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

1 Answer
Mar 19, 2018

Answer:

#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)}#