`x=a (t-sin t) , y =a (1-cos t) ` find `dy/dx`

Answer

To find \( \frac{dy}{dx} \) for the given parametric equations \( x = a(t - \sin t) \) and \( y = a(1 - \cos t) \), we will follow these steps:

### Step 1: Differentiate \( x \) with respect to \( t \)

Given:
\[ x = a(t - \sin t) \]

Differentiate \( x \) with respect to \( t \):
\[ \frac{dx}{dt} = a\left( \frac{d}{dt}(t) - \frac{d}{dt}(\sin t) \right) \]
\[ \frac{dx}{dt} = a(1 - \cos t) \]

### Step 2: Differentiate \( y \) with respect to \( t \)

Given:
\[ y = a(1 - \cos t) \]

Differentiate \( y \) with respect to \( t \):
\[ \frac{dy}{dt} = a\left( \frac{d}{dt}(1) - \frac{d}{dt}(\cos t) \right) \]
\[ \frac{dy}{dt} = a(0 + \sin t) = a \sin t \]

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

Using the chain rule, we can find \( \frac{dy}{dx} \) as follows:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
Substituting the values from Steps 1 and 2:
\[ \frac{dy}{dx} = \frac{a \sin t}{a(1 - \cos t)} \]
The \( a \) cancels out:
\[ \frac{dy}{dx} = \frac{\sin t}{1 - \cos t} \]

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

We can simplify \( \frac{dy}{dx} \) further:
\[ \frac{dy}{dx} = \frac{\sin t}{1 - \cos t} \]
Using the identity \( 1 - \cos t = 2 \sin^2\left(\frac{t}{2}\right) \) and \( \sin t = 2 \sin\left(\frac{t}{2}\right) \cos\left(\frac{t}{2}\right) \):
\[ \frac{dy}{dx} = \frac{2 \sin\left(\frac{t}{2}\right) \cos\left(\frac{t}{2}\right)}{2 \sin^2\left(\frac{t}{2}\right)} \]
The \( 2 \) cancels out:
\[ \frac{dy}{dx} = \frac{\cos\left(\frac{t}{2}\right)}{\sin\left(\frac{t}{2}\right)} = \cot\left(\frac{t}{2}\right) \]

### Final Answer
Thus, the final result is:
\[ \frac{dy}{dx} = \cot\left(\frac{t}{2}\right) \]

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