Question

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?

Answer 1

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`

Answer 2

`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`

Answer 3

`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