Find the equation of the tangent at `t=(pi)/(4)` to the curve : `x=sin 3t, y = cos 2t.`

Answer

To find the equation of the tangent at \( t = \frac{\pi}{4} \) to the curve defined by the parametric equations \( x = \sin(3t) \) and \( y = \cos(2t) \), we will follow these steps:

### Step 1: Find the coordinates of the point on the curve at \( t = \frac{\pi}{4} \).

1. Calculate \( x \):
\[ x = \sin(3t) = \sin\left(3 \cdot \frac{\pi}{4}\right) = \sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2} \]

2. Calculate \( y \):
\[ y = \cos(2t) = \cos\left(2 \cdot \frac{\pi}{4}\right) = \cos\left(\frac{\pi}{2}\right) = 0 \]

Thus, the point on the curve is \( \left(\frac{\sqrt{2}}{2}, 0\right) \).

### Step 2: Find the derivatives \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \).

1. Differentiate \( y \):
\[ \frac{dy}{dt} = \frac{d}{dt}(\cos(2t)) = -2\sin(2t) \]

2. Differentiate \( x \):
\[ \frac{dx}{dt} = \frac{d}{dt}(\sin(3t)) = 3\cos(3t) \]

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

Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-2\sin(2t)}{3\cos(3t)} \]

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

1. Calculate \( \sin(2t) \) and \( \cos(3t) \):
\[ \sin(2t) = \sin\left(2 \cdot \frac{\pi}{4}\right) = \sin\left(\frac{\pi}{2}\right) = 1 \]
\[ \cos(3t) = \cos\left(3 \cdot \frac{\pi}{4}\right) = \cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2} \]

2. Substitute into \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{-2 \cdot 1}{3 \cdot \left(-\frac{\sqrt{2}}{2}\right)} = \frac{-2}{-\frac{3\sqrt{2}}{2}} = \frac{4}{3\sqrt{2}} = \frac{4\sqrt{2}}{6} = \frac{2\sqrt{2}}{3} \]

### Step 5: Write the equation of the tangent line.

Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
where \( (x_1, y_1) = \left(\frac{\sqrt{2}}{2}, 0\right) \) and \( m = \frac{2\sqrt{2}}{3} \):
\[ y - 0 = \frac{2\sqrt{2}}{3}\left(x - \frac{\sqrt{2}}{2}\right) \]

### Step 6: Simplify the equation.

1. Distributing:
\[ y = \frac{2\sqrt{2}}{3}x - \frac{2\sqrt{2}}{3} \cdot \frac{\sqrt{2}}{2} \]
\[ y = \frac{2\sqrt{2}}{3}x - \frac{2 \cdot 2}{3 \cdot 2} = \frac{2\sqrt{2}}{3}x - \frac{2}{3} \]

Thus, the equation of the tangent line is:
\[ y = \frac{2\sqrt{2}}{3}x - \frac{2}{3} \]