If `x=3sint-sin3t ,y=3cos t-cos3t ,"f i n d"(dy)/(dx)"a t"t=pi/3dot`

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 \(t\)
We start with the equation for \(x\):
\[ x = 3\sin t - \sin 3t \]
Differentiating both sides with respect to \(t\):
\[ \frac{dx}{dt} = 3\cos t - 3\cos 3t \]
This can be simplified to:
\[ \frac{dx}{dt} = 3(\cos t - \cos 3t) \]

### Step 2: Differentiate \(y\) with respect to \(t\)
Now, we differentiate \(y\):
\[ y = 3\cos t - \cos 3t \]
Differentiating both sides with respect to \(t\):
\[ \frac{dy}{dt} = -3\sin t + 3\sin 3t \]
This can be simplified to:
\[ \frac{dy}{dt} = 3(\sin 3t - \sin t) \]

### Step 3: Find \(\frac{dy}{dx}\)
Using the chain rule, we can express \(\frac{dy}{dx}\) as:
\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3(\sin 3t - \sin t)}{3(\cos t - \cos 3t)} \]
The \(3\) cancels out:
\[ \frac{dy}{dx} = \frac{\sin 3t - \sin t}{\cos t - \cos 3t} \]

### Step 4: Evaluate \(\frac{dy}{dx}\) at \(t = \frac{\pi}{3}\)
Now we substitute \(t = \frac{\pi}{3}\):
\[ \frac{dy}{dx} \bigg|_{t = \frac{\pi}{3}} = \frac{\sin(3 \cdot \frac{\pi}{3}) - \sin(\frac{\pi}{3})}{\cos(\frac{\pi}{3}) - \cos(3 \cdot \frac{\pi}{3})} \]
Calculating the sine and cosine values:
\[ \sin(\pi) = 0, \quad \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]
\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}, \quad \cos(\pi) = -1 \]
Substituting these values:
\[ \frac{dy}{dx} \bigg|_{t = \frac{\pi}{3}} = \frac{0 - \frac{\sqrt{3}}{2}}{\frac{1}{2} - (-1)} = \frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2} + 1} \]
This simplifies to:
\[ \frac{dy}{dx} \bigg|_{t = \frac{\pi}{3}} = \frac{-\frac{\sqrt{3}}{2}}{\frac{3}{2}} = -\frac{\sqrt{3}}{3} \]

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