Parametric Curves and Motion

One parameter, two coordinated outputs

The usual equation $y=f(x)$ describes a curve by giving one vertical output for each horizontal input. That arrangement is powerful, but it cannot naturally describe a vertical line, a loop traced more than once, or an object whose position changes with time. Parametric equations begin from a different question: where is the point at each value of a shared parameter? The horizontal and vertical coordinates are each functions, synchronized by that parameter.

A parametric curve has the form

$$x=f(t),\qquad y=g(t).$$

The parameter $t$ might be time, but it does not have to be. It is simply the input that determines both coordinates. As $t$ changes, the point $(f(t),g(t))$ moves and traces a path. A graph of the path alone does not tell you the direction of travel or the timing, so a careful parametric sketch includes arrows and selected parameter values.

For example, let $x(t)=2\cos t$ and $y(t)=\sin t$ for $0\le t\le2\pi$. Eliminating the parameter gives $x^2/4+y^2=1$, an ellipse. At $t=0$, the point is $(2,0)$. For small positive $t$, $y$ increases while $x$ decreases slightly, so the curve is traced counterclockwise. The Cartesian equation identifies the shape; the parametric equations preserve the motion information that the shape alone discarded.

Velocity is coordinate-by-coordinate change

When $t$ is time in seconds, $x(t)$ and $y(t)$ are position coordinates. Their derivatives form the velocity vector

$$\mathbf{v}(t)=\left\langle\frac{dx}{dt},\frac{dy}{dt}\right\rangle.$$

For the ellipse, $\mathbf{v}(t)=\langle-2\sin t,\cos t\rangle$. At $t=0$, the velocity is $\langle0,1\rangle$, pointing straight upward, exactly as the direction arrow on the ellipse predicted. The components do not need to be positive together. A negative horizontal component means the object is moving left, while a positive vertical component means it is moving up.

Speed is the magnitude of velocity, not a signed component:

$$\text{speed}=\lVert\mathbf{v}(t)\rVert =\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}.$$

For the ellipse, speed is $\sqrt{4\sin^2t+\cos^2t}$. It varies with $t$, so equal steps in time do not correspond to equal distances along the ellipse. At the top and bottom, the horizontal component has magnitude $2$ and the speed is $2$; at the left and right ends, the speed is $1$. The picture may look symmetric, but its parameterization moves faster in some locations than in others.

Slope comes from a ratio of rates

The slope of a parametric curve is obtained through the chain rule:

$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt},$$

provided $dx/dt\ne0$. This is not an independent shortcut. It says that vertical change per horizontal change can be found by comparing their changes per unit parameter. For the ellipse,

$$\frac{dy}{dx}=\frac{\cos t}{-2\sin t}.$$

At $t=0$, $dx/dt=0$ while $dy/dt=1$, so the tangent is vertical. The ratio is undefined there for a geometric reason, not because the curve is broken. At $t=\pi/2$, $dy/dt=0$ and $dx/dt=-2$, so the tangent is horizontal. Always interpret a zero component before mechanically dividing; the components show which direction of movement has temporarily stopped.

Parameter elimination can help, but it has limits. If $x=t^2$ and $y=t$, eliminating $t$ gives $x=y^2$, a sideways parabola. The values $t=1$ and $t=-1$ produce different points with the same $x$ but different $y$, so no single function $y=f(x)$ captures the entire curve. Parametric notation does not make the curve exotic; it preserves information that the function format excludes.

Distance and displacement are not the same question

Displacement is the change in position from the initial point to the final point. For a parametric path it is the vector

$$\Delta\mathbf{r}=\langle x(b)-x(a),y(b)-y(a)\rangle.$$

Distance traveled is the accumulated speed,

$$\int_a^b\sqrt{\left[x'(t)\right]^2+\left[y'(t)\right]^2}\,dt.$$

An object that completes one circuit of the ellipse has zero displacement because it returns to its starting point, yet it travels a positive distance. This distinction is the two-dimensional version of the difference between net change and total amount. The integral uses speed, which is nonnegative, so reverse direction does not cancel previously traveled path length.

Retrieval check

Let $x(t)=t^2-1$ meters and $y(t)=2t$ meters for $-1\le t\le2$ seconds. Find the initial and final positions, predict the direction at $t=0$, and compute the velocity vector there. Then determine whether the tangent at $t=0$ is horizontal, vertical, or neither by examining the coordinate rates. Explain why a position graph without parameter arrows would not be enough to answer all of those questions. The next lesson develops the derivative and concavity tools that let a parametric path be analyzed as carefully as a conventional graph.