 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-1/15/c/solution

Differentiate

x = r cos θ

y = r sin θ

d x = cos θ d r - r sin θ d θ, d y = sin θ d r + r cos θ d θ .

Squaring and adding cancels the mixed terms and uses

cos^{2} θ + sin^{2} θ = 1

d x^{2} + d y^{2} = d r^{2} + r^{2} d θ^{2} .

sin^{2} θ + (1 + cos θ)^{2} = 2 + 2 cos θ = 2∣ cos (θ /2) ∣.

The cosine inside the absolute value changes sign at

θ = π

. Therefore the perimeter is

ℓ = 2 \int_{0}^{π} cos (θ /2) d θ - 2 \int_{π}^{2 π} cos (θ /2) d θ = 8

in coordinate length units. The isolated cusp does not obstruct this piecewise smooth arc length integral.

 New to topics? Read the docs here!