Describes a curve by giving both coordinates as functions of a third variable, which adds direction, timing and repetition that an equation cannot record. Eliminates the parameter to identify a shape, preserves the domain restriction that elimination would otherwise lose, and parametrises projectile motion.
Subject: Precalculus · 65 slides · symbolic lesson
Open the interactive version of this deck
Title
Precalculus · Chapter 8 — Further Applications of Trigonometry
§8.6 Parametric Equations, pp. 992-1007
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §8.6 Parametric Equations §8.6, pp. 992-1007 — the pages these objectives are drawn from
Warm-up
A circle's equation says which points are on it. That is all it says.
Discussion prompt
Two runners go round the same circular track, one clockwise and one anticlockwise. How do their paths differ as equations?
Hint: What does the equation record?
Answer:
Their equations are identical. Both trace exactly the same set of points, so the same circle equation describes both.
But the runs are obviously different — direction, starting point and speed all differ, and none of that appears in the equation.
So an equation describes a shape and not a journey. Parametric equations record the journey, which is why they are needed whenever motion rather than shape is the subject.
Concept
A parametrisation gives each coordinate as a function of a parameter, so the pair describes where a moving point is at each value of that parameter.
parameter — an independent variable, often representing time, that both coordinates are expressed in terms of
\[ x=f(t), \quad y=g(t) \]
The parameter is usually time, but need not be. In §8.3's polar curves it was an angle, and in geometry it is often a length — what matters is that one variable drives both coordinates.
Figure (svg): A curve traced by a moving point, with arrows showing the direction of travel and marked positions at successive parameter values
OpenStax, Precalculus, §8.6 Parametric Equations §8.6, pp. 1003-1008
Section
Section 1
Concept
Each parameter value produces one point. Tabulating several and joining them in order gives both the curve and the direction of travel.
Joining the points in parameter order rather than left to right is the whole difference from ordinary plotting. A parametrised curve may double back or cross itself, and only the parameter order reveals that.
Figure (svg): A curve traced by a moving point, with arrows showing the direction of travel and marked positions at successive parameter values
OpenStax, Precalculus, §8.6 Parametric Equations §8.6, pp. 1003-1009
Picture it
Marked positions at three parameter values.
Figure (svg): A curve traced by a moving point, with arrows showing the direction of travel and marked positions at successive parameter values
The marked points show the schedule: at zero the point is at the right, at a quarter turn at the top. Reversing the sine's sign would trace the same ellipse the other way round.
Worked example
Five parameter values are usually enough.
\[ \text{Plot } x=t^2, \; y=t+1 \text{ for } -2\le t\le 2. \]
Tabulate at five values
Why: Both coordinates each time.
\[ t = -2, -1, 0, 1, 2 \]
Compute the pairs
Why: Square and add one.
\[ (4, -1), (1, 0), (0, 1), (1, 2), (4, 3) \]
Plot in parameter order
Why: Not left to right.
Add the direction
Why: As t increases.
Figure (svg): A curve traced by a moving point, with arrows showing the direction of travel and marked positions at successive parameter values
\[ \text{opens rightward, vertex }(0,1) \]
Verify: check the doubling back
Why: The horizontal coordinate falls to zero and then rises again, so the curve doubles back on itself in that coordinate. Plotting left to right would have joined the points in the wrong order and hidden that entirely.
OpenStax, Precalculus, §8.6 Parametric Equations §8.6, pp. 1004-1007
Prediction
A parametrisation uses the cosine for the horizontal and the negative sine for the vertical.
Predict first
Which way does the point travel?
Correct: Clockwise.
Why: At the start the point is at the right, and just after, the vertical coordinate is negative — so it moves downward first, which is clockwise. Negating the sine reverses the direction without changing the circle.
Worked example
Two parametrisations of the same circle.
\[ \text{Compare } (\cos t,\sin t) \text{ with } (\cos t,-\sin t). \]
Check the shape
Why: Both satisfy the unit circle equation.
Check the start
Why: Both begin at the right.
Check a later value
Why: At a quarter turn.
Conclude
Why: Opposite directions.
Figure (svg): The solution to Worked example read the direction shown as a ladder of expressions, one row per legal move
\[ \text{anticlockwise versus clockwise} \]
Verify: confirm with the equation
Why: Eliminating the parameter from either gives the same circle equation, since squaring destroys the sign. So the equation genuinely cannot distinguish them — which is precisely the information the parametrisation adds.
OpenStax, Precalculus, §8.6 Parametric Equations §8.6, pp. 1007-1009
Trap
\[ \text{plot the pairs, then join them from left to right} \]
Order the plotted points by their horizontal coordinate
Why: The points are joined as though the curve were a function graph.
A curve that doubles back is drawn as a single sweep, in the wrong shape.
Join the points in parameter order, which is the order the moving point visits them.
A parametrised curve may double back, loop, or cross itself, none of which a left-to-right sweep can represent.
Number the plotted points as you go. The numbering is the parameter order, and joining them in sequence is then automatic.
Faded example
At the parameter value of 2.
Fill in the blanks
x=t^2=4, \quad y=t+1=3 \quad(t=2)
Why: Each parameter value is substituted into both equations to produce one point. Building a small table this way is the reliable first step before any plotting.
Sorting
Some information survives and some does not.
Sort into buckets
Sort each feature.
Step zero
You are given a parametric pair to graph.
Discussion prompt
What do you do first?
Hint: What produces a point?
Answer:
Build a table of parameter values with both coordinates computed at each. One parameter value gives one point.
Five or six values across the interval usually suffice, and including the endpoints matters when the interval is bounded.
Then plot and join in parameter order, numbering the points as you go. The numbering is what prevents the left-to-right error, which is the one mistake that changes the shape rather than just the annotation.
Section
Section 2
Concept
To identify the shape, solve whichever equation is easier for the parameter and substitute into the other, leaving a single relation between the coordinates.
When both coordinates are trigonometric, solving for the parameter would require inverse functions and a quadrant discussion. Squaring both and adding is far cleaner, and it is the standard route for circles and ellipses.
Figure (svg): A diagram showing the parameter being eliminated from two equations to leave a single relation between the coordinates
OpenStax, Precalculus, §8.6 Parametric Equations §8.6, pp. 1009-1013
Picture it
One substitution collapses two equations into one.
Figure (svg): A diagram showing the parameter being eliminated from two equations to leave a single relation between the coordinates
The red line at the bottom is the caveat. The shape is recovered exactly and everything about the journey is gone.
Worked example
One equation is linear, so solve that one.
\[ \text{Eliminate the parameter from } x=t+1, \; y=t^2. \]
Choose the simpler equation
Why: The linear one.
\[ x = t + 1 \]
Solve for the parameter
Why: Subtract one.
\[ t = x - 1 \]
Substitute
Why: Into the other.
\[ y = (x - 1) ^{2} \]
Identify
Why: A shifted parabola.
\[ \text{vertex at } (1, 0) \]
Figure (svg): A diagram showing the parameter being eliminated from two equations to leave a single relation between the coordinates
\[ y=(x-1)^2 \]
Verify: check one point
Why: At a parameter value of 2 the point is (3, 4), and substituting 3 into the equation gives 4 — matching. The equation is correct, though it says nothing about the point having been at (3, 4) at that particular moment.
OpenStax, Precalculus, §8.6 Parametric Equations §8.6, pp. 1010-1012
Prediction
Both coordinates are given by sines and cosines.
Predict first
How should you eliminate the parameter?
Correct: Isolate, square and add, using a Pythagorean identity.
Why: The identity handles both equations at once with no inverse function and no range restriction. Using an inverse would lose the half of the curve outside its range and produce an awkward composition.
Worked example
Both coordinates are trigonometric.
\[ \text{Eliminate the parameter from } x=3\cos t, \; y=2\sin t. \]
Isolate each function
Why: Divide by the coefficients.
\[ \cos = \frac{x}{3}, \sin = \frac{y}{2} \]
Square both
Why: Ready for the identity.
\[ x ^{2} / 9\text{ and } y ^{2} / 4 \]
Add them
Why: The Pythagorean identity.
\[ \sum\text{ is } 1 \]
Identify
Why: An ellipse.
\[ \text{semi-axes } 3\text{ and } 2 \]
Figure (svg): The solution to Worked example eliminate with an identity shown as a ladder of expressions, one row per legal move
\[ \frac{x^2}{9}+\frac{y^2}{4}=1 \]
Verify: check the extremes
Why: The horizontal coordinate ranges over three either side and the vertical over two, matching the semi-axes read off the equation. Solving for the parameter with an inverse cosine would have worked but required a quadrant discussion the identity avoided entirely.
OpenStax, Precalculus, §8.6 Parametric Equations §8.6, pp. 1012-1013
Error analysis
A student eliminates from a trigonometric parametrisation.
Annotate
On: \( x=3\cos t \;\Longrightarrow\; t=\arccos\tfrac{x}{3}, \; y=2\sin\!\left(\arccos\tfrac{x}{3}\right) \)
Whenever both coordinates are trigonometric, the identity route is both shorter and complete. The inverse route silently discards the half of the curve outside the inverse function's range.
Faded example
From a linear and a cubic equation.
Fill in the blanks
x=2t \;\Longrightarrow\; t=\frac28}, \; y=t^3=\frac______}
Why: Solving the linear equation and substituting gives a cubic in the coordinate, with the coefficient cubed in the denominator. Choosing the linear equation to solve is what keeps the algebra simple.
Sorting
The form of the equations decides.
Sort into buckets
Sort each pair.
Explain it to yourself
Elimination produces a familiar equation.
Discussion prompt
Explain what is lost in the process.
Hint: What could two different parametrisations share?
Answer:
The equation records which points are on the curve and nothing else. The direction of travel, the speed, and the number of times the curve is retraced all disappear.
Two parametrisations tracing the same circle in opposite directions eliminate to the identical equation, so the equation genuinely cannot tell them apart.
So elimination is the right move when the shape is the question and the wrong move when the motion is. A good understanding here is knowing which question is being asked before deciding to eliminate.
Section
Section 3
Concept
If the parameter runs over a restricted interval, or if a coordinate function has a limited range, the curve is only part of what the eliminated equation describes.
The second point is the one that catches people. A parametrisation using a square root produces only non-negative horizontal coordinates, so even with an unbounded parameter the curve is half of what the eliminated equation shows.
Figure (svg): A contrast between what a parametrisation records and what survives elimination of the parameter
OpenStax, Precalculus, §8.6 Parametric Equations §8.6, pp. 1013-1016
Picture it
The right column is what an equation can hold.
Figure (svg): A contrast between what a parametrisation records and what survives elimination of the parameter
The domain is the one item in the left column that can be preserved — but only by writing it down explicitly alongside the equation.
Worked example
The interval bounds the arc.
\[ \text{Eliminate from } x=t, \; y=t^2 \text{ for } 0\le t\le 3. \]
Eliminate
Why: The first equation is trivial.
\[ y = x ^{2} \]
Find the horizontal range
Why: From the parameter's interval.
\[ 0\text{ to } 3 \]
State both
Why: Equation and restriction.
\[ y = x ^{2}, 0 \le x \le 3 \]
Describe
Why: Only the right half, to a point.
Figure (svg): A contrast between what a parametrisation records and what survives elimination of the parameter
\[ y=x^2,\; 0\le x\le 3 \]
Verify: check the endpoints
Why: At the parameter values zero and three the points are (0,0) and (9 above 3) — the arc runs from the vertex up to (3, 9). Without the restriction the equation would include the entire left branch, which the parametrisation never visits.
OpenStax, Precalculus, §8.6 Parametric Equations §8.6, pp. 1014-1015
Prediction
The horizontal coordinate is the square root of the parameter.
Predict first
What restriction does that impose?
Correct: The horizontal coordinate is non-negative.
Why: A square root never returns a negative value, so no parameter value produces a point to the left of the vertical axis. The eliminated equation would show a full curve, and half of it is never traced.
Worked example
The parameter is unbounded but a coordinate is not.
\[ \text{Eliminate from } x=\sqrt{t}, \; y=t-1 \text{ for } t\ge 0. \]
Solve the first for the parameter
Why: Square it.
\[ t = x ^{2} \]
Substitute
Why: Into the second.
\[ y = x ^{2} - 1 \]
Find the horizontal range
Why: A square root is non-negative.
\[ x \ge 0 \]
State both
Why: Equation and restriction.
Figure (svg): The solution to Worked example a restriction from the function itself shown as a ladder of expressions, one row per legal move
\[ y=x^2-1,\; x\ge 0 \]
Verify: check why the left half is absent
Why: The horizontal coordinate is a square root, which is never negative — so no parameter value produces a point on the left. The eliminated equation would suggest a full parabola, and stating the restriction is what corrects it.
OpenStax, Precalculus, §8.6 Parametric Equations §8.6, pp. 1015-1016
Trap
\[ x=\sqrt{t},\; y=t-1 \;\Longrightarrow\; y=x^2-1 \]
Report the eliminated equation alone
Why: The restriction implied by the square root is not carried through.
The answer describes a full parabola where the curve is only its right half.
State the restriction with the equation. Here the horizontal coordinate is non-negative, so only the right half is traced.
The restriction comes from two sources: the parameter's interval, and the ranges of the coordinate functions themselves.
Check both sources every time. The second is the one that is missed, since it is not written anywhere in the problem.
Faded example
The parameter runs from 0 to 4 with the horizontal coordinate equal to it.
Fill in the blanks
y=x^2 \text0 4\le x\le___
Why: The horizontal coordinate equals the parameter, so its range is the parameter's interval directly. Stating it alongside the equation is what distinguishes the arc from the whole parabola.
Sorting
Two sources, and one is easier to miss.
Sort into buckets
Sort each restriction.
Explain it
A restriction can come from more than one place.
Discussion prompt
Explain to a classmate what to check after eliminating.
Hint: One source is written down and one is not.
Answer:
First, the parameter's interval if one was given. That is written in the problem and translates directly into a range for the coordinates.
Second, the ranges of the coordinate functions themselves. A square root gives only non-negative values; a cosine gives only values between negative one and one.
The second is the one that is missed, because nothing in the problem statement mentions it. A good explanation suggests the habit of asking what each coordinate function can actually produce, which takes a second and catches every case.
Section
Section 4
Concept
A projectile's horizontal and vertical positions each follow a simple law of their own, and writing them as a parametric pair produces the parabolic path directly.
The independence of the two motions is the physical insight the parametrisation makes visible. A dropped ball and one thrown horizontally from the same height hit the ground at the same moment, which the vertical equation says directly and the parabola's equation does not say at all.
Figure (svg): A projectile's parabolic path, with the horizontal and vertical motions shown as separate functions of time
OpenStax, Precalculus, §8.6 Parametric Equations §8.6, pp. 1016-1018
Picture it
Two equations, each with its own physical meaning.
Figure (svg): A projectile's parabolic path, with the horizontal and vertical motions shown as separate functions of time
The horizontal equation has no acceleration term because nothing pushes the projectile horizontally. The vertical one has gravity, and that asymmetry is what makes the path a parabola.
Worked example
Resolve the initial speed into components.
\[ \text{A ball is thrown at } 20 \text{ m/s at } 40^\circ. \text{ Write its path parametrically.} \]
Resolve the horizontal component
Why: Speed times the cosine.
\[ 20 \cos 40 = 15.3 \]
Write the horizontal equation
Why: Constant speed.
\[ x = 15.3 t \]
Resolve the vertical component
Why: Speed times the sine.
\[ 20 \sin 40 = 12.9 \]
Write the vertical equation
Why: With the gravity term.
\[ y = 12.9 t - 4.9 t ^{2} \]
Figure (svg): A projectile's parabolic path, with the horizontal and vertical motions shown as separate functions of time
\[ x=15.3t,\; y=12.9t-4.9t^2 \]
Verify: check the initial speed
Why: The two components combine to a speed of the root of 15.3 squared plus 12.9 squared, which is 20 — recovering the stated initial speed. That check confirms both components were resolved with the right function.
OpenStax, Precalculus, §8.6 Parametric Equations §8.6, pp. 1016-1017
Prediction
A projectile with no air resistance.
Predict first
Which coordinate's equation is quadratic in time?
Correct: The vertical, because of gravity.
Why: Nothing pushes the projectile horizontally, so that motion is at constant speed and linear in time. Gravity acts downward only, producing the squared term in the vertical equation and making the path a parabola.
Worked example
Only the parametric form can.
\[ \text{For that ball, when does it land, and how far away?} \]
Set the vertical position to zero
Why: Landing height.
\[ 12.9 t - 4.9 t ^{2} = 0 \]
Factor
Why: Time is a common factor.
\[ t(12.9 - 4.9 t) = 0 \]
Solve
Why: Discard the launch instant.
\[ t = 2.63 s \]
Find the distance
Why: Substitute into the horizontal.
\[ 15.3 \times 2.63 \]
Figure (svg): The solution to Worked example answer a timing question shown as a ladder of expressions, one row per legal move
\[ t\approx 2.63\text{ s},\; x\approx 40.3\text{ m} \]
Verify: check the two roots
Why: The factored equation gave zero and 2.63 seconds, and zero is the launch moment when the ball was also at ground level. Discarding it on physical grounds is correct, and the surviving root is the landing. The eliminated parabola would have given the distance but never the time.
OpenStax, Precalculus, §8.6 Parametric Equations §8.6, pp. 1017-1018
Error analysis
A student eliminates the parameter and then tries to find a landing time.
Annotate
On: \( y=x\tan 40^\circ-\frac{4.9x^2}{(20\cos 40^\circ)^2}, \text{ set } y=0 \text{ to find } t \)
Elimination discards exactly the variable this question was about. Choosing whether to eliminate should depend on whether the question is about the shape of the path or about the timing along it.
Faded example
A speed of 30 at an angle of 60 degrees.
Fill in the blanks
v_x=30\cos 60^\circ=15, \quad v_y=30\sin 60^\circ=15\sqrt___
Why: The cosine gives the horizontal component and the sine the vertical one, which is §5.4's right-triangle resolution. Checking that the two components recombine to 30 confirms the resolution.
Sorting
Some questions need time and some do not.
Sort into buckets
Sort each question.
Explain it to yourself
The two motions are independent.
Discussion prompt
Explain what that means and one consequence of it.
Hint: What does each equation depend on?
Answer:
Each coordinate's equation involves only time, not the other coordinate. So the horizontal motion is unaffected by what the vertical is doing and vice versa.
One consequence: a ball dropped and a ball thrown horizontally from the same height hit the ground at the same moment, since their vertical equations are identical.
That is a genuinely surprising physical fact, and the parametrisation makes it obvious where the parabola's equation makes it invisible. A good explanation notes that this is the main reason physics uses the parametric form rather than the path equation.
Section
Section 5
Concept
Any curve can be parametrised in infinitely many ways, differing in direction, speed and starting point. Choosing one is a modelling decision.
Setting the parameter equal to the horizontal coordinate always works for a function graph and gives the obvious left-to-right traversal. It is the default worth knowing, though it is rarely the most useful choice for a curve describing motion.
Figure (svg): A contrast between what a parametrisation records and what survives elimination of the parameter
OpenStax, Precalculus, §8.6 Parametric Equations §8.6, pp. 1005-1018
Picture it
Each item in the left column is a free choice.
Figure (svg): A contrast between what a parametrisation records and what survives elimination of the parameter
Because those choices are free, infinitely many parametrisations describe one curve. Which to pick depends on what the parameter is meant to represent.
Worked example
The parameter runs from zero to one.
\[ \text{Parametrise the segment from } (1,2) \text{ to } (5,10). \]
Set the start at parameter zero
Why: The first point.
\[ (1, 2)\text{ at } t = 0 \]
Set the end at parameter one
Why: The second point.
\[ (5, 10)\text{ at } t = 1 \]
Write each coordinate
Why: Start plus the change times t.
\[ x = 1 + 4 t, y = 2 + 8 t \]
State the interval
Why: Zero to one.
\[ 0 \le t \le 1 \]
Figure (svg): The solution to Worked example parametrise a line segment shown as a ladder of expressions, one row per legal move
\[ x=1+4t,\; y=2+8t,\; 0\le t\le 1 \]
Verify: check both endpoints
Why: At zero the point is (1, 2) and at one it is (5, 10), as required. And at one half it is (3, 6), the midpoint — which confirms the traversal is at constant speed, since the parameter's midpoint gives the segment's midpoint.
OpenStax, Precalculus, §8.6 Parametric Equations §8.6, pp. 1006-1010
Prediction
The parametrisation is changed so the parameter is doubled inside both functions.
Predict first
What changes?
Correct: The speed doubles; the curve is unchanged.
Why: Doubling the parameter inside means the same positions are reached in half the time, so the point moves twice as fast along exactly the same path. Only a change to the coordinate functions' outputs would change the curve itself.
Worked example
One substitution reverses it.
\[ \text{Reverse the direction of } x=\cos t, \; y=\sin t. \]
Identify the current direction
Why: Anticlockwise from the right.
Negate the parameter
Why: Replace t by its negative.
\[ \cos(-t), \sin(-t) \]
Apply the even-odd identities
Why: From §7.1.
\[ \cos t, -\sin t \]
State the result
Why: Same circle, other way.
Figure (svg): A curve traced by a moving point, with arrows showing the direction of travel and marked positions at successive parameter values
\[ x=\cos t,\; y=-\sin t \]
Verify: check just after the start
Why: Just after the start the vertical coordinate is now negative, so the point moves downward first — which is clockwise. Eliminating the parameter from either version gives the same circle equation, confirming the shape is unchanged.
OpenStax, Precalculus, §8.6 Parametric Equations §8.6, pp. 1010-1014
Trap
\[ \text{the parametrisation of this circle is } (\cos t,\sin t) \]
Refer to one parametrisation as the parametrisation
Why: The definite article implies there is only one.
The freedom to choose direction, speed and starting point is not recognised.
A curve has infinitely many parametrisations. Direction, speed and starting point are all free choices.
Scaling the parameter changes the speed; negating it reverses the direction; adding a constant shifts the start. The curve is unchanged by all three.
The choice is a modelling decision, made by what the parameter is meant to represent — usually time, in which case the physics fixes it.
Matching
Each modification affects one aspect.
Match the pairs
Why: The first three change only the journey, leaving the set of points identical. Only the fourth changes the curve, since it changes what the coordinate functions produce rather than when they produce it.
Faded example
From the point (2, 3) to the point (6, 11).
Fill in the blanks
x=2+4t, \quad y=3+8t, \quad 0\le t\le 1
Why: Each coordinate starts at its initial value and changes by the total difference over the parameter's unit interval. At the parameter value one, both coordinates reach the endpoint exactly.
Explain it
One curve, many parametrisations.
Discussion prompt
Explain to a classmate why there is no single correct parametrisation.
Hint: What is being chosen?
Answer:
A parametrisation describes a journey along a curve, and the same curve can be travelled in many ways — different directions, speeds and starting points.
All of those produce the same set of points, so they all eliminate to the same equation. The curve does not determine the journey.
So the choice is made by what the parameter represents. If it is time and the situation is physical, the physics fixes it; otherwise any convenient choice will do. A good explanation stresses that this freedom is a feature — it is exactly what lets a parametrisation carry information an equation cannot.
Comparison
Fill the blanks from memory. Each records something the other does not.
Comparison matrix
| an equation | a parametrisation | |
|---|---|---|
| records the shape | yes | yes |
| records the direction | no | yes |
| records the timing | no | yes |
| how many describe one curve | essentially one | infinitely many |
The last row is why the previous three come out as they do. The extra freedom in choosing a parametrisation is exactly the extra information it can carry.
Pattern
Five steps, and the fourth is the one that is forgotten.
Step 4 has two sources and only one of them is written in the problem. Asking what each coordinate function can produce catches the other.
OpenStax Algebra and Trigonometry 2e, §10.6 Parametric Equations §10.6
Check
What elimination costs.
Check your understanding
What information is lost when the parameter is eliminated?
Answer: A
Why: An equation records which points make up the curve and nothing about how it is traced. Two parametrisations going opposite ways round the same circle eliminate to the identical equation.
Check
Elimination method.
Check your understanding
Both coordinates are trigonometric. What is the best elimination method?
Answer: A
Why: Squaring and adding invokes a Pythagorean identity, which removes the parameter in one step with no inverse function and no range restriction. The inverse route would silently lose half the curve.
Check
Projectile motion.
Check your understanding
Which question can only the parametric form answer?
Answer: A
Why: Time does not appear in the eliminated equation at all, so any question about when something happens requires returning to the parametric form. The other three concern the shape and extent of the path.
Real world
Animation and computer graphics are built on parametrisations.
Discussion prompt
Why does animation software describe motion parametrically rather than by a path equation?
Hint: What does the software need to know at each frame?
Answer:
Each frame needs a position at a specific moment, which is exactly what a parametrisation supplies — substitute the frame's time and get a point.
A path equation would say which points the object passes through but not when, so the software would have no way to decide where to draw it in any given frame.
And the freedom in choosing the parametrisation is used deliberately: easing an animation means reparametrising the same path so the object accelerates and decelerates rather than moving uniformly. The path stays identical and only the schedule changes, which is precisely the freedom this section describes.
Commit first
State your confidence along with your answer.
Predict first
Why can one curve have infinitely many parametrisations?
Correct: Direction, speed and starting point are all free choices.
Why: A parametrisation describes a journey along the curve, and the same set of points can be traversed in many ways. All of them eliminate to the same equation, which is why the curve does not determine the journey.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
Explain to a classmate why parametric equations are worth having when an equation already describes the curve.
Hint: What can two identical equations describe?
Answer:
An equation describes a shape — which points are on the curve. It says nothing about how the curve is traced.
Two runners going opposite ways round the same track have identical equations, so the equation cannot distinguish them. Direction, speed and timing are all invisible to it.
A parametrisation records all of that, because it gives a position for each moment rather than a condition on positions. Whenever the subject is motion rather than shape, the parametrisation is the right description — and a good explanation gives the projectile as the case where the timing is the whole question.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The third is where the marks are lost, since the restriction from a function's own range is nowhere stated in the problem. The fourth is where the whole idea earns its place.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Write one parametric pair and tabulate five points, plotting them with arrows for direction. Beside it, eliminate the parameter and write the resulting equation with its restriction, noting where the restriction came from. Underneath, write the two projectile equations and one question that only the parametric form can answer.
If your restriction names its source and your projectile question is about timing, the two things this section adds beyond ordinary graphing are both on the page.
Recap
Five things, and the third is the one to check every time.
| if you remember one thing | it should be this |
|---|---|
| about plotting | join in parameter order, not left to right |
| about elimination | the shape survives; the journey does not |
| about restrictions | check the interval and each function's range |
| about projectiles | two independent motions, and only time links them |
Section 8.7 graphs parametric equations directly, adding the orientation arrows and simultaneous-motion questions that the parametric description makes possible.
OpenStax, Precalculus, §8.6 Parametric Equations §8.6, pp. 992-1007 — everything on these slides traces back here
Want this taught 1-on-1? Alexander tutors Precalculus — $55/session, free consultation.