Question

Find the arc length of the curve on the interval \([0,2 \pi]\). Hypocycloid perimeter: \(x=a \cos ^{3} \theta, y=a \sin ^{3} \theta\)

Short answer

The arc length of the curve on the interval \([0,2 \pi]\) is 0.

Step-by-step solution

Find derivatives of x and y with respect to theta

Find the derivatives of \(x=a \cos ^{3} \theta\) and \(y=a \sin ^{3} \theta\) with respect to \(\theta\). We can use the chain rule for differentiation, which results in: \[(dx/d\theta) = -3a\cos^{2}(\theta)\sin(\theta)\] and \[(dy/d\theta) = 3a\sin^{2}(\theta)\cos(\theta)\]

Square the derivatives and sum them

Square each of the derivatives from the step 1 and add them. This gives us: \[g(\theta)=(dx/d\theta)^{2} +(dy/d\theta)^{2} = 9a^{2}\cos^{4}(\theta)\sin^{2}(\theta) + 9a^{2}\sin^{4}(\theta)\cos^{2}(\theta)\]

Simplify the expression

Now simplify the expression calculated in the previous step. We can factor out \((9a^{2}\cos^{2}(\theta)\sin^{2}(\theta))\) out of the equation obtained in step 2 to get: \[g(\theta) = 9a^{2}\cos^{2}(\theta)\sin^{2}(\theta) (\cos^{2}(\theta)+\sin^{2}(\theta))= 9a^{2}\cos^{2}(\theta)\sin^{2}(\theta)\] considering the trigonometric identity \(\cos^{2}(\theta)+\sin^{2}(\theta)=1\)

Find the square root of g(θ) and integrate

Take the square root of the function \(g(\theta)\) obtained in step 3 and integrate it over the given interval [0, 2π] using the arc length formula previously mentioned i.e., \[L = \int_{\alpha}^{\beta} \sqrt {(dx/d\theta)^{2} +(dy/d\theta)^{2}} d\theta = \int_{0}^{2\pi} \sqrt {g(\theta)}d\theta = \int_{0}^{2\pi} 3a\cos(\theta)\sin(\theta)\ d\theta\]. Save the integral part for the next step.

Solve the integral

The integral calculated in step 4 simplifies to 0 because \(\cos(\theta)\sin(\theta)\) is an odd function and we are integrating over the interval \([0,2\pi]\), which is symmetric about zero.