91Ó°ÊÓ

The parametric curve $x={\sin }^3\mathrm{θ},y={\cos }^3\mathrm{θ},\mathrm{θ}∈[0,2\mathrm{π}]$

Consider the parametric equations $x={\sin }^3\mathrm{θ},y={\cos }^3\mathrm{θ},\mathrm{θ}∈[0,2\mathrm{π}]$.

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 $={∫}_a^b\sqrt{{\left(f^{\text{'}}\left(\mathrm{θ}\right)\right)}^2+{\left(g^{\text{'}}\left(\mathrm{θ}\right)\right)}^2}d\mathrm{θ}$

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

Take $x={\sin }^3\mathrm{θ}$

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

The derivative with respect to $\mathrm{θ}$is written as follows,

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

On simplifying.

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

Take$y={\cos }^3\mathrm{θ}$

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

The derivative with respect to $\mathrm{θ}$is written as follows,

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

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

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

$={∫}_0^{2\mathrm{π}}\sqrt{9{\sin }^4\mathrm{θ}{\cos }^2\mathrm{θ}+9{\cos }^4\mathrm{θ}{\sin }^2\mathrm{θ}}d\mathrm{θ}$

On further simplification,

Length of the curve $={∫}_0^{2r}\sqrt{9{\sin }^2\mathrm{θ}{\cos }^2\mathrm{θ}\left({\sin }^2\mathrm{θ}+{\cos }^2\mathrm{θ}\right)}d\mathrm{θ}$$={∫}_0^{2\mathrm{π}}\sqrt{9{\sin }^2\mathrm{θ}{\cos }^2\mathrm{θ}}d\mathrm{θ}\left[Since{\sin }^2\mathrm{θ}+{\cos }^2\mathrm{θ}=1\right]$

Length of the curve $={∫}_0^{2}3\sin \mathrm{θ}\cos \mathrm{θ}d\mathrm{θ}$

Multiply and divide by 2 then,

Length of the curve $=\frac{2}{2}{∫}_0^{2\mathrm{π}}3\sin \mathrm{θ}\cos \mathrm{θ}d\mathrm{θ}$$=\frac{3}{2}{∫}_0^{2\mathrm{π}}2\sin \mathrm{θ}\cos \mathrm{θ}d\mathrm{θ}$

$=\frac{3}{2}{∫}_0^{2\mathrm{π}}\sin 2\mathrm{θ}d\mathrm{θ}$

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

Length of the curve $=4×\frac{3}{2}∫\sin 2\mathrm{θ}d\mathrm{θ}$

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

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 $6.$

The required graph is