`x= 2 cos t -cos 2t ,y=2 sin t-sin 2t ` Find d y / d x

Answer

To find \(\frac{dy}{dx}\) for the given parametric equations \(x = 2 \cos t - \cos 2t\) and \(y = 2 \sin t - \sin 2t\), we will follow these steps:

### Step 1: Differentiate \(x\) with respect to \(t\)
We start by differentiating \(x\):
\[ x = 2 \cos t - \cos 2t \]
Using the chain rule, we differentiate:
\[ \frac{dx}{dt} = \frac{d}{dt}(2 \cos t) - \frac{d}{dt}(\cos 2t) \]
Calculating each derivative:
\[ \frac{d}{dt}(2 \cos t) = -2 \sin t \]
\[ \frac{d}{dt}(\cos 2t) = -2 \sin 2t \]
Thus, we have:
\[ \frac{dx}{dt} = -2 \sin t + 2 \sin 2t \]

### Step 2: Differentiate \(y\) with respect to \(t\)
Next, we differentiate \(y\):
\[ y = 2 \sin t - \sin 2t \]
Using the chain rule again, we differentiate:
\[ \frac{dy}{dt} = \frac{d}{dt}(2 \sin t) - \frac{d}{dt}(\sin 2t) \]
Calculating each derivative:
\[ \frac{d}{dt}(2 \sin t) = 2 \cos t \]
\[ \frac{d}{dt}(\sin 2t) = 2 \cos 2t \]
Thus, we have:
\[ \frac{dy}{dt} = 2 \cos t - 2 \cos 2t \]

### Step 3: Find \(\frac{dy}{dx}\)
Now we can find \(\frac{dy}{dx}\) using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{2 \cos t - 2 \cos 2t}{-2 \sin t + 2 \sin 2t} \]
We can simplify this by factoring out the common factor of 2:
\[ \frac{dy}{dx} = \frac{2(\cos t - \cos 2t)}{2(-\sin t + \sin 2t)} = \frac{\cos t - \cos 2t}{-\sin t + \sin 2t} \]

### Step 4: Final Expression
Thus, the final expression for \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{\cos t - \cos 2t}{\sin 2t - \sin t} \]