Visighta

Vector Kinematics in Two and Three Dimensions

3.3 Parametric Descriptions of Trajectories

Shape vs. Timing

A roller coaster traces a beautiful loop-the-loop. Stand back and photograph it --- you see a curve in space, a shape. Now watch a car ride that loop. It crawls up slowly, whips through the top, and accelerates down the far side. The shape of the track tells you nothing about that timing. A car that takes the loop in four seconds and a car that takes it in ten follow the same curve, but their motions are completely different.

Here is the question this section is about: when we write $\vec{r}(t) = (x(t), y(t))$, are we describing a shape or a motion?

The answer is both --- and learning to separate the two is the key idea.

The Same Path, Different Motions

Before you read on: Two parametric curves are defined as follows.

Curve A: $x(t) = \cos(t)$, $\quad y(t) = \sin(t)$, $\quad 0 \le t \le 2\pi$

Curve B: $x(t) = \cos(3t)$, $\quad y(t) = \sin(3t)$, $\quad 0 \le t \le 2\pi/3$

Both trace a unit circle. Are these the same trajectory? Think carefully about what "the same" means before you continue.

If "trajectory" means the path shape, then yes --- both trace the same circle. But if "trajectory" means the full motion, including when the moving point is where, then no. Curve A completes the circle in $2\pi$ time units. Curve B completes it three times faster, in $2\pi/3$ time units. At $t = 1$, the two points are at different locations on the circle, moving at different speeds.

This is not a trick. It reveals a genuine ambiguity in how we talk about motion, and resolving it is the purpose of parametric descriptions.

What a Parametric Description Gives You

Let's build the idea from the ground up.

In Section 3.1, you worked with the position function $\vec{r}(t)$, which tells you the location of a particle at every instant. You may not have noticed, but that description was already parametric --- time $t$ was the parameter, and as $t$ advanced, the point traced out a curve.

Now we are being explicit about what this buys us.

A parametric description $\vec{r}(t) = (x(t), y(t))$ encodes two distinct pieces of information:

  1. The path --- the geometric curve the point traces out in the $xy$-plane.
  2. The schedule --- how the point moves along that path: where it starts, how fast it goes, where it speeds up or slows down.

These are genuinely independent. You can change the schedule without changing the path, and you can change the path without changing the schedule. The parametric form keeps both in one package.

Exploring Parametric Motion

[Interactive: Parametric Curve Tracer. A slider controls the parameter $t$. As students drag the slider, a dot moves along a curve in the $xy$-plane, leaving a trail. Below the curve, two readouts display the current values of $x(t)$ and $y(t)$. The curve is an ellipse: $x(t) = 3\cos(t)$, $y(t) = 2\sin(t)$. A small velocity arrow is drawn tangent to the path at the dot's current position, with its length proportional to the speed $|\vec{v}|$. The following guided prompts appear one at a time:]

Prompt 1: "Drag the slider slowly from $t = 0$ to $t = 2\pi$. Watch the dot trace the ellipse. Where on the ellipse is the dot moving fastest? Where is it moving slowest?"

Prompt 2: "Look at the velocity arrow. At the top and bottom of the ellipse, is the arrow longer or shorter than at the sides?"

Prompt 3: "Now switch to a different parametrization of the same ellipse (click 'Alternative'). The new parametrization is $x(t) = 3\cos(t^2/6)$, $y(t) = 2\sin(t^2/6)$. Drag the slider again. The ellipse is the same shape. Is the speed pattern the same?"

[After completing the prompts, an annotation appears: "Same shape, different speed. The path is a property of geometry. The speed is a property of the parametrization. The parametric form captures both."]

If you worked through that interactive, you discovered something important. The first parametrization $(3\cos t, 2\sin t)$ moves fastest along the flatter parts of the ellipse (near the $x$-axis) and slowest near the sharper curves (near the $y$-axis). The second parametrization rearranges the speed entirely --- the dot starts slowly and accelerates as $t$ increases, even though the path is identical.

The path did not change. The motion along the path changed completely.

Eliminating the Parameter

Here is the simplest way to see the distinction between path and schedule.

Start with the parametric description of a circle: $x(t) = \cos t$, $y(t) = \sin t$. You can eliminate $t$ by using the identity $\cos^2 t + \sin^2 t = 1$:

$$x^2 + y^2 = 1$$

This is the equation of a unit circle. It tells you the shape of the path --- every point $(x, y)$ on the curve satisfies this equation. But the parameter $t$ is gone, and with it, all the timing information. You no longer know when the point reaches the top of the circle, how fast it moves along the bottom, or in which direction it traverses the curve.

Eliminating the parameter strips the motion down to pure geometry. What remains is the path shape and nothing else.

Pause and think: Given the parametric equations $x(t) = t$, $y(t) = t^2$, eliminate $t$ to find the path equation. What familiar curve is this?

Check your answer Since $x = t$, we can substitute directly: $y = x^2$. The path is a parabola opening upward. The parametrization tells you that the point starts at the origin (when $t = 0$) and moves to the right, speeding up as $t$ increases. The equation $y = x^2$ tells you nothing about that timing.

Why Not Just Use $y = f(x)$?

You might wonder why we bother with parametric descriptions at all. Why not just write $y$ as a function of $x$ and be done with it?

Here is the structural reason: $y = f(x)$ cannot describe every path.

Consider a circle. For any $x$ between $-1$ and $1$, there are two values of $y$ on the unit circle. The upper half has $y = \sqrt{1 - x^2}$ and the lower half has $y = -\sqrt{1 - x^2}$. No single function $f(x)$ gives both. You would need to split the circle into pieces, each described by a separate function, and stitch them together. This is clumsy and gets worse for more complex curves.

A parametric description handles the full circle effortlessly: $x(t) = \cos t$, $y(t) = \sin t$, $0 \le t \le 2\pi$. One pair of functions, one continuous sweep, no stitching.

But there is a deeper reason, too. The function $y = f(x)$ is purely geometric --- it describes a shape. The parametric form $\vec{r}(t)$ describes a motion. In physics, we almost always care about the motion, not just the shape. We want to know velocities, accelerations, and timing. The parametric form gives us all of that. The function $y = f(x)$ gives us none of it.

Historically, parametric curves became essential in the seventeenth century when mathematicians like Isaac Newton and Gottfried Leibniz needed to describe the paths of planets and projectiles --- paths where the timing mattered as much as the shape.

From Parametric Form to Velocity

This is where the parametric description pays off most directly. If you know $\vec{r}(t) = (x(t), y(t))$, then the velocity is:

$$\vec{v}(t) = \frac{d\vec{r}}{dt} = \left(\frac{dx}{dt}, \frac{dy}{dt}\right)$$

Each component of the velocity comes from differentiating the corresponding component of position. This is component independence in action --- the same idea from this chapter's big theme.

And the speed is the magnitude of the velocity vector:

$$|\vec{v}(t)| = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}$$

Let's compute this for the circular motion $x(t) = \cos t$, $y(t) = \sin t$:

$$\vec{v}(t) = (-\sin t, \cos t)$$

$$|\vec{v}(t)| = \sqrt{\sin^2 t + \cos^2 t} = 1$$

The speed is constant. The point moves around the circle at the same rate everywhere. Now compare this with $x(t) = \cos(3t)$, $y(t) = \sin(3t)$:

$$\vec{v}(t) = (-3\sin(3t), 3\cos(3t))$$

$$|\vec{v}(t)| = 3$$

Same circle, three times faster. The parameter controls the schedule; the derivative of the parameter controls the speed.

Before you read on: For the parametric path $x(t) = t$, $y(t) = t^2$, compute the velocity vector and speed at $t = 0$, $t = 1$, and $t = 2$. At which of these times is the point moving fastest?

Check your answer $$\vec{v}(t) = (1, 2t)$$ $$|\vec{v}(t)| = \sqrt{1 + 4t^2}$$ | $t$ | $\vec{v}(t)$ | Speed $|\vec{v}|$ | |:---|:---|:---| | 0 | $(1, 0)$ | 1 | | 1 | $(1, 2)$ | $\sqrt{5} \approx 2.24$ | | 2 | $(1, 4)$ | $\sqrt{17} \approx 4.12$ | The point moves fastest at $t = 2$. The $x$-component of velocity is constant (always 1), but the $y$-component grows linearly with $t$, so the speed increases over time. This makes sense: the parabola $y = x^2$ curves more steeply as $x$ grows, so the point covers more vertical distance per unit of horizontal progress.

Two Parametrizations, Same Path

To cement the distinction between path and schedule, let's construct two different parametrizations of the same curve and compare them side by side.

Path: The upper half of the unit circle, from $(1, 0)$ to $(-1, 0)$.

Parametrization 1 (constant speed):

$$x(t) = \cos t, \quad y(t) = \sin t, \quad 0 \le t \le \pi$$

Speed: $|\vec{v}| = 1$ everywhere.

Parametrization 2 (varying speed):

$$x(s) = 1 - \frac{2s}{\pi}, \quad y(s) = \sqrt{1 - \left(1 - \frac{2s}{\pi}\right)^2}, \quad 0 \le s \le \pi$$

Here the $x$-coordinate decreases linearly from 1 to $-1$. Let's compute the speed. Set $u = 1 - 2s/\pi$, so $du/ds = -2/\pi$. Then:

$$\frac{dx}{ds} = -\frac{2}{\pi}, \quad \frac{dy}{ds} = \frac{-u}{\sqrt{1 - u^2}} \cdot \left(-\frac{2}{\pi}\right) = \frac{2u}{\pi\sqrt{1 - u^2}}$$

The speed is:

$$|\vec{v}| = \frac{2}{\pi}\sqrt{1 + \frac{u^2}{1 - u^2}} = \frac{2}{\pi} \cdot \frac{1}{\sqrt{1 - u^2}}$$

This speed is not constant. When $u \approx 0$ (at the top of the semicircle), the speed is $2/\pi \approx 0.64$. As $u \to \pm 1$ (near the endpoints), the speed grows without bound.

What changed? What stayed the same? The path is identical --- both trace the upper semicircle. The speed profile is entirely different. The first parametrization moves at a steady pace. The second rushes through the endpoints and crawls at the top.

The Concept, Stated Plainly

A parametric description $\vec{r}(t) = (x(t), y(t))$ encodes both the shape of the path and the timing along it.

  • Eliminating $t$ gives the path shape alone --- the geometry.
  • Keeping $t$ gives the full motion --- positions, velocities, accelerations, and all their time dependence.

Different parametrizations of the same path describe different motions along the same curve.

This is why physicists almost always work with parametric descriptions. The path shape alone cannot tell you how fast something moves, where it accelerates, or when it arrives. The parameter $t$ carries all of that information.

Connection to What You Already Know

Every position function $\vec{r}(t)$ from Section 3.1 was already parametric. When you wrote $\vec{r}(t) = (x(t), y(t))$ and differentiated to find velocity and acceleration, you were working with a parametric curve. The parameter was time, and you took it for granted.

What is new here is the awareness that the parametrization is a choice. You could describe the same path with a different parameter and get a different velocity profile. In physics, nature makes the choice for you --- objects move along their paths according to the forces acting on them, and the resulting $\vec{r}(t)$ is the one true parametrization that describes the actual motion. But understanding that other parametrizations exist sharpens your sense of what $\vec{r}(t)$ is actually telling you.

Practice

Layer 1: Concrete

A particle moves with position $\vec{r}(t) = (2t, 3 - t^2)$, where distances are in meters and time is in seconds.

(a) Plot the trajectory by computing $(x, y)$ at $t = 0, 1, 2, 3$.

(b) Compute the velocity $\vec{v}(t)$.

(c) Find the speed at $t = 0$ and $t = 2$.

(d) At what time is the particle at its highest point?

Check your answer **(a)** Evaluating: | $t$ | $x = 2t$ | $y = 3 - t^2$ | |:---|:---|:---| | 0 | 0 | 3 | | 1 | 2 | 2 | | 2 | 4 | $-1$ | | 3 | 6 | $-6$ | The trajectory curves downward --- this looks like a projectile path. **(b)** $\vec{v}(t) = (2, -2t)$ **(c)** Speed: $|\vec{v}(t)| = \sqrt{4 + 4t^2}$ - At $t = 0$: $|\vec{v}| = \sqrt{4} = 2$ m/s - At $t = 2$: $|\vec{v}| = \sqrt{4 + 16} = \sqrt{20} \approx 4.47$ m/s **(d)** The highest point occurs when $y$ is maximized, i.e., when $dy/dt = -2t = 0$, which gives $t = 0$. The particle starts at its highest point and descends from there.

Layer 2: Pattern

Consider the path that is a straight line from $(0, 0)$ to $(4, 3)$.

(a) Write a parametrization that traverses this line at constant speed, reaching $(4, 3)$ at $t = 1$.

(b) Write a different parametrization that traverses the same line but starts slowly and finishes quickly. (Hint: try using $t^2$ instead of $t$.)

(c) Compute the speed at $t = 0.5$ for both parametrizations.

Check your answer **(a)** Constant-speed parametrization: $\vec{r}(t) = (4t, 3t)$, $0 \le t \le 1$. Velocity: $\vec{v} = (4, 3)$. Speed: $|\vec{v}| = 5$, constant throughout. **(b)** Varying-speed parametrization: $\vec{r}(t) = (4t^2, 3t^2)$, $0 \le t \le 1$. Velocity: $\vec{v} = (8t, 6t)$. Speed: $|\vec{v}| = 10t$, which starts at zero and increases linearly. **(c)** At $t = 0.5$: - Parametrization (a): speed $= 5$ m/s - Parametrization (b): speed $= 10 \times 0.5 = 5$ m/s Interestingly, the speeds match at $t = 0.5$ for these particular parametrizations, but the speed *profiles* are very different: one is constant, the other is linearly increasing. Both parametrizations trace the same line from $(0,0)$ to $(4,3)$. The geometry is identical. The kinematics are not.

Layer 3: Structure

Why can't the function $y = f(x)$ describe a complete circle?

Answer in terms of what it means for $f$ to be a function, and explain how the parametric form resolves the problem.

Check your answer A function $f(x)$ assigns exactly *one* output $y$ to each input $x$. But on a circle, most $x$-values correspond to *two* points --- one on the upper half and one on the lower half. For example, at $x = 0$ on the unit circle, both $y = 1$ and $y = -1$ are valid. No single function can produce both values for the same input. You can describe the upper or lower half of the circle separately, but not both at once as $y = f(x)$. The parametric form $x(t) = \cos t$, $y(t) = \sin t$ avoids this problem entirely. Here, $x$ and $y$ are each functions of a separate variable $t$. As $t$ advances, the point sweeps around the full circle. There is no requirement that $y$ be a single-valued function of $x$ --- only that $x(t)$ and $y(t)$ are each single-valued functions of $t$, which they are. More broadly, any path that doubles back on itself --- where the same $x$-value is visited more than once --- cannot be described as $y = f(x)$. Parametric descriptions handle such paths naturally.

Layer 4: Creation

Design a parametric path that traces a figure-eight.

Your parametrization should: - Return to the starting point after one full period. - Cross itself exactly once (at the center of the eight). - Be smooth (no sharp corners).

Write down $x(t)$ and $y(t)$ and verify that the path crosses itself by finding two different values of $t$ that give the same $(x, y)$.

Check your answer One classic figure-eight is the **lemniscate-style** parametrization: $$x(t) = \sin(t), \quad y(t) = \sin(t)\cos(t) = \frac{1}{2}\sin(2t), \quad 0 \le t \le 2\pi$$ Let's verify the self-crossing. The curve passes through the origin $(0, 0)$ when $\sin(t) = 0$, which happens at $t = 0$, $t = \pi$, and $t = 2\pi$. At $t = 0$ and $t = 2\pi$, the point is at the same location (start and end), and at $t = \pi$ it returns to the origin --- this is the crossing point. Checking the shape: for $0 < t < \pi$, $x > 0$ (the right lobe), and for $\pi < t < 2\pi$, $x < 0$ (the left lobe). The $y$-component oscillates up and down within each lobe. Another clean option is $x(t) = \sin(2t)$, $y(t) = \sin(t)$, which produces a similar figure-eight but oriented vertically. There is no single "correct" answer --- any smooth parametrization that produces a self-crossing closed curve shaped like an eight is valid. The point of the exercise is that parametric descriptions can represent paths that would be impossible to write as $y = f(x)$.

Reflection

What information do you lose when you eliminate the parameter $t$ from a parametric description?

Think beyond just "the timing." Consider: if you are given only the path equation $y = f(x)$ (or an implicit equation like $x^2 + y^2 = 1$), what questions about the motion can you no longer answer?

Try listing at least three specific things that $\vec{r}(t)$ tells you but the path equation does not.

You might note that the path equation cannot tell you the velocity at any point, the speed at any point, which direction the particle traverses the curve, where the particle starts, when it arrives at a given location, or whether it speeds up or slows down along the way. All of this requires the parameter.

Looking Ahead

You now understand that a parametric description $\vec{r}(t)$ carries more information than a path equation, and you know how to extract velocity and speed from it. You also know that the same path can be traversed with very different timing.

In the next section, we put parametric thinking to its most famous use. Projectile motion is a parametric curve: $x(t)$ describes the horizontal motion, $y(t)$ describes the vertical motion, and the two components are independent. The parabolic trajectory that a projectile follows is just what you get when you eliminate $t$ --- the pure geometry. But the full parametric form is where all the physics lives: the timing, the velocity, the peak, the landing. You already have every tool you need. Section 3.4 is about putting them together.