01 · Substitute a positive value
Find the point at t = 2 for x = 4 − t and y = 3t².
Hint
Compute x and y separately.
Worked solution
x = 2, y = 12, so the point is (2, 12).
Understand · explore · practise
Understand parametric equations as two coordinate rules for one parameter. Build a table, trace a curve manually and distinguish a point, its parameter and the direction of travel.
Before you startCoordinates, functions and basic graph shapes
01 / Use one parameter in both coordinate rules
A usual graph rule gives y from x. Parametric equations instead give both x and y from another variable, often called t. Substitute the same value into both rules to locate one point.
x = p(t), y = q(t)
Point on the curve: (p(t), q(t))
The parameter is a label for a position. It may represent time, an angle or another quantity, but no particular meaning is automatic.
x = t + 1, y = t² − 1; −2 ≤ t ≤ 2.
t = −2.00 gives (x, y) = (−1.00, 3.00).
The next larger parameter traces the curve down and right.
Grey shows the whole permitted curve. Blue shows the part traced up to your chosen parameter. Retraced portions overlap. Move t yourself; nothing advances automatically.
02 / Keep the parameter separate from the coordinates
x = 3(−2) + 2 = −4
Use the x-rule for the first coordinate.
y = (−2)² + (−2) = 2
Use the same parameter in the y-rule.
The point is (−4, 2)
The parameter −2 is not either coordinate in this example.
Parentheses matter when substituting a negative value into a square. (−2)² is 4; −2² is −4.
03 / Make a table of paired coordinates
For x = t + 1 and y = t² − 1, with −2 ≤ t ≤ 2, a short table locates useful points.
The five points are (−1, 3), (0, 0), (1, −1), (2, 0) and (3, 3). Join them in increasing parameter order with the smooth curve implied by the rules. A small table helps sketch the shape; it does not prove that no features occur between sampled points.
Pause, replay or seek freely. The notes explain the same idea and stay in view.
04 / Trace in increasing parameter order
In the first model, x increases as t increases. Starting at t = −2, the point travels down and right to (1, −1), then up and right to (3, 3). The turning point comes from y reaching its minimum at t = 0.
Increasing t describes an order along the curve.
It describes physical motion only if t has a time meaning.
In a sketch, a small arrow can show this order. Do not assume every curve is traced from left to right: the x-rule may decrease or change direction.
05 / The permitted parameter values determine the actual curve
−2 ≤ t ≤ 2
Both endpoints (−1, 3) and (3, 3) are included.
0 ≤ t ≤ 2
Only the branch from (1, −1) to (3, 3) is traced.
0 < t ≤ 2
The point (1, −1) is approached but excluded; (3, 3) remains included.
Open or closed endpoint symbols describe the actual point set. Sometimes another permitted parameter produces an endpoint too, so check all possible parameter values before declaring a point excluded.
06 / A parametric curve need not be a function of x
At t = −1, the point is (1, −1)
The square makes x positive.
At t = 1, the point is (1, 1)
The same x now has a different y.
The whole curve is not a single-valued y = f(x)
It fails the vertical line test.
It can still be described by x = y², −2 ≤ y ≤ 2
A Cartesian relation need not have y alone on one side.
Restricting to 0 ≤ t ≤ 2 gives y = √x on 0 ≤ x ≤ 4. Restricting to −2 ≤ t ≤ 0 gives y = −√x on the same x-range. The parameter domain selects the branch.
07 / Different parameter values may produce the same point
For x = t² and y = t² with −2 ≤ t ≤ 2, the points at t = −1 and t = 1 are both (1, 1). As t increases from −2 to 0, the point moves from (4, 4) to (0, 0). It then travels back along the same segment.
The locus is y = x, 0 ≤ x ≤ 4.
The segment is retraced; there are not two different copies of it.
A curve can cross itself at a point, retrace an entire portion, or return to its starting point. These are different behaviours; the coordinate rules and parameter values tell you which is happening.
08 / The same curve can have different parameterisations
In B, set t = 2u
Both rules become the coordinates in A.
Both trace y = x² for 0 ≤ x ≤ 2
The point sets are identical.
If t and u are times in the same units, B completes the path in half the time
That timing conclusion depends on giving the parameters a physical time meaning.
A reversed parameter can also reverse the direction while leaving the locus unchanged. A Cartesian equation generally loses this timing and direction information.
09 / Your turn
Find the point at t = 2 for x = 4 − t and y = 3t².
Compute x and y separately.
x = 2, y = 12, so the point is (2, 12).
Find the point at t = −3 for x = 2t + 5 and y = t² − 4.
Square the whole negative value.
x = −1 and y = 5, so (−1, 5).
Does (7, 9) lie on x = 2t + 1, y = t²?
Use x to find t, then check y.
x = 7 requires t = 3; then y = 9. Yes, provided t = 3 is permitted.
Does (5, 9) lie on x = 2t + 1, y = t²?
Both coordinates must come from the same t.
x = 5 forces t = 2, giving y = 4 rather than 9. No. Using t = 3 just for y would be invalid.
For x = t − 4, y = 2t, identify the point at t = 3. Is it (3, 6)?
The first coordinate is x, not t.
The point is (−1, 6), not (3, 6).
For x = t, y = t² with −1 < t ≤ 2, which of (−1, 1) and (2, 4) is included?
x = t makes the parameter unique.
(−1, 1) is excluded; (2, 4) is included.
For x = 5 − 2t, y = t with 0 ≤ t ≤ 2, describe the direction as t increases.
Check whether x and y increase or decrease.
The point moves up and left, from (5, 0) to (1, 2).
For x = t², y = 2t with −1 ≤ t ≤ 1, give two points with x = 1.
t can be −1 or 1.
(1, −2) and (1, 2). The whole curve is not one function y = f(x).
For x = t² + 1, y = 3t² with −2 ≤ t ≤ 2, compare t = −2 and t = 2.
Both rules depend only on t².
Both give (5, 12). The path goes from (5, 12) to (1, 0) and then retraces that segment.
Does the letter t prove that a parametric equation describes motion?
A variable name does not assign units.
No. The context must state that t represents time and give its units. It could be an angle or simply a parameter.
Compare x = t, y = 3t for 0 ≤ t ≤ 4 with x = 4u, y = 12u for 0 ≤ u ≤ 1.
Put t = 4u.
Both trace the segment y = 3x from (0, 0) to (4, 12), in the same direction as their parameters increase.
Why is plotting only a few parameter values not enough to prove the complete shape of an unfamiliar curve?
Think about what can happen between samples.
Extra turns, crossings or excluded inputs may lie between samples. Use the rules, domains, identities and further analysis to justify the sketch, rather than treating a sparse table as proof.
10 / Recap
Section 1 of 10 · Use one parameter in both coordinate rules