Question

Sketch the plane curve and find its length over the given interval. $$ \mathbf{r}(t)=a \cos ^{3} t \mathbf{i}+a \sin ^{3} t \mathbf{j}, \quad[0,2 \pi] $$

Short answer

The sketch of the plane curve represents a tear drop shape. The length of the curve over the interval \(0 \leq t \leq 2\pi\) is \(6a\).

Step-by-step solution

Sketch the Plane Curve

By setting \(x=a \cos ^{3} t\) and \(y=a \sin ^{3} t\), find the values of \(x\) and \(y\) at different values of \(t\) in the given interval. Then, plot these points on a graph to sketch the shape of the plane curve.

Calculate the Derivative of the Vector Function

The derivative of the vector function \(\mathbf{r}(t) = a \cos ^{3} t \mathbf{i} + a \sin ^{3} t \mathbf{j}\) is \(\mathbf{r}'(t) = -3a \cos^2 t \sin t \mathbf{i} + 3a \sin^2 t \cos t \mathbf{j}\).

Find the Magnitude of the Derivative

The magnitude \(||\mathbf{r}'(t)||\) is \(\sqrt{(-3a \cos^2 t \sin t)^2 + (3a \sin^2 t \cos t)^2}\), which simplifies to \(3a|\sin 2t|\).

Find the Length of the Curve

The length \(L\) of a curve over an interval \([a, b]\) is given by \(L = \int_{a}^{b} ||\mathbf{r}'(t)|| dt\). Substituting the calculated derivative magnitude and the given interval into this formula gives us \(L = \int_{0}^{2\pi} 3a|\sin 2t| dt\). Solving this integral yields \(L = 6a\).