example 3 sketch the graph of each of the following vector functions.\n\n(b) $\\vec{r}(t) = \\langle t - 2…

Explanation:

Step1: Identify parametric equations

The vector function $\vec{r}(t) = \langle t - 2\sin t, t^2 \rangle$ gives the parametric equations: $$x(t) = t - 2\sin t$$ $$y(t) = t^2$$

Step2: Analyze the y-component

Since $y = t^2$, the curve is symmetric about the y-axis (as $y(-t) = y(t)$) and exists only for $y \ge 0$.

Step3: Analyze the x-component

The function $x(t) = t - 2\sin t$ is an odd function. As $t \to \infty$, $x \to \infty$. The derivative $x'(t) = 1 - 2\cos t$ indicates oscillations.

Step4: Determine critical points for x

Set $x'(t) = 0$: $$1 - 2\cos t = 0 \implies \cos t = \frac{1}{2} \implies t = \pm \frac{\pi}{3} + 2n\pi$$ At $t = \frac{\pi}{3}$, $x \approx 0.52 - 1.73 = -1.21$. At $t = -\frac{\pi}{3}$, $x \approx 1.21$.

Step5: Describe the graph shape

The graph is a curve starting from the origin $(0,0)$ at $t=0$. As $t$ increases, $y$ increases quadratically while $x$ oscillates but generally increases, creating a "wavy" parabolic path.

Answer:

The graph is a curve in the $xy$-plane symmetric about the $y$-axis. It starts at the origin $(0,0)$ when $t=0$. For $t > 0$, the curve moves upward as $y = t^2$ increases. Because $x = t - 2\sin t$, the curve oscillates horizontally, moving left to a local minimum $x \approx -1.21$ near $y \approx 1.1$ ($t = \pi/3$), then swings back to the right, crossing the $y$-axis again and continuing to oscillate with an overall trend toward positive $x$ as $y$ grows. The left side ($t < 0$) is a mirror image across the $y$-axis.