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Parametric Curves

Calculus 2 · Chapter 9: Parametric Equations & Polar CoordinatesSection 9.1 · lesson 45 of 549 min · free

← Applications of SeriesNext lesson: Calculus with Parametric Curves →

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In this lesson

When both coordinates are driven by a third variable.

  • Describe motion with x(t) and y(t)
  • Build a parameter table and orient the curve
  • Eliminate the parameter by substitution or identity
  • Carry the parameter restriction into the Cartesian form

Transcript

Everything said in the video, section by section (1,350 words). Read along, or search it.

MAT-265 - Calculus 2 · Section 9.1

Parametric Curves

A ball is kicked at eighty feet per second, at thirty degrees above the horizontal. Where is it? And when? A single equation relating y to x can give you the shape of the path. But it says nothing about timing. It can't tell you when the ball reaches its peak, or how long it stays airborne. And there's a deeper limitation. A function assigns exactly one y to each x. So no function of x can describe a path that doubles back on itself. Or closes into a loop. The resolution is to introduce a third variable. A parameter, usually called t. And let it drive both coordinates independently.

The Idea

A Function of x Cannot Describe This

So a parametric curve is given by two equations. x equals f of t, and y equals g of t. With t running over some interval. As t advances, the point x comma y traces a path in the plane. Let's work the kicked ball. First resolve the launch velocity into components. The speed is eighty, at thirty degrees. Horizontal: eighty cosine thirty. Which is forty root three. About sixty-nine point two eight feet per second. Vertical: eighty sine thirty. Which is forty feet per second. Now the horizontal position. Neglecting air resistance, no horizontal force acts. So the horizontal motion has constant velocity, and position is velocity times time. x equals forty root three, t. And the vertical position. Gravity produces a constant downward acceleration of thirty-two feet per second squared. Constant-acceleration motion from height zero gives y equals forty t, minus sixteen t squared. Now, what's the parameter interval? The ball is airborne while y is positive. Set y equal to zero and factor. Eight t, times five minus two t, equals zero. So t equals zero, or t equals two point five. The kick happens at t equals zero, landing at two point five seconds. That's the interval. And now extract the apex. The vertical component is a downward-opening parabola in t. Its maximum is at the vertex, t equals minus b over two a. With a equal to minus sixteen and b equal to forty. That gives t equals one point two five seconds. Substitute back: y at one point two five is fifty minus twenty-five. Twenty-five feet. So the ball peaks at twenty-five feet, one and a quarter seconds after the kick. And the horizontal range follows from x at t equals two point five. One hundred root three. About one hundred seventy-three feet.

Figure 9.1.1

Timing Is Part of the Answer

Look at the path. It's a parabola, as you'd expect. But notice what the parametric form gave us that the shape alone never could. The time to apex. The total flight duration. The moment of landing. Every one of those facts comes from the parameter. None of them is directly visible in a plain equation relating y to x. That's the central advantage. Parametric equations carry timing, not just geometry.

Sketching

Table, Interpret, Then Plot

To sketch a parametric curve by hand, build a table. Take x equals t squared minus two t, and y equals t plus one, on t from minus two to three. For each value of t, compute both coordinates. And record the point. At t equals minus two: x is four plus four, which is eight. And y is minus one. So the point is eight comma minus one. At t equals minus one: x is one plus two, three. y is zero. At t equals zero: x is zero, y is one. At t equals one: x is one minus two, which is minus one. And y is two. At t equals two: x is zero, y is three. At t equals three: x is three, y is four. Now here's the important step. Interpret the pattern before you plot. Look at the y column. Minus one, zero, one, two, three, four. Rising by exactly one each step. So the vertical motion is steady and upward, throughout. Now the x column. Eight, three, zero, minus one, zero, three. It falls, then rises. Reaching a minimum of minus one at t equals one. So the curve opens to the right, with its leftmost point at minus one comma two. You know the shape before drawing a single point. Then plot, and join smoothly in the order listed. And add arrowheads. Those arrowheads record that the particle enters at eight comma minus one, sweeps left and upward, turns at minus one comma two, and exits at three comma four. That orientation is information a Cartesian equation simply discards.

Figure 9.1.2

Two Curves, Read With Their Direction

Here are two parametric curves, drawn with their direction. On the left, the parabola from the table. Traced from the bottom right, around the vertex at minus one comma two, and up to the top. The arrows show the order of travel, which the shape alone would never reveal. On the right, x equals three cosine t, y equals three sine t. As t runs from zero to two pi, that point traverses a circle of radius three. Counter-clockwise. Exactly once. And this is a closed path. No single function of x could describe it — because a vertical line through the middle crosses it twice. That's precisely the limitation parametric equations were introduced to overcome.

Eliminating the Parameter

Two Routes to a Cartesian Equation

Sometimes you want the Cartesian equation — the direct relationship between x and y, with t removed. There are two standard routes. First, elimination by substitution. Use it when one equation is easy to solve for t. Take the parabola. y equals t plus one is linear in t. So t equals y minus one. Substitute into the other equation. x equals t squared minus two t becomes y minus one, squared, minus two times y minus one. Expand fully. y minus one squared is y squared minus two y plus one. And minus two times y minus one is minus two y plus two. Combining: x equals y squared minus four y plus three. Complete the square to identify it. x equals y minus two, squared, minus one. A parabola opening to the right, with vertex at minus one comma two. Matching the gold point in the figure exactly. Second route. Elimination by identity. Use it when t sits inside trigonometric functions. For x equals three cosine t and y equals three sine t, isolate the trig functions by dividing by three. Cosine t is x over three. Sine t is y over three. Now apply the Pythagorean identity. Cosine squared plus sine squared equals one. So x over three squared, plus y over three squared, equals one. Simplify: x squared over nine, plus y squared over nine, equals one. Multiply by nine. x squared plus y squared equals nine. A circle of radius three, centered at the origin. Now — one critical final step that students skip constantly. Restore the restriction. The Cartesian equation for the parabola describes the entire infinite parabola. Extending forever in both directions. But the parameter runs only over minus two to three. Since y equals t plus one, adding one throughout gives y between minus one and four. So the complete answer is x equals y minus two squared minus one, with y between minus one and four. Without that restriction, the Cartesian form describes a curve extending forever — which is not what the parametric equations trace.

Key Takeaways

What to Carry Forward

Five takeaways. Parametric equations let a third variable drive both coordinates. Which lets them describe paths that double back, or close into loops — things no function y equals f of x can do. They carry timing and orientation, not just shape. The apex, the duration and the range all come from the parameter. Sketch by tabulating x and y against t. Interpret the pattern of each column before plotting. Then mark the direction with arrowheads. Eliminate the parameter by substitution when one equation is easy to solve for t. Or by a Pythagorean identity when t sits inside trig functions. And always carry the parameter interval across into a restriction on x or y. The Cartesian equation alone describes too much. Next up is section nine point two — Calculus with Parametric Curves. Where we differentiate without eliminating the parameter at all.

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FAQ

Questions students ask us

About Parametric Curves

What is a parametric curve?
A parametric curve is a path given by two equations, x = f(t) and y = g(t), where a third variable t runs over an interval and drives both coordinates. As t advances, the point (x, y) traces the path in order. Unlike y = f(x), parametric equations can describe loops and paths that double back, such as the circle x = 3\cos t, y = 3\sin t. They also record timing and direction.
How do you eliminate the parameter from parametric equations?
Eliminate the parameter by substitution when one equation is easy to solve for t, or by a trigonometric identity when t sits inside sine and cosine. For x = t^2 - 2t, y = t + 1, solve t = y - 1 and substitute to get x = (y-2)^2 - 1. For x = 3\cos t, y = 3\sin t, use \cos^2 t + \sin^2 t = 1 to get x^2 + y^2 = 9.
How do you find the maximum height and range of a projectile with parametric equations?
Find the maximum height by evaluating y at the vertex time t = -b/(2a), and find the range by solving y = 0 for the landing time and evaluating x there. For a ball kicked at 80 ft/s at 30^\circ, x = 40\sqrt{3}\,t and y = 40t - 16t^2. The vertex time is t = 1.25, giving a height of 25 ft. Landing occurs at t = 2.5, so the range is 100\sqrt{3} \approx 173 ft.
What is the most common mistake when converting parametric equations to Cartesian form?
The most common mistake is dropping the parameter restriction, so the Cartesian equation describes more than the parametric curve actually traces. For x = t^2 - 2t, y = t + 1 with -2 \le t \le 3, the equation x = (y-2)^2 - 1 alone describes an infinite parabola. Since y = t + 1, the complete answer must include -1 \le y \le 4. The Cartesian form also loses orientation, which arrowheads on a sketch preserve.

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