Question

If the parametric equation of curve is given by x = cosθ + log tan \(\frac{θ}{2}\) and y = sinθ, then the points for which \(\frac{dy}{dx}=0\) are given by

• A. \(θ=\frac{n\pi}{2},n ∈z\)
• B. \(θ=(2n+1)\frac{\pi}{2},n ∈z\)
• C. \(θ=(2n+1)\pi,n ∈z\)
• D. \(θ=n\pi,n ∈z\)

Answer 1

The Correct Option is D

We are given the parametric equations: \[ x = \cos(\theta) + \log \left( \tan \frac{\theta}{2} \right) \] \[ y = \sin(\theta) \] We need to find the points where \( \frac{dy}{dx} = 0 \). We know that: \[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \] #### Step 1: Derivatives of \( x \) and \( y \) with respect to \( \theta \) The derivative of \( y = \sin(\theta) \) is: \[ \frac{dy}{d\theta} = \cos(\theta) \] The derivative of \( x = \cos(\theta) + \log \left( \tan \frac{\theta}{2} \right) \) is: \[ \frac{dx}{d\theta} = -\sin(\theta) + \frac{1}{2 \tan \frac{\theta}{2}} \] #### Step 2: Setting \( \frac{dy}{dx} = 0 \) For \( \frac{dy}{dx} = 0 \), we need \( \frac{dy}{d\theta} = 0 \), which occurs when \( \cos(\theta) = 0 \). This happens when: \[ \theta = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z} \] Thus, the points for which \( \frac{dy}{dx} = 0 \) are at: \[ \theta = n\pi, \quad n \in \mathbb{Z} \] 

The correct answer is (D) : \(θ=n\pi,n ∈z\).

Answer 2

Given \( x = \cos\theta + \log\left(\tan\left(\frac{\theta}{2}\right)\right) \) and \( y = \sin\theta \), we want to find the values of \( \theta \) for which \( \frac{dy}{dx} = 0 \).

First, we find \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \):

\[ \frac{dx}{d\theta} = -\sin\theta + \frac{1}{\tan\left(\frac{\theta}{2}\right)} \cdot \sec^2\left(\frac{\theta}{2}\right) \cdot \frac{1}{2} = -\sin\theta + \frac{\cos\left(\frac{\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)} \cdot \frac{1}{2 \cos^2\left(\frac{\theta}{2}\right)} = -\sin\theta + \frac{1}{\sin\theta} \]

\[ \frac{dx}{d\theta} = \frac{1 - \sin^2\theta}{\sin\theta} = \frac{\cos^2\theta}{\sin\theta} \]

\[ \frac{dy}{d\theta} = \cos\theta \]

Now, we find \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{\cos\theta}{\frac{\cos^2\theta}{\sin\theta}} = \frac{\sin\theta}{\cos\theta} = \tan\theta \]

We want to find when \( \frac{dy}{dx} = 0 \), so we need to solve \( \tan\theta = 0 \).

\[ \tan\theta = 0 \]

\(\tan\theta = 0\) when \( \theta = n\pi \), where \( n \) is an integer.

However, we need to consider the domain of the original parametric equations. Since \( \log\left(\tan\frac{\theta}{2}\right) \) is defined, we must have \( \tan\frac{\theta}{2} > 0 \). This means that:

\[ 0 < \frac{\theta}{2} < \frac{\pi}{2} \quad \text{or} \quad \pi < \frac{\theta}{2} < \frac{3\pi}{2}, \quad \text{etc.} \]

Thus, \( 2n\pi < \theta < (2n+1)\pi \), where \( n \) is an integer.

Therefore, the values of \( \theta \) for which \( \frac{dy}{dx} = 0 \) are given by \( \theta = n\pi \), \( n \in \mathbb{Z} \), with the additional constraint that \( 2n\pi < \theta < (2n+1)\pi \).