If `x=3sint-sin3t` , `y=3cost-cos3t` , find `(dy)/(dx)` at `t=pi/3` .

Explanation

To find \(\frac{dy}{dx}\) at \(t = \frac{\pi}{3}\) given the parametric equations \(x = 3\sin t - \sin 3t\) and \(y = 3\cos t - \cos 3t\), we will follow these steps:

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

Given:
\[ x = 3\sin t - \sin 3t \]

Differentiating \(x\) with respect to \(t\):
\[ \frac{dx}{dt} = 3\cos t - \frac{d}{dt}(\sin 3t) \]
Using the chain rule for \(\sin 3t\):
\[ \frac{d}{dt}(\sin 3t) = 3\cos 3t \]
Thus:
\[ \frac{dx}{dt} = 3\cos t - 3\cos 3t \]

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

Given:
\[ y = 3\cos t - \cos 3t \]

Differentiating \(y\) with respect to \(t\):
\[ \frac{dy}{dt} = 3(-\sin t) - \frac{d}{dt}(\cos 3t) \]
Using the chain rule for \(\cos 3t\):
\[ \frac{d}{dt}(\cos 3t) = -3\sin 3t \]
Thus:
\[ \frac{dy}{dt} = -3\sin t + 3\sin 3t \]

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

Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]

Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{-3\sin t + 3\sin 3t}{3\cos t - 3\cos 3t} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{-\sin t + \sin 3t}{\cos t - \cos 3t} \]

### Step 4: Evaluate at \(t = \frac{\pi}{3}\)

Now we will substitute \(t = \frac{\pi}{3}\):
1. Calculate \(\sin\) and \(\cos\) values:
- \(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\)
- \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\)
- \(\sin\left(3 \cdot \frac{\pi}{3}\right) = \sin(\pi) = 0\)
- \(\cos\left(3 \cdot \frac{\pi}{3}\right) = \cos(\pi) = -1\)

Substituting these values into \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{-\frac{\sqrt{3}}{2} + 0}{\frac{1}{2} - (-1)} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2} + 1} = \frac{-\frac{\sqrt{3}}{2}}{\frac{3}{2}} = -\frac{\sqrt{3}}{3} \]

### Final Answer
Thus, the value of \(\frac{dy}{dx}\) at \(t = \frac{\pi}{3}\) is:
\[ \frac{dy}{dx} = -\frac{1}{\sqrt{3}} \]