Graphs parametric equations with their orientation marked, handles curves that loop or revisit points, and distinguishes paths that cross from objects that collide. Uses the parametric form to answer simultaneous-motion questions an equation cannot state.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 8 — Further Applications of Trigonometry
§8.7 Parametric Equations: Graphs, pp. 1008-1021
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §8.7 Parametric Equations: Graphs §8.7, pp. 1008-1021 — the pages these objectives are drawn from
Warm-up
Two roads crossing does not mean two cars collide.
Discussion prompt
Two cars travel along paths that intersect. Under what condition do they actually collide?
Hint: What else besides position matters?
Answer:
They must be at the intersection point at the same time. Passing through the same place at different moments is not a collision.
So the question is about schedules, not just about paths. The paths crossing is necessary but nowhere near sufficient.
An equation records only the paths, so it cannot express the question. The parametric form gives a position for each moment, which is exactly what a collision question needs.
Concept
A parametric graph is a curve with arrows on it. The arrows record the order in which points are visited, which the curve alone does not.
\[ \text{same curve}, \text{ different orientation} \;\Longrightarrow\; \text{different motion} \]
Marking the orientation is not an optional decoration. It is half the content of the graph, and a parametric graph without arrows has thrown away the information that distinguishes it from an ordinary one.
Figure (svg): A parametrised curve with arrowheads along it marking the direction of increasing parameter
OpenStax, Precalculus, §8.7 Parametric Equations: Graphs §8.7, pp. 1019-1024
Section
Section 1
Concept
As the parameter increases the point moves along the curve, and arrows drawn in that direction record the order of the journey.
Deciding the direction takes one comparison: compute two nearby parameter values and see which way the point moved. That is more reliable than reasoning about the functions, especially when both coordinates are changing.
Figure (svg): A parametrised curve with arrowheads along it marking the direction of increasing parameter
OpenStax, Precalculus, §8.7 Parametric Equations: Graphs §8.7, pp. 1019-1025
Picture it
Four arrows record the direction all the way round.
Figure (svg): A parametrised curve with arrowheads along it marking the direction of increasing parameter
Negating the sine would give an identical ellipse with all four arrows reversed. The two graphs are different even though the curves are identical.
Worked example
Compare two nearby parameter values.
\[ \text{Find the orientation of } x=t^2, \; y=t^3. \]
Evaluate at a negative value
Why: Take the parameter as negative one.
\[ (1, -1) \]
Evaluate at zero
Why: The origin.
\[ (0, 0) \]
Evaluate at a positive value
Why: Take the parameter as one.
\[ (1, 1) \]
Read the order
Why: The point moves left then right, always upward.
Figure (svg): A parametrised curve with arrowheads along it marking the direction of increasing parameter
\[ \text{cusp at the origin, upward throughout} \]
Verify: check the vertical direction
Why: The vertical coordinate is the parameter cubed, which increases steadily — so the point always moves upward regardless of what the horizontal is doing. The horizontal reverses at the origin, which is what produces the cusp there.
OpenStax, Precalculus, §8.7 Parametric Equations: Graphs §8.7, pp. 1020-1023
Prediction
The parameter is replaced by its negative throughout.
Predict first
What changes?
Correct: The orientation reverses; the curve is unchanged.
Why: The same parameter values are visited in the opposite order, so the point traces the identical set of points backwards. The set of points is unaffected, which is why elimination gives the same equation.
Worked example
Negating the parameter does it.
\[ \text{Reverse the orientation of } x=2t, \; y=t^2-1 \text{ on } [-2,2]. \]
Replace the parameter by its negative
Why: Throughout.
\[ x = -2 t, y = t ^{2} - 1 \]
Check the curve is unchanged
Why: Eliminate to compare.
Check the interval
Why: Symmetric, so unchanged.
\[ [-2, 2] \]
Confirm the direction
Why: Now rightward becomes leftward.
Figure (svg): The solution to Worked example reverse an orientation shown as a ladder of expressions, one row per legal move
\[ x=-2t,\; y=t^2-1 \]
Verify: check by eliminating
Why: Both versions eliminate to the same parabola, since the horizontal coordinate is squared in the substitution and the sign disappears. So the curve is genuinely identical and only the orientation differs, which is exactly what was wanted.
OpenStax, Precalculus, §8.7 Parametric Equations: Graphs §8.7, pp. 1023-1025
Trap
\[ \text{plot the points, join them, done} \]
Draw the curve with no direction marked
Why: The graph is treated as an ordinary curve sketch.
The distinguishing feature of a parametric graph has been discarded.
Mark the orientation with arrows. It is half the content, not an annotation.
Two parametrisations of the same curve differ only in their arrows, so a graph without them cannot distinguish them.
Determine the direction by comparing two nearby parameter values, which is one substitution and more reliable than reasoning about the functions.
Faded example
Comparing two parameter values on a parametrisation.
Fill in the blanks
t=0:\;(0,0) \quad t=1:\;(1,1) \quad(x=t^2,\;y=t^3)
Why: Substituting the two parameter values gives two points, and the order they come in is the orientation. Comparing nearby values is the most reliable way to settle the direction.
Sorting
Some changes affect the points and some do not.
Sort into buckets
Sort each modification.
Step zero
You have plotted the points of a parametric curve.
Discussion prompt
What must you do before the graph is complete?
Hint: What distinguishes it from an ordinary sketch?
Answer:
Mark the orientation with arrows in the direction of increasing parameter. Without them the graph is just a curve.
Determine the direction by comparing two nearby parameter values, which is one substitution and settles it unambiguously.
And mark the start and end if the interval is bounded, since those are points the journey actually has. A parametric graph without arrows has discarded exactly what made it parametric.
Section
Section 2
Concept
A parametrised curve may pass through the same point more than once, either by looping or by retracing. The parameter values at which it does are what matter.
Finding where a curve crosses itself requires solving for two different parameter values giving the same point — both coordinates must match. That is a system rather than a single equation, which is why it takes more work than finding a curve's intersections with a line.
Figure (svg): A parametrised curve that crosses itself, with the crossing point reached at two different parameter values
OpenStax, Precalculus, §8.7 Parametric Equations: Graphs §8.7, pp. 1025-1029
Picture it
The marked point is reached at two parameter values.
Figure (svg): A parametrised curve that crosses itself, with the crossing point reached at two different parameter values
The crossing is invisible in the eliminated equation, which simply describes a curve that happens to pass through that point. Only the parametrisation records that it does so twice.
Worked example
Both coordinates must match at two different parameter values.
\[ \text{Where does } x=t^2-2, \; y=t^3-3t \text{ cross itself?} \]
Require equal horizontal coordinates
Why: At two parameter values.
\[ a ^{2} = b ^{2},\text{ so } b = -a \]
Require equal vertical coordinates
Why: With that substitution.
\[ a ^{3} - 3 a = -a ^{3} + 3 a \]
Solve
Why: Collect and factor.
\[ 2 a(a ^{2} - 3) = 0 \]
Discard the trivial root
Why: Zero gives one value, not two.
\[ a = \sqrt{3} \]
Figure (svg): A parametrised curve that crosses itself, with the crossing point reached at two different parameter values
\[ (1,0) \text{ at } t=\pm\sqrt{3} \]
Verify: substitute both parameter values
Why: At root three the point is (1, 0) since three minus two is 1 and the cube root three times three minus three root three is zero. At negative root three both coordinates come out the same. So the curve genuinely visits that point twice.
OpenStax, Precalculus, §8.7 Parametric Equations: Graphs §8.7, pp. 1026-1028
Prediction
A curve crosses itself once.
Predict first
What does the eliminated equation record about the crossing?
Correct: Nothing; it just describes a curve through that point.
Why: The equation records which points are on the curve, and the crossing point is on it once as far as the equation is concerned. That the journey visits it twice is parametric information the equation cannot carry.
Worked example
The interval covers the curve more than once.
\[ \text{Describe } x=\cos t, \; y=\sin t \text{ on } 0\le t\le 4\pi. \]
Identify the curve
Why: The unit circle.
Find the period
Why: One full turn.
\[ 2 \pi \]
Compare with the interval
Why: Twice the period.
Describe
Why: The circle traced twice.
Figure (svg): The solution to Worked example a retraced curve shown as a ladder of expressions, one row per legal move
\[ \text{two complete laps} \]
Verify: check what the equation would show
Why: Eliminating gives the unit circle equation, identical to what a single lap would give. The equation cannot record that the curve was traced twice, which is exactly the information the interval carries.
OpenStax, Precalculus, §8.7 Parametric Equations: Graphs §8.7, pp. 1028-1029
Error analysis
A student describes a looping parametric curve.
Annotate
On: \( \text{the curve is in two pieces, meeting at }(1,0) \)
A crossing point is where the journey returns to somewhere it has already been. Following the orientation arrows through the crossing shows the continuity that the static picture obscures.
Faded example
Two parameter values giving the same point.
Fill in the blanks
x(a)=x(b) \textb y(a)=y(b), \text2 a\ne___ \text______\text___
Why: Both coordinates must agree at two different parameter values, which is a system rather than a single equation. That is why finding a self-crossing takes more work than finding an intersection with a line.
Sorting
Some features survive elimination and some do not.
Sort into buckets
Sort each feature.
Explain it to yourself
A curve that crosses itself looks like two pieces.
Discussion prompt
Explain why it is one continuous curve.
Hint: What does the parameter do?
Answer:
The parameter runs continuously through its interval, and both coordinate functions are continuous — so the point never jumps.
At the crossing it simply arrives somewhere it has been before, having travelled continuously in between. Nothing is broken.
Following the orientation arrows through the crossing shows this directly: the arrows lead in and out along the four branches in a single connected order. The static picture is what suggests two pieces, and the arrows are what correct it.
Section
Section 3
Concept
Two parametrised objects collide only if they are at the same position at the same parameter value. Their paths crossing is a weaker condition entirely.
The difference in the number of unknowns explains why collisions are rare. Two curves generally cross somewhere, but requiring the crossing to happen at a single shared parameter value is one condition more than the system has freedom to satisfy.
Figure (svg): Two parametrised paths crossing, with the two moving points shown at the same parameter value in different places
OpenStax, Precalculus, §8.7 Parametric Equations: Graphs §8.7, pp. 1029-1032
Picture it
Both points shown at the same parameter value.
Figure (svg): Two parametrised paths crossing, with the two moving points shown at the same parameter value in different places
The paths clearly intersect, and the two objects are nowhere near each other at that moment. The intersection happens, but not simultaneously.
Worked example
Allow different parameter values.
\[ \text{Do the paths } (t,2t) \text{ and } (s+1,s^2) \text{ cross?} \]
Set the horizontal coordinates equal
Why: Different parameters allowed.
\[ t = s + 1 \]
Set the vertical coordinates equal
Why: A second equation.
\[ 2 t = s ^{2} \]
Substitute
Why: Eliminate one unknown.
\[ 2(s + 1) = s ^{2} \]
Solve
Why: A quadratic.
\[ s = 1\text{ plus or minus } \sqrt{3} \]
Figure (svg): Two parametrised paths crossing, with the two moving points shown at the same parameter value in different places
\[ s=1\pm\sqrt{3} \]
Verify: check one intersection
Why: At s equal to about 2.73 the second path is at about (3.73, 7.46), and the first path reaches that point when t is 3.73, giving a vertical coordinate of 7.46 — they match. So the paths genuinely cross there, though at different parameter values.
OpenStax, Precalculus, §8.7 Parametric Equations: Graphs §8.7, pp. 1030-1031
Prediction
Two moving objects with parametrised paths.
Predict first
Which is the rarer event?
Correct: A collision, since it needs the same time as well as place.
Why: Path crossing is a system in two unknowns and generally has solutions. A collision imposes the extra condition that the two parameter values are equal, which is one more condition than the system has freedom to satisfy.
Worked example
Require the same parameter value.
\[ \text{For the same two paths, do the objects collide?} \]
Use one parameter for both
Why: The same moment.
\[ s = t \]
Set the horizontal coordinates equal
Why: One equation.
\[ t = t + 1 \]
Observe
Why: This is impossible.
Conclude
Why: No collision.
Figure (svg): The solution to Worked example do they collide shown as a ladder of expressions, one row per legal move
\[ \text{no collision} \]
Verify: check why the paths still cross
Why: The crossings found earlier occurred at different parameter values for the two objects — one arrived before the other. Requiring a single shared parameter value imposes one more condition than the system can satisfy, which is why collisions are the rarer event.
OpenStax, Precalculus, §8.7 Parametric Equations: Graphs §8.7, pp. 1031-1032
Trap
\[ \text{set } t \text{ equal in both parametrisations to find where the paths cross} \]
Impose a shared parameter value
Why: The same symbol is used for both objects' parameters.
The stronger collision condition is imposed, and genuine crossings are missed.
Use different symbols for the two parameters when asking whether the paths cross.
The paths cross wherever some value of one parameter and some value of the other give the same point — they need not be equal.
Reserve a shared parameter for the collision question, which is the stronger condition. Using the wrong one gives a wrong answer in either direction.
Sorting
Shared parameter or separate ones.
Sort into buckets
Sort each question.
Faded example
Two objects, one shared parameter.
Fill in the blanks
x_1(t)=x_2}(t) \text2 y_1(t)=y____}(t) \text___t
Why: Both coordinate conditions must hold at the same parameter value, which makes this two equations in one unknown. Over-determination is why collisions are rare.
Explain it
Crossing paths and colliding objects are different.
Discussion prompt
Explain to a classmate what distinguishes them, and why it matters.
Hint: Think about two roads.
Answer:
Two roads crossing does not mean two cars crash. The paths intersect but the cars may pass through at different moments.
Mathematically, a path crossing allows each object its own parameter value; a collision requires one shared value. That is two equations in one unknown rather than in two.
So a collision is genuinely rarer, and only the parametric form can even state the question — an equation records paths and not schedules. A good explanation notes that this is the clearest case where parametrisation is not just convenient but necessary.
Section
Section 4
Concept
Negating the parameter reverses the orientation, scaling it changes the speed, and restricting its interval truncates the curve. Only the third changes the set of points.
Only the last two rows change the eliminated equation or its restriction. The first three are invisible to elimination, which is exactly why they are the ones a parametrisation exists to record.
| change | effect on the curve | effect on the journey |
|---|---|---|
| negate the parameter | none | orientation reverses |
| scale the parameter | none | speed changes |
| shift the parameter | none | starting point moves |
| restrict the interval | truncated to an arc | start and end appear |
| scale the coordinate functions | the curve is stretched | unchanged in kind |
Figure (svg): Three cards giving the standard modifications to a parametrisation and what each does to the graph
OpenStax, Precalculus, §8.7 Parametric Equations: Graphs §8.7, pp. 1024-1030
Picture it
Each card names one change and its effect.
Figure (svg): Three cards giving the standard modifications to a parametrisation and what each does to the graph
The caption is the key observation: two of the three leave the set of points untouched, so they show up only in the arrows and not in any equation.
Worked example
Doubling the parameter inside.
\[ \text{Compare } (\cos t,\sin t) \text{ on } [0,2\pi] \text{ with } (\cos 2t,\sin 2t) \text{ on the same interval.} \]
Identify the curve
Why: Both are the unit circle.
Compare the periods
Why: The second completes in half the interval.
Count the laps
Why: Over the same interval.
Describe
Why: Same circle, traced twice.
Figure (svg): Three cards giving the standard modifications to a parametrisation and what each does to the graph
\[ \text{two laps instead of one} \]
Verify: check what elimination shows
Why: Both eliminate to the identical unit circle equation, since the doubling is entirely inside the parameter. The difference is only in the schedule, which confirms that scaling the parameter is a journey change rather than a curve change.
OpenStax, Precalculus, §8.7 Parametric Equations: Graphs §8.7, pp. 1025-1027
Prediction
A constant is added to the parameter inside both functions.
Predict first
What changes?
Correct: The starting point; the curve is unchanged.
Why: Adding a constant inside means the journey begins at a different place along the same path. Since every point is still visited, the curve is identical and elimination gives the same equation.
Worked example
This one does change the curve.
\[ \text{Describe } (\cos t,\sin t) \text{ on } 0\le t\le\pi. \]
Identify the full curve
Why: The unit circle.
Note the interval
Why: Half a turn.
Find the endpoints
Why: At zero and at a half turn.
\[ (1, 0)\text{ and } (-1, 0) \]
Describe
Why: The upper semicircle.
Figure (svg): The solution to Worked example restrict to an arc shown as a ladder of expressions, one row per legal move
\[ \text{upper semicircle}, (1,0)\to(-1,0) \]
Verify: check the vertical coordinate's sign
Why: Over the interval from zero to a half turn the sine is non-negative, so the vertical coordinate never goes below zero — confirming the upper half. Eliminating would give the full circle equation, so the restriction must be stated separately.
OpenStax, Precalculus, §8.7 Parametric Equations: Graphs §8.7, pp. 1027-1030
Error analysis
A student compares two parametrisations of a circle.
Annotate
On: \( (\cos 2t,\sin 2t) \text{ must be a different curve, since it moves faster} \)
Scaling the parameter changes when each point is reached, never which points are reached. The distinction between the curve and the journey along it is the section's central idea.
Matching
Four changes, four effects.
Match the pairs
Why: The first two are invisible to elimination and the last two are not. That split is what distinguishes changes to the journey from changes to the curve itself.
Faded example
A circle parametrisation with the parameter tripled, over one full turn.
Fill in the blanks
\text3=\frac3___}, \text______\text___
Why: Tripling the parameter divides the period by three, so a full-turn interval contains three complete periods. The curve is unchanged and only the number of traversals differs.
Explain it
Some modifications change the curve and some do not.
Discussion prompt
Explain to a classmate how to tell which is which.
Hint: What test would distinguish them?
Answer:
Ask whether the set of points visited changes. If the same points are reached, only in a different order or at a different rate, the curve is unchanged.
Negating, scaling and shifting the parameter all leave the point set alone. Restricting the interval removes points, and scaling the coordinate functions moves them.
A quick test: eliminate the parameter in both versions. If the equations agree, only the journey changed. A good explanation notes that this is a genuine use for elimination even when the timing is what you care about.
Section
Section 5
Concept
Problems involving two moving objects are set up as two parametrisations sharing a parameter, after which the questions are about when their positions relate in some way.
The distance between the two objects is itself a function of the parameter, and questions about closest approach are questions about its minimum. That connection to optimisation is why these problems recur in calculus with the tools to answer them fully.
Figure (svg): Two parametrised paths crossing, with the two moving points shown at the same parameter value in different places
OpenStax, Precalculus, §8.7 Parametric Equations: Graphs §8.7, pp. 1031-1034
Picture it
Both points are shown at the same parameter value.
Figure (svg): Two parametrised paths crossing, with the two moving points shown at the same parameter value in different places
The distance between the two dots is a function of the parameter, and watching it change as the parameter runs is what a simultaneous-motion question is about.
Worked example
The distance formula, with both positions parametrised.
\[ \text{Two objects follow } (t,2t) \text{ and } (4-t,t). \text{ Write their separation.} \]
Find the horizontal difference
Why: Subtract.
\[ t - (4 - t) = 2 t - 4 \]
Find the vertical difference
Why: Subtract.
\[ 2 t - t = t \]
Apply the distance formula
Why: Root of the sum of squares.
\[ \sqrt{(2 t - 4) ^{2} + t ^{2}} \]
Expand
Why: Collect terms.
\[ \sqrt{5 t ^{2} - 16 t + 16} \]
Figure (svg): Two parametrised paths crossing, with the two moving points shown at the same parameter value in different places
\[ d(t)=\sqrt{5t^2-16t+16} \]
Verify: check at a convenient value
Why: At the parameter value zero the objects are at (0,0) and (4,0), four apart — and the formula gives the root of 16, which is 4. The setup is correct, and the expression under the root being a quadratic means the closest approach is at its vertex.
OpenStax, Precalculus, §8.7 Parametric Equations: Graphs §8.7, pp. 1032-1033
Prediction
You want the closest approach of two objects.
Predict first
What is easiest to minimise?
Correct: The expression under the square root.
Why: The square root is an increasing function, so whatever minimises the expression inside also minimises the root. Working with the polynomial inside avoids differentiating or manipulating a radical.
Worked example
Minimise the quadratic under the root.
\[ \text{For those objects, when are they closest?} \]
Note the root is increasing
Why: So minimise what is inside.
Find the vertex
Why: Negative b over twice a.
\[ \frac{16}{10} \]
Compute the parameter value
Why: The closest moment.
\[ t = 1.6 \]
Compute the distance
Why: Substitute back.
\[ \sqrt{5(2.56) - 25.6 + 16} \]
Figure (svg): The solution to Worked example find the closest approach shown as a ladder of expressions, one row per legal move
\[ t=1.6,\; d\approx 1.79 \]
Verify: check the endpoints
Why: At zero the separation is 4 and at parameter value 3 it is the root of 45 minus 48 plus 16, about 3.6 — both larger than 1.79. The vertex is genuinely the minimum, and minimising the quadratic rather than the root was legitimate since the square root preserves order.
OpenStax, Precalculus, §8.7 Parametric Equations: Graphs §8.7, pp. 1033-1034
Trap
\[ \text{differentiate the square root directly to find the minimum} \]
Work with the root throughout
Why: The square root is carried through every step of the minimisation.
The algebra becomes far messier for no benefit.
Minimise what is under the root instead. The square root is increasing, so it preserves which value is smallest.
Here the expression under the root is a quadratic whose vertex is one formula away.
Take the root only at the end, to report the distance. The location of the minimum is unaffected and the work is much shorter.
Faded example
From two parametrised positions.
Fill in the blanks
d(t)=\sqrt2}})^2+(y_1-y_2})^2}
Why: The distance formula applied to the two parametrised positions gives a function of the parameter. Every simultaneous-motion question about proximity is a question about this function.
Sorting
Some questions involve time and some do not.
Sort into buckets
Sort each question.
Explain it to yourself
Two objects, one parameter symbol.
Discussion prompt
Explain why simultaneous-motion problems share the parameter.
Hint: What does the parameter represent?
Answer:
The parameter represents time, and both objects experience the same time. Using two symbols would allow them to be at different moments.
So sharing the symbol is what encodes simultaneity, which is exactly what makes a collision different from a path crossing.
And it makes the separation a function of one variable, which is what allows questions about closest approach to be answered by minimisation. The shared parameter is the whole mechanism by which motion questions become algebra questions.
Comparison
Fill the blanks from memory. The difference is one shared symbol.
Comparison matrix
| paths cross | objects collide | |
|---|---|---|
| parameters | one for each object | one shared between them |
| unknowns | two | one |
| equations | two | two |
| how common | usually happens | rare, over-determined |
The second row explains the fourth: two equations in two unknowns generally have solutions, and two equations in one unknown generally do not.
Pattern
Five steps, and the fourth is what makes it a parametric graph.
Step 5 confirms the shape but never the orientation, so it is a partial check — the arrows have to be got right from the table.
OpenStax Algebra and Trigonometry 2e, §10.7 Parametric Equations: Graphs §10.7
Check
Orientation.
Check your understanding
What do the arrows on a parametric graph record?
Answer: A
Why: The arrows record the order in which points are visited as the parameter increases. That is information the curve alone does not carry, and it is what distinguishes a parametric graph from an ordinary sketch.
Check
Collisions.
Check your understanding
What distinguishes a collision from a path crossing?
Answer: A
Why: A path crossing allows each object to reach the shared point at its own moment. A collision requires simultaneity, which is one shared parameter value — making it two equations in one unknown rather than two.
Check
Modifications.
Check your understanding
Which modification changes the set of points on the curve?
Answer: A
Why: Restricting the interval removes points from the curve, truncating it to an arc. The other three change only the order, rate or starting point of the journey, leaving every point still visited.
Real world
Air traffic control is a simultaneous-motion problem solved continuously.
Discussion prompt
Why is it not enough for a controller to check that two aircraft's routes do not cross?
Hint: What would that check miss, and what would it over-report?
Answer:
Routes crossing is almost unavoidable in busy airspace, and it is harmless if the aircraft pass through the crossing at different times. Flagging every crossing would be useless.
What matters is the separation as a function of time — the distance between the two parametrised positions — and whether it ever drops below the required minimum.
So the system computes a closest-approach distance and the time it occurs, exactly the calculation in this section's last worked example. The path question and the schedule question give completely different answers, and only the second is the one that matters.
Commit first
State your confidence along with your answer.
Predict first
Why are collisions rarer than path crossings?
Correct: A collision is two equations in one unknown rather than two.
Why: Requiring a shared parameter value removes one unknown while keeping both coordinate equations, so the system is over-determined. Path crossing keeps two unknowns and generally has solutions.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
A classmate drew a parametric curve with no arrows. Explain what is missing.
Hint: What could two identical drawings describe?
Answer:
The orientation — which way the point travels along the curve as the parameter increases.
Without it, a parametrisation and its reverse produce identical drawings, even though they describe opposite motions. The drawing has thrown away the only thing that distinguished them.
Adding arrows takes seconds: compare two nearby parameter values and see which way the point moved. A good explanation stresses that this is not a presentational nicety — it is half of what the graph is meant to communicate.
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 the conceptual heart of the section and the one that is most often confused. The fourth is where the whole parametric apparatus proves itself necessary rather than merely convenient.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Graph one parametric curve with its arrows and endpoints marked. Beside it, write the two setups for the crossing question and the collision question, showing which uses one parameter and which uses two. Underneath, list the four parameter modifications and mark which change the curve and which change only the journey.
If your two setups differ only in the parameter symbols and your modification list has the split marked, the section's two ideas are on the page rather than its examples.
Recap
Five things, and the third is the one worth stating carefully.
| if you remember one thing | it should be this |
|---|---|
| about graphs | arrows are half the content, not decoration |
| about crossings | separate parameters for paths, one shared for collisions |
| about modifications | only restricting the interval changes the set of points |
| about separation | minimise what is under the root, not the root |
Section 8.8 introduces vectors, which give the same directed-quantity idea an algebra of its own — and turn the component resolution used throughout this chapter into a general operation.
OpenStax, Precalculus, §8.7 Parametric Equations: Graphs §8.7, pp. 1008-1021 — everything on these slides traces back here
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