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Parametric Curves and Derivatives · Tutorial 900 of 998

Parameter Domains and Curve Traces

Determine the trace of a parametrized curve from its allowed parameter interval, including omitted portions, included endpoints, and repeated points.

Beginner 10 min read

What You'll Learn

  • Define the trace of a parametrized curve as the set of points produced by allowed parameter values.
  • Find endpoint points for linear parametrizations on closed intervals.
  • Use trigonometric ranges to identify a restricted circular arc.
  • Recognize how a parameter interval can omit part of a full curve.
  • Use periodicity to detect points traced more than once.
  • Distinguish the curve’s full possible shape from the trace on a specified interval.

Why the Parameter Interval Matters

A pair of parametric equations describes points in the plane as a parameter \(t\) varies. But the equations alone do not determine which points are included in a particular problem: the allowed values of \(t\) matter too. Restricting those values can leave out part of the curve, while allowing a parameter to run through more than one period can make some points occur repeatedly.

For example, \(x(t)=\cos t\) and \(y(t)=\sin t\) describe a circle when \(t\) runs through a full period. If the allowed interval is only \([0,\pi]\), the points produced lie on only half of that circle. If the interval is longer than one period, points can be produced more than once.

To understand a parametrized curve accurately, therefore, identify both the coordinate formulas and the domain of the parameter.

Parameter Domain and Trace

Definition: A parametrization consists of coordinate functions \(x(t)\) and \(y(t)\), together with an allowed parameter set \(I\). Its trace on \(I\) is the set of all points \((x(t),y(t))\) produced as \(t\) ranges over \(I\): $$ \{(x(t),y(t)):t\in I\}. $$

The parameter set \(I\) is often an interval such as \([a,b]\), \((a,b)\), or \([a,b)\). Square brackets include an endpoint; parentheses exclude it. A point belongs to the trace if at least one allowed parameter value produces it. If two different allowed parameter values produce the same point, that point is still one point in the trace, although it has been produced more than once.

It is useful to distinguish the equations from the trace. The equations specify how coordinates depend on \(t\); the interval specifies which inputs are permitted. Changing the interval can change the trace even when the coordinate formulas stay exactly the same.

Key takeaway: The trace is determined by the coordinate formulas and the parameter domain. Do not assume that the equations trace a whole line or curve unless the allowed parameter values actually produce all of it.

A Reliable Way to Read a Parameter Interval

For an interval with finite endpoints, begin by evaluating the coordinates at each included endpoint. Then use the formulas and the permitted parameter values to determine which other points occur. For trigonometric parametrizations, pay particular attention to the range of angles and to periodicity. Finally, check whether different allowed parameter values give the same point.

1
Record the permitted values.
Note whether each endpoint is included and whether the interval spans a full period of any trigonometric functions.
2
Evaluate boundary values.
Substitute included endpoints into both coordinate functions to find the boundary points they produce.
3
Determine what the formulas produce inside the interval.
Use the coordinate functions’ ranges and any relevant identities to identify the trace and any omitted portions.
4
Check for repetition.
Look for distinct allowed parameter values that give the same coordinate pair, especially when the functions are periodic.

Worked Example: A Linear Parametrization on a Restricted Interval

Worked Example: A Linear Parametrization on a Restricted Interval

Find the trace of \(x(t)=t\), \(y(t)=2t+1\) for \(1\le t\le 3\), and identify the points produced at the endpoints and at \(t=2\).

At the left endpoint, \(t=1\), the coordinates are

$$ x(1)=1,\qquad y(1)=2(1)+1=3. $$

Thus the endpoint parameter value produces \((1,3)\).

At the right endpoint, \(t=3\), the coordinates are

$$ x(3)=3,\qquad y(3)=2(3)+1=7. $$

This endpoint produces \((3,7)\). At the intermediate value \(t=2\),

$$ x(2)=2,\qquad y(2)=2(2)+1=5, $$

so the point \((2,5)\) is also in the trace.

As \(t\) ranges from \(1\) to \(3\), the first coordinate \(x=t\) takes every value from \(1\) to \(3\), and the second coordinate is fixed by \(y=2t+1\). The trace is the line segment whose endpoints are \((1,3)\) and \((3,7)\); values of \(t\) outside \([1,3]\), which would extend the line, are not allowed.

Check: The listed points satisfy the coordinate formulas: at \(t=1\) they give \((1,3)\), at \(t=2\) they give \((2,5)\), and at \(t=3\) they give \((3,7)\). Each of these parameter values lies in the specified interval.

Worked Example: A Trigonometric Interval That Omits Part of a Circle

Worked Example: A Trigonometric Interval That Omits Part of a Circle

Find the trace of \(x(t)=\cos t\), \(y(t)=\sin t\) for \(0\le t\le\pi\). State which familiar points occur at the endpoints and determine which half of the circle is traced.

At \(t=0\),

$$ x(0)=\cos 0=1,\qquad y(0)=\sin 0=0, $$

so the point is \((1,0)\). At \(t=\pi\),

$$ x(\pi)=\cos\pi=-1,\qquad y(\pi)=\sin\pi=0, $$

so the point is \((-1,0)\).

For every allowed \(t\), the identity \(\cos^2 t+\sin^2 t=1\) gives

$$ x(t)^2+y(t)^2 =\cos^2 t+\sin^2 t =1. $$

Thus every produced point lies on the unit circle. Also, \(\sin t\ge 0\) for \(0\le t\le\pi\), so every produced point lies on or above the \(x\)-axis. As the angle ranges from \(0\) to \(\pi\), the parametrization produces the upper semicircle, including both endpoints \((1,0)\) and \((-1,0)\).

The lower half of the unit circle is not in this trace. It would require parameter values in an interval where \(\sin t<0\), and no such values are permitted here.

Check: At \(t=\pi/2\), the coordinates are \(\cos(\pi/2)=0\) and \(\sin(\pi/2)=1\), so the point \((0,1)\) is on the trace. It satisfies \(0^2+1^2=1\) and lies above the \(x\)-axis, as required.

Why Periodic Formulas Can Repeat Points

A function is periodic with period \(P>0\) if its value is unchanged when its input is increased by \(P\): \(f(t+P)=f(t)\) wherever the formula is defined. Sine and cosine both have period \(2\pi\). Consequently, for \(x(t)=\cos t\) and \(y(t)=\sin t\), increasing \(t\) by \(2\pi\) leaves both coordinates unchanged.

Result: For the unit-circle parametrization, if \(t\) and \(t+2\pi\) are both allowed, then they produce the same point: $$ (\cos(t+2\pi),\sin(t+2\pi))=(\cos t,\sin t). $$

This follows by applying the periodicity identities separately to each coordinate. The parameter interval can therefore include distinct values that produce the same point. Periodicity can also help identify the trace: once a full period has been included, additional parameter values may repeat points already produced.

Worked Example: A Full Circle Followed by a Repeated Arc

Worked Example: A Full Circle Followed by a Repeated Arc

Find the trace of \(x(t)=\cos t\), \(y(t)=\sin t\) for \(0\le t\le 5\pi/2\). Identify a portion that is produced more than once and check a specific repeated point.

The interval \([0,2\pi]\) covers a full period, so it produces the full unit circle. The remaining parameter values run from \(2\pi\) to \(5\pi/2\). Write those values as \(t=2\pi+u\), where \(0\le u\le\pi/2\). By the \(2\pi\)-periodicity of sine and cosine,

$$ \begin{aligned} x(2\pi+u)&=\cos(2\pi+u)=\cos u,\\ y(2\pi+u)&=\sin(2\pi+u)=\sin u. \end{aligned} $$

As \(u\) ranges from \(0\) to \(\pi/2\), these are the points produced by the first-quadrant portion of the unit circle. Therefore, the trace is still the full unit circle, but its first-quadrant arc is produced again. Its endpoints, \((1,0)\) and \((0,1)\), are included in that repeated portion.

For a numerical check, take \(u=\pi/3\). At \(t=\pi/3\),

$$ (x(t),y(t)) = \left(\cos\frac{\pi}{3},\sin\frac{\pi}{3}\right) = \left(\frac12,\frac{\sqrt3}{2}\right). $$

The corresponding later parameter is \(t=2\pi+\pi/3=7\pi/3\), which is within the allowed interval because \(7\pi/3<5\pi/2\). At this value,

$$ (x(t),y(t)) = \left(\cos\frac{7\pi}{3},\sin\frac{7\pi}{3}\right) = \left(\frac12,\frac{\sqrt3}{2}\right). $$

The two distinct allowed parameter values, \(\pi/3\) and \(7\pi/3\), produce the same point.

Check: Both coordinates agree because \(7\pi/3=\pi/3+2\pi\). The repeated point lies on the unit circle since

$$ \left(\frac12\right)^2+ \left(\frac{\sqrt3}{2}\right)^2 = \frac14+\frac34 =1. $$

A Common Mistake: Ignoring the Allowed Interval

A common error is to recognize a familiar pair of coordinate formulas and immediately claim the full curve, without checking the parameter domain. For the unit-circle formulas with \(0\le t\le\pi\), one might incorrectly say, “The trace is the entire circle because sine and cosine parametrize a circle.” That statement overlooks the restriction on \(t\).

For a concrete check, the lower point \((0,-1)\) would require an angle such as \(t=3\pi/2\), since

$$ \cos\frac{3\pi}{2}=0,\qquad \sin\frac{3\pi}{2}=-1. $$

But \(3\pi/2>\pi\), so this parameter value is not allowed. The correction is to use the specified interval: on \([0,\pi]\), \(\sin t\ge 0\), and the trace is only the upper semicircle, not the entire circle.

Why Identifying the Trace Matters

The trace tells you exactly which points a parametrization includes. This matters whenever an interval is restricted, endpoints are excluded, or periodic functions are allowed to run for more than one period. The same coordinate formulas can describe different traces on different parameter intervals, so the interval is part of the mathematical information, not an optional detail.

When analyzing a parametrized curve, keep three questions separate: which points can the formulas produce, which of those points occur for permitted parameter values, and whether a point is produced by more than one permitted value. Answering these questions prevents both missing parts of a trace and incorrectly adding points that the parameter domain does not allow.

Check Your Understanding

Use the coordinate formulas and the stated parameter domains to answer each question.

  1. For \(x(t)=t\), \(y(t)=t+2\), and \(0\le t\le 4\), what points are produced at the two endpoints?
  2. For \(x(t)=\cos t\), \(y(t)=\sin t\), and \(0\le t\le\pi\), which sign condition on \(y(t)\) shows that the lower semicircle is omitted?
  3. For the unit-circle parametrization, why do \(t=\pi/4\) and \(t=9\pi/4\) produce the same point?
  4. For \(x(t)=\cos t\), \(y(t)=\sin t\), and \(0\le t\le 2\pi\), is the endpoint point at \(t=2\pi\) also produced at another allowed parameter value? Identify it.
  5. Explain why the coordinate formulas alone are not enough to determine the trace.

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