Choose the parameter interval
How far around does the interval go?
A point moves along \(x = 3\cos t\) and \(y = 2\sin t\). At \(t = 0\), it is at \((3, 0)\), the faint ring. The faint dashed curve is the whole ellipse. The solid curve is the part traced for \(0 \le t \le b\), and the orange point is at \(t = b\).
Set \(b = \pi\). The point is at \((-3, 0)\), and the solid curve is the upper half of the ellipse.
Set \(b = 2\pi\). The point is back at \((3, 0)\), and the whole ellipse is traced once.
Move \(b\) past \(2\pi\). The point goes over the ellipse again, so the part it traces a second time is red.
The interval \(0 \le t \le 2\pi\) traces the ellipse once, because \(\cos t\) and \(\sin t\) repeat after \(2\pi\).
An interval that matches the piece
This curve has no numbers. The orange rings mark the ends of the piece that you want, and the interval \(a \le t \le b\) decides which part the point traces.
Move a and b. The solid part is traced once, the red part is traced twice, and the faint dashed part is not traced.
The interval fits the piece when the point starts at the first ring and stops at the second. No part of the piece is red or missing.
The two rings do not fix the interval, because many values of t give the same point. The interval also depends on how often the equations repeat.
Notation
Find the parameter value at each end of the piece. Set \(x(t)\) and \(y(t)\) equal to the coordinates of that end.
\[x(t) = x_{\text{end}},\qquad y(t) = y_{\text{end}}\]One equation can have several solutions. The other equation decides between them.
Use the two values as the limits, with \(a < b\), so the point traces the piece as \(t\) increases.
\[\cy{a} \le t \le \cy{b}\]If \(x\) and \(y\) use \(\cos kt\) and \(\sin kt\), the motion repeats after the period \(T\). An interval longer than \(T\) traces part of the curve again, as the red part shows.
\[T = \dfrac{2\pi}{k}\]If the interval is given, divide its length by the period to find the number of times around.
\[\dfrac{b - a}{T}\]
When it fails
The equations \(x = 2\sin t\) and \(y = 4\sin^2 t\) trace the arc of the parabola \(y = x^2\) from \((-2, 4)\) to \((2, 4)\). During one period, \(0 \le t \le 2\pi\), the point goes out to \((2, 4)\) and back to the origin, then out to \((-2, 4)\) and back.
The point passes along each part of the arc twice, so one period does not trace the curve once. The interval \(\frac{\pi}{2} \le t \le \frac{3\pi}{2}\) traces the arc once, because \(x = 2\sin t\) decreases from 2 to \(-2\) on it.
Example 1: The first arch
A point moves with \(x = t - \sin t\) and \(y = 1 - \cos t\). Find the interval that traces the first arch of the curve, from \((0, 0)\) to \((2\pi, 0)\), once.
The arch starts at the origin, and both coordinates are 0 at t = 0.
\[x(0) = 0,\qquad y(0) = 0\]The arch ends on the x-axis, where \(y = 0\), so \(\cos t = 1\).
\[\begin{aligned} 1 - \cos t &= 0 \\ t &= 0,\ 2\pi,\ 4\pi,\dots \end{aligned}\]The x-coordinate decides which time. At \(t = 2\pi\), \(x = 2\pi\), the end of the first arch, not the second.
\[x(2\pi) = 2\pi - \sin 2\pi = 2\pi\]Because \(\frac{dx}{dt}\) is never negative, the point never moves to the left, so the interval \(0 \le t \le 2\pi\) traces the arch once.
\[\dfrac{dx}{dt} = 1 - \cos t \ge 0\]
Example 2: A quarter of the ellipse
A point moves with \(x = 3\cos t\) and \(y = 2\sin t\). Find an interval that traces once the arc from \((0, -2)\) to \((3, 0)\), with \(t\) increasing.
For the start \((0, -2)\), the equation for \(x\) gives more than one time.
\[3\cos t = 0\ \Rightarrow\ t = \tfrac{\pi}{2},\ \tfrac{3\pi}{2},\dots\]A tempting wrong move: Take the smaller time as the start. The end \((3, 0)\) is at \(t = 2\pi\).
\[\dfrac{\pi}{2} \le t \le 2\pi\]At \(t = \frac{\pi}{2}\), the point is at the red ring \((0, 2)\), not at \((0, -2)\). This interval traces three quarters of the ellipse instead of the last quarter.
The equation for \(y\) decides. The start has \(y = -2\), so \(\sin t = -1\).
\[2\sin t = -2\ \Rightarrow\ t = \tfrac{3\pi}{2}\]The end \((3, 0)\) comes next, at \(t = 2\pi\). The interval is \(\frac{3\pi}{2} \le t \le 2\pi\).
\[x(2\pi) = 3,\qquad y(2\pi) = 0\]
Practice
A point moves with \(x = 5\cos\pi t\) and \(y = 2\sin\pi t\). Find an interval that traces once the upper half of the ellipse, from \((5, 0)\) to \((-5, 0)\).
Move b until the point first reaches (−5, 0), and see which half of the ellipse is solid.
Solution
The start is at \(t = 0\), because \(\cos 0 = 1\) and \(\sin 0 = 0\).
\[x(0) = 5,\qquad y(0) = 0\]The end \((-5, 0)\) needs \(\cos\pi t = -1\). The first time is \(t = 1\).
\[\cos\pi t = -1\ \Rightarrow\ \pi t = \pi\ \Rightarrow\ t = 1\]On \(0 \le t \le 1\), \(y\) is not negative, so the point traces the upper half. The period is \(\frac{2\pi}{\pi} = 2\), so this interval is half a period.
\[\cy{0 \le t \le 1}\]
Questions
A point moves along \(x = 2\cos 3t\) and \(y = 2\sin 3t\), starting at \(t = 0\). The solid curve is the part traced for \(0 \le t \le b\), and the red part is traced a second time. The figure shows \(b = \pi\).
The interval is \(0 \le t \le \pi\). How many times around the circle does the point go?
- 1.5
- 1
- 3
Answer
1.5
What is the largest \(b\) for which the interval \(0 \le t \le b\) has no red part?
- \(\frac{2\pi}{3}\)
- \(2\pi\)
- \(\pi\)
Answer
\(\frac{2\pi}{3}\)
A point moves with \(x = 4\cos t\) and \(y = 4\sin t\). Which interval traces the arc from \((0, 4)\) to \((-4, 0)\) once, with \(t\) increasing?
- \(\frac{\pi}{2} \le t \le \pi\)
- \(0 \le t \le \pi\)
- \(\pi \le t \le \frac{3\pi}{2}\)
Answer
\(\frac{\pi}{2} \le t \le \pi\)
Summary
You find the parameter value at each end of the piece, and you check how many times the interval goes around.
This works because the motion repeats after one period, so a longer interval traces the curve again.
You use it when a question needs a part of the curve traced once, such as its length.
This is one of four skills in topic 9.3 of Which & Why: AP Calculus.