What is the arclength of the polar curve #f(theta) = -3sin(3theta)-2cot4theta # over #theta in [0,pi/8] #?

1 Answer
Jul 25, 2017

Answer:

# L= ∞#

Explanation:

Let us start with the derivation of the formula to find the arclength of a polar curve, where #r = f(theta)#

From basic geometry and trigonometry,
#x = r*cos(theta)# and #y = r*sin(theta)#

since #r = f(theta)#,
#x = f(theta) * cos(theta)# and #y = f(theta) * sin(theta)#

Starting with #x# first and differentiating both sides with respect to #theta#using the product rule,
#dx/(d theta) = d/(d theta) (f(theta) * cos(theta)) = f'(theta)cos(theta) - f(theta)sin(theta)#
#= (dr)/(d theta) cos(theta)-rsin(theta)#

Doing the same with the #y# equation,
#dy/(d theta) = d/(d theta) (f(theta) * sin(theta)) = f'(theta)sin(theta) + f(theta)cos(theta)#
# = (dr)/(d theta) sin(theta)+rcos(theta)#

To find #dl#, a differential of the arclength, each infinitesimal piece of the arclength can be seen as the hypotenuse (or simply a straight line segment) whose length can be figured out by finding

#dl = sqrt((dx/(d theta))^2 + (dy/(d theta))^2) d(theta)#

let us concentrate on the #(dx/(d theta))^2 + (dy/(d theta))^2# bit first.

#(dx/(d theta))^2 + (dy/(d theta))^2 = ((dr)/(d theta)cos(theta) - rsin(theta))^2 + ((dr)/(d theta)sin(theta) + rcos(theta))^2#

#=((dr)/(d theta))^2cos^2(theta)-2r(dr)/(d theta)cos(theta)sin(theta)+r^2sin^2(theta) + ((dr)/(d theta))^2sin^2(theta)+2r(dr)/(d theta)cos(theta)sin(theta)+r^2cos^2(theta)#

simplifying,
#= ((dr)/(d theta))^2 (cos^2(theta) + sin^2(theta))+r^2(cos^2(theta) + sin^2(theta)) #

and since #cos^2(theta) + sin^2(theta) = 1# (trigonometric identity),

#(dx/(d theta))^2 + (dy/(d theta))^2 = ((dr)/(d theta))^2 +r^2 #

therefore,

#sqrt((dx/(d theta))^2 + (dy/(d theta))^2) = sqrt(((dr)/(d theta))^2 +r^2 )#

#dl = (sqrt(((dr)/(d theta))^2 +r^2 )) d(theta)#

integrating all the infinitesimal pieces of length, say from #theta = a# to #theta = b#

#int_a^b dl = int_a^b(sqrt(((dr)/(d theta))^2 +r^2 ))d(theta)#

note that #r = f(theta)#
in your case, #theta# ranges in the interval #[0,pi/8]#
#f'(theta) = (dr)/(d theta) = d/(d theta) (-3sin(3 theta)-2cot(4 theta))#
# = -9cos(3 theta)+8csc ^2(4)#

Arc-length =
#int_0^(pi/8) (sqrt((-9cos(3 theta)+8csc ^2(4 theta))^2) + (-3sin(3 theta)-2cot(4 theta))^2) d(theta)#

You could use a calculator to evaluate the integral, but by even graphing, the arc length reaches up and down towards infinity, so that there is not a closed curve to calculate its length, which can be seen graphically on a Cartesian plane:
graph{-3sin(3x)-2cot(4x) [-14.24, 14.24, -7.12, 7.12]}

There are discontinuous segments throughout its domain, which are infinite, so its arc length cannot be finite.

Therefore, # L= ∞#