If `x = a sin2t(1+cos2t)` and `y =b cos 2t(1-cos2t)`, then show that `((dy)/(dx))_(t=pi//4) = (b)/(a)`.

Explanation

To solve the problem, we start with the given equations for \( x \) and \( y \):

\[ x = a \sin(2t)(1 + \cos(2t)) \]
\[ y = b \cos(2t)(1 - \cos(2t)) \]

We need to find \( \frac{dy}{dx} \) at \( t = \frac{\pi}{4} \) and show that it equals \( \frac{b}{a} \).

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

Using the product rule, we differentiate \( x \):

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

Calculating the derivatives:

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

Substituting these back into the equation:

\[ \frac{dx}{dt} = a \left( \sin(2t)(-2 \sin(2t)) + (1 + \cos(2t))(2 \cos(2t)) \right) \]

Simplifying:

\[ \frac{dx}{dt} = a \left( -2 \sin^2(2t) + 2 \cos(2t)(1 + \cos(2t)) \right) \]

Factoring out the common terms:

\[ \frac{dx}{dt} = 2a \left( \cos(2t)(1 + \cos(2t)) - \sin^2(2t) \right) \]

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

Now we differentiate \( y \):

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

Calculating the derivatives:

- \( \frac{d}{dt}(1 - \cos(2t)) = 2 \sin(2t) \)
- \( \frac{d}{dt}(\cos(2t)) = -2 \sin(2t) \)

Substituting these back into the equation:

\[ \frac{dy}{dt} = b \left( \cos(2t)(2 \sin(2t)) + (1 - \cos(2t))(-2 \sin(2t)) \right) \]

Simplifying:

\[ \frac{dy}{dt} = b \left( 2 \cos(2t) \sin(2t) - 2 \sin(2t)(1 - \cos(2t)) \right) \]

Factoring out the common term:

\[ \frac{dy}{dt} = 2b \sin(2t) \left( \cos(2t) - (1 - \cos(2t)) \right) \]
\[ \frac{dy}{dt} = 2b \sin(2t)(2\cos(2t) - 1) \]

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

Now we can find \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2b \sin(2t)(2\cos(2t) - 1)}{2a \left( \cos(2t)(1 + \cos(2t)) - \sin^2(2t) \right)} \]

### Step 4: Evaluate at \( t = \frac{\pi}{4} \)

At \( t = \frac{\pi}{4} \):

\[ \sin(2t) = \sin\left(\frac{\pi}{2}\right) = 1, \quad \cos(2t) = \cos\left(\frac{\pi}{2}\right) = 0 \]

Substituting these values:

\[ \frac{dy}{dx} = \frac{2b(1)(2 \cdot 0 - 1)}{2a \left( 0(1 + 0) - 1 \right)} = \frac{2b(-1)}{2a(-1)} = \frac{b}{a} \]

### Conclusion

Thus, we have shown that:

\[ \left( \frac{dy}{dx} \right)_{t=\frac{\pi}{4}} = \frac{b}{a} \]