If `x=a cos^3 t " and "y=a sin ^3 t "then find " dy/dx`

Options

A. `cot t `
B. `-tan t `
C. `tan t `
D. `-cot t `

Explanation

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

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

We start by differentiating \( x \) with respect to \( t \):

\[ x = a \cos^3 t \]

Using the chain rule:

\[ \frac{dx}{dt} = a \cdot 3 \cos^2 t \cdot (-\sin t) = -3a \cos^2 t \sin t \]

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

Next, we differentiate \( y \) with respect to \( t \):

\[ y = a \sin^3 t \]

Using the chain rule:

\[ \frac{dy}{dt} = a \cdot 3 \sin^2 t \cdot \cos t = 3a \sin^2 t \cos t \]

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

Now we can find \( \frac{dy}{dx} \) using the formula:

\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]

Substituting the derivatives we found:

\[ \frac{dy}{dx} = \frac{3a \sin^2 t \cos t}{-3a \cos^2 t \sin t} \]

### Step 4: Simplify the expression

We can simplify this expression:

\[ \frac{dy}{dx} = \frac{3a \sin^2 t \cos t}{-3a \cos^2 t \sin t} = \frac{\sin t}{-\cos t} = -\tan t \]

### Final Answer

Thus, we have:

\[ \frac{dy}{dx} = -\tan t \]