Find te slope of the normal to the curve `x= a cos^(2) theta and y= a sin^(3) theta" at "theta =(pi)/(4)`

Answer

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

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

We need to find \(\frac{dy}{d\theta}\) and \(\frac{dx}{d\theta}\).

1. Differentiate \( y = a \sin^3 \theta \):
\[ \frac{dy}{d\theta} = a \cdot 3 \sin^2 \theta \cdot \cos \theta = 3a \sin^2 \theta \cos \theta \]

2. Differentiate \( x = a \cos^2 \theta \):
\[ \frac{dx}{d\theta} = a \cdot 2 \cos \theta \cdot (-\sin \theta) = -2a \cos \theta \sin \theta \]

### Step 2: Find the slope of the tangent line

The slope of the tangent line \( m_1 \) is given by:
\[ m_1 = \frac{dy/d\theta}{dx/d\theta} = \frac{3a \sin^2 \theta \cos \theta}{-2a \cos \theta \sin \theta} \]

### Step 3: Simplify the expression for \( m_1 \)

Cancel out \( a \) and \( \cos \theta \) (assuming \( \cos \theta \neq 0 \)):
\[ m_1 = \frac{3 \sin \theta}{-2} = -\frac{3}{2} \sin \theta \]

### Step 4: Evaluate \( m_1 \) at \( \theta = \frac{\pi}{4} \)

Substituting \( \theta = \frac{\pi}{4} \):
\[ m_1 = -\frac{3}{2} \sin\left(\frac{\pi}{4}\right) = -\frac{3}{2} \cdot \frac{1}{\sqrt{2}} = -\frac{3}{2\sqrt{2}} \]

### Step 5: Find the slope of the normal line

The slope of the normal line \( m_2 \) is the negative reciprocal of the slope of the tangent line:
\[ m_2 = -\frac{1}{m_1} = -\frac{1}{-\frac{3}{2\sqrt{2}}} = \frac{2\sqrt{2}}{3} \]

### Final Answer

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