The slope of tangent to the curve `x =a sin ^(3) t , y = b cos ^(3)t   at = (pi)/(2)` is :

Options

A. `- (b)/(a)`
B. `(b)/(a)`
C. `0`
D. `1`

Answer

To find the slope of the tangent to the curve defined by the parametric equations \( x = a \sin^3 t \) and \( y = b \cos^3 t \) at \( t = \frac{\pi}{2} \), we will follow these steps:

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

1. Differentiate \( x = a \sin^3 t \):
\[ \frac{dx}{dt} = a \cdot 3 \sin^2 t \cdot \cos t = 3a \sin^2 t \cos t \]

2. Differentiate \( y = b \cos^3 t \):
\[ \frac{dy}{dt} = b \cdot 3 \cos^2 t \cdot (-\sin t) = -3b \cos^2 t \sin t \]

### Step 2: Find the slope of the tangent \( \frac{dy}{dx} \)

Using the chain rule, the slope of the tangent can be expressed as:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{-3b \cos^2 t \sin t}{3a \sin^2 t \cos t} \]

### Step 3: Simplify the expression

We can simplify the expression:
\[ \frac{dy}{dx} = \frac{-b \cos^2 t \sin t}{a \sin^2 t \cos t} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{-b}{a} \cdot \frac{\cos t}{\sin t} = \frac{-b}{a} \cot t \]

### Step 4: Evaluate at \( t = \frac{\pi}{2} \)

Now, we need to evaluate the slope at \( t = \frac{\pi}{2} \):
\[ \frac{dy}{dx} \bigg|_{t = \frac{\pi}{2}} = \frac{-b}{a} \cot\left(\frac{\pi}{2}\right) \]
Since \( \cot\left(\frac{\pi}{2}\right) = 0 \):
\[ \frac{dy}{dx} \bigg|_{t = \frac{\pi}{2}} = \frac{-b}{a} \cdot 0 = 0 \]

### Final Answer
The slope of the tangent to the curve at \( t = \frac{\pi}{2} \) is \( 0 \).

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