Find `dy/dx`, if x and y are connected parametrically by the equations, given below without eliminating the parameter:
`x=2cos^(3)theta,y=2sin^(3)theta`

Answer

To find \(\frac{dy}{dx}\) given the parametric equations \(x = 2\cos^3\theta\) and \(y = 2\sin^3\theta\), we will follow these steps:

### Step 1: Differentiate \(x\) with respect to \(\theta\)
We start by differentiating \(x\) with respect to \(\theta\):
\[ x = 2\cos^3\theta \]
Using the chain rule:
\[ \frac{dx}{d\theta} = 2 \cdot 3\cos^2\theta \cdot \frac{d}{d\theta}(\cos\theta) = 2 \cdot 3\cos^2\theta \cdot (-\sin\theta) = -6\cos^2\theta\sin\theta \]

### Step 2: Differentiate \(y\) with respect to \(\theta\)
Next, we differentiate \(y\) with respect to \(\theta\):
\[ y = 2\sin^3\theta \]
Using the chain rule:
\[ \frac{dy}{d\theta} = 2 \cdot 3\sin^2\theta \cdot \frac{d}{d\theta}(\sin\theta) = 2 \cdot 3\sin^2\theta \cdot \cos\theta = 6\sin^2\theta\cos\theta \]

### Step 3: Find \(\frac{dy}{dx}\)
Now, we can find \(\frac{dy}{dx}\) using the formula:
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{6\sin^2\theta\cos\theta}{-6\cos^2\theta\sin\theta} \]

### Step 4: Simplify the expression
We can simplify this expression:
\[ \frac{dy}{dx} = \frac{6\sin^2\theta\cos\theta}{-6\cos^2\theta\sin\theta} = \frac{\sin\theta}{-\cos\theta} = -\tan\theta \]

Thus, the final result is:
\[ \frac{dy}{dx} = -\tan\theta \]