Parametric rates of change

Over an interval of parameter values, each coordinate function has its own average rate of change: change in that coordinate divided by change in t. If a particle is at (3, 2) when t = 1 and at (7, 10) when t = 3, then over that interval

1. x changes at $(7-3)/(3-1) = 2$ units per unit of t, and
2. y changes at $(10-2)/(3-1) = 4$ units per unit of t.

Two rates, computed separately, never averaged together. The x-rate says how fast the particle drifts horizontally; the y-rate how fast it climbs. Signs carry direction: a negative x-rate is net leftward drift, a negative y-rate net downward.

The slope of the secant between two positions on the path is rise over run in the plane: $\Delta y / \Delta x$. Using the same two positions, slope $= (10-2)/(7-3) = 2$. Notice the parameter has vanished: both changes happened over the same t interval, so the t's cancel, and the slope is also the ratio of the two component rates, $4/2 = 2$.

That gives two equivalent recipes and one warning. Recipes: divide the coordinate changes directly, or divide the y-rate by the x-rate. Warning: the y-rate alone is NOT the slope. A y-rate of 4 with an x-rate of 2 gives slope 2; the same y-rate with an x-rate of 8 gives slope 1/2. Slope always needs both components.

Read the two signs first, the ratio second. A particle moving left and up has $\Delta x < 0$ and $\Delta y > 0$, so its secant slope is negative, even though the particle is rising. A negative slope on a parametric path does not mean "going down"; it means the two components disagree in sign, and which one is negative matters. Rising-while-moving-left and falling-while-moving-right both produce negative slopes from opposite motions.

This is the sharpest difference from y = f(x) graphs, where x always increases and slope alone tells the vertical story. On a path, direction is two independent signs; the slope is only their quotient.

Comparing the two rates in size predicts the path's local character. When $|\Delta x|$ dwarfs $|\Delta y|$, travel is nearly horizontal and the slope is near 0. When $|\Delta y|$ dwarfs $|\Delta x|$, travel is nearly vertical and the slope is large; in the limit of a zero x-change the secant is vertical and the slope is undefined, which a parametric path handles without complaint.

And remember these are average rates over the interval: an x-rate of zero across [2, 5] means the particle ended at the horizontal position where it started, not that it never moved horizontally in between. The interval's interior is invisible to an average computed from endpoints.