Question

The derivative \(\frac{\mathrm dy}{\mathrm d x}\) ,if \(x=a(\theta -sin\theta),y=a(1+cos\theta)\) is :

• A. \(cos \frac{\theta}{2}\)
• B. \(-cot \frac{\theta}{2}\)
• C. \(cot \frac{\theta}{2}\)
• D. \(tan \frac{\theta}{2}\)

Answer 1

The Correct Option is B

To find the derivative \(\frac{\mathrm dy}{\mathrm dx}\), we need to use the parametric equations given: \(x=a(\theta - \sin \theta)\) and \(y=a(1 + \cos \theta)\).

First, we find the derivatives \(\frac{\mathrm{d}x}{\mathrm{d}\theta}\) and \(\frac{\mathrm{d}y}{\mathrm{d}\theta}\):

\(\frac{\mathrm{d}x}{\mathrm{d}\theta} = a(1 - \cos \theta)\)

\(\frac{\mathrm{d}y}{\mathrm{d}\theta} = a(-\sin \theta)\)

Next, we use the chain rule to find \(\frac{\mathrm{d}y}{\mathrm{d}x}\):

\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}\theta}}{\frac{\mathrm{d}x}{\mathrm{d}\theta}} = \frac{a(-\sin \theta)}{a(1 - \cos \theta)} = \frac{-\sin \theta}{1 - \cos \theta}\)

Notice that \(\frac{-\sin \theta}{1 - \cos \theta}\) can be simplified using the identity:

\(1 - \cos \theta = 2 \sin^2 \frac{\theta}{2}\)

and

\(\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\)

Substituting in, we get:

\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin^2 \frac{\theta}{2}} = \frac{-\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}} = -\cot \frac{\theta}{2}\)

Thus, the derivative \(\frac{\mathrm dy}{\mathrm dx}\) is \(-\cot \frac{\theta}{2}\).