Examine and plot a graph of a function given in parametric form

Question

I am having trouble understanding how to examine and plot a graph of a function given in parametric form.

I know how to examine and plot a graph of a real function with real variable in explicit form given by $y=f(x)$: domain, parity, null points, asymptotes, extreme points, inflection points. From this data, we can plot a graph $y=f(x)$.

For example, what are the steps to examine and plot a graph of real function with real variable given in parametric form, $$x=t^2-2t$$ $$y=t^2+2t,t\in\mathbb R?$$

Answer 1

Think about each of the terms. If we had only $$\pmatrix{x \\ y}=\pmatrix{t^2 \\ t^2}$$ we would simply have a straight line with slope $1$. This tells us that when $t$ gets large (when the squared terms dominate the linear), the graph will look like such a line. This additionally tells us that the graph is not a closed curve.

Now, what do the "anti-symmetric" terms $\pm 2t$ contribute with? Well, the $-2t$ for $x$ makes the otherwise straight line grow slower in the $x$-direction, and likewise with $y$, only with an increase in growth instead, so the graph should have a slope that generally larger than $1$ in the beginning.

Lastly, let's plug in some numbers! Your function starts at the origin, and $y$ always increases. $x$, on the other hand, actually dabbles in negativity between $0$ and $2$ (as can be easily visualized by imagining the quadratic function $f(x)=x^2-2x$). However, from then, $x$ is also always increasing. With this knowledge, combined with the fact that the graph should resemble a straight line after some time, you can kind of get an idea of what the whole graph would look like, namely:

Answer 2

Not a general answer, but any time you see a pair of equations of the form \begin{align*} u &= a + b, \\ v &= a - b, \end{align*} a little voice in your head should ask, "What happens if I add and subtract (i.e., solve for $a$ and $b$ in terms of $u$ and $v$)?"

Here, you get the equivalent system \begin{align*} u &:= \tfrac{1}{2}(y + x) = t^{2}, \\ v &:= \tfrac{1}{2}(y - x) = 4t, \end{align*} and your curve clearly traces the entire parabola $u = \bigl(\frac{1}{4}v\bigr)^{2}$.

Now, $(u, v)$ is a rotated and scaled Cartesian coordinate system:

• The $u$-axis has equation $v = 0$, i.e., $y = x$;

• The $v$-axis has equation $u = 0$, i.e., $y = -x$;

• Both coordinates increase with $y$;

• The point $(x, y) = (1, 1)$ satisfies $(u, v) = (1, 0)$, so one unit of length in $(u, v)$ equals $\sqrt{2}$ units of length in $(x, y)$.

Your parabola consequently looks like: