If `x=asin2t(1+cos2t)` and `y=bcos2t(1-cos2t)` , show that at `t=pi/4` , `(dy)/(dx)=b/a` .

Explanation

To solve the problem, we need to find \(\frac{dy}{dx}\) when \(t = \frac{\pi}{4}\) given the equations:

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

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

First, we differentiate \(x\):

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

Using the product rule, we have:

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

Calculating the derivatives:

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

Now substituting these into the equation:

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

This simplifies to:

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

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

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

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

Next, we differentiate \(y\):

\[ y = b \cos(2t)(1 - \cos(2t)) \]

Using the product rule again:

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

Calculating the derivatives:

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

Now substituting these into the equation:

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

This simplifies to:

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

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

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

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

This simplifies to:

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

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

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

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

Now substituting these values into \(\frac{dy}{dx}\):

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

### Conclusion

Thus, we have shown that at \(t = \frac{\pi}{4}\):

\[ \frac{dy}{dx} = \frac{b}{a} \]