Question

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

$x=\cos \theta +\theta \sin \theta ,y=\sin \theta -\theta \cos \theta ,\theta \in [0,2\pi ]$

Short answer

The graphical representation of the points $(1,0)(1,-6.2)$ is as follows,

Therefore, the length of the curve is equals to $2{\pi }^2$or 19.76.

Step-by-step solution

Given information

The parametric curve $x=\cos \theta +\theta \sin \theta ,y=\sin \theta -\theta \cos \theta ,\theta \in [0,2\pi ]$

Calculation

Consider the parametric equations $x=\cos \theta +\theta \sin \theta ,y=\sin \theta -\theta \cos \theta ,\theta \in [0,2\pi ]$.

The objective is to draw the parametric curve and find the arc length of the curve. The formula to find the are 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 $x=\cos \theta +\theta \sin \theta ,y=\sin \theta -\theta \cos \theta$

Take $x=\cos \theta +\theta \sin \theta$

That is $f\left(\theta \right)=\cos \theta +\theta \sin \theta$

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

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

Simplification

On simplifying further,

$$\begin{gathered} f^l\left(\theta \right)=\frac{d}{d\theta }\left(\cos \theta \right)+\frac{d}{d\theta }\left(\theta \sin \theta \right) \\ {f}^{\text{'}}\left(\theta \right)=-\sin \theta +\left(\theta \frac{d}{d\theta }\left(\sin \theta \right)+\sin \theta \frac{d\theta }{d\theta }\right)[sinced(uv)=udv+vdu] \\ {f}^{\text{'}}\left(\theta \right)=-\sin \theta +(\theta \cos \theta +\sin \theta ) \\ {f}^l\left(\theta \right)=\theta \cos \theta \end{gathered}$$

Take $y=\sin \theta -\theta \cos \theta$.

That is $g\left(\theta \right)=\sin \theta -\theta \cos \theta$

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

$$\begin{gathered} g^1\left(\theta \right)=\frac{d}{d\theta }(\sin \theta -\theta \cos \theta ) \\ {g}^1\left(\theta \right)=\frac{d}{d\theta }\left(\sin \theta \right)-\frac{d}{d\theta }\left(\theta \cos \theta \right) \end{gathered}$$

On simplifying further,

$$\begin{gathered} g^1\left(\theta \right)=\cos \theta -\left(\theta \frac{d}{d\theta }\left(\cos \theta \right)+\cos \theta \frac{d\theta }{d\theta }\right) \\ {g}^1\left(\theta \right)=\cos \theta -(-\theta \sin \theta +\cos \theta ) \\ {g}^1\left(\theta \right)=\cos \theta +\theta \sin \theta -\cos \theta \\ {g}^1\left(\theta \right)=\theta \sin \theta \end{gathered}$$

Further calculation

Take$y=\sin \theta -\theta \cos \theta$.

That is $g\left(\theta \right)=\sin \theta -\theta \cos \theta$

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

$$\begin{gathered} g^1\left(\theta \right)=\frac{d}{d\theta }(\sin \theta -\theta \cos \theta ) \\ {g}^1\left(\theta \right)=\frac{d}{d\theta }\left(\sin \theta \right)-\frac{d}{d\theta }\left(\theta \cos \theta \right) \end{gathered}$$

On simplifying further,

$$\begin{gathered} g^1\left(\theta \right)=\cos \theta -\left(\theta \frac{d}{d\theta }\left(\cos \theta \right)+\cos \theta \frac{d\theta }{d\theta }\right) \\ {g}^1\left(\theta \right)=\cos \theta -(-\theta \sin \theta +\cos \theta ) \\ {g}^1\left(\theta \right)=\cos \theta +\theta \sin \theta -\cos \theta \\ {g}^1\left(\theta \right)=\theta \sin \theta \end{gathered}$$

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

Length of curve $={\int }_0^{2\pi }\sqrt{(\theta \cos \theta {)}^2+(\theta \sin \theta {)}^2}d\theta$

$\left[since{f}^{-1}\left(\theta \right)=\theta \cos \theta ,g^1\left(\theta \right)=\theta \sin \theta \right]$

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

$={\int }_0^2\sqrt{{\theta }^2\left({\cos }^2\theta +{\sin }^2\theta \right)}d\theta$

On further simplification,

$$\begin{gathered} ={\int }_0^{2\pi }\sqrt{{\theta }^2}d\theta \\ ={\int }_0^{2\pi }\theta d\theta \end{gathered}$$

Length of the curve $={\left[\frac{{\theta }^2}{2}\right]}_0^{2n}$

By substituting the limits,

$$\begin{gathered} \text{Length of the curve}=\left[\frac{(2\pi {)}^2}{2}\right] \\ =\frac{3^{2}4{\pi }^2}{2} \end{gathered}$$

Length of the curve $=2{\pi }^2$or$19.75$

Simplification

To draw the curve for the parametric equations, first find the points when $\theta \in [0,2\pi ]$.

Given that role="math" localid="1653043766172" $\theta \in [0,2\pi ]$

When $\theta =0$,

role="math" localid="1653043840592" $$\begin{gathered} (x,y)=(\cos \theta +\theta \sin \theta ,\sin \theta -\theta \cos \theta ) \\ (x,y)=(\cos 0+0\sin 0,\sin 0-0\cos 0) \\ (x,y)=(1,0) \end{gathered}$$

When $\theta =2\pi$.

$(x,y)=(\cos \theta +\theta \sin \theta ,\sin \theta -\theta \cos \theta )$

Then .

$$\begin{gathered} (x,y)=(\cos 2\pi +2\pi \sin 2\pi ,\sin 2\pi -2\pi \cos 2\pi ) \\ (x,y)=(1+0,0-2\pi ·1) \\ (x,y)=(1,-2\pi ) \\ (x,y)=(1,-6.2) \end{gathered}$$

The graphical representation of the points$(1,0)(1,-6.2)$is as follows,

Therefore, the length of the curve is equals to 19.76.