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

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

... can someone check my answer?

I know the length formula is

... can someone check my answer?

##### 3 Answers

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

#### Explanation:

Considering that this is for a semicircle perimeter of a full cycle..

#### Explanation:

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

A more analytic solution would go as follows

So, for

and hence

Thus, the arc length is

# 2pi #

#### Explanation:

We seek the arc length of

We calculate polar arc length using the formula:

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

Then, given that

# (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

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