Parametric Equations — Print

Let a third variable drive both coordinates, and curves that were impossible become easy.

What the practice questions cover

These questions use 7 ideas from the course. Follow any of them for a definition, where it is taught, and what it leads to.

Throw a ball and photograph its flight.

The path is a neat arc, and y = f(x) describes it perfectly well.

the path does not say when t=0 t=1 t=2 t=3 t=4 same curve, but the dots say where it was at each second

But the photograph has lost the timing. Was the ball fast at the top, or slow? The curve cannot say.

Worse, a curve that loops back — a circle, a figure of eight — fails the vertical line test entirely.

So let time drive both coordinates separately: x = f(t) and y = g(t). Now the path carries its schedule, and loops are no trouble at all.

A circle traced by a parameter t one parameter, two coordinates x = r cos t y = r sin t t runs from 0 to 2π no y = f(x) can do this

Building a table

Choose values of t, work out x and y for each, and plot the points in order.

The order is the extra information: it tells you the direction of travel, which a plain graph cannot.

Mark a few points with their t values and the curve carries its own schedule.

The arrow matters

Two parametrisations can trace the same curve in opposite directions. Same picture, different motion, and physics problems care which.

Eliminating the parameter

To recover a plain x–y equation, solve one equation for t and substitute into the other.

From x = t+1 and y = t2: t = x-1, so y = (x-1)2 — an ordinary parabola.

You lose the timing and the direction, but gain a familiar shape.

Watch the domain

If t ≥ 0 then only part of the curve is traced. The eliminated equation usually describes more than the parametric one does, so state the restriction.

When trigonometry is the parameter

x = rcos t, y = rsin t traces a circle of radius r, because cos2 t + sin2 t = 1.

That identity is how you eliminate t: square both, add, and the parameter cancels itself.

Change the two radii and you get x2a2 + y2b2 = 1, an ellipse.

Square and add

Whenever you see cos t in one equation and sin t in the other, the Pythagorean identity is the tool for removing t.

Projectile motion

Horizontally nothing pushes the ball, so x = v0cosθ · t — steady speed, straight line.

Vertically gravity pulls, so y = v0sinθ · t - 16t2 in feet, or -4.9t2 in metres.

The two behave completely differently, and keeping them apart is exactly what parametric form allows.

Landing time

Set y = 0 and solve for t. The positive root is when it lands, and substituting that into x gives the range.

Curves no function can draw

A circle fails the vertical line test, so no y = f(x) can describe it. Parametrically it takes one line.

The same is true of loops, spirals and anything that crosses its own path.

That is the real reason this form exists: it removes the requirement that each x have only one y.

One input, two outputs

The parameter is the input, and it returns a pair. Nothing about that breaks the definition of a function.

Worked examples

Eliminating the parameter

Eliminate t from x = 2t, y = t2 + 1.

  1. Solve the simpler one: t = x2. Pick whichever is easier to rearrange.
  2. Substitute: y = (x2)2 + 1. Replace every t.
  3. y = x24 + 1. A parabola with vertex (0,1).

y = x24+1

A circle

What curve is traced by x = 3cos t, y = 3sin t?

  1. Square both: x2 = 9cos2 t and y2 = 9sin2 t. Setting up the identity.
  2. Add: x2+y2 = 9(cos2 t + sin2 t) = 9. The identity removes t.
  3. That is a circle of radius 3 centred at the origin. Traced anticlockwise as t increases.

A circle of radius 3

A projectile

A ball is thrown at 40 ft/s at 30°. Using y = 20t - 16t2, when does it land?

  1. It lands when y = 0: 20t - 16t2 = 0. Ground level.
  2. Factor: 4t(5 - 4t) = 0, so t = 0 or t = 1.25. t=0 is the throw itself.
  3. So it lands after 1.25 seconds. The positive root is the one that means anything.

1.25 seconds

Words to know

Parameter
The extra variable, usually t, that drives both coordinates.
Parametric equations
A pair x = f(t), y = g(t) describing a curve.
Eliminating the parameter
Combining the two equations to recover a relation in x and y alone.
Orientation
The direction the curve is traced as t increases.
Plane curve
The set of points traced out, with the timing stripped away.

Mistakes to watch for

Losing the domain when eliminating t. If t ≥ 0 then x = 2t only reaches x ≥ 0, so y = x24+1 describes twice as much curve as the original. Say what x is restricted to.

Ignoring the direction of travel. x=cos t and x=cos(-t) trace the same circle opposite ways. The picture is identical and the motion is not.

Trying to eliminate t by dividing the two equations. From x=2t and y=t2, dividing gives yx=t2, which still has t in it. Solve one equation for t and substitute.

Forgetting the identity route with trig. Squaring and adding removes t in one step, while solving x = 3cos t for t drags an arccosine through the whole problem.

Taking t=0 as the landing time. It is the moment of release. The other root of the quadratic is the answer.

Every animation and every CNC machine runs on parametric equations, because a machine needs to know where the tool is at each moment, not merely which line it traces.

A cycloid — the path of a point on a rolling wheel — has a horrible implicit equation and a beautiful parametric one. It was called the Helen of geometers, for all the quarrels it caused.

Bézier curves, which draw every font on your screen, are parametric and cubic.

Problem 1

For x=t+1, y=2t, find x when t=3.

Worked answer

4

3+1 = 4

Problem 2

For the same equations, find y when t=3.

Worked answer

6

2(3) = 6

Problem 3

Eliminate t from x=t, y=t2. Give y in terms of x.

Worked answer

x^2

t = x, so y = x2

Problem 4

Eliminate t from x=t-2, y=t. Give y in terms of x.

Worked answer

x+2

t = x+2, so y = x+2

Problem 5

Why can a circle not be written as y=f(x)?

  1. It fails the vertical line test
  2. It has no equation
  3. It is not continuous
  4. It has no slope
Worked answer

It fails the vertical line test

Two, for most x on a circle

Problem 6

Eliminate t from x=2t, y=t2+1. Give y in terms of x.

Worked answer

x^2/4+1

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Problem 7

What curve is x=5cos t, y=5sin t?

  1. A circle of radius 5
  2. An ellipse
  3. A line
  4. A parabola
Worked answer

A circle of radius 5

x2+y2 = 25(cos2 t + sin2 t) = 25

Problem 8

For x=6cos t, y=6sin t, find x2+y2.

Worked answer

36

36(cos2 t+sin2 t) = 36

Problem 9

For x=3cos t, y=3sin t, find x when t=0.

Worked answer

3

3(1) = 3

Problem 10

A ball has y = 30t - 16t2. At what t does it return to the ground? Give the positive value.

Worked answer

1.875

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Problem 11

With x = 20t, how far has that ball travelled horizontally at t=1.5?

Worked answer

30

20(1.5) = 30

Problem 12

Eliminate t from x=t2, y=t4. Give y in terms of x.

Worked answer

x^2

y = (t2)2 = x2

Note x ≥ 0, since it is a square

Problem 13

For x=4cos t, y=3sin t, find the largest value x can take.

Worked answer

4

4(1) = 4

Problem 14

What curve is x=4cos t, y=3sin t?

  1. An ellipse
  2. A circle
  3. A hyperbola
  4. A line
Worked answer

An ellipse

x216+y29 = cos2 t + sin2 t = 1

Problem 15

For x=t2 with t any real number, what is the smallest value x can take?

Worked answer

0

t2 ≥ 0

Problem 16

A ball has y = 48t - 16t2. Find the time at which it reaches its highest point.

Worked answer

1.5

Roots at t=0 and t=3

Halfway is 1.5

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