The last idea of the course: let both coordinates depend on a third variable, so a curve becomes the path of a moving point. Covers sketching from a parametrization by reading each coordinate's behaviour separately, eliminating the parameter by substitution or by a trigonometric identity, the standard recipes for parametrizing graphs, segments and ellipses, and the two adjustments that reverse an orientation or delay a start. Closes with the cycloid, a curve that has no usable equation in x and y at all.
Subject: Trigonometry · 65 slides · symbolic lesson
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Title
Trigonometry · Chapter 11 — Applications of Trigonometry
§11.10 Parametric Equations, pp. 1048-1059
Objectives
Five outcomes, and the last one shows why the whole idea was worth building.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.10 Parametric Equations §11.10, pp. 1048-1059 — the pages these objectives are drawn from
Warm-up
Every curve so far has been described by an equation relating x and y directly.
Discussion prompt
A bug crawls across a table, looping back to cross its own path. Its route fails both the vertical and the horizontal line test. Is there any way to describe it with functions?
Hint: How many places can the bug be at one moment?
Answer:
The bug is in exactly one place at each moment, so its x-coordinate is a function of time, and so is its y-coordinate. Two ordinary functions, of a third variable.
The route crossing itself is no obstacle, because the bug reaches the crossing point at two different times. Neither coordinate function is one-to-one, and neither needs to be.
So the curve is described by a pair of functions of a parameter rather than by one equation in x and y. That is the idea of this lesson, and it also supplies something the bare curve never had — a direction of travel.
Concept
Let a point move through the plane, and record its two coordinates separately as functions of a parameter. The set of positions is the curve, and the parameter supplies an order in which they are visited.
parametrization — A pair of equations giving each coordinate as a function of a parameter, together with the interval over which the parameter runs. The curve is the set of resulting points; the parametrization also gives it an orientation.
\[ \begin{cases} x = f(t) \\ y = g(t) \end{cases} \]
The article is indefinite on purpose: a curve has infinitely many parametrizations, differing in speed, starting point and direction, all tracing the same set of points.
Figure (svg): A curve traced by a moving point, with arrows showing the direction of travel and the point's two coordinates each written as a function of time
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.10 Parametric Equations §11.10, pp. 1048-1048
Section
Section 1
Concept
The curve itself is a set of points and has no direction. The parametrization adds one, by saying which points are visited first, and different parametrizations of the same curve can disagree about it.
The parameter is usually written as t and thought of as time, but it need not be. For a circle the natural parameter is the angle, and for the cycloid it is the rotation of the rolling wheel.
Figure (svg): A curve traced by a moving point, with arrows showing the direction of travel and the point's two coordinates each written as a function of time
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.10 Parametric Equations §11.10, pp. 1048-1049
Picture it
The arrows are the new information.
Figure (svg): A curve traced by a moving point, with arrows showing the direction of travel and the point's two coordinates each written as a function of time
Everything about the shape was already expressible with an equation in x and y. The orientation was not, and it is the parametrization's own contribution.
Worked example
The unit circle, met long before this lesson.
\[ \text{Identify the curve traced by } x = \cos(t), \; y = \sin(t). \]
Recognise the pair
Why: The definition of the circular functions.
Note the distance from the origin
Why: The Pythagorean Identity.
\[ \text{always } 1 \]
Note the starting point
Why: At t equal to zero.
\[ (1, 0) \]
Note the direction
Why: Increasing t sweeps counter-clockwise.
Figure (svg): The solution to Worked example a familiar parametrization shown as a ladder of expressions, one row per legal move
\[ x^2 + y^2 = \cos^2 t + \sin^2 t = 1 \]
Verify: check a second point
Why: At t equal to pi over 2 the point is 0 comma 1, straight up. Coming from 1 comma 0, that is a counter-clockwise quarter turn, confirming the orientation. This parametrization has been in use since Lesson 10.2 without being named as one.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.10 Parametric Equations §11.10, pp. 1048-1049
Prediction
A parametric curve crosses itself at a point.
Predict first
What does that say about the coordinate functions?
Correct: Neither is one-to-one.
Why: A crossing means the point is reached at two different parameter values, so both coordinate functions take the same value at two different inputs. That is exactly the failure of one-to-one. It causes no problem at all, because a parametrization only requires each coordinate to be a function of t, not an invertible one — which is precisely why parametrizations can describe self-crossing curves and equations in x and y cannot describe them as functions.
Worked example
Example 11.10.1. Points plus arrows.
\[ \text{Sketch } x = t^2 - 3, \; y = 2t - 1 \text{ for } t \ge -2. \]
Tabulate a few values of t
Why: Starting at negative 2.
Plot them in order
Why: Each t gives one point.
\[ (1, -5), (-2, -3), (-3, -1),... \]
Draw arrows for increasing t
Why: The direction of travel.
Join smoothly
Why: The shape suggests a parabola.
Figure (svg): The solution to Worked example sketching from a table shown as a ladder of expressions, one row per legal move
\[ \text{vertex } (-3,-1), \text{ opening right, traced from } (1,-5) \text{ upward} \]
Verify: check the extreme point
Why: The x-coordinate is smallest when t is zero, giving x equal to negative 3, and y is then negative 1. That is the leftmost point of the curve, which for a rightward-opening parabola is the vertex. The table and the shape agree.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.10 Parametric Equations §11.10, pp. 1049-1049
Trap
\[ x^2 + y^2 = 1 \;\Longrightarrow\; \text{traced counter-clockwise} \]
Read a direction off the equation
Why: The parametrization had one, so the curve should too.
But the equation describes only the set of points. Nothing in it says anything about order, and the pair with sine negated traces the same circle clockwise.
The curve has no orientation; the parametrization does. Two parametrizations of one curve can run in opposite directions and both be correct.
So orientation must be stated separately, and it is lost the moment the parameter is eliminated.
The practical consequence: mark the arrows before eliminating, because no amount of work on the resulting equation will recover them.
Sorting
The set of points is the curve; the schedule is the parametrization.
Sort into buckets
Sort each pair by whether it traces the unit circle.
Faded example
For x = t squared minus 3 and y = 2t minus 1, find the point at t equal to 3.
Fill in the blanks
x = 3^2 - 3 = 6, \quad y = 2(3) - 1 = 5
Why: Each coordinate is computed independently from the same value of t, which is what makes the two equations a pair of ordinary functions. The resulting point is 6 comma 5, further right and further up than the earlier points, consistent with the curve heading into the first quadrant.
Socratic
Some of these curves could be described by an equation in x and y instead.
Discussion prompt
What does the parametrization give that such an equation cannot?
Hint: Think about what a physical problem needs to know.
Answer:
It gives the order and the timing. An equation says where the curve is; a parametrization says where the point is at each moment, which is what any question about motion requires.
It also handles curves for which no usable equation in x and y exists. The cycloid at the end of this lesson is the standard example: its parametrization is two short lines and its rectangular form is unusable.
And it composes. Two parametrized paths can be joined end to end by shifting one parameter, producing a single description of a route with corners in it — which no single equation in x and y could do.
Section
Section 2
Concept
Graph each coordinate against the parameter separately. Those two graphs say, over each subinterval, whether the point is moving left or right and up or down, which is enough to draw the curve.
This is the same two-plane method as the polar graphing lesson, with the theta-r graph replaced by a pair of graphs. The reasoning is identical and so is the payoff: it works on a curve you have never seen.
Figure (svg): Three graphs side by side: x plotted against the parameter, y plotted against the parameter, and the resulting curve in the plane
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.10 Parametric Equations §11.10, pp. 1050-1052
Picture it
The two on the left are ordinary function graphs.
Figure (svg): Three graphs side by side: x plotted against the parameter, y plotted against the parameter, and the resulting curve in the plane
Choosing the subinterval breaks at the turning points of either graph is what makes the transfer straightforward, since each coordinate then has a single direction throughout.
Worked example
Example 11.10.2, part 1.
\[ \text{Sketch } x = t^3, \; y = 2t^2 \text{ for } -1 \le t \le 1. \]
Find the ranges
Why: The cube ranges over negative 1 to 1; the square over 0 to 2.
Note the behaviour of x
Why: Increasing throughout.
Note the behaviour of y
Why: Decreasing then increasing, with a minimum at zero.
Transfer
Why: Right and down to the origin, then right and up.
Figure (svg): The solution to Worked example a curve with a cusp shown as a ladder of expressions, one row per legal move
\[ y = 2x^{2/3}, \quad -1 \le x \le 1 \]
Verify: check the shape against the eliminated equation
Why: Solving the first for t and substituting gives y equal to twice the cube root of x, squared — a curve with a cusp at the origin, exactly as the transfer predicted. Note the curve is not smooth there, which a purely algebraic approach might not have anticipated.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.10 Parametric Equations §11.10, pp. 1050-1050
Sorting
Translate the behaviour of each coordinate function.
Sort into buckets
Sort each observation by what the point does.
Worked example
Example 11.10.2, part 4. The interval is what makes it interesting.
\[ \text{Sketch } x = 1 + 3\cos t, \; y = 2\sin t \text{ for } 0 \le t \le \tfrac{3\pi}{2}. \]
Find the ranges
Why: X between negative 2 and 4; y between negative 2 and 2.
Evaluate at the quarter points
Why: Four convenient values of t.
\[ (4, 0), (1, 2), (-2, 0), (1, -2) \]
Track the directions on each quarter
Why: One coordinate turns at each.
Stop at three quarters
Why: The interval ends there.
Figure (svg): The solution to Worked example three quarters of an ellipse shown as a ladder of expressions, one row per legal move
\[ \frac{(x-1)^2}{9} + \frac{y^2}{4} = 1, \quad \text{three quarters traced} \]
Verify: check the ellipse's features
Why: The eliminated equation is an ellipse centred at 1 comma 0 with semi-axes 3 and 2, so its rightmost point is 4 comma 0 and its topmost is 1 comma 2 — matching the tabulated points. The parametrization covers all but the fourth quarter, which the interval's endpoint confirms.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.10 Parametric Equations §11.10, pp. 1052-1052
Error analysis
A student sketches a parametrization of an ellipse.
Annotate
On: \( x = 1 + 3\cos t, \; y = 2\sin t, \; 0 \le t \le \tfrac{3\pi}{2} \;\Longrightarrow\; \text{the whole ellipse} \)
A parametrization is the equations together with the interval, and the interval is doing real work. Eliminating the parameter discards it entirely, which is why the extent has to be recorded separately.
Faded example
For x = 1 + 3 cosine t and y = 2 sine t, describe the first quarter turn.
Fill in the blanks
As t runs from 0 to pi over 2, x decreases from 4 to 1 while y increases from 0 to 2.
Why: At t equal to pi over 2 the cosine is zero and the sine is 1, giving the point 1 comma 2. The point has moved left and up, which is counter-clockwise motion in the first quadrant relative to the ellipse's centre.
Prediction
On a subinterval, x is increasing and y is decreasing.
Predict first
Which way is the curve heading?
Correct: Down and to the right.
Why: A rising x-coordinate moves the point right and a falling y-coordinate moves it down. The two are read independently, which is the whole reason for graphing them separately. Combining the two readings is the only step in the transfer, and doing it subinterval by subinterval keeps each answer unambiguous.
Edge cases
The method breaks the parameter interval at the turning points of each coordinate.
Discussion prompt
Why those points, and what would go wrong without the breaks?
Hint: What is true of each coordinate between two consecutive turning points?
Answer:
Between two consecutive turning points, each coordinate is monotone — always rising or always falling — so the direction of travel is a single unambiguous statement over the whole subinterval.
Without the breaks, a subinterval could contain a turn, and the description would have to be moving right then left, which cannot be drawn as one stroke.
The same principle governed the polar graphing method, where the interval was broken at the zeros of r and the quadrant boundaries. Subdivide until each piece has one consistent behaviour, and then each piece draws itself. That is the general technique, and it is what makes both methods work on unfamiliar curves.
Section
Section 3
Concept
Solving one equation for the parameter and substituting into the other gives an equation in x and y. With trigonometric parametrizations, solving for the trigonometric functions and using an identity is easier.
Sometimes the extent cannot be fixed by any restriction, because the parametrization covers part of the curve twice and no equation in x and y can record a multiplicity.
Figure (svg): Two columns contrasting what is kept and what is lost when the parameter is eliminated from a pair of parametric equations
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.10 Parametric Equations §11.10, pp. 1049-1052
Picture it
The shape survives; the schedule does not.
Figure (svg): Two columns contrasting what is kept and what is lost when the parameter is eliminated from a pair of parametric equations
The right column is worth reading before eliminating rather than after, since two of the four items can be preserved by writing them down first.
Worked example
Example 11.10.1 revisited, with the restriction.
\[ \text{Eliminate } t \text{ from } x = t^2-3, \; y = 2t-1, \; t \ge -2. \]
Solve the easier equation for t
Why: The linear one.
\[ t = \frac{y + 1}{2} \]
Substitute into the other
Why: Squaring the fraction.
\[ x = (\frac{y + 1}{2}) ^{2} - 3 \]
Rearrange to standard form
Why: Multiply out.
\[ (y + 1) ^{2} = 4(x + 3) \]
Restrict to match the extent
Why: The smallest y occurs at t equal to negative 2.
\[ y \ge - 5 \]
Figure (svg): The solution to Worked example an algebraic elimination shown as a ladder of expressions, one row per legal move
\[ (y+1)^2 = 4(x+3), \quad y \ge -5 \]
Verify: state what the restriction cannot fix
Why: The restriction recovers the extent correctly, but nothing in the equation records that the curve is traced upward from the bottom. The orientation is unrecoverable, which is why it must be noted before eliminating.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.10 Parametric Equations §11.10, pp. 1049-1049
Prediction
You eliminate the parameter from a parametrization and obtain a familiar equation.
Predict first
What can you no longer determine from that equation?
Correct: The direction it is traced.
Why: An equation in x and y describes a set of points, and a set has no order. So the orientation is lost permanently, along with the speed and any doubling back. The shape, position and type all survive, since they are properties of the point set. This is why the arrows should be marked on the sketch before any elimination is attempted.
Worked example
Example 11.10.2, part 4. The identity does the work.
\[ \text{Eliminate } t \text{ from } x = 1 + 3\cos t, \; y = 2\sin t. \]
Isolate the cosine
Why: From the first equation.
\[ \cos t = \frac{x - 1}{3} \]
Isolate the sine
Why: From the second.
\[ \sin t = \frac{y}{2} \]
Substitute into the Pythagorean Identity
Why: Both squared.
\[ (\frac{x - 1}{3}) ^{2} + (\frac{y}{2}) ^{2} = 1 \]
Read the conic
Why: An ellipse in standard form.
\[ \text{centre } (1, 0) \]
Figure (svg): The solution to Worked example a trigonometric elimination shown as a ladder of expressions, one row per legal move
\[ \frac{(x-1)^2}{9} + \frac{y^2}{4} = 1 \]
Verify: note why solving for t would have been worse
Why: Solving x equals 1 plus 3 cosine t for t would give an arccosine, whose range is only half a turn, and substituting that into the sine would need a further identity. Isolating the trigonometric functions instead avoids the inverse entirely, which is why the identity route is the standard one.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.10 Parametric Equations §11.10, pp. 1052-1052
Trap
\[ x = \sin t, \; y = \csc t, \; 0 < t < \pi \;\Longrightarrow\; y = \tfrac{1}{x} \]
Eliminate and report the resulting equation
Why: The reciprocal identity gives it immediately, and it is correct as far as it goes.
But the parametrization covers only the part with x between 0 and 1, and it covers that part twice — out as t rises to pi over 2 and back as t continues to pi.
Report the equation with its restriction: y equals one over x, for x between 0 and 1 inclusive of 1.
And note separately that the arc is traced twice, since no equation in x and y can record a multiplicity.
This is the case the book warns about: sometimes no restriction can make the eliminated equation faithful, because the information lost is not about extent at all. Knowing which situation you are in requires looking at the parameter graphs, not at the final equation.
Faded example
Eliminate t from x = 2 cosine t and y = 3 sine t.
Fill in the blanks
\cos t = \frac49, \; \sin t = \frac______ \;\Longrightarrow\; \frac______} + \frac______} = 1
Why: Squaring each isolated function and adding gives the Pythagorean Identity's left side, which equals 1. The denominators are the squares of the two coefficients, so the curve is an ellipse with semi-axes 2 and 3 centred at the origin.
Sorting
Algebraic parametrizations and trigonometric ones want different strategies.
Sort into buckets
Sort each parametrization.
Explain it to yourself
Eliminating the parameter is sometimes helpful and sometimes misleading.
Discussion prompt
Explain when it helps, and what to record before doing it.
Hint: What is the eliminated equation good for?
Answer:
It helps when you want to identify the curve — to recognise that a mess of parametric equations is really a parabola or an ellipse, and to read off its centre and axes. That is genuinely useful and often the fastest route to a sketch.
Before eliminating, record the orientation and the extent, because the first cannot be recovered and the second only sometimes can, by adding a restriction.
And check whether the parametrization covers any part more than once, because that too is unrecoverable. The eliminated equation answers what curve is this; the parametrization answers how is it traced, and only one of the two questions survives the elimination.
Section
Section 4
Concept
For a graph, let the parameter be the independent variable. For a segment, use the starting point plus the displacement times a parameter running from 0 to 1. For a circle or ellipse, use the cosine and sine.
The ellipse recipe is the Pythagorean Identity read backwards, which is why it can be reconstructed rather than memorised. The segment recipe is the vector idea of starting point plus displacement, from Lesson 11.8a.
Figure (svg): The four standard parametrization recipes, each with the form of curve it applies to
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.10 Parametric Equations §11.10, pp. 1053-1055
Picture it
Each applies to a shape, not to an equation.
Figure (svg): The four standard parametrization recipes, each with the form of curve it applies to
Recognising which shape you have is the whole difficulty; once that is settled the recipe writes itself, and the ellipse one can be rederived from the identity if forgotten.
Worked example
Example 11.10.3, part 3. Starting point plus displacement.
\[ \text{Parametrize the segment from } (2,-3) \text{ to } (1,5). \]
Find the displacement in x
Why: One minus 2.
\[ -1 \]
Find the displacement in y
Why: Five minus negative 3.
\[ 8 \]
Write each coordinate
Why: Start plus displacement times t.
\[ x = 2 - t, y = -3 + 8 t \]
Set the interval
Why: So that t equal to 0 is the start and 1 the end.
\[ 0 \le t \le 1 \]
Figure (svg): The solution to Worked example a line segment shown as a ladder of expressions, one row per legal move
\[ \begin{cases} x = 2 - t \\ y = -3 + 8t \end{cases}, \quad 0 \le t \le 1 \]
Verify: check both endpoints
Why: At t equal to zero the point is 2 comma negative 3, the start. At t equal to 1 it is 1 comma 5, the end. And eliminating t gives y equal to negative 8x plus 13, a line — so the parametrization traces exactly the segment between those two points.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.10 Parametric Equations §11.10, pp. 1054-1054
Faded example
Parametrize the segment from (0,0) to (3,4).
Fill in the blanks
x = 0 + (3 - 0)t = 3t, \quad y = 0 + (4-0)t = 4t, \quad 0 \le t \le 1
Why: Starting at the origin makes the recipe especially simple: each coordinate is just the displacement times t. At t equal to 1 the point is 3 comma 4, the endpoint, as required.
Worked example
Example 11.10.3, part 5. The recipe, then a restriction.
\[ \text{Parametrize the left half of } \frac{x^2}{4} + \frac{y^2}{9} = 1. \]
Apply the ellipse recipe
Why: Semi-axes 2 and 3, centred at the origin.
\[ x = 2 \cos t, y = 3 \sin t \]
Identify which t give the left half
Why: Where the cosine is non-positive.
Set the interval
Why: From a quarter turn to three quarters.
\[ \frac{\pi}{2} \le t \le 3 \pi / 2 \]
Check the endpoints
Why: Top and bottom of the ellipse.
\[ (0, 3)\text{ to } (0, -3) \]
Figure (svg): The solution to Worked example half an ellipse shown as a ladder of expressions, one row per legal move
\[ \begin{cases} x = 2\cos t \\ y = 3\sin t \end{cases}, \quad \tfrac{\pi}{2} \le t \le \tfrac{3\pi}{2} \]
Verify: substitute back
Why: Putting the two expressions into the ellipse equation gives four cosine squared over 4 plus nine sine squared over 9, which is the Pythagorean Identity and equals 1. So every point lies on the ellipse, and the interval restricts to the half where x is non-positive.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.10 Parametric Equations §11.10, pp. 1055-1055
Error analysis
A student parametrizes the right half of an ellipse.
Annotate
On: \( x = 2\cos t, \; y = 3\sin t, \quad 0 \le t \le \pi \)
Which half an interval produces depends on which trigonometric function controls the coordinate being restricted. Checking one endpoint against the intended half settles it in a second.
Matching
Recognise the shape first.
Match the pairs
Why: The first and last are graphs, so the parameter is set equal to whichever variable is the independent one — x in the first case and y in the last, which is why the fourth recipe exists at all. The circle uses the cosine and sine recipe with the centre added to each coordinate.
Prediction
A curve has been parametrized in one way.
Predict first
How many other parametrizations does it have?
Correct: Infinitely many.
Why: Any change of parameter that covers the same points gives another parametrization: run at twice the speed, start at a different place, run backwards, or compose with any increasing function of a new parameter. The curve is a set of points and the parametrization is one way of scheduling a walk through it, of which there are endlessly many. This is why the book says a parametrization rather than the parametrization.
Two truths and a lie
Three of these are true and one is false.
Eliminate the wrong options
One of these statements about parametrizations is wrong.
Survives elimination: C
Why: Statement C is false. The eliminated equation usually describes more of the curve than the parametrization traced — the whole parabola rather than a piece of it — and a restriction on x or y is needed to fix it. Sometimes even that is not enough, as when the parametrization covers a piece twice.
Section
Section 5
Concept
Replacing the parameter by its negative reverses the orientation; replacing it by the parameter minus a constant delays the start. Both substitutions must be applied to the bounds as well as to the equations.
Both adjustments leave the set of points completely unchanged. Only the schedule of the walk is altered, which is precisely the information the parametrization adds over a bare equation.
Figure (svg): The two ways to adjust a parametrization: replacing t by its negative to reverse the orientation, and by t minus a constant to delay the start
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.10 Parametric Equations §11.10, pp. 1055-1057
Picture it
The path of a point on the rim of a rolling wheel.
Figure (svg): A cycloid, the curve traced by a point on the rim of a rolling circle, with the generating circle drawn at one position
Its rectangular equation exists but cannot be solved for either variable and is of no practical use. The parametrization is two short lines, and it is the reason this topic is worth having.
Worked example
Example 11.10.4, part 1. Both adjustments in turn.
\[ \text{Parametrize } y = x^2 \text{ from } (2,4) \text{ to } (-1,1), \text{ starting at } t = 0. \]
Start from the standard recipe
Why: Which runs the wrong way.
\[ x = t, y = t ^{2}, -1 \le t \le 2 \]
Reverse: replace t by its negative
Why: Bounds included, then tidy.
\[ x = -t, y = t ^{2}, -2 \le t \le 1 \]
Delay by 2: replace t by t minus 2
Why: Bounds included again.
\[ -2 \le t - 2 \le 1 \]
Simplify
Why: Expand the square.
\[ x = 2 - t, y = t ^{2} - 4 t + 4, 0 \le t \le 3 \]
Figure (svg): The solution to Worked example reverse and delay shown as a ladder of expressions, one row per legal move
\[ \begin{cases} x = 2 - t \\ y = t^2 - 4t + 4 \end{cases}, \quad 0 \le t \le 3 \]
Verify: check both ends
Why: At t equal to zero the point is 2 comma 4, the required start. At t equal to 3 it is negative 1 comma 1, the required end. And eliminating t gives y equal to x squared, so the path is on the parabola throughout.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.10 Parametric Equations §11.10, pp. 1055-1055
Faded example
Delay the parametrization with bounds 0 to 1 so that it starts at t equal to 1.
Fill in the blanks
\text1 t \text2 (t-1): \quad 0 \le t - 1 \le 1 \;\Longrightarrow\; ___ \le t \le ___
Why: Adding 1 throughout the inequality shifts the interval to 1 through 2. This is the step that lets a second path pick up exactly where a first one left off, which is how a piecewise route is assembled.
Worked example
The derivation, from a rolling circle.
\[ \text{Find parametric equations for a point on the rim of a circle of radius } r \text{ rolling along the } x\text{-axis.} \]
Parametrize the point on a stationary circle
Why: Clockwise, starting at the bottom.
\[ x = -r \sin \theta, y = -r \cos \theta \]
Find where the centre is at angle theta
Why: It has rolled an arc length r theta.
\[ (r \theta, r) \]
Shift both coordinates by the centre
Why: Add the centre's position.
Simplify
Why: Factor out r.
\[ x = r(\theta - \sin \theta), y = r(1 - \cos \theta) \]
Figure (svg): The solution to Worked example the cycloid shown as a ladder of expressions, one row per legal move
\[ \begin{cases} x = r(\theta - \sin\theta) \\ y = r(1 - \cos\theta) \end{cases}, \quad \theta \ge 0 \]
Verify: check the special angles
Why: At theta equal to zero the point is at the origin, on the ground. At theta equal to pi it is at r pi comma 2r, the top of the wheel and half an arch along. At theta equal to 2 pi it is back on the ground at 2 pi r, one full circumference along. All three match a wheel rolling one revolution.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.10 Parametric Equations §11.10, pp. 1056-1057
Trap
\[ x = t, \; y = t^2, \; -1 \le t \le 2 \;\Longrightarrow\; x = -t, \; y = t^2, \; -1 \le t \le 2 \]
Replace t by its negative in the two equations
Why: That is what the rule says to do.
But the rule says everywhere, and the bounds contain a t as well. Left unchanged, they now describe a different arc of the parabola than intended.
\[ x = -t, \; y = t^2, \quad -1 \le -t \le 2 \;\Longrightarrow\; -2 \le t \le 1 \]
Substitute into the inequality too, then solve it
Why: Dividing by negative 1 flips both inequality signs.
The bounds are part of the parametrization, not an afterthought. Any substitution in the equations must be made in the bounds, and for the reversal that means flipping the inequality, which is the step most often dropped.
Prediction
You replace t by negative t throughout a parametrization.
Predict first
What changes about the curve?
Correct: Only its orientation.
Why: The substitution reaches exactly the same points, just in the reverse order, so the set of points is untouched and only the direction of travel flips. This is the cleanest illustration of the distinction the lesson keeps making: the curve is the set of points and the parametrization is the schedule, and this substitution changes only the second.
Sorting
Two substitutions, each fixing one thing.
Sort into buckets
Sort each requirement.
Real world
The cycloid was studied intensely in the seventeenth century, and turned out to answer two famous questions: the shape of the fastest slide between two points, and the shape of a pendulum path whose period does not depend on its amplitude.
Discussion prompt
Explain why a parametric description was essential to that work, and what this lesson has added to the course as a whole.
Hint: What does the rectangular equation of a cycloid look like?
Answer:
The cycloid's equation in x and y cannot be solved for either variable and is of essentially no use for calculation. Its parametrization is two short expressions in the wheel's rotation angle, and every classical result about it was obtained from that form.
More than that, both questions are about motion — how long a bead takes to slide, how long a pendulum takes to swing. A question about time needs a description that mentions time, and that is exactly what a parameter supplies.
What this lesson adds to the course is the ability to describe a point moving along a path, which is the form nearly every physical application actually takes. Projectile motion, orbits, the position of a machine part: all are naturally two functions of time and unnaturally one equation in x and y.
It is a fitting end to the chapter. The Laws of Sines and Cosines measured static triangles; polar coordinates and complex numbers described position and rotation; vectors described magnitude and direction; and the parameter finally puts all of it in motion. Everything in this course was building toward describing things that move, which is what calculus takes up next.
Comparison
Fill the blanks from memory. The last two rows are what the parameter buys.
Comparison matrix
| An equation in x and y | A parametrization | |
|---|---|---|
| what it describes | a set of points | a point moving along that set |
| curves that are not functions | handled, but not as a function | each coordinate is a genuine function of t |
| orientation | none; a set has no order | supplied by increasing t |
| speed and timing | not recorded | recorded exactly |
| uniqueness | essentially one equation | infinitely many parametrizations |
The two descriptions answer different questions. Which curve is this is answered by the equation; how is it traced is answered only by the parametrization.
Pattern
Five moves, covering both directions between a curve and a parametrization.
Check every parametrization by substituting back into the curve's equation and testing the two endpoints.
OpenStax Algebra and Trigonometry 2e, §10.6 Parametric Equations §10.6
Check
Eliminating the parameter.
Check your understanding
What curve is traced by x = 3 cosine t and y = 3 sine t?
Answer: A
Why: Isolating the cosine and sine gives x over 3 and y over 3, and substituting into the Pythagorean Identity gives x squared plus y squared equal to 9. That is a circle of radius 3, traced counter-clockwise from the point 3 comma 0.
Check
Parametrizing a segment.
Check your understanding
Which parametrizes the segment from (1, 2) to (5, -6) with t running from 0 to 1?
Answer: A
Why: The displacement is 5 minus 1, namely 4, in x, and negative 6 minus 2, namely negative 8, in y. Starting point plus displacement times t gives the answer, and at t equal to 1 it lands on 5 comma negative 6 as required.
Check
Adjusting a parametrization.
Check your understanding
A parametrization has bounds from -3 to 1. After replacing t by negative t, what are the bounds?
Answer: B
Why: The condition becomes negative 3 at most negative t at most 1, and multiplying through by negative 1 flips both inequality signs, giving negative 1 at most t at most 3. The interval keeps its length of 4 but is reflected about zero.
Real world
An animator needs a character's hand to move along a specific arc, arriving at a target at a specific moment, and slowing as it arrives. The path is drawn once and stored.
Discussion prompt
Explain what the animator is manipulating, and which parts of this lesson correspond to which parts of the task.
Hint: Which is the path and which is the timing?
Answer:
The stored arc is the curve — a set of points, with no timing in it. The motion along it is the parametrization, and the animator adjusts the two independently: change the arc without changing the timing, or change the timing without touching the arc.
Arriving at a specific moment is a shift of parameter, exactly the substitution from this section. Reversing the gesture is the reversal substitution. Slowing on approach is a reparametrization — a different schedule through the same points, which animators call easing.
The reason animation software separates the two is precisely the distinction this lesson draws. A path and a schedule are different objects, and mixing them makes either impossible to adjust without disturbing the other.
The same separation runs through robotics, where a planned trajectory is stored as a path plus a timing law; through computer graphics, where curves are parametrized to be evaluated at arbitrary points; and through physics, where a trajectory is position as a function of time and nothing else. The parameter is what turns a shape into a motion, and that is what this lesson has been about.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
You eliminate the parameter and get the equation of a full circle. What must you check?
Correct: Whether the parametrization traces the whole circle, and in which direction.
\[ x = 3\cos t, \; y = 3\sin t, \; 0 \le t \le \pi \;\Longrightarrow\; \text{only the upper half} \]
Why: The eliminated equation always describes the full circle, but the parameter's interval may cover only part of it — as the three-quarter ellipse did — and the equation records no direction at all. Both must be determined from the parametrization and stated separately, since neither survives elimination.
Explain it
They have just met parametric equations and think they are a needlessly complicated way to write down a curve they could have written directly.
Discussion prompt
In no more than five sentences, give them a case where the direct route is not available.
Hint: Ask them to write down the path of a point on a rolling wheel.
Answer:
Ask them for the equation of the path traced by a spot of paint on a bicycle tyre as it rolls. The curve is easy to picture and its equation in x and y is unusable — it cannot be solved for either variable and nobody works with it.
With a parameter it takes two short lines: x is r times the angle minus its sine, and y is r times 1 minus its cosine. The rotation angle of the wheel is the natural variable, and using it makes the description simple.
More generally, a parametrization records when the point is where, which any question about motion needs and no equation in x and y can supply.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: Sketching is fixed by graphing both coordinates against t and breaking the interval at every turning point. Elimination is fixed by substituting for algebraic pairs and using an identity for trigonometric ones, then adding the restriction. Writing one is fixed by recognising the shape and applying the matching recipe. Adjusting is fixed by applying the substitution to the bounds as well as the equations. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of the page draw a self-crossing curve with arrows on it, and write beside it why each coordinate is a function of the parameter even though the curve is a function of neither variable. Underneath, take x equal to 1 plus 3 cosine t and y equal to 2 sine t over the interval from 0 to three halves of pi, and draw the three graphs: x against t, y against t, and the resulting curve, marking the orientation. Beside them, eliminate the parameter and write the ellipse's equation, noting in one line what the elimination lost. In the middle of the page write the four parametrization recipes. In the bottom left, parametrize the segment from (2, -3) to (1, 5) and check both endpoints. In the bottom right, write the two adjustment rules and derive the cycloid equations for a wheel of radius r. Finally, circle the one piece of information that no equation in x and y can ever carry.
The circled item is the orientation. A curve is a set of points and a set has no order, so the direction of travel exists only in the parametrization — which is exactly the thing this last lesson adds to everything that came before it.
Recap
Five things, and together they close the course.
| If the question says | Your first move is |
|---|---|
| Sketch this parametrization | Graph x against t and y against t first |
| Identify the curve | Eliminate the parameter |
| The parametrization uses sine and cosine | Isolate them and use the Pythagorean Identity |
| Parametrize this segment | Start plus displacement times t, on [0, 1] |
| Reverse it, or make it start later | Substitute in the bounds as well as the equations |
That is the last section of the course. It began by measuring a static triangle and ends by describing a point in motion, and every tool along the way — the identities, the polar system, the complex numbers, the vectors and the dot product — was a step toward describing something that moves. What comes next is calculus, which asks how fast.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.10 Parametric Equations §11.10, pp. 1048-1059 — everything on these slides traces back here
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