Calculate distance traveled
Adding up the speed
A particle has velocity \(\langle 6t,\ 3 - 3t^2\rangle\). The chalk curve at the top is its path, and the orange arrow is its velocity. The graph below shows the speed, which is the length of that arrow, and the orange ruler shows the distance traveled so far.
Drag the slider. The teal area under the speed graph grows with the ruler. A thin strip is as high as the speed and a short time wide, so its area is about the distance covered in that time.
At \(t = 1\), the speed is 6 and the teal area is 4, the distance traveled so far. At \(t = 2\), the speed is 15 and the area is 14.
The area under the speed graph
Drag the time slider. The area of each teal strip is the speed times a short time, which is about the length of a short piece of the path.
The speed graph never goes below the axis, so no strip subtracts and the total never decreases.
Add strips with the second slider. Their total comes closer to the area under the graph, so in the limit the distance traveled is that area.
Notation
The speed is the length of the velocity.
\[\cy{|\vec v(t)|} = \sqrt{x'(t)^2 + y'(t)^2}\]In a short time \(dt\), the particle moves about \(|\vec v|\,dt\) along its path. The integral adds these pieces from \(a\) to \(b\).
\[\begin{aligned} &\cb{\text{distance}} \\ &= \int_a^b |\vec v(t)|\,dt \\ &= \int_a^b \sqrt{x'(t)^2 + y'(t)^2}\,dt \end{aligned}\]The integrand is the speed, which is never negative, so the distance is never negative.
If the sum of the squares is a perfect square \(u^2\), the speed is \(\sqrt{u^2} = |u|\), not \(u\). If \(u\) changes sign, split the integral at that time.
The distance traveled is also the length of the path, but only when the particle moves along each part of the path once.
When it fails
After one lap, the orange ruler and the faint bar under it both reach \(4\pi\), the length of the circle.
On the second lap, the particle covers the circle again, and the part it covers twice turns red. The ruler grows to \(8\pi\), but the faint bar stops at \(4\pi\), because the distance counts every pass.
Example 1: A curved path
A particle has velocity \(\vec v(t) = \langle \sqrt{t},\ 1\rangle\). Find the distance traveled from \(t = 0\) to \(t = 8\).
Find the speed. Square each component and add, using \((\sqrt{t})^2 = t\).
\[|\vec v| = \sqrt{(\sqrt{t})^2 + 1^2} = \sqrt{1 + t}\]Integrate the speed from 0 to 8. An antiderivative is \(\tfrac23 (1 + t)^{3/2}\).
\[\int_0^8 \sqrt{1 + t}\,dt = \tfrac23\left[(1 + t)^{3/2}\right]_0^8\]Substitute the limits. The value at 8 is \(9^{3/2} = 27\), and the value at 0 is 1.
\[\tfrac23(27 - 1) = \cb{\tfrac{52}{3}}\]As a check, \(52/3 \approx 17.3\). The straight line from \((0, 0)\) to the end point, about \((15.1, 8)\), is about 17.1 long, and a curved path is longer than the straight line.
Example 2: Dropping the square root
A particle has velocity \(\vec v(t) = \langle 4t^2 - 1,\ 4t\rangle\). Find the distance traveled from \(t = 0\) to \(t = 1\).
Square each component and add. The sum is a perfect square.
\[\begin{aligned} &(4t^2 - 1)^2 + (4t)^2 \\ &= 16t^4 + 8t^2 + 1 \\ &= (4t^2 + 1)^2 \end{aligned}\]A tempting wrong move: Integrate the sum of the squares from 0 to 1.
\[\int_0^1 (4t^2 + 1)^2\,dt = \tfrac{103}{15}\]The red curve is the speed squared, not the speed. It climbs far above the speed curve, so the area under it, about 6.9, is far too large.
Take the square root first. Because \(4t^2 + 1 > 0\), the square root of its square is \(4t^2 + 1\) itself.
\[|\vec v| = 4t^2 + 1\]Integrate the speed.
\[\int_0^1 (4t^2 + 1)\,dt = \left[\tfrac43 t^3 + t\right]_0^1 = \cb{\tfrac73}\]
Practice
A particle has velocity components \(x'(t) = e^t(\cos t - \sin t)\) and \(y'(t) = e^t(\cos t + \sin t)\). Find the distance traveled from \(t = 0\) to \(t = \ln 3\).
Square each component and add. The cross terms cancel, and the cosine and sine terms combine to a constant. Then integrate the speed.
Solution
Square each component and add. Factor out \(e^{2t}\). The cross terms \(-2\sin t\cos t\) and \(2\sin t\cos t\) cancel, and \(\cos^2 t + \sin^2 t = 1\) appears twice, so the bracket is 2.
\[\begin{aligned} |\vec v|^2 &= e^{2t}\Bigl[(\cos t - \sin t)^2 \\ &\qquad + (\cos t + \sin t)^2\Bigr] \\ &= 2e^{2t} \end{aligned}\]The speed is \(\sqrt{2}\,e^t\), because \(e^t > 0\). Integrate it from 0 to \(\ln 3\).
\[\begin{aligned} \int_0^{\ln 3} \sqrt{2}\,e^t\,dt &= \sqrt{2}\,(e^{\ln 3} - e^0) \\ &= \sqrt{2}\,(3 - 1) \\ &= \cb{2\sqrt{2}} \end{aligned}\]As a check, \(2\sqrt{2} \approx 2.83\).
Questions
The graph shows the speed of a particle for \(0 \le t \le 6\). The speed rises from 0 to 4 during the first 2 seconds, stays at 4 until \(t = 5\), and then falls to 0 at \(t = 6\). The teal area is the distance traveled so far.
What is the distance traveled from \(t = 0\) to \(t = 6\)?
- 18
- 24
- 12
Answer
18
What is the distance traveled from \(t = 0\) to \(t = 2\)?
- 4
- 8
- 2
Answer
4
A particle has velocity \(\langle t^2 - 4,\ 4t\rangle\). What is the distance traveled from \(t = 0\) to \(t = 3\)?
- 21
- \(3\sqrt{37}\)
- 18
Answer
21
Summary
To find the distance traveled, you integrate the speed, which is the length of the velocity, over the time interval.
This works because each short time adds about the speed times that time, and the speed is never negative, so no piece cancels another.
It answers how far an object travels, which is not the same as where it ends up.
This is one of five skills in topic 9.6 of Which & Why: AP Calculus.