If `x=a (theta-sin theta) and y=a(1-cos theta)`, find `(d^(2)y)/(dx^(2))` at `theta=pi.`

Answer

To solve the problem, we need to find the second derivative of \( y \) with respect to \( x \) at \( \theta = \pi \), given the equations:

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

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

We differentiate \( x \) with respect to \( \theta \):

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

**Hint:** Use the derivative of \( \sin \theta \) which is \( \cos \theta \).

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

Next, we differentiate \( y \) with respect to \( \theta \):

\[ \frac{dy}{d\theta} = a\left(0 + \sin \theta\right) = a \sin \theta \]

**Hint:** Remember that the derivative of \( \cos \theta \) is \( -\sin \theta \).

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

Now we can find \(\frac{dy}{dx}\) using the chain rule:

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

**Hint:** The \( a \) cancels out in the fraction.

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

To find the second derivative, we need to differentiate \(\frac{dy}{dx}\) with respect to \( \theta \) and then multiply by \(\frac{d\theta}{dx}\):

\[ \frac{d^2y}{dx^2} = \frac{d}{d\theta}\left(\frac{\sin \theta}{1 - \cos \theta}\right) \cdot \frac{d\theta}{dx} \]

First, we need to compute \(\frac{d\theta}{dx}\):

\[ \frac{d\theta}{dx} = \frac{1}{\frac{dx}{d\theta}} = \frac{1}{a(1 - \cos \theta)} \]

Now we differentiate \(\frac{\sin \theta}{1 - \cos \theta}\) using the quotient rule:

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

This simplifies to:

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

### Step 5: Substitute and simplify

Now we substitute back into the equation for \(\frac{d^2y}{dx^2}\):

\[ \frac{d^2y}{dx^2} = \frac{-\sin^2 \theta}{(1 - \cos \theta)^2} \cdot \frac{1}{a(1 - \cos \theta)} \]

This gives:

\[ \frac{d^2y}{dx^2} = \frac{-\sin^2 \theta}{a(1 - \cos \theta)^3} \]

### Step 6: Evaluate at \(\theta = \pi\)

Now we evaluate at \(\theta = \pi\):

\[ \sin(\pi) = 0 \quad \text{and} \quad 1 - \cos(\pi) = 1 - (-1) = 2 \]

Thus:

\[ \frac{d^2y}{dx^2} = \frac{-0^2}{a(2)^3} = 0 \]

### Final Answer

\[ \frac{d^2y}{dx^2} \text{ at } \theta = \pi \text{ is } 0. \]

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