Find arc length of r=2\cos\theta in the range 0\le\theta\le\pi?

I know the length formula is \int_a^b\sqrt(1+(y')^2)dx

... can someone check my answer?

3 Answers
May 23, 2018

This is an answer from a tutor but I don't get their work.

Explanation:

L=\intds=\int_(\theta_1)^(\theta_2)rd\theta

s=r\theta so ds=rd\theta, thus
2\int_0^\pi\cos\theta=2(\sin\theta)|_0^\pi=2(\sin\pi-\sin0)=...

Considering that this is for a semicircle perimeter of a full cycle.. ...4(\sin\theta)|_0^(\pi/2)=4(\sin(\pi/2)-\sin0)=4(1-0)=4

May 23, 2018

2pi

Explanation:

It is easy to see that the curve is a circle of radius 1. It's length is obviously 2pi

A more analytic solution would go as follows

ds^2 = dr^2+r^2d theta^2

So, for r = 2 cos theta, we have

dr = -2 sin theta d theta

and hence

ds^2 = (-2 sin theta d theta)^2+(2 cos theta)^2 d theta^2 = 4d theta^2 implies

ds = 2 d theta

Thus, the arc length is

L = int_{theta = 0}^{theta=pi}ds = 2pi

May 23, 2018

2pi

Explanation:

We seek the arc length of r=2cos theta for theta in [0,pi]

We calculate polar arc length using the formula:

l = int_alpha^beta \ sqrt(r^2 + ((dr)/(d theta))^2 ) \ d theta

Then, given that r=2cos theta then differentiating wrt theta we get:

(dr)/(d theta) = -2sin theta

So then:

l = int_(0)^(pi) \ sqrt((2cos theta)^2 + (-2sin theta)^2 ) \ d theta

\ = int_(0)^(pi) \ sqrt(4(cos^2 theta + sin^2 theta) ) \ d theta

\ = int_(0)^(pi) \ sqrt(4 ) \ d theta \ \ \ \ \ \ \ \ \ (because cos^2 theta + sin^2 theta -= 1)

\ = int_(0)^(pi) \ 2 \ d theta

\ = 2[ \ theta \ ]_(0)^(pi)

\ = 2(pi - 0)

\ = 2pi

Notes:

The observant reader will note that that r=2cos theta represents a circle of radius 1 centred on (1,0), so we can calculate the arc length as the perimeter of a unit circle, thus>

P = (2pi)(1) = 2pi, as above