sketching the parametric curve $(5\sin t, t^2)$ for $t\in [-2\pi, 2\pi]$ I want to sketch the parametric curve on $t\in[-2\pi, 2\pi]$ defined by $$x=5\sin t, y=t^2$$
The first thing I did was find the domain and range. Since $-1 \leq \sin t \leq 1$ then $x\in[-5, 5]$. And of course $y\in\mathbb R$.
My second step was finding a min/max. I differentiated y with respect to x given by $$dx/dt = 5\cos t \\ dy/dt = 2t \\ \implies dy/dx = \cfrac{2t}{5\cos t}$$ and found only one critical point at $t=0$, which is located on the origin. I searched for vertical asymptotes, which appear when $t=\pm(\pi/2, 3\pi/2)$. There are 3 total y-intercepts given by $t=0, \pi, 2\pi$.
In total I have the points $(\pm5, \pi^2/4), (\pm5, 9\pi^2/4), (0,0), (0, \pi^2), (0, 4\pi^2)$.
But how do I most quickly use my variations in x and y to accurately sketch the curve that passes through these points?  I have a good guess what the curve looks like based on these points, but I want to ensure my directions are correct
 A: The points that you got are a good start. Let me give you a couple of guidelines that will help you get a good sketch.
Let us write $f(t) = (5\sin t, t^2)$.
The first useful thing to notice is that $f(-t) = (5\sin(-t), (-t)^2) = (-5\sin t,t^2)$, so the graph will be symmetric with respect to the $y$-axis. Thus, it is enough to sketch the curve on $[0,2\pi]$ and to get it on $[-2\pi, 0]$ we only need to reflect the curve with respect to the $y$-axis.
Now, let us first think of what $(5\sin t,t)$ would look like and how we would sketch it. Quite obviously, it is a sinusoid. It looks just like sine wave, only its amplitude is $5$ and it is reflected with respect to the $y=x$ line. So, we would sketch it as usual, we want to plot some points on which we know the values of sine explicitly:
\begin{array}{c | c c c c c}
t & 0 & \frac\pi 6 & \frac\pi 4 & \frac\pi 3 & \frac\pi 2 & \cdots \\ \hline
\sin t & 0 & \frac 12 & \frac {\sqrt 2}2 & \frac{\sqrt 3}2 & 1 & \cdots
\end{array}
Of course, we'd multiply the values of sine by $5$ to get points on $(5\sin t,t)$ and we get the rest of values from the trigonometric circle.
The next step is to realise that $(5\sin t, t^2)$ is quite similar to the sine wave, we can think of it as a sinusoid that gets more streched as we go further along the $y$-axis. This can help visualise the general behaviour. But to get more explicit, we can use the above table to produce the following one:
\begin{array}{c | c c c c c}
t^2 & 0 & \left(\frac\pi 6\right)^2 & \left(\frac\pi 4\right)^2 & \left(\frac\pi 3\right)^2 & \left(\frac\pi 2\right)^2 & \cdots \\ \hline
5\sin t & 0 & \frac 52 & \frac {5\sqrt 2}2 & \frac{5\sqrt 3}2 & 5 & \cdots
\end{array}
and this will get you all the points you need for an accurate sketch. It will be as precise as your sketch of sine would be.
Remembering the symmetry, just reflect this sketch with respect to the $y$-axis and we are done. The result will look kind of like a sketch of fish.
