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

Answer

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

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

Given:
\[ x = a(\theta + \sin \theta) \]
\[ y = a(1 - \cos \theta) \]

Now, we differentiate \(x\) and \(y\) with respect to \(\theta\):

\[ \frac{dx}{d\theta} = a\left(1 + \cos \theta\right) \]
\[ \frac{dy}{d\theta} = a\sin \theta \]

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

Using the chain rule, we find \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a\sin \theta}{a(1 + \cos \theta)} = \frac{\sin \theta}{1 + \cos \theta} \]

### Step 3: Differentiate \(\frac{dy}{dx}\) with respect to \(\theta\)

Now we differentiate \(\frac{dy}{dx}\) with respect to \(\theta\):

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

Simplifying this gives:
\[ = \frac{(1 + \cos \theta)\cos \theta + \sin^2 \theta}{(1 + \cos \theta)^2} \]
\[ = \frac{\cos \theta + \cos^2 \theta + \sin^2 \theta}{(1 + \cos \theta)^2} \]
Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\):
\[ = \frac{\cos \theta + 1}{(1 + \cos \theta)^2} = \frac{1 + \cos \theta}{(1 + \cos \theta)^2} = \frac{1}{1 + \cos \theta} \]

### Step 4: Find \(\frac{d^2y}{dx^2}\)

Now we can find \(\frac{d^2y}{dx^2}\):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{d\theta}\left(\frac{dy}{dx}\right) \cdot \frac{d\theta}{dx} \]
We already have \(\frac{d}{d\theta}\left(\frac{dy}{dx}\right) = \frac{1}{1 + \cos \theta}\).

Now we need \(\frac{d\theta}{dx}\):
\[ \frac{d\theta}{dx} = \frac{1}{\frac{dx}{d\theta}} = \frac{1}{a(1 + \cos \theta)} \]

Combining these:
\[ \frac{d^2y}{dx^2} = \frac{1}{1 + \cos \theta} \cdot \frac{1}{a(1 + \cos \theta)} = \frac{1}{a(1 + \cos \theta)^2} \]

### Step 5: Evaluate at \(\theta = \frac{\pi}{2}\)

Now we substitute \(\theta = \frac{\pi}{2}\):
\[ \cos\left(\frac{\pi}{2}\right) = 0 \implies 1 + \cos\left(\frac{\pi}{2}\right) = 1 \]
Thus,
\[ \frac{d^2y}{dx^2} = \frac{1}{a(1)^2} = \frac{1}{a} \]

### Final Answer
\[ \frac{d^2y}{dx^2} \text{ at } \theta = \frac{\pi}{2} = \frac{1}{a} \]