`x=2 cos^2 t, y= 6 sin ^2 t ` find `dy/dx`

Answer

To find \(\frac{dy}{dx}\) given the parametric equations \(x = 2 \cos^2 t\) and \(y = 6 \sin^2 t\), we will follow these steps:

### Step 1: Differentiate \(x\) with respect to \(t\)
We start with the equation for \(x\):
\[ x = 2 \cos^2 t \]
Using the chain rule, we differentiate \(x\) with respect to \(t\):
\[ \frac{dx}{dt} = 2 \cdot \frac{d}{dt}(\cos^2 t) = 2 \cdot 2 \cos t \cdot \frac{d}{dt}(\cos t) = 2 \cdot 2 \cos t \cdot (-\sin t) = -4 \cos t \sin t \]
This can also be expressed using the double angle identity:
\[ \frac{dx}{dt} = -2 \sin(2t) \]
Let’s label this as Equation (1).

### Step 2: Differentiate \(y\) with respect to \(t\)
Now we differentiate \(y\):
\[ y = 6 \sin^2 t \]
Using the chain rule again, we get:
\[ \frac{dy}{dt} = 6 \cdot \frac{d}{dt}(\sin^2 t) = 6 \cdot 2 \sin t \cdot \frac{d}{dt}(\sin t) = 6 \cdot 2 \sin t \cdot \cos t = 12 \sin t \cos t \]
This can also be expressed using the double angle identity:
\[ \frac{dy}{dt} = 6 \sin(2t) \]
Let’s label this as Equation (2).

### Step 3: Find \(\frac{dy}{dx}\)
To find \(\frac{dy}{dx}\), we use the relationship:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
Substituting the values from Equations (1) and (2):
\[ \frac{dy}{dx} = \frac{12 \sin(2t)}{-2 \sin(2t)} \]
The \(\sin(2t)\) terms cancel out (assuming \(\sin(2t) \neq 0\)):
\[ \frac{dy}{dx} = \frac{12}{-2} = -6 \]

### Final Answer
Thus, the value of \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = -6 \]