If `x=k(theta+sintheta)` and `y=k(1+costheta)`. then what is the derivative of y with respect to x at `theta=pi//2` ?

Options

A. A) `-1`
B. B) 0
C. C) 1
D. D) 2

Answer

To find the derivative of \( y \) with respect to \( x \) at \( \theta = \frac{\pi}{2} \), we start with the given equations:

\[ x = k(\theta + \sin \theta) \]
\[ y = k(1 + \cos \theta) \]

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

We need to find \( \frac{dy}{d\theta} \):

\[ \frac{dy}{d\theta} = k \cdot \frac{d}{d\theta}(1 + \cos \theta) \]
\[ \frac{dy}{d\theta} = k \cdot (0 - \sin \theta) = -k \sin \theta \]

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

Now, we find \( \frac{dx}{d\theta} \):

\[ \frac{dx}{d\theta} = k \cdot \frac{d}{d\theta}(\theta + \sin \theta) \]
\[ \frac{dx}{d\theta} = k \cdot (1 + \cos \theta) \]

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

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

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

The \( k \) cancels out:

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

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

Now we substitute \( \theta = \frac{\pi}{2} \):

\[ \frac{dy}{dx} \bigg|_{\theta = \frac{\pi}{2}} = \frac{-\sin\left(\frac{\pi}{2}\right)}{1 + \cos\left(\frac{\pi}{2}\right)} \]

We know:

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

So:

\[ \frac{dy}{dx} \bigg|_{\theta = \frac{\pi}{2}} = \frac{-1}{1 + 0} = -1 \]

### Final Answer

Thus, the derivative of \( y \) with respect to \( x \) at \( \theta = \frac{\pi}{2} \) is:

\[ \boxed{-1} \]

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