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

1 Answer
Jul 25, 2017

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 thetausing 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= ∞