Question

The length of the normal at the point $t$ of the curve $x=a\left(t+\sin t\right)$, $y=a\left(1-\cos t\right)$ is

• A. $a\sin \left(t\right)$
• B. $2a{\sin }^3\left(\frac{t}{2}\right)\sec \left(\frac{\mathrm{t}}{2}\right)$
• C. $2a\sin \left(\frac{t}{2}\right)\tan \left(\frac{\mathrm{t}}{2}\right)$
• D. $2a\sin \left(\frac{t}{2}\right)$

Answer 1

The correct option is C

$2a\sin \left(\frac{t}{2}\right)\tan \left(\frac{\mathrm{t}}{2}\right)$

The parametric form of the given curve is as follows,

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

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

Differentiate both sides of the equation with respect to $t$.

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

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

Differentiate both sides of the equation with respect to $t$.

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

$$\begin{gathered} \therefore \frac{dy}{dx}=\frac{{\displaystyle \frac{dy}{dt}}}{{\displaystyle \frac{dx}{dt}}} \\ \Rightarrow \frac{dy}{dx}=\frac{a\sin t}{2a{\cos }^2\left({\displaystyle \frac{t}{2}}\right)} \\ \Rightarrow \frac{dy}{dx}=\frac{2a\sin \left({\displaystyle \frac{t}{2}}\right)\cos \left({\displaystyle \frac{t}{2}}\right)}{2a{\cos }^2\left({\displaystyle \frac{t}{2}}\right)}...\left[\because sin2t=2sint·cost\right] \\ \Rightarrow \frac{dy}{dx}=\frac{\sin \left({\displaystyle \frac{t}{2}}\right)}{\cos \left({\displaystyle \frac{t}{2}}\right)} \\ \Rightarrow \frac{dy}{dx}=\tan \left(\frac{t}{2}\right) \end{gathered}$$

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

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

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