The slope of the tangent to the curve x=2 ` sin ^(3) t, y=3 cos ^(3) t " at " t= (pi)/(4) ` is

Options

A. ` (3)/(2)`
B. ` -(3)/(2)`
C. ` ( 2)/(3)`
D. ` -(2)/(3)`

Answer

To find the slope of the tangent to the curve defined by the parametric equations \( x = 2 \sin^3 t \) and \( y = 3 \cos^3 t \) at \( t = \frac{\pi}{4} \), we will follow these steps:

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

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

1. **For \( y = 3 \cos^3 t \)**:
- Using the chain rule, we differentiate:
\[ \frac{dy}{dt} = 3 \cdot 3 \cos^2 t \cdot (-\sin t) = -9 \cos^2 t \sin t \]

2. **For \( x = 2 \sin^3 t \)**:
- Again, using the chain rule:
\[ \frac{dx}{dt} = 2 \cdot 3 \sin^2 t \cdot \cos t = 6 \sin^2 t \cos t \]

### Step 2: Find \( \frac{dy}{dx} \)

Now, we can find the slope of the tangent \( \frac{dy}{dx} \) using the formula:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{-9 \cos^2 t \sin t}{6 \sin^2 t \cos t} \]

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

We can simplify the expression:
\[ \frac{dy}{dx} = \frac{-9 \cos^2 t \sin t}{6 \sin^2 t \cos t} = \frac{-9}{6} \cdot \frac{\cos t}{\sin t} = -\frac{3}{2} \cdot \cot t \]

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

Now, we substitute \( t = \frac{\pi}{4} \):
\[ \frac{dy}{dx} \bigg|_{t = \frac{\pi}{4}} = -\frac{3}{2} \cdot \cot\left(\frac{\pi}{4}\right) \]
Since \( \cot\left(\frac{\pi}{4}\right) = 1 \):
\[ \frac{dy}{dx} \bigg|_{t = \frac{\pi}{4}} = -\frac{3}{2} \cdot 1 = -\frac{3}{2} \]

### Final Result

Thus, the slope of the tangent to the curve at \( t = \frac{\pi}{4} \) is:
\[ \boxed{-\frac{3}{2}} \]

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