AP Precalculus 4.3: Parametric Functions and Rates of Change

A parametric function gives the position of a moving point as $(x(t),y(t))$. To tell which way the point moves as $t$ increases, you do not need the picture. Look at the two coordinates separately.

• If $x(t)$ is increasing, the point moves right. If $x(t)$ is decreasing, it moves left.
• If $y(t)$ is increasing, the point moves up. If $y(t)$ is decreasing, it moves down.

Then put the two together. A beetle on graph paper at $(6-2t,t^2)$ for $0\le t\le 3$ has $x(t)=6-2t$, which decreases, and $y(t)=t^2$, which increases when $t\ge 0$. It crawls left and up the whole time.

A coordinate can turn around partway through. Take $x(t)=t^2-4t$ and $y(t)=2t$ on $0\le t\le 4$. The graph of $x(t)$ against $t$ is a parabola with its vertex at $t=2$, so $x(t)$ decreases on $0\le t\le 2$ and increases on $2\le t\le 4$. Meanwhile $y(t)$ increases the whole time. The point moves left and up, then right and up. Split the interval wherever either coordinate changes direction.

A path can pass through the same point more than once, and it does not have to be going the same way each time. The path $(3\sin t,2\sin (2t))$ for $0\le t\le 2\pi$ is the figure eight in the diagram. It is at the origin when $t=0$ and again when $t=\pi$.

Check each visit on a short interval of $t$, using what you know about sine.

On $0\le t\le \frac{\pi }{4}$, $\sin t$ increases, so $x(t)$ increases. The angle $2t$ runs from 0 to $\frac{\pi }{2}$, where sine increases, so $y(t)$ increases too. The point leaves the origin moving right and up.

On $\pi \le t\le \frac{5\pi }{4}$, $\sin t$ decreases, so $x(t)$ decreases. The angle $2t$ runs from $2\pi$ to $\frac{5\pi }{2}$, where sine increases, so $y(t)$ increases. This time the point leaves the origin moving left and up.

One point, two directions. So when you are asked which way a particle moves at a point, you need the value of $t$, not only the point.

Back to the two walkers. Their maps matched, but their walks did not have to. The same curve in the plane can be parametrized in different ways, and different parametric functions can trace it in different directions.

Each of these traces the piece of $y=x^2$ from $(-2,4)$ to $(2,4)$:

• $(t,t^2)$ for $-2\le t\le 2$. Here $x(t)=t$ increases, so the point goes left to right.
• $(2t,4t^2)$ for $-1\le t\le 1$. Here $4t^2=(2t{)}^2$, so $y=x^2$ again, and $x(t)=2t$ runs from $-2$ to 2 while increasing. Left to right again, over a shorter interval of $t$.
• $(-t,t^2)$ for $-2\le t\le 2$. Here $x(t)=-t$ decreases, so the point starts at $(2,4)$ and goes right to left.

When the curve is the graph of a function like $y=x^2$, and each $x(t)$ only increases or only decreases, the check is short. Show that both parametric functions satisfy $y=x^2$ and cover the same stretch of $x$; then they draw the same piece. Compare directions by whether each $x(t)$ increases or decreases. For other curves, such as a circle, check that both give the same set of points, then compare directions interval by interval. A curve drawn without arrows cannot tell you.

Over an interval $t_1\le t\le t_2$, each coordinate has its own average rate of change, worked out the way you did in Unit 1:

Take $(t^2,10-3t)$ on $1\le t\le 4$. The points are $(1,7)$ and $(16,-2)$. The average rate of change of $x$ is $\frac{15}{3}=5$ and the average rate of change of $y$ is $\frac{-9}{3}=-3$, each in units per unit of $t$.

Divide the rate of $y$ by the rate of $x$: $\frac{-3}{5}=-0.6$. That is the slope of the line through $(1,7)$ and $(16,-2)$, because the $t_2-t_1$ in both denominators cancels and leaves $\frac{-9}{15}$.

This only works when the average rate of change of $x$ is not 0. On $(t^2-4t,2t)$ from $t=0$ to $t=4$, $x$ starts and ends at 0, so its average rate of change is 0. The points $(0,0)$ and $(0,8)$ sit one straight above the other, and the slope between them is not defined. The point did move sideways in between, left and then right. An average rate of change sees only the two ends.