`x=a sin 2t(1+cccos 2t ),y=b cos2 t (1-cos 2t)` find `dy/dx`

Answer

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

### Step 1: Differentiate \(x\) with respect to \(t\)
Given:
\[ x = a \sin 2t (1 + \cos 2t) \]

Using the product rule:
\[ \frac{dx}{dt} = a \left( \frac{d}{dt}(\sin 2t) \cdot (1 + \cos 2t) + \sin 2t \cdot \frac{d}{dt}(1 + \cos 2t) \right) \]

Calculating the derivatives:
- \(\frac{d}{dt}(\sin 2t) = 2 \cos 2t\)
- \(\frac{d}{dt}(1 + \cos 2t) = -2 \sin 2t\)

Substituting these into the equation:
\[ \frac{dx}{dt} = a \left( 2 \cos 2t (1 + \cos 2t) + \sin 2t (-2 \sin 2t) \right) \]
\[ = a \left( 2 \cos 2t + 2 \cos^2 2t - 2 \sin^2 2t \right) \]
Using the identity \(\cos^2 2t - \sin^2 2t = \cos 4t\):
\[ \frac{dx}{dt} = 2a \cos 2t \cdot \cos 4t \]

### Step 2: Differentiate \(y\) with respect to \(t\)
Given:
\[ y = b \cos 2t (1 - \cos 2t) \]

Using the product rule:
\[ \frac{dy}{dt} = b \left( \frac{d}{dt}(\cos 2t) \cdot (1 - \cos 2t) + \cos 2t \cdot \frac{d}{dt}(1 - \cos 2t) \right) \]

Calculating the derivatives:
- \(\frac{d}{dt}(\cos 2t) = -2 \sin 2t\)
- \(\frac{d}{dt}(1 - \cos 2t) = 2 \sin 2t\)

Substituting these into the equation:
\[ \frac{dy}{dt} = b \left( -2 \sin 2t (1 - \cos 2t) + \cos 2t (2 \sin 2t) \right) \]
\[ = b \left( -2 \sin 2t + 2 \sin 2t \cos 2t + 2 \sin 2t \cos 2t \right) \]
\[ = 2b \sin 2t \cos 2t \]

### Step 3: Find \(\frac{dy}{dx}\)
Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{2b \sin 2t \cos 2t}{2a \cos 2t \cos 4t} \]
Simplifying:
\[ \frac{dy}{dx} = \frac{b \sin 2t}{a \cos 4t} \]

### Final Result
Thus, the derivative \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{b \sin 2t}{a \cos 4t} \]

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