Consider the curve `x=a(cos theta+thetasintheta)` and `y=a(sin theta-thetacostheta)`.
What is `(dy)/(dx)` equal to ?

Options

A. `tan theta`
B. `cot theta`
C. `sin 2 theta`
D. `cos 2 theta`

Answer

To find \(\frac{dy}{dx}\) for the given parametric equations \(x = a(\cos \theta + \theta \sin \theta)\) and \(y = a(\sin \theta - \theta \cos \theta)\), we will use the formula:

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

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

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

Using the product rule for differentiation, we differentiate \(y\) with respect to \(\theta\):

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

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

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

Again using the product rule for differentiation:

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

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

Now we can substitute \(\frac{dy}{d\theta}\) and \(\frac{dx}{d\theta}\) into the formula for \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = \frac{a\theta \sin \theta}{a\theta \cos \theta} \]

The \(a\) and \(\theta\) terms cancel out (assuming \(\theta \neq 0\)):

\[ \frac{dy}{dx} = \frac{\sin \theta}{\cos \theta} = \tan \theta \]

### Final Answer

Thus, we find that:

\[ \frac{dy}{dx} = \tan \theta \]

---