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 need to find \(\frac{dy}{dx}\) at \(t = \frac{\pi}{4}\) given the equations:

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

### 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 for \(\frac{dx}{dt}\):

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

Using the identity \(\cos^2(2t) - \sin^2(2t) = \cos(4t)\):

\[ \frac{dx}{dt} = 2a \left( \cos(2t) + \frac{1}{2}\cos(4t) \right) \]

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

Similarly, 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 for \(\frac{dy}{dt}\):

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

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

Using the chain rule:

\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]

Substituting the expressions we derived:

\[ \frac{dy}{dx} = \frac{2b \sin(2t)(2\cos(2t) - 1)}{2a \left( \cos(2t) + \frac{1}{2}\cos(4t) \right)} \]

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

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

\[ \sin(2t) = \sin\left(\frac{\pi}{2}\right) = 1 \]
\[ \cos(2t) = \cos\left(\frac{\pi}{2}\right) = 0 \]
\[ \cos(4t) = \cos(\pi) = -1 \]

Thus, we have:

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

### Conclusion

We have shown that:

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