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?
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