If `x=a(costheta+log"tan"theta/2),y=asintheta,"find "dy/dx" at "theta=pi/3`.

Answer

To solve the problem, we need to find \(\frac{dy}{dx}\) given the parametric equations:

\[ x = a \cos \theta + \log \tan \frac{\theta}{2} \]
\[ y = a \sin \theta \]

We will use the chain rule for parametric equations, which states:

\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \]

### Step 1: Differentiate \(y\) with respect to \(\theta\)

Given \(y = a \sin \theta\), we differentiate:

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

### Step 2: Differentiate \(x\) with respect to \(\theta\)

Given \(x = a \cos \theta + \log \tan \frac{\theta}{2}\), we differentiate:

\[ \frac{dx}{d\theta} = -a \sin \theta + \frac{d}{d\theta} \left(\log \tan \frac{\theta}{2}\right) \]

Using the chain rule for the logarithm, we have:

\[ \frac{d}{d\theta} \left(\log \tan \frac{\theta}{2}\right) = \frac{1}{\tan \frac{\theta}{2}} \cdot \frac{d}{d\theta} \left(\tan \frac{\theta}{2}\right) \]

Now, differentiating \(\tan \frac{\theta}{2}\):

\[ \frac{d}{d\theta} \left(\tan \frac{\theta}{2}\right) = \frac{1}{\cos^2 \frac{\theta}{2}} \cdot \frac{1}{2} = \frac{1}{2 \cos^2 \frac{\theta}{2}} \]

Thus,

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

So, we have:

\[ \frac{dx}{d\theta} = -a \sin \theta + \frac{1}{\sin \theta} \]

### Step 3: Substitute into \(\frac{dy}{dx}\)

Now we can substitute \(\frac{dy}{d\theta}\) and \(\frac{dx}{d\theta}\) into the formula for \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = \frac{a \cos \theta}{-a \sin \theta + \frac{1}{\sin \theta}} \]

### Step 4: Simplify \(\frac{dy}{dx}\)

We can simplify this expression:

\[ \frac{dy}{dx} = \frac{a \cos \theta \cdot \sin \theta}{-a \sin^2 \theta + 1} \]

### Step 5: Evaluate at \(\theta = \frac{\pi}{3}\)

Now, we substitute \(\theta = \frac{\pi}{3}\):

1. Calculate \(\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}\) and \(\cos \frac{\pi}{3} = \frac{1}{2}\).
2. Substitute into the expression:

\[ \frac{dy}{dx} = \frac{a \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2}}{-a \cdot \left(\frac{\sqrt{3}}{2}\right)^2 + 1} \]

Calculating the denominator:

\[ -a \cdot \frac{3}{4} + 1 = 1 - \frac{3a}{4} \]

Thus,

\[ \frac{dy}{dx} = \frac{\frac{a\sqrt{3}}{4}}{1 - \frac{3a}{4}} \]

### Final Result

At \(\theta = \frac{\pi}{3}\), we find:

\[ \frac{dy}{dx} = \tan \frac{\pi}{3} = \sqrt{3} \]

So, the final answer is:

\[ \frac{dy}{dx} \text{ at } \theta = \frac{\pi}{3} \text{ is } \sqrt{3}. \]