Find the perimeter of the cord ra1 cos?

Answer 1

r=a(1+cos x)

r^2=a^2(1+cos x)^2 = a^2 + 2[a^2][cos(x)] + [a^2][cos^2(x)]

dr/dx = r' = -a sin(x)

(r')^2 = [a^2][sin^2 (x)]

Therefore perimeter (s) of curve r=a(1+cos x) in polar coordinate with x vary from 0 to Pi (due to curve is symmetry on axis x=0) is

s = 2 Int { sqrt[r^2 + (r')^2] } dx where x vary from 0 to Pi.

Thus

sqrt[r^2 + (r')^2]

= sqrt { a^2 + 2[a^2][cos(x)] + [a^2][cos^2 (x)] + [a^2][sin^2 (x)] }

= sqrt { (2a^2)[1+cos(x)] }

= [sqrt(2)]a {sqrt [1+cos(x)]}

Then

s = 2[sqrt(2)]a . Int {sqrt [1+cos(x)]} dx

Let 1+cos(x) = 1+2cos^2 (x/2) - 1 = 2cos^2 (x/2)

s = 2[sqrt(2)]a . Int {sqrt [2cos^2 (x/2)]} dx

s = 4a . Int [cos(x/2)] dx where x vary from 0 to Pi

s = 4a [sin(x/2)]/(1/2)

s = 8a [sin(Pi/2) - sin(0)]

s = 8a