## Multivariable Calculus, 7th Edition

$(1,1), (\frac{3}{4},\frac{1}{2}), (\frac{1}{2}, 1-\frac{\sqrt 2}{2}), (\frac{1}{4},1-\frac{\sqrt 3}{2}),$ and $(0,0)$ are consecutive points on the curve. The arrow points from $(1,1)$ to $(0,0)$.
Since $0 \leq t \leq \frac{\pi}{2},$ plot the points where $t=0,$ $t=\frac{\pi}{6},$ $t=\frac{\pi}{4},$ $t=\frac{\pi}{3},$ and $t=\frac{\pi}{2}$. You can find the $x$ and $y$ coordinates of each point by plugging the value of $t$ into the given formulas for $x$ and $y$. For instance, when $t=0,$ $$x = (\cos0)^2=1$$$$y = 1 - \sin0 = 1$$ Therefore, $t=0$ corresponds to the point $(1,1)$. The same calculation gives: $(\frac{3}{4},\frac{1}{2})$ for $t=\frac{\pi}{6}$ $(\frac{1}{2}, 1-\frac{\sqrt 2}{2})$ for $t=\frac{\pi}{4}$ $(\frac{1}{4},1-\frac{\sqrt 3}{2})$ for $t=\frac{\pi}{3}$ $(0,0)$ for $t=\frac{\pi}{2}$ Plot these five points and connect them with a curve (in order of increasing $t$), as shown. As $t$ increases from $0$ to $\frac{\pi}{2}$, the curve is traced from $(1,1)$ to $(0,0)$, so an arrow should be drawn on the curve in that direction, as shown.