Question

The length of the normal to the curve $x=a\left(\theta +\sin \theta \right)$, $y=a\left(1-\cos \theta \right)$ at $\theta =\frac{\mathrm{\pi }}{2}$ is

• A. $2a$
• B. $\frac{a}{2}$
• C. $\frac{a}{\sqrt{2}}$
• D. $\sqrt{2}a$

Answer 1

The correct option is D

$\sqrt{2}a$

Step 1 : Write the parametric form of the given curve

The parametric form of the given curve is as follows,

$x=a\left(\theta +\sin \theta \right)$ and $y=a\left(1-\cos \theta \right)$.

Consider the equation: $x=a\left(\theta +\sin \theta \right)$.

Step 2 : Differentiate both sides of the equation with respect to $\theta$.

$$\begin{gathered} \frac{d}{d\theta }\left(x\right)=\frac{d}{d\theta }\left[a\left(\theta +\sin \theta \right)\right] \\ \Rightarrow \frac{dx}{d\theta }=a\left[\frac{d\theta }{d\theta }+\frac{d}{d\theta }\left(\sin \theta \right)\right] \\ \Rightarrow \frac{dx}{d\theta }=a\left(1+\cos \theta \right) \\ \Rightarrow \frac{dx}{d\theta }=2a{\cos }^2\left(\frac{\theta }{2}\right)...\left[\because 1+cos\theta =2co{s}^2\left(\frac{\theta }{2}\right)\right] \end{gathered}$$

Consider the equation: $y=a\left(1-\cos \theta \right)$.

Step 3: Differentiate both sides of the equation with respect to $\theta$.

$$\begin{gathered} \frac{d}{d\theta }\left(y\right)=\frac{d}{d\theta }\left[a\left(1-\cos \theta \right)\right] \\ \Rightarrow \frac{dy}{d\theta }=a\left[\frac{d}{d\theta }\left(1\right)-\frac{d}{d\theta }\left(\cos \theta \right)\right] \\ \Rightarrow \frac{dy}{d\theta }=a\sin \theta \end{gathered}$$

$$\begin{gathered} \therefore \frac{dy}{dx}=\frac{{\displaystyle \frac{dy}{d\theta }}}{{\displaystyle \frac{dx}{d\theta }}} \\ \Rightarrow \frac{dy}{dx}=\frac{a\sin \theta }{2a{\cos }^2\left({\displaystyle \frac{\theta }{2}}\right)} \\ \Rightarrow \frac{dy}{dx}=\frac{2a\sin \left({\displaystyle \frac{\theta }{2}}\right)\cos \left({\displaystyle \frac{\theta }{2}}\right)}{2a{\cos }^2\left({\displaystyle \frac{\theta }{2}}\right)}...\left[\because sin2\theta =2sin\theta ·cos\theta \right] \\ \Rightarrow \frac{dy}{dx}=\frac{\sin \left({\displaystyle \frac{\theta }{2}}\right)}{\cos \left({\displaystyle \frac{\theta }{2}}\right)} \\ \Rightarrow \frac{dy}{dx}=\tan \left(\frac{\theta }{2}\right) \end{gathered}$$

Step 4 : Find the length of normal to the curve

The length of a normal to the curve can be given by, $L=y\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}$.

$$\begin{gathered} \Rightarrow L=a\left(1-\cos \theta \right)\sqrt{1+{\tan }^2\left(\frac{\theta }{2}\right)} \\ \Rightarrow L=2a{\sin }^2\left(\frac{\theta }{2}\right)\sqrt{{\sec }^2\left(\frac{\theta }{2}\right)}...\left[\because \left(1-cos2\theta \right)=2si{n}^2\theta and\left(1+ta{n}^2\theta \right)=se{c}^2\theta \right] \\ \Rightarrow L=2a\sin \left(\frac{\theta }{2}\right)\frac{\sin \left(\frac{\theta }{2}\right)}{\cos \left(\frac{\theta }{2}\right)} \\ \Rightarrow L=2a\sin \left(\frac{\theta }{2}\right)\tan \left(\frac{\theta }{2}\right)...\left[\because tan\theta =\frac{sin\theta }{cos\theta }\right] \end{gathered}$$

Now, at point $\theta =\frac{\mathrm{\pi }}{2}$, $L\left(\frac{\mathrm{\pi }}{2}\right)=2a\sin \left(\frac{\mathrm{\pi }}{4}\right)\tan \left(\frac{\mathrm{\pi }}{4}\right)=2a\times \frac{1}{\sqrt{2}}\times 1=\sqrt{2}a$.

Therefore, the length of the normal at the point $\theta =\frac{\mathrm{\pi }}{2}$ of the curve $x=a\left(\theta +\sin \theta \right)$, $y=a\left(1-\cos \theta \right)$ is $\sqrt{2}a$.