Distinguish a vector rate from a scalar rate
Three rates from one arrow
A point has position \(\mathbf r(t) = \langle 16 - 4t,\ 3t\rangle\), in meters, with \(t\) in seconds. At \(t = 2\), it is at \((8, 6)\), and the dashed line from the origin to the point has length \(D = 10\).
The velocity is \(\mathbf r'(t) = \langle -4,\ 3\rangle\), the orange arrow. Its red part is 4 units to the left, because the point moves to the left, and its green part is 3 units up.
Three different numbers describe this motion. The horizontal rate \(\frac{dx}{dt} = -4\) is one part of the velocity. The speed is the length of the arrow, \(\sqrt{(-4)^2 + 3^2} = 5\) m/s.
The third number is the rate of change of \(D\). Press the button to show the part of the arrow along the dashed line. It is red because the point moves toward the origin, so \(D\) falls, here at 1.4 m/s.
Drag the time. The speed stays 5, but the rate of change of \(D\) changes, because the dashed line turns while the arrow does not.
The part of the arrow along the line
The orange arrow is the velocity, and its length is the speed. The figure splits the arrow into a part along the dashed line from the origin and a part across it.
The part along the line is the rate at which \(D\) changes. It is green when the point moves away from the origin and red when the point moves toward it.
The part across the line moves the point around the origin, so it does not change \(D\) at that instant.
Move the point. The arrow keeps its length, but the part along the line changes.
Notation
The velocity is a vector. Its horizontal part \(x'\) is one number, and it can be negative.
\[\mathbf v = \langle x',\ y'\rangle\]The speed is the length of the velocity vector, so it is never negative.
\[\cy{|\mathbf v|} = \sqrt{(x')^2 + (y')^2}\]The distance from the origin is \(D = \sqrt{x^2 + y^2}\). The chain rule gives its rate of change.
\[\begin{aligned} D' &= \dfrac{2xx' + 2yy'}{2\sqrt{x^2 + y^2}} \\ &= \dfrac{xx' + yy'}{D} \end{aligned}\]\(D'\) is positive when the point moves away from the origin and negative when it moves toward it.
The formula divides by \(D\), so it needs \(D \ne 0\).
When it fails
Take \(\mathbf r(t) = \langle 2t,\ t\rangle\), which passes through the origin at \(t = 0\). The arrow has length \(\sqrt5\) and lies along the dashed line, so \(D' = -\sqrt5\) before \(t = 0\) and \(D' = \sqrt5\) after it.
The distance \(D = \sqrt5\,|t|\) has a corner at \(t = 0\), so \(D'\) does not exist there. The formula divides by \(D = 0\) there.
Example 1: All three rates
A point has position \(\mathbf r(t) = \langle 5t + 1,\ 12t - 4\rangle\). At \(t = 1\), find the horizontal rate, the speed, and the rate of change of the distance from the origin.
Find the velocity. The horizontal rate is 5.
\[\mathbf r'(t) = \langle 5,\ 12\rangle\]The figure draws the arrow at half its length.
The speed is the length of the velocity.
\[|\mathbf v| = \sqrt{5^2 + 12^2} = 13\]At \(t = 1\), the point is at \((6, 8)\), so the distance from the origin is 10.
\[D = \sqrt{6^2 + 8^2} = 10\]Use the formula for \(D'\).
\[D' = \dfrac{6\cdot 5 + 8\cdot 12}{10} = 12.6\]The three numbers 5, 13 and 12.6 are different. \(D'\) is positive because the point moves away from the origin.
Example 2: A boat that moves across
A boat has position \(\mathbf r(t) = \langle 3t,\ 4\rangle\), in kilometers, with \(t\) in hours. How fast is its distance from the origin changing at \(t = 0\)?
The speed is 3 km/h, but the question asks for a different rate.
A tempting wrong move: The boat moves at 3 km/h, so the distance changes at 3 km/h.
\[D'(0) = 3\]At \(t = 0\), the arrow points to the right, across the dashed line to the origin. After 0.1 hour, the boat is at \((0.3, 4)\), only about 0.011 km farther from the origin. That is an average of about 0.11 km/h, not 3 km/h.
Write the distance from the origin and differentiate it.
\[\begin{aligned} D &= \sqrt{9t^2 + 16} \\ D' &= \dfrac{9t}{\sqrt{9t^2 + 16}} \end{aligned}\]At \(t = 0\), the rate is 0.
\[D'(0) = 0\]At \(t = 1\), the rate is \(\frac95 = 1.8\) km/h. The speed is 3 km/h at both times.
Practice
A point has position \(\mathbf r(t) = \langle 2t,\ 9 - t^2\rangle\). At \(t = 2\), find the horizontal rate, the speed, and the rate of change of the distance from the origin.
On the figure, compare the direction of the arrow with the dashed line from the origin.
Solution
Find the velocity and substitute \(t = 2\). The horizontal rate is 2.
\[\begin{aligned} \mathbf r'(t) &= \langle 2,\ -2t\rangle \\ \mathbf r'(2) &= \langle 2,\ -4\rangle \end{aligned}\]The speed is the length of the velocity.
\[|\mathbf v| = \sqrt{2^2 + (-4)^2} = 2\sqrt5\]The point is at \((4, 5)\), so \(D = \sqrt{41}\). Use the formula for \(D'\).
\[D' = \dfrac{4\cdot 2 + 5\cdot(-4)}{\sqrt{41}} = -\dfrac{12}{\sqrt{41}}\]As a check, \(D' \approx -1.87\). It is negative, so the point moves toward the origin.
Questions
Questions 1 and 2 use the path \(\mathbf r(t) = \langle 4 + 3t,\ 3 - 4t\rangle\). The orange point is at \(t = 0\) and the white point is at \(t = 1\). Each point has its velocity arrow, and the dashed line joins the origin to the orange point.
At \(t = 0\), the orange arrow is perpendicular to the dashed line. At what rate does the distance from the origin change?
- 0
- 5
- 3
Answer
0
Which of these numbers is different at \(t = 0\) and at \(t = 1\)?
- The rate of change of the distance from the origin
- The speed
- The horizontal rate
Answer
The rate of change of the distance from the origin
A point has position \(\mathbf r(t) = \langle 2t,\ t^2\rangle\). At \(t = 1\), what is the rate of change of its distance from the origin?
- \(\dfrac{6}{\sqrt5}\)
- \(2\sqrt2\)
- 2
- 6
Answer
\(\dfrac{6}{\sqrt5}\)
Summary
You tell three rates apart: a component rate such as \(x'\), the speed \(|\mathbf v|\), and the rate \(D'\) of the distance from the origin.
The first is one part of the velocity, the second is its length, and the third is its part along the line from the origin.
The three numbers usually differ, so you use the one the question names.
This is one of four skills in topic 9.4 of Which & Why: AP Calculus.