:

#L=(35pi)/4-24/5((7pi)/16+2sin(7/16pi)cos(11/8pi))+40sum_(n=2)^oo((1/2),(n))(-6/25)^n{((2n),(n))(7pi)/32+2sum_(k=0)^(n-1)((2n),(k))(sin((n-k)(7pi)/16)cos((n-k)(11pi)/8))/(n-k)}#

Simplify:

#L=(133pi)/20-48/5sin(7/16pi)cos(11/8pi)+40sum_(n=2)^oo((1/2),(n))(-6/25)^n{((2n),(n))(7pi)/32+2sum_(k=0)^(n-1)((2n),(k))(sin((n-k)(7pi)/16)cos((n-k)(11pi)/8))/(n-k)}#