Interpret direction and repeated points
One circle, two ways to travel it
Two points move on the same circle of radius 3. The orange point has \(x = 3\cos t\), \(y = 3\sin t\), and the white point has \(x = 3\cos 2t\), \(y = -3\sin 2t\).
Move t forward. Both points start at (3, 0), but the orange point moves counterclockwise and the white point moves clockwise twice as fast.
At \(t = \pi\), the orange point is halfway around, at \((-3, 0)\), and the white point is back at \((3, 0)\) after one full turn.
Both points trace the same circle, so only the equations show the direction, the speed and how many times each point is reached.
A path and a clock
The path is the set of all points that the equations reach. As t increases, the orange point moves along the path in the direction of the arrowheads.
The value of \(t\) works as a clock: \(\cy{t = a}\) and \(\cy{t = b}\) are the times when the point reaches those places. At the crossing, two different times, \(\cy{t_1}\) and \(\cy{t_2}\), give the same point.
Press the buttons. The arrowheads and the times change, but the path stays the same.
Notation
On an interval \(a \le t \le b\), the path is the set of points \((x(t), y(t))\), so it starts at \(t = a\) and ends at \(t = b\).
\[\begin{aligned} \text{start }&\cy{\big(x(a),\,y(a)\big)} \\ \text{end }&\cy{\big(x(b),\,y(b)\big)} \end{aligned}\]To find the direction, make a table of points for increasing values of t. Plot them in this order and draw arrows.
The signs of the rates also give the direction. If \(\frac{dx}{dt} > 0\), the point moves to the right. If \(\frac{dy}{dt} < 0\), it moves down. Green marks a positive rate and red a negative one.
\[\begin{aligned} \cg{\dfrac{dx}{dt} > 0}&:\ \text{right} \\ \cr{\dfrac{dy}{dt} < 0}&:\ \text{down} \end{aligned}\]A point is repeated if two different values of \(t\) give it, as \(\cy{t_1}\) and \(\cy{t_2}\) do at the crossing. Both coordinates must match.
\[\begin{aligned} x(t_1) &= x(t_2) \\ y(t_1) &= y(t_2) \\ t_1 &\neq t_2 \end{aligned}\]The interval, not the equations alone, decides which part of the curve is traced, so the point never reaches the faint part.
When it fails
The circle \(x = 3\cos t\), \(y = 3\sin t\) is traced once while \(t\) goes from 0 to \(2\pi\). Use the circle and let \(t\) run to \(4\pi\).
The point goes around twice and reaches every point at least twice, but the path is still the same circle. A repeated point adds no new part to the curve.
Example 1: The part of a circle that is traced
A point moves with \(x = 2\sin t\) and \(y = 2\cos t\) for \(0 \le t \le \pi\). Describe its path and its direction.
Make a table of points for increasing values of t.
\[\begin{aligned} t = 0&:\ (0,\,2) \\ t = \tfrac{\pi}{2}&:\ (2,\,0) \\ t = \pi&:\ (0,\,-2) \end{aligned}\]Plot the points in this order. The point starts at the top, passes through (2, 0) and ends at the bottom, so it moves clockwise.
All points satisfy \(x^2 + y^2 = 4\), so they lie on a circle of radius 2. The path is only the right half, because \(x = 2\sin t\) is not negative when \(0 \le t \le \pi\).
\[x^2 + y^2 = 4\sin^2 t + 4\cos^2 t = 4\]
Example 2: A path that doubles back
A point moves with \(x = 1 + \cos t\) and \(y = 2 + 2\cos t\) for \(0 \le t \le 2\pi\). Describe its path.
Compare \(y\) with \(x\). Because \(y = 2(1 + \cos t)\), every point satisfies \(y = 2x\).
\[y = 2 + 2\cos t = 2x\]A tempting wrong move: The path is the whole line \(y = 2x\).
\[\cr{\text{the line }y = 2x}\]Because \(\cos t\) is between \(-1\) and 1, \(x\) is between 0 and 2. The point never reaches the red parts of the line beyond \((0, 0)\) and \((2, 4)\).
The path is only the segment from (0, 0) to (2, 4). To find the direction, make a table of points.
\[\begin{aligned} t = 0&:\ (2,\,4) \\ t = \tfrac{\pi}{2}&:\ (1,\,2) \\ t = \pi&:\ (0,\,0) \\ t = \tfrac{3\pi}{2}&:\ (1,\,2) \\ t = 2\pi&:\ (2,\,4) \end{aligned}\]While t goes from 0 to π, the point moves down to (0, 0). It then turns around and moves back up to (2, 4), so it passes (1, 2) twice, once moving down and once moving up.
Practice
For \(0 \le t \le 2\), point \(P\) has \(x = 2t\), \(y = 4 - 2t\), and point \(Q\) has \(x = 4 - 2t\), \(y = 2t\). Show that both move along one segment. Find their directions, and where and when they meet.
The orange point is P and the white point is Q. Move the slider to compare their positions at t = 0 and t = 2. Then find the t that makes their x-coordinates equal.
Solution
For both points, \(x + y = 4\) with \(x\) between 0 and 4, so both move along the segment from \((0, 4)\) to \((4, 0)\).
\[\begin{aligned} 2t + (4 - 2t) &= 4 \\ (4 - 2t) + 2t &= 4 \end{aligned}\]\(P\) starts at \((0, 4)\) and ends at \((4, 0)\). \(Q\) starts at \((4, 0)\) and ends at \((0, 4)\). They move in opposite directions.
Because both points lie on \(x + y = 4\), they meet when their \(x\)-coordinates are equal at the same \(t\). This gives \(t = 1\) and the point \((2, 2)\).
\[2t = 4 - 2t\ \Rightarrow\ t = 1,\qquad \cy{(2,\,2)}\]
Questions
A point moves around the square, one side for each unit of \(t\). The labels give its position at \(t = 0, 1, 2\) and 3, and the orange point is its position at \(t = 1.5\).
In which direction does the point move between t = 1 and t = 2?
- Up along the right side
- Down along the right side
- Left along the top
Answer
Up along the right side
The point continues around the square with the same timing. At which time is it next at the corner (2, 2) after t = 2?
- t = 6
- t = 4
- t = 5
Answer
t = 6
A point has \(x = 5\cos 3t\) and \(y = -5\sin 3t\). Where is it at \(t = \pi\)?
- (−5, 0)
- (5, 0)
- (0, −5)
Answer
(−5, 0)
Summary
You find the direction from points in order of increasing \(t\), and a repeated point from two values of \(t\) that give the same \((x, y)\).
This works because the equations give a path and a clock, and the direction, the speed and the number of visits come from the clock.
Eliminating t gives only the path, so you need t to find the direction, the part traced and how often a point is reached.
This is one of four skills in topic 9.1 of Which & Why: AP Calculus.