Find \( \frac{d y}{d x} \) if \( x=a \sin ^{2} t \) and \( y=a \cos ^{2} t, 9 \) is a constant

To find \( \frac{d y}{d x} \), we can use the chain rule. We will first find \( \frac{d y}{d t} \) and \( \frac{d x}{d t} \), and then use the relationship: \[ \frac{d y}{d x} = \frac{d y/d t}{d x/d t} \] ### Step 1: Find \( \frac{d y}{d t} \) Given \( y = a \cos^2 t \), we can differentiate it with respect to \( t \): \[ \frac{d y}{d t} = a \cdot 2 \cos t \cdot (-\sin t) = -2a \cos t \sin t \] ### Step 2: Find \( \frac{d x}{d t} \) Given \( x = a \sin^2 t \), we can differentiate it with respect to \( t \): \[ \frac{d x}{d t} = a \cdot 2 \sin t \cdot \cos t = 2a \sin t \cos t \] ### Step 3: Find \( \frac{d y}{d x} \) Now we can substitute \( \frac{d y}{d t} \) and \( \frac{d x}{d t} \) into the formula for \( \frac{d y}{d x} \): \[ \frac{d y}{d x} = \frac{-2a \cos t \sin t}{2a \sin t \cos t} \] ### Step 4: Simplify The \( 2a \sin t \cos t \) terms cancel out: \[ \frac{d y}{d x} = -1 \] Thus, the final result is: \[ \frac{d y}{d x} = -1 \]