If `x=acos^(3)t and y=asin^(3)t`, then `(dy)/(dx)` is equal to

Options

A. cot t
B. sec t
C. `cosect`
D. `-tant`

Answer

To find \(\frac{dy}{dx}\) given \(x = a \cos^3 t\) and \(y = a \sin^3 t\), we will use the chain rule for differentiation.

### Step-by-Step Solution:

1. **Differentiate \(x\) with respect to \(t\)**:
\[ x = a \cos^3 t \]
Using the chain rule:
\[ \frac{dx}{dt} = 3a \cos^2 t \cdot \frac{d}{dt}(\cos t) = 3a \cos^2 t \cdot (-\sin t) = -3a \cos^2 t \sin t \]

2. **Differentiate \(y\) with respect to \(t\)**:
\[ y = a \sin^3 t \]
Again using the chain rule:
\[ \frac{dy}{dt} = 3a \sin^2 t \cdot \frac{d}{dt}(\sin t) = 3a \sin^2 t \cdot \cos t \]

3. **Find \(\frac{dy}{dx}\)** using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
Substitute the derivatives we found:
\[ \frac{dy}{dx} = \frac{3a \sin^2 t \cos t}{-3a \cos^2 t \sin t} \]

4. **Simplify the expression**:
\[ \frac{dy}{dx} = \frac{3a \sin^2 t \cos t}{-3a \cos^2 t \sin t} = \frac{\sin t}{-\cos t} = -\tan t \]

Thus, we conclude that:
\[ \frac{dy}{dx} = -\tan t \]