Find `(d^(2)y)/(dx^(2))" at "theta=pi/2` when :
`x=a(1-costheta),y=a(theta+sintheta)`

Answer

To find \(\frac{d^2y}{dx^2}\) at \(\theta = \frac{\pi}{2}\) given the parametric equations \(x = a(1 - \cos \theta)\) and \(y = a(\theta + \sin \theta)\), we can follow these steps:

### Step 1: Find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\)

1. **Differentiate \(x\) with respect to \(\theta\)**:
\[ x = a(1 - \cos \theta) \implies \frac{dx}{d\theta} = a \sin \theta \]

2. **Differentiate \(y\) with respect to \(\theta\)**:
\[ y = a(\theta + \sin \theta) \implies \frac{dy}{d\theta} = a(1 + \cos \theta) \]

### Step 2: Find \(\frac{dy}{dx}\)

Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a(1 + \cos \theta)}{a \sin \theta} = \frac{1 + \cos \theta}{\sin \theta} \]

### Step 3: Find \(\frac{d^2y}{dx^2}\)

To find \(\frac{d^2y}{dx^2}\), we need to differentiate \(\frac{dy}{dx}\) with respect to \(\theta\) and then divide by \(\frac{dx}{d\theta}\):

1. **Differentiate \(\frac{dy}{dx}\)**:
\[ \frac{dy}{dx} = \frac{1 + \cos \theta}{\sin \theta} \]
Using the quotient rule:
\[ \frac{d}{d\theta}\left(\frac{1 + \cos \theta}{\sin \theta}\right) = \frac{\sin \theta(-\sin \theta) - (1 + \cos \theta)\cos \theta}{\sin^2 \theta} \]
Simplifying:
\[ = \frac{-\sin^2 \theta - (1 + \cos \theta)\cos \theta}{\sin^2 \theta} \]

2. **Now, find \(\frac{d^2y}{dx^2}\)**:
\[ \frac{d^2y}{dx^2} = \frac{d}{d\theta}\left(\frac{dy}{dx}\right) \cdot \frac{1}{\frac{dx}{d\theta}} = \frac{-\sin^2 \theta - (1 + \cos \theta)\cos \theta}{\sin^2 \theta} \cdot \frac{1}{a \sin \theta} \]

### Step 4: Evaluate at \(\theta = \frac{\pi}{2}\)

1. **Calculate \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\)**:
\[ \frac{dx}{d\theta} = a \sin\left(\frac{\pi}{2}\right) = a \]
\[ \frac{dy}{d\theta} = a\left(1 + \cos\left(\frac{\pi}{2}\right)\right) = a(1 + 0) = a \]

2. **Substituting \(\theta = \frac{\pi}{2}\)** into \(\frac{d^2y}{dx^2}\):
\[ \frac{d^2y}{dx^2} = \frac{-\sin^2\left(\frac{\pi}{2}\right) - (1 + \cos\left(\frac{\pi}{2}\right))\cos\left(\frac{\pi}{2}\right)}{a \sin^2\left(\frac{\pi}{2}\right)} = \frac{-1 - (1 + 0) \cdot 0}{a \cdot 1} = \frac{-1}{a} \]

### Final Result:
\[ \frac{d^2y}{dx^2} \bigg|_{\theta = \frac{\pi}{2}} = -\frac{1}{a} \]