If `x=costheta(1+costheta),y=sintheta(1+costheta)," then "((dy)/(dx))" at "theta=(pi)/(2)` is

Options

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

Answer

To find \(\frac{dy}{dx}\) at \(\theta = \frac{\pi}{2}\), we start with the given equations:

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

Since \(x\) and \(y\) are both functions of \(\theta\), we can use the chain rule to find \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \]

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

Using the product rule for differentiation:

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

Let \(u = \cos \theta\) and \(v = 1 + \cos \theta\). Then:

\[ \frac{dx}{d\theta} = u \frac{dv}{d\theta} + v \frac{du}{d\theta} \]

Calculating \(\frac{du}{d\theta}\) and \(\frac{dv}{d\theta}\):

\[ \frac{du}{d\theta} = -\sin \theta \]
\[ \frac{dv}{d\theta} = -\sin \theta \]

Now substituting back:

\[ \frac{dx}{d\theta} = \cos \theta (-\sin \theta) + (1 + \cos \theta)(-\sin \theta) \]
\[ = -\cos \theta \sin \theta - \sin \theta - \cos \theta \sin \theta \]
\[ = -2\cos \theta \sin \theta - \sin \theta \]

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

Using the product rule for \(y\):

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

Let \(u = \sin \theta\) and \(v = 1 + \cos \theta\):

\[ \frac{dy}{d\theta} = u \frac{dv}{d\theta} + v \frac{du}{d\theta} \]

Calculating \(\frac{du}{d\theta}\) and \(\frac{dv}{d\theta}\):

\[ \frac{du}{d\theta} = \cos \theta \]
\[ \frac{dv}{d\theta} = -\sin \theta \]

Now substituting back:

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

### Step 3: Evaluate at \(\theta = \frac{\pi}{2}\)

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

For \(\frac{dx}{d\theta}\):

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

For \(\frac{dy}{d\theta}\):

\[ \cos\left(\frac{\pi}{2}\right) = 0 \quad \text{and} \quad \sin\left(\frac{\pi}{2}\right) = 1 \]
\[ \frac{dy}{d\theta} = 0 + 1 - 1^2 = 0 \]

### Step 4: Calculate \(\frac{dy}{dx}\)

Now substituting into the formula:

\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{0}{-1} = 0 \]

### Final Answer

Thus, the value of \(\frac{dy}{dx}\) at \(\theta = \frac{\pi}{2}\) is:

\[ \frac{dy}{dx} = 0 \]