If `x = a (theta + sin theta), y = b (1+ cos theta),` then `(dy)/(dx)" at "theta = (pi)/(2)` is:

Options

A. `-(a^(2))/(b^(2))`
B. `(a)/(b)`
C. `(b^(2))/(a^(2))`
D. `-(b)/(a)`

Answer

To find \(\frac{dy}{dx}\) at \(\theta = \frac{\pi}{2}\) for the given equations \(x = a(\theta + \sin \theta)\) and \(y = b(1 + \cos \theta)\), we will follow these steps:

### Step 1: Differentiate \(x\) with respect to \(\theta\)
Given:
\[ x = a(\theta + \sin \theta) \]
Differentiating \(x\) with respect to \(\theta\):
\[ \frac{dx}{d\theta} = a\left(1 + \cos \theta\right) \]

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

### Step 3: Find \(\frac{dy}{dx}\)
Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{b(-\sin \theta)}{a(1 + \cos \theta)} \]

### Step 4: Substitute \(\theta = \frac{\pi}{2}\)
Now, substitute \(\theta = \frac{\pi}{2}\):
\[ \frac{dy}{dx} = \frac{b(-\sin(\frac{\pi}{2}))}{a(1 + \cos(\frac{\pi}{2}))} \]
Calculating the trigonometric values:
\[ \sin\left(\frac{\pi}{2}\right) = 1 \quad \text{and} \quad \cos\left(\frac{\pi}{2}\right) = 0 \]
Thus,
\[ \frac{dy}{dx} = \frac{b(-1)}{a(1 + 0)} = \frac{-b}{a} \]

### Final Answer
The value of \(\frac{dy}{dx}\) at \(\theta = \frac{\pi}{2}\) is:
\[ \frac{dy}{dx} = -\frac{b}{a} \]

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