If `x=2(theta-sintheta) and y=2(1+costheta), " find "(d^(2)y)/(dx^(2))" at "theta=pi/3`.

Answer

To find \(\frac{d^2y}{dx^2}\) at \(\theta = \frac{\pi}{3}\), we will follow these steps:

### Step 1: Find \(\frac{dx}{d\theta}\)
Given:
\[ x = 2(\theta - \sin \theta) \]
We differentiate \(x\) with respect to \(\theta\):
\[ \frac{dx}{d\theta} = 2\left(1 - \cos \theta\right) \]

### Step 2: Find \(\frac{dy}{d\theta}\)
Given:
\[ y = 2(1 + \cos \theta) \]
We differentiate \(y\) with respect to \(\theta\):
\[ \frac{dy}{d\theta} = 2(-\sin \theta) = -2\sin \theta \]

### Step 3: Find \(\frac{dy}{dx}\)
Using the chain rule, we can find \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{-2\sin \theta}{2(1 - \cos \theta)} = \frac{-\sin \theta}{1 - \cos \theta} \]

### Step 4: Simplify \(\frac{dy}{dx}\)
We can simplify \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{-\sin \theta}{1 - \cos \theta} = \frac{-\sin \theta}{2\sin^2(\theta/2)} = -\cot(\theta/2) \]

### Step 5: Find \(\frac{d^2y}{dx^2}\)
To find the second derivative, we need to differentiate \(\frac{dy}{dx}\) with respect to \(\theta\):
\[ \frac{d^2y}{dx^2} = \frac{d}{d\theta}\left(-\cot(\theta/2)\right) \cdot \frac{1}{\frac{dx}{d\theta}} \]
The derivative of \(-\cot(\theta/2)\) is:
\[ \frac{d}{d\theta}\left(-\cot(\theta/2)\right) = \frac{1}{2}\csc^2(\theta/2) \]
Thus,
\[ \frac{d^2y}{dx^2} = \frac{\frac{1}{2}\csc^2(\theta/2)}{2(1 - \cos \theta)} = \frac{\csc^2(\theta/2)}{4(1 - \cos \theta)} \]

### Step 6: Substitute \(\theta = \frac{\pi}{3}\)
Now we substitute \(\theta = \frac{\pi}{3}\):
\[ 1 - \cos\left(\frac{\pi}{3}\right) = 1 - \frac{1}{2} = \frac{1}{2} \]
\[ \csc^2\left(\frac{\pi}{6}\right) = \left(\frac{1}{\sin\left(\frac{\pi}{6}\right)}\right)^2 = \left(\frac{1}{\frac{1}{2}}\right)^2 = 4 \]
Thus,
\[ \frac{d^2y}{dx^2} = \frac{4}{4 \cdot \frac{1}{2}} = \frac{4}{2} = 2 \]

### Final Answer:
\[ \frac{d^2y}{dx^2} \text{ at } \theta = \frac{\pi}{3} \text{ is } 2. \]