Curves described by a parameter rather than as graphs, sketching with orientation, eliminating the parameter and what that operation loses, and why a parametrisation is a choice rather than a property of the curve.
Subject: Calculus II · 66 slides · symbolic lesson
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Title
Calculus II · Section 7.1
Curves as journeys: where the point is, and when
Objectives
Until now a curve has meant the graph of a function: one height for each x. Circles, orbits and the path of a point on a rolling wheel do not fit that mould. This lesson describes a curve by telling you where a moving point is at each moment.
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, p. 606 — learning objectives 7.1.1 to 7.1.4
Every curve you have worked with in calculus so far has been the graph of a function: feed in x, get out one y. That covers a lot, but it leaves out some of the most natural curves there are. A circle fails. So does the orbit of a planet, a figure eight, or the path of a pebble stuck in a bicycle tyre.
This lesson changes what you record. Instead of a relation between x and y, you record a journey: where a moving point is at each moment t. Both coordinates become functions of that one parameter, and suddenly circles, loops and cusps are no harder to write down than a straight line.
You will learn to sketch such curves with their direction of travel, to turn them back into ordinary equations when that helps, and to notice exactly what gets lost when you do. The last part meets the famous cycloid, the curve a point on a rolling wheel traces, which is the reason this whole chapter is worth having.
Warm-up
Discussion prompt
The circle of radius 4 about the origin has equation x squared plus y squared equal to 16. Is it the graph of a function y = f(x)? Say which test decides it, and what you would have to do to describe the circle with functions of x.
Write your answer before you reveal it. The vertical line test is the tool you want: a curve is the graph of a function of x exactly when no vertical line meets it more than once.
The circle fails, and the fix you know is a patch: solve for y, get a plus-or-minus, and treat the circle as two separate functions, an upper semicircle and a lower one, glued at the far left and far right. That works for a circle. It gets ugly fast for anything more complicated, and for a curve that loops back over itself, like the ones later in this lesson, there may be no sensible way to cut it into function pieces at all.
Keep this warm-up in mind. By the end of Part 1 you will have a description of the whole circle in one line, with no plus-or-minus and nothing glued.
Section
Part 1
Concept
Number the days of the year 1 to 365. On each day the Earth is at one definite point of its elliptical orbit, so the day number decides both of its coordinates.
\[ \text{day } t \;\longmapsto\; \big(x(t),\, y(t)\big) \]
Neither coordinate is a function of the other: the Earth passes the same x twice a year, once on each side of the Sun. But each coordinate is a perfectly good function of time. That is the whole idea of a parametric description.
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, pp. 606-607 — Figures 7.2 and 7.3
The textbook opens with the Earth for a reason. Its orbit is an ellipse, and if you put axes on it, there is no way to write the orbit as y equals a function of x: for most x-values the Earth is at that x twice a year, once on the near side and once on the far side.
But ask a different question and everything becomes simple. On day 100, where is the Earth? That has one answer. So the day number t determines the x-coordinate, and it determines the y-coordinate. Each coordinate is an honest function of time.
This is the shift in point of view that the whole chapter rests on. Instead of relating x and y to each other directly, you relate each of them to a third quantity, the parameter, and let the curve be whatever the pair traces out as the parameter moves.
Concept
Parametric equations — If x and y are continuous functions of t on an interval I, the equations x = x(t) and y = y(t) are parametric equations and t is the parameter. The set of points (x, y) obtained as t runs over I is the parametric curve, or plane curve, C.
\[ x = x(t), \qquad y = y(t), \qquad t \in I \]
\[ C = \{\,\big(x(t),\, y(t)\big) : t \in I\,\} \]
The letters x and y do double duty: as the functions x(t) and y(t), and as the coordinates of the points they produce. Keep the two roles apart.
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, pp. 607-608 — definition of parametric equations
Read the definition slowly, because it has two layers. The first layer is two ordinary continuous functions of t on an interval. The second layer is the set of points they produce together, and that set is what the book calls the curve C.
The book makes a point that is worth repeating: the letters x and y are being used in two ways at once. In x of t they are names of functions, things you can graph against t. In the pair (x, y) they are coordinates of a point in the plane. When you write x equals t squared minus 3, you are saying that the coordinate x of the moving point is given by the function of t on the right.
Notice also that the interval I is part of the definition. Two parametric descriptions with the same formulas but different intervals are different curves, which is a fact you will use again and again in this lesson.
Notation
Annotate
On: \( x = t^2 - 3, \quad y = 2t + 1, \quad -2 \le t \le 3 \)
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, p. 608 — Example 7.1b
Step through the notes one at a time. The first two tell you that each formula, taken alone, is a function you already know how to graph: a parabola in t, and a line in t. Neither of those graphs is the curve.
The third note is the one people skip. The interval from minus 2 to 3 is what makes the curve start at one point and stop at another. Stretch the interval and the curve gets longer; shrink it and a piece disappears.
The last note is about what you will not see. When you draw the curve in the xy-plane, t is nowhere on the axes. The only traces it leaves are the labels you choose to write beside a few points and the arrows that show which way the point moves. That is why those labels and arrows are not decoration; they are the only record of the parameter in the picture.
Picture it
Figure (svg): Three panels. Left: x of t equals t squared minus 3, a parabola in t. Middle: y of t equals 2t plus 1, a straight line in t. Right: the points (x of t, y of t) for t from minus 2 to 3, forming a parabola lying on its side, with arrows showing the direction of travel.
Read the left two panels at the same t and plot the pair: at t equal to minus 2, x is 1 and y is minus 3, which is the first dot on the right. Step t along and the dots walk up a parabola lying on its side.
Look at the three panels from left to right. The first is x as a function of t: a parabola with its lowest point at t equal to 0. The second is y as a function of t: a straight line. Neither of them looks like the third panel.
To build the third panel, pick a value of t, read x from the first panel and y from the second, and plot the pair. At t equal to minus 2 that gives the point (1, minus 3). At t equal to 0 it gives (minus 3, 1), the leftmost point, because x of t is smallest there. At t equal to 3 it gives (6, 7).
The result is a parabola lying on its side. It is sideways because the quadratic is in x rather than in y. Keep this in mind: which coordinate carries the square decides which way the parabola opens.
Concept
Figure (svg): A circle of radius 4 centred at the origin, drawn as x equals 4 cos t, y equals 4 sin t, with a dashed vertical line x equals 2 meeting it at two points, (2, 3.46) at t equals pi over 3 and (2, minus 3.46) at t equals 5 pi over 3.
\[ x = 4\cos t, \quad y = 4\sin t, \quad 0 \le t \le 2\pi \]
The whole circle comes from one pair of formulas, with no plus-or-minus and no gluing. Two points above the same x are simply visited at two different times.
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, p. 610 — Example 7.1c
Here is the payoff promised in the warm-up. The two formulas four cos t and four sin t describe the complete circle, top and bottom, with no plus-or-minus anywhere.
Look at the dashed vertical line at x equal to 2. It meets the circle twice, at heights of about 3.46 and minus 3.46, which is exactly why the circle is not a function of x. Parametrically there is no conflict at all: the upper point is reached when t is a third of pi, and the lower one when t is five thirds of pi. Two points with the same x, visited at two different times.
The vertical line test is a question about functions of x. A parametric curve is not claiming that y is a function of x, so the test simply does not apply to it. It is not that parametric curves pass the test; the test is asking the wrong question.
Prediction
\[ x = t^2 - 3, \quad y = 2t + 1, \quad -2 \le t \le 3 \]
Predict first
How many points of this curve have x-coordinate −2?
Correct: Two
Why: Setting t squared minus 3 equal to minus 2 gives t squared equal to 1, so t is 1 or minus 1. Both are in the interval, and they give different heights, y equal to 3 and y equal to minus 1. So the vertical line x equal to minus 2 meets the curve twice, and the curve is not the graph of any function of x.
\[ t^2 - 3 = -2 \iff t = \pm 1 \;\Longrightarrow\; (-2, -1),\; (-2, 3) \]
Make your prediction from the formulas, not from memory of the earlier picture. The question is really: how many values of t in the interval make x of t equal to minus 2?
Setting t squared minus 3 equal to minus 2 gives t squared equal to 1, so t is 1 or minus 1, and both are inside the interval from minus 2 to 3. Two values of t, and since y equals 2t plus 1 gives different heights for them, two different points.
This is the algebraic version of the vertical line test failing. Whenever x of t takes the same value at two different times, and y does not, the curve has two points stacked vertically, and it cannot be the graph of a function of x.
Prediction
Figure (svg): The Lissajous curve x equals sin 4t, y equals sin 3t for t from 0 to 2 pi: a woven closed curve inside the square from minus 1 to 1, crossing itself many times, with the origin marked as the point reached at t equals 0 and t equals pi.
Predict first
For x = sin 4t, y = sin 3t with 0 ≤ t < 2π, how many values of t put the point at the origin?
Correct: Two
Why: You need sine of 4t and sine of 3t both zero. The first holds when t is a multiple of a quarter pi, the second when t is a multiple of a third pi; both together only when t is a multiple of pi. In the interval that means t equal to 0 and t equal to pi: one point, two parameter values.
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, p. 624 — Exercise 56
This picture is Exercise 56 from the section, and it is worth staring at. Two very plain functions, sine of 4t and sine of 3t, together trace a woven figure that crosses itself over and over. No function of x could ever do that.
For the prediction, you need both sines to be zero at once. Sine of 4t is zero when 4t is a whole multiple of pi, so when t is a multiple of a quarter of pi. Sine of 3t is zero when t is a multiple of a third of pi. Both happen together only when t is a whole multiple of pi, which in the interval means t equal to 0 and t equal to pi.
So the origin is one point with two parameter values. That is what every self-crossing is: the moving point comes back to a place it has already been, at a later time. The curve as a set of points does not know it was visited twice; the parametrisation does.
Section
Part 2
Concept
The reliable way to sketch any parametric curve is a table with t in the first column. Each row is one point, and the order of the rows is the order of travel.
| step | what you do | what it records |
|---|---|---|
| 1 | choose t values across the interval, including both ends | where the journey starts and stops |
| 2 | compute x(t) and y(t) for each | the points |
| 3 | plot, and join them in order of t | the shape |
| 4 | put arrows along the curve in the direction t increases | the orientation |
Orientation — The direction in which the point moves along the curve as the parameter increases. It is part of the answer to every sketch.
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, p. 608 — Example 7.1a
This procedure is not glamorous, but it never fails, and on an exam it is the method that earns full marks. Put t in the first column, because t is the independent variable. The x and y columns are then just evaluations.
Two details matter. Always include both ends of the interval, because they are where the curve starts and stops, and they are the points most likely to be asked about. And join the points in the order of the rows, not in the order of increasing x. The order of the rows is the order in which the point actually visits them.
The fourth step is what makes a sketch of a parametric curve different from a sketch of a graph. The arrows showing the orientation are part of the answer. Two sketches with the same shape but opposite arrows are sketches of different parametric curves.
Worked example
Sketch the curve and mark its orientation.
\[ x(t) = t - 1, \quad y(t) = 2t + 4, \quad -3 \le t \le 2 \]
Start the table at the left end of the interval
Why: At t equal to minus 3.
\[ t = -3: \quad x = -3 - 1 = -4, \quad y = 2(-3) + 4 = -2 \]
Fill in the remaining rows
Why: Each unit step in t moves x by 1 and y by 2.
| t | −3 | −2 | −1 | 0 | 1 | 2 |
|---|---|---|---|---|---|---|
| x(t) | −4 | −3 | −2 | −1 | 0 | 1 |
| y(t) | −2 | 0 | 2 | 4 | 6 | 8 |
Plot in order of t and add arrows
Why: The points lie on a line and are reached bottom-left first.
Figure (svg): The line segment from (minus 4, minus 2) to (1, 8), with dots at t equals minus 3, minus 2, minus 1, 0, 1, 2 labelled, and arrows pointing up and to the right.
Check by eliminating t
Why: From the first equation t equals x plus 1; substituting gives a line, and both endpoints satisfy it.
\[ y = 2(x+1) + 4 = 2x + 6: \quad 2(-4)+6 = -2, \quad 2(1)+6 = 8 \]
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, pp. 608-609 — Example 7.1a and Figure 7.4
Start by computing the first row carefully, because it fixes the method. At t equal to minus 3, x is minus 3 minus 1, which is minus 4, and y is 2 times minus 3 plus 4, which is minus 2. The rest of the table follows the same pattern: every step of one in t adds one to x and two to y.
Because x and y both change by fixed amounts for each step in t, the points lie on a straight line, and the curve is a segment, not a whole line, because t only runs from minus 3 to 2. The arrows point up and to the right, since both coordinates increase with t.
The check eliminates t, which is quick here: t is x plus 1, so y is 2x plus 6. Then you test both endpoints against that line. The segment sits on the line y equals 2x plus 6, from x equal to minus 4 to x equal to 1.
Worked example
\[ x(t) = t^2 - 3, \quad y(t) = 2t + 1, \quad -2 \le t \le 3 \]
Tabulate
Why: x depends on t squared, so t and minus t give the same x.
| t | −2 | −1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|---|
| x(t) | 1 | −2 | −3 | −2 | 1 | 6 |
| y(t) | −3 | −1 | 1 | 3 | 5 | 7 |
Find the turning point
Why: x is smallest when t is 0, where the point stops moving left.
\[ t = 0: \quad (x, y) = (-3, 1) \]
Plot in order, with arrows
Why: The point runs left and up to the vertex, then right and up.
Figure (svg): A parabola opening to the right, x equals t squared minus 3, y equals 2t plus 1, from (1, minus 3) at t equals minus 2 up to (6, 7) at t equals 3, vertex (minus 3, 1) at t equals 0, with arrows pointing upward along it.
Check an intermediate value
Why: Halfway between two table rows, at t equal to one half, the point must lie on the same curve between them.
\[ t = \tfrac12: \; (x,y) = \left(-\tfrac{11}{4},\, 2\right), \quad y = 2 \text{ lies between } 1 \text{ and } 3 \]
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, p. 609 — Example 7.1b and Figure 7.5
The table is the same one you saw in the three-panel figure, and the thing to notice is the x column: 1, minus 2, minus 3, minus 2, 1, 6. It goes down and comes back up, because x depends on t squared.
That turn is the vertex of the sideways parabola, and it happens at t equal to 0, where x reaches its smallest value, minus 3. Before that moment the point is moving left; after it, the point is moving right. It is moving up the whole time, because y equals 2t plus 1 always increases.
The check uses a value you did not put in the table. At t equal to one half, the point is (minus eleven quarters, 2). Its height, 2, is between the heights of the neighbouring rows, 1 and 3, and its x-value is just to the right of the vertex, which is exactly where the curve in the figure passes. A table can hide surprises between its rows; testing a point in a gap is a cheap way to catch them.
Worked example
\[ x(t) = 4\cos t, \quad y(t) = 4\sin t, \quad 0 \le t \le 2\pi \]
Use multiples of pi over 6
Why: The standard angles give exact values.
\[ t = \tfrac{\pi}{6}: \quad x = 4\cdot\tfrac{\sqrt3}{2} = 2\sqrt3 \approx 3.46, \quad y = 4\cdot\tfrac12 = 2 \]
Continue round the quarter-turns
Why: Each quarter of the interval is a quarter-turn.
\[ t = 0, \tfrac{\pi}{2}, \pi, \tfrac{3\pi}{2}: \quad (4,0),\; (0,4),\; (-4,0),\; (0,-4) \]
Plot and read the direction
Why: Increasing t moves the point from the positive x-axis up to the positive y-axis: counterclockwise.
Figure (svg): A circle of radius 4 about the origin with twelve dots at t equal to multiples of pi over 6, labels at t equals 0, pi over 6, pi over 2, pi and 3 pi over 2, and arrows running counterclockwise.
Check that every point is on the circle
Why: Square and add; the Pythagorean identity leaves 16.
\[ (4\cos t)^2 + (4\sin t)^2 = 16(\cos^2 t + \sin^2 t) = 16 \]
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, pp. 610-611 — Example 7.1c and Figure 7.6
Use multiples of pi over 6 because they give exact values: cosine and sine of the standard angles. At pi over 6 the point is (2 root 3, 2), about (3.46, 2). The quarter-turns land on the axes: (4, 0), (0, 4), (minus 4, 0), (0, minus 4).
The direction comes straight from those quarter-turns. The point starts on the positive x-axis and a quarter of the way through the interval it is on the positive y-axis. That is counterclockwise. And it starts and finishes at the same point, (4, 0), because cosine and sine both repeat after 2 pi.
The check is the real reason this parametrisation works. Square both coordinates and add, and the factor 16 comes out in front of cos squared plus sin squared, which is always 1. So every point is at distance 4 from the origin: every point is on the circle, whatever t is.
Worked example
\[ x(t) = 3t + 2, \quad y(t) = t^2 - 1, \quad -3 \le t \le 2 \]
Tabulate
Why: Now x is the linear one and y the quadratic one.
| t | −3 | −2 | −1 | 0 | 1 | 2 |
|---|---|---|---|---|---|---|
| x(t) | −7 | −4 | −1 | 2 | 5 | 8 |
| y(t) | 8 | 3 | 0 | −1 | 0 | 3 |
Plot with arrows
Why: x increases with t, so the travel is left to right.
Figure (svg): An upward-opening parabola traced from (minus 7, 8) at t equals minus 3, down to the vertex (2, minus 1) at t equals 0, and up to (8, 3) at t equals 2, with arrows pointing left to right.
Identify the curve
Why: Solve the linear equation for t and substitute.
\[ t = \frac{x-2}{3} \;\Longrightarrow\; y = \left(\frac{x-2}{3}\right)^2 - 1, \quad -7 \le x \le 8 \]
Check both endpoints
Why: The ends of the t-interval must land on the xy-equation.
\[ x=-7: \left(\tfrac{-9}{3}\right)^2 - 1 = 8, \qquad x = 8: \left(\tfrac{6}{3}\right)^2 - 1 = 3 \]
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, p. 611 — Checkpoint 7.1
Compare this with Example 7.1(b). The roles have swapped: now x is the linear function of t and y is the quadratic one. So this time the parabola opens upward, like the graphs you are used to.
Because x equals 3t plus 2 increases steadily with t, the point moves left to right the whole way. It comes down the left side of the parabola, reaches the vertex (2, minus 1) at t equal to 0, and climbs the right side to (8, 3).
When one of the equations is linear in t, eliminating is easy: solve it for t and substitute. Here that gives y as the square of x minus 2 over 3, minus 1. The last step checks the two endpoints against that equation, and it also reminds you to attach the restriction: x only runs from minus 7 to 8, because t only runs from minus 3 to 2.
Concept
Figure (svg): Two unit circles side by side. Left: x equals cos t, y equals sin t, starting at (1, 0) with arrows counterclockwise; t equals pi over 2 is at the top. Right: x equals cos t, y equals minus sin t, also starting at (1, 0) but with arrows clockwise; t equals pi over 2 is at the bottom.
Replace t by minus t and every point is visited in the opposite order. The interval flips too, so the new journey starts where the old one ended.
\[ \big(x(t), y(t)\big),\; a \le t \le b \quad\longrightarrow\quad \big(x(-t), y(-t)\big),\; -b \le t \le -a \]
Stewart, Calculus: Early Transcendentals 8e, §10.1 Curves Defined by Parametric Equations §10.1, p. 642 — orientation of a parametric curve
Look at the two circles. They are the same set of points, both starting at (1, 0), but on the left the point first rises and on the right it first falls. On the left, a quarter of the way through, it is at the top; on the right, at the bottom.
The general trick is to replace t by minus t. The point that used to be visited at time t is now visited at time minus t, so the whole schedule runs backward. For the circle, cosine of minus t is cosine of t and sine of minus t is minus sine of t, which is exactly the right-hand panel.
Do not forget the interval. If the old parameter ran from a to b, the new one runs from minus b to minus a. Get this wrong and you trace a different piece of the curve instead of the same piece backward.
Concept
Give a parametrisation of the same segment that runs downward.
\[ x = t - 1, \quad y = 2t + 4, \quad -3 \le t \le 2 \]
Replace t by minus t
Why: Every formula gets minus t where t was.
\[ x = -t - 1, \quad y = -2t + 4 \]
Flip the interval
Why: Minus 3 to 2 becomes minus 2 to 3.
\[ -3 \le -t \le 2 \iff -2 \le t \le 3 \]
Find the new start
Why: At the new left end, t equal to minus 2.
\[ t = -2: \quad (x, y) = (2 - 1,\, 4 + 4) = (1, 8) \]
Check the end
Why: At t equal to 3 the point should be the old start.
\[ t = 3: \quad (x, y) = (-4,\, -2) \quad \checkmark \]
This is the reversal rule applied to a concrete example, and each line is one move. Replace every t by minus t in the formulas, then turn the interval inside out: minus 3 to 2 becomes minus 2 to 3.
The new starting point is where the new parameter begins, at t equal to minus 2. Plugging that in gives (1, 8), which is where the old journey ended. The new journey ends at t equal to 3, at (minus 4, minus 2), the old starting point.
So the new description traces exactly the same segment, from the top right down to the bottom left. That check, new start equals old end and new end equals old start, is the fastest way to confirm a reversal is right.
Error analysis
Annotate
On: \( x = t^2,\; y = t,\; -2 \le t \le 2 \;\Longrightarrow\; t = \sqrt{x} \;\Longrightarrow\; y = \sqrt{x} \)
Step through the notes and find the line where the argument goes wrong. The formulas are fine, and y equals the square root of x does contain some of the points. The problem is the step from t squared equals x to t equals the square root of x.
A square root returns only the non-negative answer. The equation t squared equals x has two solutions when x is positive, plus and minus the root, and choosing only the plus one throws away every point with negative t. At t equal to minus 1 the curve is at (1, minus 1), below the x-axis, and the proposed answer has no points there.
The general lesson is to solve for t from the equation that makes it easy and one-to-one. Here y equals t is already solved, so x equals y squared, a sideways parabola, with y from minus 2 to 2. Solving the harder equation introduced a choice that was not really there.
Matching
Match the pairs
Why: All four live on the unit circle. Flipping the sign of sine reverses the direction; doubling the angle doubles the speed, so the same interval goes round twice; halving the interval stops the point at the far side.
All four descriptions put the point on the unit circle, so the matching is about the journey, not the shape. Work out where each one starts and which way it goes at the start.
Flipping the sign of sine reflects the circle top to bottom, which reverses the direction of travel. Doubling the angle to 2t does not change the circle, but the angle now runs up to 4 pi, so the point goes round twice in the same interval. Cutting the interval to 0 to pi stops the point halfway round, at (minus 1, 0), so only the upper half is traced.
If you mixed up any of these, notice that you cannot tell any of them apart from the equation x squared plus y squared equals 1. The differences live only in the parametrisation.
Trap
Asked to sketch the curve below, a student draws a full circle:
\[ x = 4\cos t,\; y = 4\sin t \]
\[ 0 \le t \le \pi \]
Wrong. The interval was ignored.
The formulas are the circle's, but the interval stops at pi, the far side. Only the upper semicircle is drawn, from (4, 0) to (minus 4, 0), counterclockwise. The range of t is part of the curve.
\[ y = 4\sin t \ge 0 \text{ for } 0 \le t \le \pi \]
This is the mistake of recognising the formulas and forgetting to read the rest of the question. Four cos t and four sin t are the circle's formulas, and a full circle is what they trace when t runs through 2 pi. But the interval here stops at pi.
Check it directly: for t between 0 and pi, sine of t is never negative, so y is never below the x-axis. The point starts at (4, 0), passes over the top at (0, 4), and stops at (minus 4, 0). An upper semicircle, counterclockwise.
Make it a habit to read the interval before you sketch anything. The formulas tell you which curve the point is on; the interval tells you how much of it gets drawn.
Section
Part 3
Concept
To recognise a parametric curve, rewrite it as one equation in x and y. When one equation can be solved for t, solve it and substitute into the other. Example 7.1(b), in visible lines:
\[ y = 2t + 1 \;\Longrightarrow\; t = \frac{y-1}{2} \]
\[ x = \left(\frac{y-1}{2}\right)^2 - 3 \]
\[ x = \frac{y^2 - 2y + 1}{4} - 3 = \frac{y^2 - 2y - 11}{4} \]
This is x as a function of y: a parabola opening to the right, exactly the picture from the table. The endpoints (1, minus 3) and (6, 7) came from the interval for t, and the equation alone does not know about them.
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, p. 611 — eliminating the parameter
A table and a sketch tell you what a curve looks like, but they do not tell you its name. To recognise the curve, you want one equation in x and y, with t gone. That process is called eliminating the parameter.
Here y equals 2t plus 1 is linear, so it is easy to solve for t: subtract 1 and halve. Substituting that into the x equation and expanding the square gives x as a quadratic in y. A quadratic in y is a parabola opening sideways, which is exactly what the table showed.
Look at the last sentence on the slide, because it is the theme of this whole part. The equation you get describes an infinite parabola. The curve you started with is a finite piece of it, from (1, minus 3) to (6, 7). The endpoints came from the interval for t, and elimination does not carry them across automatically.
Notation
Annotate
On: \( x = \frac{y^2 - 2y - 11}{4}, \qquad -3 \le y \le 7 \)
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, p. 611 — the restriction on the parameter
Step through the three notes, because together they are the honest summary of what elimination does. It keeps the shape: every point of the curve satisfies the equation, so the equation tells you what kind of curve you are on.
It does not keep the restriction. The interval for t translated into y running from minus 3 to 7, and that range has to be written next to the equation by hand. Leave it off and you are describing a much bigger set of points than the curve actually contains.
And some things are simply gone. No equation in x and y can say which way the point travels, where it starts, or how fast it moves. If the question asks about any of those, you answer it from the parametric form, not from the eliminated one.
Worked example
\[ x(t) = \sqrt{2t + 4}, \quad y(t) = 2t + 1, \quad -2 \le t \le 6 \]
Square the first equation
Why: Squaring loses the sign, so record now that x cannot be negative.
\[ x^2 = 2t + 4, \qquad x \ge 0 \]
Solve for t
Why: Subtract 4 and halve.
\[ t = \frac{x^2 - 4}{2} \]
Substitute into y
Why: The 2 cancels the halving.
\[ y = 2\left(\frac{x^2-4}{2}\right) + 1 = x^2 - 4 + 1 = x^2 - 3 \]
Find the x-range from the t-range
Why: The square root increases with t.
\[ t = -2: x = \sqrt{0} = 0, \qquad t = 6: x = \sqrt{16} = 4 \]
Figure (svg): The parabola y equals x squared minus 3 drawn dashed for x from minus 4 to 4; only the piece from (0, minus 3) at t equals minus 2 to (4, 13) at t equals 6 is drawn solid, with arrows pointing up the right half.
Check the far endpoint in both forms
Why: At t equal to 6 the parametric point and the xy-equation must agree.
\[ (x,y) = (4,\,13), \qquad 4^2 - 3 = 13 \;\checkmark \]
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, pp. 612-613 — Example 7.2a and Figure 7.7
The first step squares the equation for x to get rid of the root. That is a legitimate move, but it is not reversible: squaring turns minus 2 and 2 into the same number. So at the moment you square, write down what the root was telling you, that x is never negative. The textbook makes exactly this point.
After that the algebra is routine: solve for t, substitute into y, and the 2 cancels the half to leave y equals x squared minus 3. Then translate the t-range into an x-range. The square root of 2t plus 4 increases with t, so x runs from 0 at t equal to minus 2 up to 4 at t equal to 6.
The figure shows why all of this matters. The dashed curve is everything the equation y equals x squared minus 3 describes. The solid piece is the actual curve: the right half only, and only up to height 13. The final check confirms the far endpoint satisfies the equation in both forms.
Worked example
\[ x(t) = 4\cos t, \quad y(t) = 3\sin t, \quad 0 \le t \le 2\pi \]
Do not solve for t
Why: Cosine and sine are not one-to-one, so an inverse would lose most of the curve. Isolate them instead.
\[ \cos t = \frac{x}{4}, \qquad \sin t = \frac{y}{3} \]
Use the Pythagorean identity
Why: The one relation that cosine and sine always satisfy.
\[ \cos^2 t + \sin^2 t = 1 \]
Substitute
Why: Replace each trig function by its expression in x or y.
\[ \left(\frac{x}{4}\right)^2 + \left(\frac{y}{3}\right)^2 = 1 \iff \frac{x^2}{16} + \frac{y^2}{9} = 1 \]
Figure (svg): A horizontal ellipse with semi-axes 4 and 3, x equals 4 cos t and y equals 3 sin t, with dots at t equals 0, pi over 2, pi and 3 pi over 2 on the axes, and arrows running counterclockwise.
Check a point off the axes
Why: At t equal to pi over 4 both coordinates are nonzero.
\[ \frac{(2\sqrt2)^2}{16} + \frac{(3\sqrt2/2)^2}{9} = \frac{8}{16} + \frac{4.5}{9} = 1 \;\checkmark \]
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, pp. 613-614 — Example 7.2b and Figure 7.8
Here solving for t is the wrong move. You could write t as the inverse cosine of x over 4, but inverse cosine only returns angles from 0 to pi, so you would silently lose the whole lower half of the curve. The book says it plainly: sine and cosine are not one-to-one, so do not invert them.
Instead, get cos t and sin t on their own, and then use the one equation that links them no matter what t is: cos squared plus sin squared equals 1. Substitute and the parameter is gone, leaving x squared over 16 plus y squared over 9 equals 1. That is an ellipse centred at the origin, 4 units to each side and 3 up and down.
The figure shows the direction: at t equal to 0 the point is at (4, 0), a quarter-turn later it is at (0, 3), so it runs counterclockwise, once round. The check uses a point off the axes, where a wrong answer would be most likely to show up.
Worked example
\[ x(t) = 2 + \frac{3}{t}, \quad y(t) = t - 1, \quad 2 \le t \le 6 \]
Solve the first equation for t
Why: Subtract 2, then take reciprocals.
\[ x - 2 = \frac{3}{t} \;\Longrightarrow\; t = \frac{3}{x - 2} \]
Substitute into y
Why: A shifted reciprocal: a hyperbola.
\[ y = \frac{3}{x-2} - 1 \]
Find the endpoints
Why: x decreases as t increases.
\[ t = 2: (3.5,\, 1), \qquad t = 6: (2.5,\, 5) \]
Attach the restriction
Why: Only the arc between those x values is drawn, traced right to left.
\[ y = \frac{3}{x-2} - 1, \qquad 2.5 \le x \le 3.5 \]
Figure (svg): The curve y equals 3 over x minus 2, minus 1, drawn dashed with its vertical asymptote x equals 2; the parametric piece from (3.5, 1) at t equals 2 to (2.5, 5) at t equals 6 is drawn solid, with arrows moving up and to the left.
Check a middle value
Why: At t equal to 4 the point is (2.75, 3).
\[ \frac{3}{2.75 - 2} - 1 = \frac{3}{0.75} - 1 = 3 \;\checkmark \]
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, p. 614 — Checkpoint 7.2
The equation for y is the simple one here, but the equation for x is also one-to-one in t on this interval, so either can be solved. Solving the x equation gives t equals 3 over x minus 2, and substituting gives y as a shifted reciprocal: a hyperbola with a vertical asymptote at x equal to 2.
Now the restriction. At t equal to 2 the point is (3.5, 1) and at t equal to 6 it is (2.5, 5). So only a short arc of one branch is drawn, and as t increases, x decreases. The point moves up and to the left, which is easy to get backwards if you assume curves are traced left to right.
In the figure, the dashed curves are everything the equation describes, both branches, stretching off to infinity. The solid arc is the real curve. The check at t equal to 4 confirms a middle point sits on the equation.
Trap
Eliminating t from Example 7.2(a):
\[ x = \sqrt{2t+4},\; y = 2t+1 \]
\[ \Longrightarrow\; y = x^2 - 3 \]
Wrong as a description of the curve: that is the whole parabola.
The square root can never be negative and t stops at 6, so the curve is only the arc from (0, minus 3) to (4, 13). The equation is right; the answer is incomplete without its domain.
\[ y = x^2 - 3, \quad 0 \le x \le 4 \]
This is the most common way to lose marks on elimination questions, and the algebra is not where it goes wrong. Every line of the elimination is correct. The answer is still wrong, because it describes the whole parabola, and the curve is only part of it.
The fix is to ask, every time, what values x and y can actually take. Look for the usual suspects: a square root, which is never negative; a square, which is never negative; an exponential, which is always positive; a sine or cosine, which stays between minus 1 and 1; and the ends of the t-interval.
Write the restriction next to the equation, as part of the same answer. An equation with its domain describes the curve. An equation alone describes something else.
Worked example
Eliminate the parameter, t ranging over all real numbers, and describe the motion.
\[ x = 2t^2, \qquad y = t^4 + 1 \]
Solve the first equation for t squared
Why: No need for t itself: y only uses t to the fourth, the square of t squared.
\[ t^2 = \frac{x}{2} \]
Substitute
Why: t to the fourth is t squared, squared.
\[ y = \left(\frac{x}{2}\right)^2 + 1 = \frac{x^2}{4} + 1 \]
Record the restriction
Why: x is twice a square.
\[ x = 2t^2 \ge 0 \]
Figure (svg): Two panels showing the same half-parabola from (0, 1) to (4.5, 6.06). Left: for t from minus 1.5 to 0 the point runs down the arc toward (0, 1). Right: for t from 0 to 1.5 it runs back up the same arc.
Check the double tracing
Why: Opposite values of t give the same point, so the arc is covered once coming in and once going out.
\[ t = \pm 1: \quad (x, y) = (2,\, 2), \qquad \frac{2^2}{4} + 1 = 2 \;\checkmark \]
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, p. 623 — Exercise 5
This exercise has no t-interval, so t can be any real number, and the restriction comes from the formulas instead. Notice that you never need t itself. The y equation uses t to the fourth, which is just t squared, squared, so solve for t squared and substitute.
That gives y equals x squared over 4 plus 1, but x is twice a square, so x can never be negative. Only the right half of that parabola is part of the curve.
The figure shows something the equation cannot. Both coordinates use even powers of t, so t and minus t land on the same point. As t runs from very negative up to 0, the point slides down the arc to (0, 1); as t continues from 0 upward, it climbs back out along the very same arc. The curve is traced twice, once in each direction, and the eliminated equation records neither trip.
Worked example
\[ x = t^2, \qquad y = t^3 \]
Look for a combination without t
Why: Cube x and square y; both give t to the sixth.
\[ x^3 = t^6, \qquad y^2 = t^6 \]
Set them equal
Why: The parameter is gone.
\[ y^2 = x^3, \qquad x \ge 0 \]
Solve for y
Why: Two branches, one for positive t and one for negative t.
\[ y = \pm x^{3/2} \]
Figure (svg): The curve x equals t squared, y equals t cubed for t from minus 1.6 to 1.6: two branches meeting at a sharp point at the origin, the lower branch traced leftward into the origin and the upper branch traced rightward out of it.
Check a point on each branch
Why: At t equal to 2 and to minus 2.
\[ (4, 8),\; (4, -8): \quad 8^2 = 64 = 4^3 \;\checkmark \]
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, p. 623 — Exercise 26
Neither equation is convenient to solve for t, and there is no trig identity to use. The trick is to raise the two equations to powers that make them match: x cubed and y squared are both t to the sixth, so they must be equal.
The result, y squared equals x cubed, has two branches, y equal to plus or minus x to the three halves. Positive t gives the upper branch and negative t gives the lower branch, and x, being a square, is never negative.
Look at the origin in the figure. Two perfectly smooth formulas have produced a sharp point, called a cusp. The reason is that the point momentarily stops there: at t equal to 0, both x and y have zero rate of change, and when the point sets off again it leaves in a new direction. Section 7.2 will make that precise with derivatives.
Sorting
Sort into buckets
Eliminate t in your head. Is the xy-equation on its own exactly the curve, or must a restriction be attached?
For each item, eliminate the parameter in your head, and then ask a second question: can the point actually reach every point of that equation's graph? If yes, the equation is the curve. If not, a restriction has to be attached.
The common culprits are quantities that cannot take every value. A square, like t squared, is never negative. A sine or cosine never leaves the band from minus 1 to 1. An exponential is always positive. Each of these limits x or y, and the equation you get does not know it.
The ellipse is a useful contrast. Its t-interval is finite, but it runs through a whole turn, so every point of the ellipse is reached, and the equation on its own is exactly right.
Step zero
\[ x = 1 + \cos t, \qquad y = 3 - \sin t, \qquad 0 \le t \le 2\pi \]
Discussion prompt
Exercise 27. Before writing anything, decide: will you solve for t, or use an identity? Then carry it out, name the curve, and find its direction.
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, p. 623 — Exercise 27
Decide your strategy before you write, because the wrong choice here produces an answer that looks fine and is missing most of the curve. Inverse cosine and inverse sine only return angles in a limited range, so solving for t would lose points.
Whenever both equations contain a cosine and a sine of the same angle, isolate them and use cos squared plus sin squared equals 1. Here cos t is x minus 1 and sin t is 3 minus y, and the identity gives a circle of radius 1 centred at (1, 3).
The direction needs one more thought, because of the minus sign in front of the sine. At t equal to 0 the point is at (2, 3), to the right of the centre. A quarter-turn later it is at (1, 2), below the centre. Right, then down, is clockwise.
Section
Part 4
Concept
Any graph of a function can be parametrised by letting x itself be the parameter. The curve is traced left to right, at unit speed in x.
\[ y = f(x) \quad\longrightarrow\quad x(t) = t, \quad y(t) = f(t) \]
But nothing forces that choice. Any x(t) whose values cover the whole domain will do, with y(t) then defined as f of x(t). This process is called parameterization of the curve.
\[ x(t) = g(t) \quad\Longrightarrow\quad y(t) = f\big(g(t)\big) \]
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, p. 614 — parameterization of a curve
So far you have started with parametric equations and found the curve. Now go the other way: start with a curve and find parametric equations for it.
For the graph of a function there is always one easy answer. Let x be the parameter itself, so x of t is t, and then y of t is f of t. The point moves left to right, with its x-coordinate increasing at exactly one unit per unit of t.
But there is nothing special about that choice. Any function g of t will do for x, as long as its values cover every x the curve needs. Then y is forced: it must be f of g of t, so that the point stays on the graph. The next example uses this to build a second parametrisation of a parabola.
Worked example
\[ y = 2x^2 - 3 \]
The first choice: x equals t
Why: No restriction on x, so none on t.
\[ x(t) = t, \quad y(t) = 2t^2 - 3, \quad t \in \mathbb{R} \]
A second choice for x
Why: 3t minus 2 takes every real value, so the whole parabola is still covered.
\[ x(t) = 3t - 2 \]
Substitute into the equation
Why: y must be 2 x squared minus 3 with this x.
\[ y(t) = 2(3t-2)^2 - 3 \]
Expand the square
Why: One move.
\[ y(t) = 2\left(9t^2 - 12t + 4\right) - 3 \]
Distribute and collect
Why: 18t squared minus 24t plus 8, minus 3.
\[ y(t) = 18t^2 - 24t + 5 \]
Figure (svg): Two copies of the parabola y equals 2x squared minus 3 with dots at x equals minus 2 to 2. Left, for x equals t, the dots are labelled t equals minus 2 to 2. Right, for x equals 3t minus 2, the same dots are labelled t equals 0, 1/3, 2/3, 1 and 4/3.
Check at t equal to 0, and correct the book
Why: x is minus 2, so y must be 2 times 4 minus 3, which is 5. The printed solution writes minus 2 for minus 3 in its first line and arrives at plus 6, which gives the wrong point.
\[ t = 0: \; (-2,\, 5), \quad 2(-2)^2 - 3 = 5 \ne 6 \]
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, pp. 614-615 — Example 7.3 (misprint: +6 should be +5)
The first parametrisation is the automatic one: x equals t, y equals 2t squared minus 3. The book then picks x equals 3t minus 2 as a second choice. The only condition to check is that 3t minus 2 takes every real value, which a non-constant linear function does.
Then y is not a choice any more: it must be 2 times x squared minus 3 with this x. Expand the square in one line, distribute and collect in the next, and you get 18t squared minus 24t plus 5.
Now the check, which catches a real misprint. The printed solution writes minus 2 instead of minus 3 at the start of its working and ends with plus 6. Test t equal to 0: x is minus 2, and the parabola says y must be 2 times 4 minus 3, which is 5. The printed formula gives 6, a point that is not on the curve. The correct constant is 5.
The figure shows what changed between the two parametrisations: nothing about the points, only which value of t is attached to each one.
Worked example
\[ y = x^2 + 2x \]
The standard one
Why: Let x be t.
\[ x = t, \quad y = t^2 + 2t \]
A shift that suits this curve
Why: Completing the square shows the vertex is at x equal to minus 1, so start the clock there.
\[ x = t - 1 \]
Substitute
Why: Expand both terms.
\[ y = (t-1)^2 + 2(t-1) = t^2 - 2t + 1 + 2t - 2 \]
Simplify
Why: The t terms cancel.
\[ x = t - 1, \quad y = t^2 - 1 \]
Check with a point
Why: At t equal to 3 the point is (2, 8); on the original equation 4 plus 4 is 8.
\[ x = 2: \quad 2^2 + 2(2) = 8 = 3^2 - 1 \;\checkmark \]
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, p. 615 — Checkpoint 7.3
The first answer is the standard one, x equals t. For the second, you can use any x of t that covers all real numbers, and it is worth choosing one that makes the algebra nicer.
Completing the square shows the vertex of this parabola is at x equal to minus 1. If you set x equal to t minus 1, the clock reads zero exactly at the vertex, and after expanding, the t terms cancel and y becomes t squared minus 1. That is a much simpler formula than you would get from a random choice.
The check picks t equal to 3, which gives the point (2, 8), and confirms that 2 squared plus 2 times 2 is indeed 8. The book's own answer uses a different second choice; any correct one is equally right.
Picture it
Figure (svg): Four copies of the line y equals 2x. Each shows dots at t equals minus 1, minus one half, 0, one half and 1 for a different parametrisation: (t, 2t) evenly spaced moving up; (2t, 4t) evenly spaced but twice as far apart; (minus t, minus 2t) moving down; (t cubed, 2t cubed) bunched near the origin and spreading out.
Each panel is the line y equal to 2x. The dots are at the same five values of t in every panel, and they sit in very different places: evenly spaced, twice as spread, running the other way, or bunched at the origin.
Look at the four panels as four different travellers on the same road. The dots mark where each one is at t equal to minus 1, minus a half, 0, a half and 1, and the yellow dot is where each one is at t equal to 1.
The first moves steadily up and to the right. The second covers twice the distance in the same time, so its dots are twice as far apart. The third goes the other way. The fourth, t cubed, is the strangest: it crawls through the origin, where its dots are bunched together, and then speeds up.
Let t run over all real numbers and every panel covers the whole line. As sets of points they are identical. As journeys they are all different, and the differences are exactly the things elimination throws away: starting point, direction and speed.
Intuition
A curve is a set of points. A parametrisation is a way of travelling through that set: it adds a starting place, a direction and a speed that the set itself does not have.
\[ (t, 2t), \quad (2t, 4t), \quad (-t, -2t), \quad (t^3, 2t^3) \]
So there is never one correct parametrisation of a curve. There are infinitely many, and the useful question is which one suits the job: the one whose parameter means something (time, an angle, a distance rolled), or the one that makes the algebra easy.
This is the conceptual heart of Part 4. A curve, the set of points, exists independently of how you travel along it. A parametrisation adds a travel plan: where you are at time zero, which way you are heading, how fast you go.
So it never makes sense to ask for the parametrisation of a curve. There are infinitely many, and all of them are correct. What does make sense is to ask which one is useful. Sometimes the parameter should mean something physical, like time for a projectile or the angle a wheel has turned for the cycloid. Sometimes the best choice is simply the one that makes the formulas short.
When you are asked to parametrise a curve, any choice that traces the right set of points is a right answer. When you are given a parametrisation, the extra information it carries is often exactly what the question is about.
Comparison
Comparison matrix
| parametrisation | point at t = 1 | direction as t increases | at the origin when t = |
|---|---|---|---|
| (t, 2t) | (1, 2) | up and right | 0 |
| (2t, 4t) | (2, 4) | up and right | 0 |
| (−t, −2t) | (−1, −2) | down and left | 0 |
| (t³, 2t³) | (1, 2) | up and right | 0 |
| (t + 1, 2t + 2) | (2, 4) | up and right | −1 |
Fill in each blank by substituting, not by guessing. For the point at t equal to 1, just evaluate both formulas at 1. For the direction, ask what happens to x as t increases. For when the point is at the origin, solve for the t that makes x zero.
Notice which column changes and which does not. The direction and the timing change from row to row, but every row lives on y equals 2x.
The last row is a shift in time: t plus 1 reaches the origin at t equal to minus 1, one unit earlier than the others. Same road, same speed, same direction, but the clock has been reset.
Worked example
Exercise 49 shows the equations below describe a circle; use them for radius 5 and centre (minus 2, 3).
\[ x = h + r\cos\theta, \qquad y = k + r\sin\theta \]
Isolate the trig functions
Why: Move the centre across and divide by r.
\[ \cos\theta = \frac{x-h}{r}, \qquad \sin\theta = \frac{y-k}{r} \]
Apply the identity
Why: The sum of the squares is 1.
\[ (x-h)^2 + (y-k)^2 = r^2 \]
Insert the given centre and radius
Why: h is minus 2, k is 3, r is 5.
\[ x = -2 + 5\cos\theta, \quad y = 3 + 5\sin\theta, \quad 0 \le \theta < 2\pi \]
Check two points
Why: At theta equal to 0 the point is (3, 3); at a half-turn it is (minus 7, 3). Both are 5 from the centre.
\[ (3+2)^2 + 0^2 = 25, \qquad (-7+2)^2 + 0^2 = 25 \]
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, p. 624 — Exercises 49 and 50
This generalises the circle you have been using all lesson. The cosine and sine terms describe a circle of radius r about the origin, and adding h and k slides it so the centre sits at (h, k).
To prove that, isolate the cosine and the sine, and use the identity: the squares of x minus h over r and y minus k over r add to 1, which rearranges to the familiar centre-radius equation of a circle.
Then it is substitution: centre (minus 2, 3) and radius 5. The check tests two points, the start at theta equal to 0 and the half-turn point, and confirms both are exactly 5 from the centre. This is the form to reach for whenever you need a circle as parametric equations; the direction is counterclockwise, and you can reverse it by changing the sign in front of the sine.
Counterexample
Discussion prompt
Claim: if two parametrisations eliminate to the same xy-equation, they draw the same curve. Break it, using the equation y equal to x squared.
Try to break the claim before you reveal the answer. The claim sounds plausible, since elimination is supposed to identify the curve. But you now know elimination can lose restrictions, so look for a parametrisation that cannot reach all of y equals x squared.
Sine of t is a good candidate, because it is trapped between minus 1 and 1. Set x equal to sin t and y equal to sin squared t. Then y is x squared, yet x never leaves the interval from minus 1 to 1. As t increases the point sweeps back and forth along a short arc of the parabola, forever.
So two parametrisations can share an eliminated equation and still describe different sets of points. That is precisely why the restriction belongs in the answer.
Prediction
\[ x = \cos 2t, \qquad y = \sin 2t, \qquad 0 \le t \le 2\pi \]
Predict first
How many times does the point go round the unit circle?
Correct: Twice
Why: The angle is 2t, which runs from 0 to 4 pi as t runs from 0 to 2 pi. Each 2 pi of angle is one lap, so the point laps the circle twice. The xy-equation is still x squared plus y squared equal to 1, which cannot show the second lap at all.
Predict before you compute. The formulas are the unit circle's, but with 2t inside instead of t.
The angle is what counts. As t goes from 0 to 2 pi, the angle 2t goes from 0 to 4 pi, and every 2 pi of angle is a full lap. So the point goes round twice. You could also think of it as the same trip taken at double speed: the point reaches each place in half the time, so in the same interval it has time for two laps.
This is another thing an xy-equation cannot show. Once round or a hundred times round, the equation is still x squared plus y squared equals 1.
Trap
On a quiz asking for a parametrisation of the ellipse, one answer marked wrong:
\[ \frac{x^2}{16} + \frac{y^2}{9} = 1 \]
\[ x = -4\cos t,\; y = 3\sin t \]
The grader wanted 4 cos t. The grader was wrong.
Both are correct: the minus sign starts the journey at (minus 4, 0) and runs clockwise, but every point still satisfies the equation, and the whole ellipse is covered. A question that asks for a parametrisation has infinitely many right answers.
\[ \frac{16\cos^2 t}{16} + \frac{9\sin^2 t}{9} = 1 \]
This trap is on the grader's side of the desk, and it is surprisingly common. The expectation is that the ellipse has a standard parametrisation, 4 cos t and 3 sin t, and anything else is wrong.
Check the answer that was marked wrong. Minus 4 cos t squared is still 16 cos squared t, so the equation is satisfied, and as t runs through a full turn every point of the ellipse is reached. The only differences are that the journey starts at (minus 4, 0) and goes clockwise. Neither was asked about.
Whenever a question asks for a parametrisation, check a proposed answer the way the right-hand side does: substitute into the equation, and make sure the whole curve is covered. If both hold, it is correct, however unfamiliar it looks.
Section
Part 5
Concept
Figure (svg): A cycloid for a wheel of radius a: one arch from (0, 0) up to height 2a at x equals pi a and back to the ground at 2 pi a. The wheel is drawn at t equals 0, pi over 2, pi, 3 pi over 2 and 2 pi, appearing one after another, each with a spoke from its centre to the traced point.
An ant grips the rim of a wheel of radius a that rolls without slipping along a straight road. The path it traces is a cycloid, and it is one of the curves that is far easier to describe parametrically than any other way.
\[ x(t) = a(t - \sin t), \qquad y(t) = a(1 - \cos t) \]
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, p. 615 — the cycloid and Figure 7.9
Watch the figure: the wheel appears at five moments of one full turn, and each time a spoke joins its centre to the point the ant is clinging to. The ant starts on the ground, is carried up and over the top, and comes back down to touch the ground again one full circumference later.
The path is a series of arches, and each arch meets the road in a sharp point, a cusp, because the rim point is momentarily at rest when it touches the ground. That is what rolling without slipping means: the bit of the wheel touching the road is not sliding along it.
Try writing this curve as y equals a function of x and you will get nowhere useful. As a pair of functions of the turning angle t, it is two short lines. That contrast is the best advertisement for parametric equations in the whole chapter.
Worked example
Let t be the angle the wheel has turned through, and split the ant's position into the centre plus the ant's offset from it.
Where the centre is
Why: Rolling without slipping: the ground covered equals the arc turned, a times t.
\[ C = (at,\, a) \]
Where the ant is relative to the centre
Why: The wheel turns clockwise from the bottom; the negative signs set that direction.
\[ A - C = (-a\sin t,\, -a\cos t) \]
Figure (svg): A wheel of radius a whose centre C sits at (at, a) after rolling a distance at. The ant A is on the rim, rotated clockwise by angle t from the bottom. A dashed horizontal leg of length a sin t and a vertical leg of length a cos t connect A to C. The arch traced so far runs from the origin to A.
Add the two
Why: Position of the ant is the centre plus the offset.
\[ A = \big(at - a\sin t,\; a - a\cos t\big) = \big(a(t - \sin t),\; a(1-\cos t)\big) \]
Check the top and the landing
Why: Half a turn should put the ant on top of the wheel; a full turn should put it back on the ground one circumference along.
\[ t = \pi: (\pi a,\, 2a), \qquad t = 2\pi: (2\pi a,\, 0) \]
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, pp. 615 and 620 — the cycloid; project Figure 7.13
The idea is to break a complicated motion into two simple ones: the centre of the wheel sliding steadily to the right, and the ant going round the centre. Add them and you have the ant's position.
The centre is at height a always. How far along is it? Because the wheel does not slip, the length of road it has covered equals the length of rim that has touched the road, which is the radius times the angle, a times t. So the centre is at (at, a).
In the figure, the ant is joined to the centre by the orange spoke, and the dashed legs split that spoke into a horizontal piece, a sin t, and a vertical piece, a cos t. The ant is behind and below the centre, so both are subtracted. The signs are the book's way of making the wheel turn clockwise, which is the direction a wheel rolling to the right must turn.
The check is physical: after half a turn the ant should be on top of the wheel, at height 2a, and after a full turn back on the ground, one circumference, 2 pi a, from the start.
Notation
Annotate
On: \( x = a\,t - a\sin t, \qquad y = a - a\cos t \)
Step through the notes and connect each term to the figure you just saw. The a t is the forward progress of the centre, the only part of the motion that keeps growing. The constant a is the centre's height.
The two remaining terms are the ant's offset from the centre, and they are a circle of radius a traced clockwise, starting from the bottom. At t equal to 0 the offset points straight down, so the ant is on the ground. At t equal to pi it points straight up, so the ant is on top.
Reading equations this way, as a sum of a simple motion and a circular one, is a skill you will use again. The hypocycloid and the curtate and prolate cycloids later in this part are all built the same way; only the size of the pieces changes.
Concept
Roll a wheel of radius b around the inside of a fixed circle of radius a. Its centre moves round a circle of radius a minus b, and a point on its rim traces a hypocycloid.
\[ x = (a-b)\cos t + b\cos\!\left(\tfrac{a-b}{b}t\right) \]
\[ y = (a-b)\sin t - b\sin\!\left(\tfrac{a-b}{b}t\right) \]
Figure (svg): Four hypocycloids inside dashed unit circles: a over b equal to 3 (three cusps), 5 (five cusps), 5 over 2 (a five-pointed star closing after two laps) and 7 over 3 (seven cusps, closing after three laps).
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, pp. 615-617 — Figures 7.10 and 7.11
Now the wheel rolls around the inside of a bigger circle instead of along a road. The first term of each equation is the centre of the small wheel going round a circle of radius a minus b. The second term is the point on its rim going round that centre, in the opposite direction and faster.
The figure shows the effect of the ratio a over b. When it is a whole number, that number is the count of cusps, and the curve closes up after one lap. When it is a fraction in lowest terms, like five halves, the numerator is still the number of cusps, but the curve needs as many laps as the denominator before it closes, which is how the five-pointed star gets drawn.
The textbook adds that when the ratio is irrational the curve never closes at all. It just keeps adding cusps forever, gradually filling the ring between the two circles.
Worked example
The hypocycloid of Figure 7.10. Simplify it and eliminate the parameter.
\[ x = 3\cos t + \cos 3t, \qquad y = 3\sin t - \sin 3t \]
Use the triple-angle formulas
Why: Both come from expanding cosine and sine of 2t plus t.
\[ \cos 3t = 4\cos^3 t - 3\cos t, \qquad \sin 3t = 3\sin t - 4\sin^3 t \]
Substitute and simplify
Why: The 3 cos t and 3 sin t terms cancel.
\[ x = 4\cos^3 t, \qquad y = 4\sin^3 t \]
Isolate cosine and sine
Why: Take cube roots.
\[ \cos t = \left(\frac{x}{4}\right)^{1/3}, \qquad \sin t = \left(\frac{y}{4}\right)^{1/3} \]
Apply the Pythagorean identity
Why: Square each and add.
\[ \left(\frac{x}{4}\right)^{2/3} + \left(\frac{y}{4}\right)^{2/3} = 1 \iff x^{2/3} + y^{2/3} = 4^{2/3} \]
Figure (svg): The four-cusped astroid inside a dashed circle of radius 4. A small wheel of radius 1 is drawn inside the big circle, touching it, with its centre on a dotted circle of radius 3 and a spoke to the tracing point on the astroid.
Check numerically at t equal to pi over 3
Why: The original form and the simplified form must agree.
\[ 3(0.5) + \cos\pi = 0.5 = 4(0.5)^3, \qquad 3\tfrac{\sqrt3}{2} - \sin\pi \approx 2.598 \approx 4\left(\tfrac{\sqrt3}{2}\right)^3 \]
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, p. 616 — Figure 7.10, a = 4 and b = 1
With a equal to 4 and b equal to 1, the general formulas give 3 cos t plus cos 3t for x. That is correct, but it hides a much nicer form. The triple-angle formulas, which come from expanding cosine and sine of 2t plus t, turn cos 3t and sin 3t into cubes, and the 3 cos t and 3 sin t terms cancel exactly.
So x is 4 cos cubed t and y is 4 sin cubed t. Take cube roots to get cos t and sin t on their own, and the Pythagorean identity eliminates t: the two-thirds powers of x and y add up to the two-thirds power of 4. That is the astroid, the four-cusped star of the figure.
The figure also shows the mechanism: the small wheel's centre runs round the dotted circle of radius 3, and the wheel touches the dashed outer circle at each cusp. The check evaluates both forms at pi over 3 and gets the same point, which confirms the simplification.
Concept
Figure (svg): A circle of radius 1 with its bottom at the origin O and top at (0, 2). A ray from O at angle theta meets the circle at A and the line y equals 2 at B. The point P sits directly below B at the height of A. As theta varies, P traces a bell-shaped curve, the witch of Agnesi, y equals 8 over x squared plus 4.
The construction fixes everything by one angle theta: the ray from the origin at angle theta meets the circle at A and the top line at B, and P takes its x from B and its y from A.
\[ x = 2a\cot\theta, \qquad y = 2a\sin^2\theta, \qquad 0 < \theta < \pi \]
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, pp. 618-619 — Student Project: the witch of Agnesi
This curve comes from one of the section's student projects, and it is a good example of a parameter chosen for its geometric meaning. Everything in the construction is fixed once you choose the angle theta of the ray from the origin.
Follow the figure: the ray meets the circle at A and the horizontal line at height 2a at B. Drop straight down from B until you are level with A, and that is P. As theta varies, P sweeps out the bell-shaped curve.
The project leads you through the trigonometry that gives the formulas on the slide: x is 2a cot theta, the horizontal distance to B, and y is 2a sin squared theta, the height of A. You do not need to reproduce every step, but notice that no equation in x and y was needed to build the curve; the angle did all the work.
Worked example
Project step 9: eliminate theta to show the witch is the graph of a function.
\[ x = 2a\cot\theta, \qquad y = 2a\sin^2\theta \]
Solve the first equation for cot theta
Why: Divide by 2a.
\[ \cot\theta = \frac{x}{2a} \]
Write sin squared in terms of cot
Why: Divide the Pythagorean identity by sine squared.
\[ 1 + \cot^2\theta = \csc^2\theta \;\Longrightarrow\; \sin^2\theta = \frac{1}{1 + \cot^2\theta} \]
Substitute the cotangent
Why: Clear the fraction inside.
\[ \sin^2\theta = \frac{1}{1 + \frac{x^2}{4a^2}} = \frac{4a^2}{x^2 + 4a^2} \]
Multiply by 2a
Why: That is y.
\[ y = 2a\cdot\frac{4a^2}{x^2 + 4a^2} = \frac{8a^3}{x^2 + 4a^2} \]
Check with a equal to 1, theta equal to 1
Why: The parametric point, then the formula.
\[ (x, y) = (2\cot 1,\, 2\sin^2 1) \approx (1.2842,\, 1.4161), \quad \frac{8}{1.2842^2 + 4} \approx 1.4161 \]
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, p. 619 — project step 9
The last step of the project eliminates theta, and it is a good test of choosing the right identity. The x equation gives cot theta directly. The y equation needs sin squared theta, so you want an identity linking the two.
Divide cos squared plus sin squared equals 1 by sin squared, and you get 1 plus cot squared equals csc squared. Flip it, and sin squared is 1 over 1 plus cot squared. Now substitute x over 2a for the cotangent and clear the small fraction.
Multiplying by 2a gives y equals 8a cubed over x squared plus 4a squared, an ordinary function of x. So this curve, unlike the circle, is a graph; the parametric form was simply the natural way to build it. The check takes a equal to 1 and theta equal to 1, computes the point from the parametric form, and confirms the formula gives the same height to four decimals.
Tweak it
Parameter explorer
The curve is the witch for a circle of radius a. Slide a. How does the peak height depend on a, and how does the width of the bell change?
\[ y = \frac{8 \cdot {a}^{3}}{x^2 + 4 \cdot {a}^{2}} \]
Move the slider and watch two things: the height of the peak and the width of the bell. Before you move it, predict how the peak depends on a, using the formula with x equal to 0.
At x equal to 0 the formula gives 8a cubed over 4a squared, which is 2a. That makes sense geometrically: the peak of the witch is the top of the circle, at height 2a. Doubling a doubles the peak.
The width also scales with a, because the whole construction is scaled: a circle twice as big gives the same shape, twice as large in every direction. Far from the centre, the curve decays like 8a cubed over x squared, which is why it has such long, slowly thinning tails, the feature that makes it the Cauchy distribution in probability.
Concept
Now let the ant sit a distance b from the centre instead of on the rim. The centre still moves along at height a; only the length of the offset changes.
\[ x = a\,t - b\sin t, \qquad y = a - b\cos t \]
Figure (svg): Three curves for a wheel of radius 1 rolling two full turns: a curtate cycloid with b equals one half, a gentle wave between heights one half and one and a half; the ordinary cycloid with b equals 1, touching the ground in cusps; and a prolate cycloid with b equals 1.6, dipping below the ground in small loops.
With b smaller than a (climbing a spoke) the path is curtate, a smooth wave. With b larger than a (a train wheel's flange) it is prolate, and the ant moves backwards for part of each turn.
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, pp. 620-622 — Student Project: curtate and prolate cycloids
The second project asks what happens if the ant is not on the rim. The centre's motion is exactly the same as before, so the a t and the a stay. Only the offset changes, from length a to length b, the ant's distance from the centre.
Compare the three curves in the figure. With b equal to one half, inside the rim, the ant never reaches the ground and never stops, so the path is a smooth wave with no cusps. With b equal to 1, the ordinary cycloid, the ant touches the ground in cusps. With b equal to 1.6, beyond the rim, as on the flange of a train wheel, the path dips below the road and loops.
Those loops mean the ant is moving backward for part of every turn, even though the train is moving forward. The project's final question asks you to notice that parts 3 and 4 give the same formula; the only difference between curtate and prolate is whether b is smaller or larger than a.
Estimation
\[ x = 100t, \qquad y = -4.9t^2 + 4000, \qquad t \ge 0 \]
Predict first
A plane flying at 100 m/s at 4000 m releases a package. Guess first: roughly how far before the target must it be released?
Correct: About 3 km
Why: The fall takes the time for 4.9 t squared to reach 4000, about 28.6 seconds, and all that time the package keeps the plane's 100 metres per second forward. That is roughly 2860 metres, close to 3 km.
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, p. 624 — Exercise 54
Make an honest guess before you calculate. The trap is to think the package drops straight down; it does not, because it keeps the plane's forward speed the whole way down.
A rough estimate is quick. The fall time is the time for 4.9 t squared to reach 4000, and t squared is about 800, so t is a bit under 30 seconds. At 100 metres per second, that is just under 3000 metres forward.
This is the parametric view doing real work. The horizontal and vertical motions are separate functions of the same time t, so you can solve the vertical one for when, and put that when into the horizontal one to get where.
Worked example
\[ x = 100t, \qquad y = -4.9t^2 + 4000, \qquad t \ge 0 \]
The package lands when y is zero
Why: Set the height to zero.
\[ -4.9t^2 + 4000 = 0 \iff t^2 = \frac{4000}{4.9} \approx 816.33 \]
Take the positive root
Why: Time after release is positive.
\[ t \approx 28.57 \text{ s} \]
Find the horizontal distance at that time
Why: x is 100 times t.
\[ x = 100(28.57) \approx 2857 \text{ m} \]
Figure (svg): The trajectory x equals 100t, y equals minus 4.9 t squared plus 4000 from the release point (0, 4000) down to the ground at x about 2857 metres, with dots every 10 seconds.
Check the landing height
Why: Put the time back into y.
\[ y(28.57) = -4.9(816.33) + 4000 \approx 0 \;\checkmark \]
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, p. 624 — Exercise 54
Landing means height zero, so set y equal to 0 and solve for t. The equation has no linear term, so it is just a square root: t squared is 4000 over 4.9, about 816.33, and t is about 28.57 seconds. Take the positive root, since the package is released at t equal to 0.
Then feed that time into x. The horizontal motion is uniform, 100 metres per second, so the package travels about 2857 metres forward while falling. That is how far before the target it must be released.
The figure shows the path with dots every 10 seconds: evenly spaced left to right, because the horizontal speed is constant, but falling farther in each interval, because the vertical speed keeps growing. The check substitutes the landing time back into y and gets zero.
Real world
\[ x = (v_0\cos\alpha)t, \qquad y = (v_0\sin\alpha)t - \tfrac12 g t^2 \]
Discussion prompt
A bullet leaves the gun at 500 m/s at 30 degrees above the horizontal, with g equal to 9.8 metres per second squared. When does it hit the ground, and how far away? Ignore air resistance.
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, p. 624 — Exercise 55
Write your answer before you reveal it. The horizontal speed is 500 cos 30 degrees, about 433 metres per second, and the initial vertical speed is 500 sin 30 degrees, exactly 250.
The bullet is at ground level when y is zero. Factor out t and there are two solutions: t equal to 0, the moment it leaves the gun, and t equal to 250 over 4.9, about 51 seconds, when it lands. Put that time into x and the range is about 22 kilometres.
The reason this problem is parametric is that the question is about time as well as place. A single equation for the path would tell you its shape, a parabola, but not when the bullet is anywhere. In reality air resistance would cut that range drastically; the model ignores it.
Section
Part 6
Pattern
Figure (svg): A flow diagram: from x of t and y of t with a range for t, one branch tabulates and plots points in order of t with arrows, the other eliminates t, either by solving for t or with an identity; both meet in a final box that attaches the restriction from the t range.
The flow diagram shows two branches that start from the same parametric equations and end in one complete answer. The upper branch, the table, gives you the things only the parametrisation knows: where the curve starts and stops and which way it goes. The lower branch, elimination, gives you the shape's name.
The yellow arrows meet at the step that students most often skip: attaching the restriction. It uses the t-interval, and any square, root, exponential or trig function that limits the values of x or y.
Read the numbered list as a checklist for any question on this section. A complete answer names the curve with its domain, and states the orientation with its start and end.
Ranking
Put in order
Order the steps of a complete sketch of a parametric curve.
Why: The table comes first because it carries the order of travel, which the xy-equation cannot. Elimination comes last, as a check and a name for the shape you have drawn.
Before you drag anything, ask which step needs the results of which. You cannot compute x and y until you have chosen the t values; you cannot join the points in order until you have them; and the arrows follow the order of joining.
Elimination goes last here, because in a sketching question its job is to name and check the shape you drew. In a question that only asks you to identify the curve, you might do it first, but you would still need the t-range to state the restriction.
If you put the arrows before the plotting, reconsider: the direction comes from the order of the table's rows, so it can only be drawn once the points are in place.
Check
Check your understanding
Which xy-description is exactly the curve x = t², y = t⁴, t any real number?
Answer: B
Why: Since t to the fourth is t squared, squared, y equals x squared. But x is a square, so it is never negative: only the right half of the parabola is reached, traced inward and then outward again.
Commit to an answer before you check. The elimination itself is one line: t to the fourth is the square of t squared, so y equals x squared.
The question is what values x can take. Since x is t squared, it is never negative, so only the right half of the parabola is reached. That makes the equation with the restriction the exact description.
If you chose the square root, you swapped the roles of x and y. If you chose the whole parabola, you did the algebra right and missed the restriction, which is the trap this lesson has warned about more than once.
Check
Check your understanding
What does x = 2 cos t, y = 2 sin t, 0 ≤ t ≤ π trace?
Answer: B
Why: At t equal to 0 the point is (2, 0); sine is never negative between 0 and pi, so the point stays on the upper half, reaching the top at a quarter-turn and (minus 2, 0) at the end of the interval.
Work from the interval before the formulas. At t equal to 0 the point is at (2, 0), which settles where the curve starts. Between 0 and pi, sine is never negative, so the point stays on the upper half.
At t equal to pi the point is at (minus 2, 0), so the curve is the upper semicircle from right to left, which is counterclockwise.
The wrong answers each ignore one piece of information: the length of the interval, the starting point, or the sign of sine on that interval. Any one of those is enough to decide the question.
Check
Check your understanding
A wheel of radius 3 rolls along the x-axis; a rim point starts at the origin. Where is it after half a turn (t = π)?
Answer: A
Why: With a equal to 3, x is 3 times pi minus sine of pi, which is 3 pi, and y is 3 times 1 minus cosine of pi, which is 6. After half a turn the point is at the top of the wheel, twice the radius up, and the wheel has rolled half its circumference.
Substitute a equal to 3 and t equal to pi into the cycloid equations. Sine of pi is 0, so x is 3 pi; cosine of pi is minus 1, so y is 3 times 2, which is 6.
Then check it physically: half a turn puts the rim point on top of the wheel, twice the radius above the road, and the wheel has rolled half its circumference, which is pi times the radius.
If you chose a height of 3, you found the centre, not the rim point. If you chose pi for x, you forgot that the rolled distance is the radius times the angle, not the angle alone.
Explain it to yourself
Discussion prompt
You have met two things an xy-equation cannot record: the direction of travel, and the restriction coming from the t-interval. Explain in two sentences why losing the restriction is the more serious error.
Write your two sentences before revealing the model answer. The point to make is the difference between changing how a set of points is travelled and changing which points are in the set.
Losing the orientation is a loss of information, but what you are left with is still true: the curve really is that set of points. Losing the restriction makes your answer false: it includes points that the curve never reaches, like the left half of the parabola in Example 7.2(a).
That is why this lesson keeps asking for the restriction as part of the answer, while the orientation is recorded separately with arrows.
Exit ticket
\[ x = 4 + 2\cos\theta, \qquad y = -1 + \sin\theta, \qquad 0 \le \theta \le 2\pi \]
Discussion prompt
Eliminate the parameter and name the curve, give its direction of travel, and write a parametrisation of the same curve that goes the other way.
OpenStax Calculus Volume 2, §7.1 Parametric Equations §7.1, p. 623 — Exercise 14
This is Exercise 14, and it uses everything: isolate the trig functions, apply the identity, name the curve, read the direction, then reverse it.
Cos theta is x minus 4 over 2 and sin theta is y plus 1, and the identity gives an ellipse centred at (4, minus 1), stretched 2 units horizontally and 1 vertically. The full interval means no restriction beyond the ellipse itself.
For the direction, compare theta equal to 0, at (6, minus 1), with a quarter-turn, at (4, 0): right of centre, then above it, so counterclockwise. To reverse it, flip the sign of the sine term; any other correct reversal, such as replacing theta by minus theta, is equally good.
Recap
| idea | what to do | what to remember |
|---|---|---|
| parametric curve | x = x(t), y = y(t), t in an interval | the vertical line test does not apply |
| sketching | table in order of t, plot, arrows | orientation is part of the answer |
| eliminating t | solve for t, or cos² t + sin² t = 1 | attach the restriction; direction is lost |
| parametrising | x = t, y = f(t), or any x(t) covering the domain | infinitely many choices, all correct |
| cycloid | x = a(t − sin t), y = a(1 − cos t) | centre (at, a) plus a rotating offset |
\[ x = a(t - \sin t), \qquad y = a(1 - \cos t) \]
Next, Section 7.2 does calculus on these curves: slopes, areas and arc lengths, all computed directly from x(t) and y(t) without ever eliminating the parameter.
Stewart, Calculus: Early Transcendentals 8e, §10.1 Curves Defined by Parametric Equations §10.1, pp. 640-648 — the same material in Stewart
The table collects the five ideas of the lesson. A parametric curve records a journey, not just a shape, which is why the vertical line test does not apply to it. Sketch it from a table in order of t, with arrows.
Eliminating the parameter names the curve but loses the direction, and it loses the restriction unless you put it back by hand. Going the other way, any curve has infinitely many parametrisations, and the right one is the one that suits the job.
The cycloid is the model example of why this is worth doing: the path of a point on a rolling wheel is two short formulas as a function of the turning angle, and hopeless as a function of x. In the next section you will find slopes, areas and lengths directly from x of t and y of t, without eliminating anything.
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