How can I sketch this vector function? $r(t) = \langle t-2\sin(t), t^2\rangle$

Question

So from all the resource I was able to find about graphing or visualizing vector functions, none of them could justify this problem. The sketch of $r(t) = \langle t-2\sin(t), t^2\rangle$ looks nothing like the picture above and I don't see how, when coupled with $y=t^2$, it produces that graph.

Answer 1

Some things that might get you started …

The curve certainly passes through the origin.

At the origin, the tangent is in the direction $(1,0)$.

When $t$ is large, the $\sin(t)$ term is not significant, so the curve looks like $r(t) = (t, t^2)$, which is just the parabola $y=x^2$.

The curve will cross the y-axis when $t=2 \sin(t)$.

It’s symmetric about the y-axis.

The tangents are vertical when $t= \pm\pi/3$.

Answer 2

First, notice that the graph of the path of $$\mathbf{r}(t)=\langle x(t),y(t)\rangle=\langle t-2\sin(t),t^2\rangle $$

is symmetric with resptect to the $y$-axis because the vector

\begin{eqnarray}\mathbf{r}(-t)&=&\langle x(-t),y(-t)\rangle\\ &=&\langle -t-2\sin(-t),(-t)^2\rangle\\ &=&\langle -t+2\sin(t),t^2\rangle\\ &=&\langle -x(t),y(t)\rangle\end{eqnarray} has the same $y$-component but the opposite $x$-component of $\mathbf{r}(t)$.

So the path of the equation satisfies the equation

$$ \lvert x\rvert=\sqrt{y}-2\sin\left(\sqrt{y}\right) $$

Which you can verify is the graph as shown in your question.