Question

Prove that the arc length of the polar curve `r = a(1-cos theta)` is `8a`?

Answer 1

The polar arc length of a curve is given by:

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

We have:

`r = a(1-cos theta)`
`\ \ = a-acos theta`

Thus:

`(dr)/(d theta) = asin theta`

So, the arc length is:

`L = int_0^(2pi) \ sqrt( (a-acos theta)^2 + (asin theta)^2 ) \ d theta`
`\ \ \ = int_0^(2pi) \ sqrt( (a^2-2a^2cos theta + a^2 cos^2 theta) + (asin theta)^2 ) \ d theta`
`\ \ \ = int_0^(2pi) \ asqrt( 1-2cos theta + cos^2 theta + sin^2 theta ) \ d theta`
`\ \ \ = a \ int_0^(2pi) sqrt( 2-2cos theta ) \ d theta`

`\ \ \ = a \ int_0^(2pi) sqrt( 2)sqrt(1-cos theta ) \ d theta`

Using the trig identity:

`cos2x -= 1-2sin^2x => 2sin^2x-=1-cos2x`

We then have:

`L = a \ int_0^(2pi) sqrt( 2)sqrt(2sin^2 (theta/2) ) \ d theta`
`\ \ \ = a \ int_0^(2pi) 2sin (theta/2) \ d theta`
`\ \ \ = 2a \ [ -2cos (theta/2) ]_0^(2pi)`
`\ \ \ = -4a \ (cos pi - cos 0)`
`\ \ \ = -4a \ (-1-1)`
`\ \ \ = (-4a)(-2)`
`\ \ \ = 8a` QED