Checkpoint 1 Answers

1. Solve the given equation for ${\displaystyle y}$: ${\displaystyle y=\pm 3\sqrt{1-\frac{x^2}{4}}}$. The positive solution corresponds to the top half of the ellipse.

2. ${\displaystyle \frac{dy}{dx}=3\cdot \frac{1}{2}\cdot \frac{1}{\sqrt{1-\frac{x^2}{4}}}\cdot \frac{d}{dx}(1-\frac{x^2}{4})=\frac{3}{2}\cdot \frac{1}{\sqrt{1-\frac{x^2}{4}}}\cdot \frac{-2x}{4}=-\frac{3}{4}\cdot \frac{x}{\sqrt{1-\frac{x^2}{4}}}}$.
   At ${\displaystyle x=1}$, ${\displaystyle \frac{dy}{dx}=-\frac{3}{4}\cdot \frac{1}{\sqrt{\frac{3}{4}}}=-\sqrt{\frac{3}{4}}=-\frac{\sqrt{3}}{2}.}$