Calculate speed
A bug on a plane
A bug crawls on a plane. After \(t\) seconds, its position in centimeters is \(\langle 3t,\ 2t^2\rangle\). The orange arrow is its velocity \(\langle 3,\ 4t\rangle\), drawn at 0.6 of its length so that it fits.
The horizontal leg is 3, the rate at which \(x\) changes, and the vertical leg is \(4t\), the rate at which \(y\) changes. Both legs are green because both coordinates increase.
At \(t = 1\), the legs are 3 and 4. The arrow is the long side of a right triangle, so its length is \(\sqrt{3^2 + 4^2} = 5\).
The length of the arrow is the speed, which tells how fast the bug moves along its path, in centimeters per second. At \(t = 2\), the legs are 3 and 8, so the speed is larger.
The length of the velocity
Move the first two sliders. The legs change, and the length of the arrow changes with them. A leg turns red when its component is negative.
Move the third slider. The legs change, but the length of the arrow stays the same, because turning an arrow does not make it longer. The speed depends on the length of the velocity, not on its direction.
The speed is the length of the long side of a right triangle whose legs are the two components of the velocity.
Notation
The two components of the velocity are the rates of change of the two coordinates.
\[\cy{\vec v}(t) = \langle x'(t),\ y'(t)\rangle\]The speed is the length of the velocity vector. You square each component, add the squares, and take the positive square root.
\[\text{speed} = |\vec v(t)| = \sqrt{x'(t)^2 + y'(t)^2}\]The speed is a number, not a vector. It has no direction, and it is never negative.
The velocity has a direction, which is the direction of motion.
The speed is zero only when both components are zero. If one component is zero, the particle still moves, and the speed is the size of the other component.
Example 1: Around a circle
A particle has position \(\vec r(t) = \langle 4\cos 2t,\ 4\sin 2t\rangle\). Find its speed.
Differentiate each component. The chain rule gives a factor 2.
\[\vec v(t) = \langle -8\sin 2t,\ 8\cos 2t\rangle\]Square each component, add the squares, and factor out 64.
\[\begin{aligned} |\vec v| &= \sqrt{64\sin^2 2t + 64\cos^2 2t} \\ &= \sqrt{64\left(\sin^2 2t + \cos^2 2t\right)} \end{aligned}\]The sum \(\sin^2 2t + \cos^2 2t\) is 1, so the speed is 8 at every time. The faint arrows at other times point in other directions but have the same length.
\[|\vec v| = \cy{8}\]
Example 2: A particle that speeds up
A particle has velocity \(\vec v(t) = \langle 2t,\ t^2 - 1\rangle\). Find its speed at \(t = 3\).
Substitute \(t = 3\) into each component.
\[\vec v(3) = \langle 6,\ 8\rangle\]A tempting wrong move: The speed is the sum of the sizes of the two components.
\[|6| + |8| = 14\]The red path goes up 8 and then 6 to the right, so its length is 14. The straight arrow between the same two points is shorter.
Put the lengths of the legs into the Pythagorean theorem.
\[\sqrt{6^2 + 8^2} = \sqrt{100} = \cy{10}\]The speed is 10 units per second.
For this velocity, the speed is \(t^2 + 1\) at every time, because \(4t^2 + (t^2 - 1)^2 = (t^2 + 1)^2\).
Practice
A particle has position \(\vec r(t) = \langle 2t^3,\ 3t^2\rangle\). Find its speed at \(t = 1\). Then find every time when the particle is at rest.
Find the velocity first, because the speed is its length. Then move the slider until the arrow shrinks to a dot. There both components are zero, so the particle is at rest.
Solution
Differentiate each component.
\[\vec v(t) = \langle 6t^2,\ 6t\rangle\]At \(t = 1\), both components are 6. The speed is the length of the arrow.
\[\sqrt{6^2 + 6^2} = \sqrt{72} = \cy{6\sqrt{2}}\]The speed is zero only if both components are zero. Both \(6t^2\) and \(6t\) are zero only at \(t = 0\), so the particle is at rest only at \(t = 0\).
Questions
The figure shows the velocity of a particle at two times, A and B. Each arrow starts at the origin, and its legs are the two components.
What is the speed at time A, when the legs are 5 and 12?
- 13
- 17
- 7
Answer
13
What is the speed at time B, when the x-part is \(-6\) and the y-part is 8?
- 10
- 2
- 14
Answer
10
A particle has position \(\langle t^2,\ 2t\rangle\). What is its speed at \(t = 2\)?
- \(2\sqrt{5}\)
- 6
- \(4\sqrt{2}\)
Answer
\(2\sqrt{5}\)
Summary
You find the velocity, square its two components, add the squares, and take the square root.
This works because the velocity arrow is the long side of a right triangle whose legs are its two components.
You use it to find how fast an object moves at one instant, or when it is at rest.
This is one of five skills in topic 9.6 of Which & Why: AP Calculus.