Parametric curves and polar coordinates

Begin by following one pair of coordinates

Let a marker have coordinates x = t² and y = t, with the dimensionless input restricted to −2 ≤ t ≤ 1. At t = −2, −1, 0 and 1 it visits (4, −2), (1, −1), (0, 0) and (1, 1). The two visits with x = 1 have different heights. Eliminating t gives x = y², but the permitted part still has −2 ≤ y ≤ 1. A formula for the whole sideways parabola would add locations the marker never reaches.

Prepare with plotting coordinate pairs and reading function domains. Reversing a substitution means recovering an original permitted input; inverse functions and the restrictions in rational and radical equations help with that step. Circle and polar examples also use the unit-circle values of sine and cosine, directed radian angles and the identity cos² t + sin² t = 1.

On a first reading, follow the parameter image, line segment, restricted parabola and circle-direction sections, then the signed polar conversion. Return to the rational-circle inverse and the guarded polar quotients after you can check original denominator and square-root restrictions. The polar parabola proof uses a Cartesian relation to reconstruct a positive radius; read that proof after the elementary circle and line conversions.

A parameter gives a point for each permitted input

Parametric equations x = f(t), y = g(t), with t in a stated set I, describe the image C = {(f(t), g(t)): t ∈ I}. Each permitted input must make both expressions real and defined. The parameter may represent time, an angle, a distance or an abstract number; name its meaning and units before interpreting it as physical motion. An empty permitted set gives an empty image, even when a later algebraic equation has many solutions.

The image is a set of points. The parameter map contains additional information: which point occurs at each input, the order of visits as t increases, and whether a point is revisited. Two maps can have the same image while traversing it in opposite directions or at different rates. Equal locations at two inputs do not make those inputs equal.

A curve need not be a graph y = F(x). A circle contains two different y-values at many x-values, yet x(t) and y(t) can each be ordinary functions of t. Parametric coordinates describe the pair together; separate graphs of x against t and y against t are not themselves the curve in the x-y plane.

Specify I as carefully as the formulas. Closed and open endpoints affect membership, and a removed parameter can create a missing point unless another permitted input reaches it. An excluded denominator value remains excluded after simplification. A finite table shows selected visits; proving the complete image requires an argument covering every permitted parameter and every claimed image point.

A line parameter becomes a segment when its interval is bounded

For two points A = (a₁, a₂) and B = (b₁, b₂), use P(t) = A + t(B − A): x = a₁ + t(b₁ − a₁), y = a₂ + t(b₂ − a₂). If A ≠ B and t ranges over all real numbers, the image is the complete line through them. Every point on that line has a real scalar displacement from A in the nonzero direction B − A, which proves the reverse implication.

On 0 ≤ t ≤ 1 the image is exactly the closed segment from A to B. An open endpoint removes that endpoint because the map is one-to-one when B − A ≠ 0. For t ≥ 0 the image is the ray starting at A toward B. Thus the same formulas with different parameter sets describe different geometric sets.

If one coordinate difference is zero, recover t from the other nonzero component. Do not divide by b₁ − a₁ when the line is vertical. If A = B, both differences vanish and the map is constant: every permitted input gives A. Its image is one point whenever I is nonempty, and an empty set when I is empty; an undefined slope or inverse does not turn it into a line.

If t is time and x = x₀ + vₓt, y = y₀ + vᵧt on a physical time interval, the constants vₓ and vᵧ are coordinate rates in length per time. Over a time increment Δt, the displacement length is √(vₓ² + vᵧ²)|Δt|. The geometric line equation alone does not retain those rates or the original beginning and ending times.

An implicit equation is often only a consequence

Eliminating t produces a relation that every permitted pair satisfies. To claim equality of images, also start with every point in the proposed Cartesian set and recover at least one t in I that gives that exact pair. An equation without the original restrictions can contain extra branches or endpoints.

For x = t² and y = t on −2 ≤ t ≤ 1, elimination gives x = y², but the complete image also requires −2 ≤ y ≤ 1. Conversely any point with those two conditions is reached by t = y. Taking only t = √x would discard the negative-y part; taking both signs without checking I would add points above y = 1.

For x = t, y = √t with t ≥ 0, squaring gives x = y². The original square root is nonnegative, so y ≥ 0 must remain. Conversely x = y² with y ≥ 0 gives t = x ≥ 0 and √t = y. The lower half of the sideways parabola satisfies the squared relation but has no original preimage.

Another example is x = t², y = t⁴ for all real t. The image is y = x² with x ≥ 0: every x ≥ 0 has the preimages t = ±√x, merging at t = 0. The unrestricted parabola also contains x < 0 and is too large. Likewise x = t², y = t² reaches exactly the ray y = x with x ≥ 0, twice at every positive x and once at the origin. Elimination must preserve the range of a squared coordinate even when t itself is unrestricted.

For x = t², y = t³ over all real t, the image is the cusp y² = x³. This equation itself forces x ≥ 0. At x > 0, the inverse t = y/x gives t² = y²/x² = x and t³ = tx = y, recovering every such point. At x = 0 the equation forces y = 0, and t = 0 reaches the origin. The inverse quotient is undefined there, so that point needs its separate reconstruction rather than being discarded.

A denominator hole works in the other direction: x = t, y = 1/t with t ≠ 0 gives xy = 1. Conversely every point with xy = 1 has x ≠ 0, so t = x is permitted and gives y = 1/x. This complete image has two disconnected branches. A sampled trace must not join the portions across t = 0, where the original map is undefined.

The same image can have a different journey

Suppose P(t) is defined on I and substitute t = φ(u) for u in J. The new map has image P(φ(J)), which is contained in P(I) if φ(J) ⊆ I. A sufficient condition for exactly the same image is φ(J) = I. If φ fails to reach an input, the images can still agree when that input's point has another reached preimage; compare points, not just input sets.

A strictly increasing one-to-one φ onto I preserves the order of parameter visits; a strictly decreasing one reverses it. A substitution that revisits parameter values can retrace the curve. These are statements about the original order, not claims that every curve has one global clockwise direction or a constant speed.

For the full parabola P(t) = (t, t²), t ∈ ℝ, the substitution t = u³ with u ∈ ℝ reaches every t exactly once and preserves their order. The substitution t = −u reaches every t once in reverse order. But t = u² reaches only t ≥ 0; it gives the right half of the parabola and visits each point with positive x twice, once at each sign of u.

For P(t) = (cos t, sin t), increasing t traverses the unit circle counterclockwise. Using −t reverses that direction; using 2t on 0 ≤ t < 2π traces the full circle twice. A geometric set records each location only once, so its equation cannot tell you how often it was visited. An endpoint at the same location as the start is a second visit, not a second point in the image.

Trigonometric parameters retain arcs and axis degeneracies

For r > 0, x = h + r cos t, y = k + r sin t satisfies (x − h)² + (y − k)² = r² for every real t. Conversely every point of that circle normalises to a unit-circle point and has an angle t. An interval 0 ≤ t < 2π reaches each point exactly once. The closed interval 0 ≤ t ≤ 2π has the same complete image but repeats the starting point at the end.

Restricting to 0 ≤ t ≤ π/2 gives exactly the first-quadrant arc relative to the centre, including both endpoints. The complete Cartesian description is the circle equation together with x ≥ h and y ≥ k. The signs select the unique unit-circle angle in that interval, proving the converse. Removing t = 0 from this quarter interval removes its rightmost endpoint; the same removal from an interval containing t = 2π would leave that location included.

The interval [0, π] similarly gives the closed upper semicircle, while (0, π) removes both horizontal endpoints and retains y > k. On [0, 4π] the full circle is traversed twice, but the start location has three visits at t = 0, 2π and 4π; every other point has two. A count stated for regular points must identify this endpoint exception. A repeated closed start does not become an extra geometric point.

With a > 0 and b > 0, x = h + a cos t, y = k + b sin t gives the ellipse (x − h)²/a² + (y − k)²/b² = 1. Normalising by the positive semiaxes proves completeness over a full turn. A restricted angle interval may instead give only an arc. The circle is the equal-semiaxis case; translations affect locations without changing the parameter order.

If a = 0 and b > 0, the formulas remain defined: x = h and y = k + b sin t. Over a full turn their image is the segment h fixed, k − b ≤ y ≤ k + b, with repeated visits. The standard ellipse expression with denominator a² is undefined and cannot be cleared to justify it. If both a and b are zero and the parameter set is nonempty, the image is only (h, k). Treat these maps directly rather than assigning an ordinary ellipse to them.

A rational parameter can miss a point of its implicit circle

Consider x = (1 − t²)/(1 + t²), y = 2t/(1 + t²) for every real t. The denominator is always positive. Expanding (1 − t²)² + 4t² = (1 + t²)² proves x² + y² = 1 for all t. This is an entire algebraic identity, not a conclusion from plotted samples.

However, x + 1 = 2/(1 + t²) is strictly positive for every finite real t. Thus the point (−1, 0) is never reached. The complete image is the unit circle with that one point removed, not the full implicit circle.

Conversely, take any unit-circle point other than (−1, 0). Then x > −1, so t = y/(1 + x) is defined. From y² = 1 − x², obtain t² = (1 − x)/(1 + x). Hence 1 + t² = 2/(1 + x), and substitution recovers both x and y. This proves that there are no other missing circle points and gives a unique finite preimage for every point in this image.

As t tends to positive or negative infinity, the coordinates approach (−1, 0), but infinity is not an allowed real input. A limit point need not belong to the image. To cover the full circle, use the trigonometric parameterisation or explicitly add that point as a separate part of the description. A drawing that joins the two far ends across the omitted point would conceal the hole.

A signed polar coordinate is different from a distance

A polar pair (r, θ) represents the Cartesian point x = r cos θ, y = r sin θ. We use a real signed r and a real directed radian angle θ from the positive x-axis. For r > 0, the point lies along that terminal ray; for r < 0 it lies in the opposite direction. Its actual distance from the origin is |r| because x² + y² = r². A negative signed polar coordinate is valid and is not a negative geometric distance.

For every integer k, the pairs (r, θ + 2kπ) and (−r, θ + (2k + 1)π) represent the same point as (r, θ). Full-turn periodicity gives the first equivalence. Adding an odd half-turn negates both sine and cosine, which is cancelled by negating r in the second equivalence.

For a nonzero point, these are all its representations relative to one chosen pair with r ≠ 0. Equality of the squared Cartesian lengths forces any other signed radius to be r or −r. With the same radius, equality of both unit-circle coordinates forces the angle difference to be 2kπ; with the opposite radius it forces an odd multiple of π. Equality of tangent alone would miss the distinction between opposite rays.

When r = 0, every real θ represents the origin. Its angle is not determined by the point, and it has no unique reference direction. This exception must be separated before dividing by r or using an inverse angle. A representation convention r ≥ 0 does not change the point set, but it removes negative-radius representatives and needs a separate rule for the origin.

Recover the quadrant before accepting an inverse angle

For a nonzero Cartesian point (x, y), first choose the canonical positive radius R = √(x² + y²) > 0. Its normalised pair (x/R, y/R) is on the unit circle. There is exactly one terminal angle Θ with 0 ≤ Θ < 2π having those two coordinates. This existence and uniqueness comes from the unit-circle direction, and does not require dividing by x.

If x ≠ 0, tan Θ = y/x can help, but the principal arctan output lies strictly between −π/2 and π/2. When x < 0, add π to that principal output to reach the correct quadrant. When x > 0, retain it if nonnegative and add 2π if negative. Verify both coordinate signs afterward; the tangent ratio has period π and cannot identify a terminal ray by itself.

If x = 0 and y > 0, use Θ = π/2; if x = 0 and y < 0, use Θ = 3π/2. For y = 0, the nonzero positive and negative x-axis points use 0 and π respectively. At (0, 0), R is zero and no angle can be recovered uniquely. All pairs (0, θ) represent that point. Alternatively, for a nonzero point, arccos(x/R) gives an angle in [0, π]; retain it when y ≥ 0 and use 2π − arccos(x/R) when y < 0. This also keeps both coordinate signs rather than relying on a rounded angle.

Once a nonzero canonical pair (R, Θ) is found, every signed representation is (R, Θ + 2kπ) or (−R, Θ + (2k + 1)π), with integer k. Keep the original directed angle separate from this canonical terminal representative: different complete turns can give the same point, and changing the sign of the radius also changes which ray represents it.

A polar equation describes represented points, with its domain retained

For r = f(θ), θ ∈ I, the image is {(f(θ)cos θ, f(θ)sin θ): θ ∈ I and f is defined}. This is a parametric map with θ as the parameter. A point belongs when at least one permitted representative satisfies the equation; another polar representative of that same point need not satisfy it. For example, the origin can occur at one zero of f even if r = 0 fails the equation at another angle.

For a > 0, r = a over all real θ gives the circle x² + y² = a²; r = −a gives the same full circle with opposite directed radii. Restricting the angle can change either image. The equation r = 0 gives the origin whenever its angle set is nonempty. Similarly θ = θ₀ with all signed real r gives the complete line through the origin in that direction; restricting r ≥ 0 gives only a ray.

For r = 4cos θ over all real θ, multiplying by r gives r² = 4r cos θ and hence x² + y² = 4x, the circle (x − 2)² + y² = 4. To prove the reverse and retain the origin, write x = 4cos² θ = 2 + 2cos(2θ), y = 4sin θ cos θ = 2sin(2θ). A full range of 2θ reaches the entire translated circle, including the origin at θ = π/2. Dividing by r without a zero check would conceal that point. On −π/2 ≤ θ ≤ π/2 the circle is traversed once with its endpoint repeated; on 0 ≤ θ < 2π it is traversed twice.

For r = 3sec θ = 3/cos θ, retain cos θ ≠ 0. Then x = 3 and y = 3tan θ. Every point (3, y) has a permitted angle θ = arctan(y/3) in (−π/2, π/2), with positive radius 3/cos θ, so this is the whole vertical line when all permitted real angles are allowed. On 0 ≤ θ < π/2 it is only the upper ray from (3, 0). The forbidden endpoint is a pole, not a finite point to add.

The zero coefficient behaves differently: r = 0/cos θ, still with cos θ ≠ 0, gives only the origin. Multiplying by cos θ yields x = 0, but that complete vertical line is too large; the original numerator-zero quotient has no nonzero-radius points. The domain guard remains even when the numerator is zero, and a retained Cartesian consequence is not automatically an equivalent image.

Another guarded example is r = 2/(1 + cos θ), with θ ≠ (2k + 1)π for every integer k. The denominator is strictly positive on that domain, so r > 0. Multiplication gives r + x = 2; using r² = x² + y² gives y² = 4(1 − x), a left-opening parabola. Conversely any point of this parabola has x ≤ 1 and positive distance R = 2 − x ≥ 1. Its unit direction has cos θ = x/R and sin θ = y/R, and 1 + cos θ = 2/R > 0. Thus its permitted polar representative reconstructs the original equation and proves the whole image. The origin is not on this curve.

With zero numerator instead, r = 0/(1 + cos θ) retains the same forbidden angles and reaches only the origin. Clearing the denominator would allow arbitrary r at a forbidden angle with cos θ = −1, adding points on the x-axis. Those points do not belong to the original guarded quotient. Even the origin's forbidden-angle representative remains invalid; its other permitted representatives still establish its geometric membership.

A repeated angle does not always repeat a curve point

The unit direction repeats after 2π, but a polar curve point repeats only if the radius and direction together produce the same Cartesian pair. For the spiral r = θ with θ ≥ 0, the angles θ and θ + 2π have the same terminal ray but different positive distances, so their nonzero points differ. Its permitted input starts at the origin and its radius is unbounded. A finite angular interval gives only part of that image.

For r = 4cos θ, replacing θ by θ + π negates both r and the unit direction, leaving the point unchanged. The Cartesian point map has period π, although the signed-radius function itself has period 2π. The scalar formula's period and the geometric tracing period therefore answer different questions. A closed endpoint can repeat a location without enlarging the set.

A symmetry proof must preserve the whole allowed image. For instance, if every permitted θ has a permitted −θ with the same r, the corresponding points (r cos θ, r sin θ) and (r cos θ, −r sin θ) give reflection symmetry in the x-axis. An invariant formula alone does not suffice if the selected angle interval removes its partner points. Failing a particular substitution test does not by itself prove absence of geometric symmetry, because other polar representatives may realise the reflected points.

A finite point table illustrates exact visits but does not prove a complete loop, a hole, symmetry or unbounded continuation. Use identities and reverse reconstruction for those claims. A plotting window can clip even a valid full parameter image; a sampled polyline may join points that are not joined by the actual curve, cross a pole, or fill a missing endpoint. Distinguish a true parameter boundary from an edge of the screen, and use equal length units on the coordinate axes.

Predict membership and visits before checking the construction

In elimination, compare the unrestricted sideways parabola with its nonnegative-parameter branch and its bounded [−2, 1] piece. Before reading the verdict, recover t from a proposed point: does it satisfy the formula and the original interval? Try the rational circle next and explain why (−1, 0) satisfies the circle equation but has no finite input.

In tracing, predict which endpoint locations survive on a closed or open upper semicircle. On one half-open turn, explain why t = 2π is forbidden although its location is included through t = 0. Compare increasing-parameter arrows on opposite circle directions. For two closed turns, count the start separately: it has three visits, while regular points have two. A zero semiaxis must be read from the original formulas, giving a segment or a point without an ordinary ellipse denominator.

For signed polar points, predict the quadrant before multiplying. A negative radius places the point opposite the stated angle ray. Keep the distance |r| separate from the signed r, and use both coordinate signs when recovering a canonical angle. At zero radius, changing the angle still gives the origin and assigns it no geometric direction.

For polar curves, compare positive and zero quotient numerators while preserving each forbidden denominator. Decide separately whether the formula is defined, a supplied representative satisfies it, and that representative's location belongs through some permitted input. The origin may have a failing representative and a valid one. Read the exact tables and reconstruction before interpreting a loop, a missing point or a clipped unbounded piece; a finite drawing does not establish the complete image.

Worked examples

Try it yourself

1. Parameterise the closed segment from A = (1, −2) to B = (7, 1) using 0 ≤ u ≤ 1. Give its complete Cartesian description and midpoint visit. What changes if u = 1 is excluded?

   Show the answer and reasoning

   Use x = 1 + 6u, y = −2 + 3u. The image is y = (x − 5)/2 with 1 ≤ x ≤ 7; u = 1/2 gives (4, −1/2). Excluding u = 1 removes B and changes the x-bound to x < 7.

   The nonzero horizontal difference gives u = (x − 1)/6. Substitution yields y = −2 + (x − 1)/2 = (x − 5)/2, and the parameter interval gives the exact x-bounds. Conversely those bounds recover an allowed u, proving the full segment. The map is one-to-one, so no remaining parameter reaches B after its only preimage is removed.

2. For x = t − 2, y = √t on 0 < t ≤ 9, find the complete image and identify its omitted and included endpoint locations. Explain why squaring alone adds extra points.

   Show the answer and reasoning

   The image is x = y² − 2 with 0 < y ≤ 3. The limit location (−2, 0) is omitted and (7, 3) is included. The lower square-root branch is not reached.

   The original square root requires nonnegative y, and strict t > 0 makes it positive. Solving t = y² and retaining t ≤ 9 gives 0 < y ≤ 3. Every point with these conditions is reached by that t. Without the sign and interval restrictions, x = y² − 2 contains y ≤ 0 and y > 3, whose recovered parameters or square-root signs fail the original map.

3. For x = 1 + 3cos(2t), y = −1 + 2sin(2t), 0 ≤ t < 2π, give the complete image, orientation and number of visits to each point. Find the points at t = 0, π/4 and π.

   Show the answer and reasoning

   The full ellipse is (x − 1)²/9 + (y + 1)²/4 = 1, traced counterclockwise twice. Each point has two permitted visits. The three points are (4, −1), (1, 1) and (4, −1).

   The phase 2t ranges over [0, 4π), two half-open full turns. Normalising by the positive semiaxes proves that every ellipse point occurs once per turn, hence twice in the stated interval. Increasing phase gives counterclockwise order. Substitution at phases 0, π/2 and 2π yields the exact coordinates; the repeated start location remains one point of the geometric image.

4. For x = (1 − t²)/(1 + t²), y = 2t/(1 + t²), t ∈ ℝ, recover the parameter of (3/5, 4/5). Is (−1, 0) reached at any finite input, and does its limiting appearance restore membership?

   Show the answer and reasoning

   The point (3/5, 4/5) has t = 1/2. The point (−1, 0) has no finite preimage and remains omitted, even though the coordinates approach it as t → ±∞.

   Use t = y/(1 + x) = (4/5)/(8/5) = 1/2. Substitution gives (1 − 1/4)/(1 + 1/4) = 3/5 and 1/(5/4) = 4/5. For every real t, x + 1 = 2/(1 + t²) > 0, excluding x = −1. Infinity is a limiting process rather than an allowed real parameter, so it adds no point to the original image.

5. Find the canonical positive-radius polar pair for (−√3, −1), and one negative-radius representative. Why is arctan(y/x) alone insufficient, and what are all polar pairs for the origin?

   Show the answer and reasoning

   The canonical pair is (2, 7π/6); one negative-radius pair is (−2, π/6). The principal arctan ratio gives π/6, which is on the opposite ray. Every pair (0, θ), with real θ, represents the origin.

   The squared distance is 3 + 1 = 4, so R = 2. Both normalised coordinates are negative, requiring quadrant III and terminal angle 7π/6. The tangent ratio is 1/√3, shared with the first-quadrant ray; adding π corrects the principal inverse for x < 0. Changing the signed radius to −2 and subtracting π from the direction gives the same point. At the origin there is no nonzero normalised direction, so its angle is arbitrary.

6. Determine the complete image of r = 5sec θ on −π/3 ≤ θ ≤ π/3, including the endpoints. Does this interval reach the whole vertical line x = 5?

   Show the answer and reasoning

   The image is the closed segment x = 5, −5√3 ≤ y ≤ 5√3, with endpoints (5, −5√3) and (5, 5√3). It does not reach the whole vertical line.

   Cosine is positive and nonzero throughout this interval, so the quotient is defined. The coordinates are x = 5 and y = 5tan θ. Tangent is strictly increasing and continuous on this interval and has endpoint values −√3 and √3. Conversely θ = arctan(y/5) lies in the required interval for every y between those bounds, proving the whole segment. Removing the angular restriction would be a different parameter image.

7. For r = 4sin θ on 0 ≤ θ ≤ π, find the complete image and the image obtained by restricting to 0 ≤ θ ≤ π/2. Does the origin belong even though its representative (0, π/2) fails the equation?

   Show the answer and reasoning

   The full image is x² + (y − 2)² = 4, traced once with the origin visited at both endpoints. The restricted image is its right semicircle, x ≥ 0. The origin belongs through θ = 0 or π, despite the failing representative (0, π/2).

   The coordinates are x = 2sin(2θ), y = 2 − 2cos(2θ). Their centre-relative unit pair completes one turn as θ runs from 0 to π, proving the whole circle. On [0, π/2], x is nonnegative and the pair covers exactly the right semicircle. The allowed endpoint angles have r = 0 and reach the origin. At θ = π/2 the formula instead requires r = 4, so the pair (0, π/2) does not satisfy it; membership of a geometric point needs one permitted satisfying representative, not every equivalent pair.

What to carry forward

• A parametric curve is the complete image of its defined coordinate pair on the original parameter set. The image, order of visits and number of visits are different pieces of information.
• Eliminating a parameter proves a consequence. Equality of images also requires reverse reconstruction with every original interval, sign, square-root and denominator guard retained.
• Changing parameters can preserve, reverse, repeat or restrict a journey. Trigonometric full turns, partial arcs and zero-axis degeneracies must be interpreted from the actual formulas and input set.
• The rational unit-circle map covers every point except (−1, 0); its inverse and strict denominator argument identify that hole, while an infinite-parameter limit does not fill it.
• A real signed polar pair represents (r cos θ, r sin θ) with distance |r|. Nonzero points have full-turn and sign-reversed half-turn representation families; the origin has every angle with r = 0.
• Recover a polar direction using both coordinate signs and the axis cases. A principal arctan ratio cannot distinguish opposite rays, and the origin has no uniquely recoverable angle.
• A polar equation needs a permitted satisfying representative, with zero-radius visits and poles checked separately. Complete identities and inverse arguments prove images; samples, tracing intervals and viewing windows describe how much is displayed.