Question

In Exercises 49-53 sketch the parametric curve and find its length.

$x={\sin }^3\theta ,y={\cos }^3\theta ,\theta \in [0,2\pi ]$

Short answer

The length of the curve is equals to $6$.

Step-by-step solution

Given information

The parametric curve $x={\sin }^3\theta ,y={\cos }^3\theta ,\theta \in [0,2\pi ]$

Calculation

Consider the parametric equations $x={\sin }^3\theta ,y={\cos }^3\theta ,\theta \in [0,2\pi ]$.

The objective is to draw the parametric curve and find the arc length of the curve.

The formula to find the arc length of the curve is .

Length of the curve $={\int }_a^b\sqrt{{\left(f^{\text{'}}\left(\theta \right)\right)}^2+{\left(g^{\text{'}}\left(\theta \right)\right)}^2}d\theta$

First find the derivative of the parametric equations$x={\sin }^3\theta ,y={\cos }^3\theta$.

Take $x={\sin }^3\theta$

That is$f\left(\theta \right)={\sin }^3\theta$

The derivative with respect to $\theta$is written as follows,

$f^{\text{'}}\left(\theta \right)=\frac{d}{d\theta }\left({\sin }^3\theta \right)$

On simplifying.

$$\begin{gathered} f^{\text{'}}\left(\theta \right)=3{\sin }^2\theta \\ \frac{d}{d\theta }\left(\sin \theta \right)f^{\text{'}}\left(\theta \right)=3{\sin }^2\theta \cos \theta \end{gathered}$$

Take$y={\cos }^3\theta$

That is $g\left(\theta \right)={\cos }^3\theta$

The derivative with respect to $\theta$is written as follows,

$$\begin{gathered} g^{\text{'}}\left(\theta \right)=\frac{d}{d\theta }\left({\cos }^3\theta \right) \\ {g}^{\text{'}}\left(\theta \right)=3{\cos }^2\theta \frac{d}{d\theta }\left(\cos \theta \right) \\ {g}^{\text{'}}\left(\theta \right)=-3{\cos }^2\theta \sin \theta \end{gathered}$$

Now by using the, Length of the curve $={\int }_0^{2r}\sqrt{{\left(f^{\text{'}}\left(\theta \right)\right)}^2+{\left(g^{\text{'}}\left(\theta \right)\right)}^2}d\theta$.

Length of the curve $={\int }_0^2\sqrt{{\left(3{\sin }^2\theta \cos \theta \right)}^2+{\left(-3{\cos }^2\theta \sin \theta \right)}^2}d\theta$

$={\int }_0^{2\pi }\sqrt{9{\sin }^4\theta {\cos }^2\theta +9{\cos }^4\theta {\sin }^2\theta }d\theta$

Further calculation

On further simplification,

Length of the curve $={\int }_0^{2r}\sqrt{9{\sin }^2\theta {\cos }^2\theta \left({\sin }^2\theta +{\cos }^2\theta \right)}d\theta$$={\int }_0^{2\pi }\sqrt{9{\sin }^2\theta {\cos }^2\theta }d\theta \left[Since{\sin }^2\theta +{\cos }^2\theta =1\right]$

Length of the curve $={\int }_0^{2}3\sin \theta \cos \theta d\theta$

Multiply and divide by 2 then,

Length of the curve $=\frac{2}{2}{\int }_0^{2\pi }3\sin \theta \cos \theta d\theta$$=\frac{3}{2}{\int }_0^{2\pi }2\sin \theta \cos \theta d\theta$

$=\frac{3}{2}{\int }_0^{2\pi }\sin 2\theta d\theta$

Thus, to find the length of the curve the limits are taken from 0 to $\frac{\pi }{2}$and multiply it with 4 since the curves are symetric .

Length of the curve $=4\times \frac{3}{2}\int \sin 2\theta d\theta$

$$\begin{gathered} =4·\frac{3}{2}{\left(-\frac{\cos 2\theta }{2}\right)}_0^{\frac{\pi }{2}} \\ =6\left(-\frac{\cos 2·\frac{\pi }{2}}{2}+\frac{\cos 0}{2}\right) \\ =6\left(-\frac{\cos \pi }{2}+\frac{\cos 0}{2}\right) \end{gathered}$$

Thus,

The arc length $=6\left(-\frac{-1}{2}+\frac{1}{2}\right)$

Arc length $=6\left(1\right)$

Therefore, the length of the curve is equals to $6.$

Plot the graph

The required graph is