Calculate vector displacement
How far has the boat moved?
A boat has the velocity \(\langle t,\ 6 - 2t\rangle\) meters per second. It leaves the start point at time 0. Drag the time slider to move the boat.
The two graphs show the two parts of the velocity. The shaded area under each graph, from 0 to the time on the slider, is the change in that coordinate. Green area counts as positive, and red area lies below the axis, so it counts as negative.
The teal arrow runs from the start to the boat. It is the displacement, which is the change in position, and its two parts are the two shaded areas.
Go to \(t = 4\). The x-area is 8. The y-area is the green 9 minus the red 1, which is 8. The displacement is \(\langle 8,\ 8\rangle\), even though the boat first went up and then moved down.
Only the two ends matter
Move the two time sliders. The teal arrow runs from the position at time \(a\) to the position at time \(b\), and the areas between \(a\) and \(b\) are shaded.
Bend the route. The path and the two graphs change, but the teal arrow stays the same, because the two end positions stay the same.
Each shaded area, with red counted as negative, is the change in one coordinate, so the two areas stay the same too. The displacement depends only on the two end positions.
Notation
The displacement from time \(a\) to time \(b\) is the change in position. It is a vector, because position is a vector.
\[\cb{\Delta\vec r} = \vec r(b) - \vec r(a)\]Each coordinate changes by the integral of its rate, so you integrate each part of the velocity over the interval.
\[\begin{aligned} \cb{\Delta\vec r} &= \int_a^b \cy{\vec v}(t)\,dt \\ &= \left\langle \int_a^b x'(t)\,dt,\ \int_a^b y'(t)\,dt\right\rangle \end{aligned}\]To evaluate, find an antiderivative \(F\) of the x-part and \(G\) of the y-part. In each part, subtract the value at \(a\) from the value at \(b\).
\[\begin{aligned} \Delta\vec r &= \langle F(b) - F(a), \\ &\qquad G(b) - G(a)\rangle \end{aligned}\]The result is one constant vector, not a function of \(t\), because both ends of the integral are numbers.
If you leave out \(F(a)\) and \(G(a)\), the answer is wrong unless the antiderivative is 0 at \(a\).
To find where the object is at time \(b\), add the displacement to the position at time \(a\).
\[\vec r(b) = \vec r(a) + \cb{\Delta\vec r}\]
Example 1: A bird's flight
A bird has the velocity \(\vec v(t) = \langle 2\sin t,\ 3\rangle\). Find its displacement from \(t = 0\) to \(t = \pi\).
Integrate the x-part from 0 to \(\pi\). An antiderivative of \(2\sin t\) is \(-2\cos t\).
\[\begin{aligned} \int_0^{\pi} 2\sin t\,dt &= \left[-2\cos t\right]_0^{\pi} \\ &= 2 + 2 = 4 \end{aligned}\]Integrate the y-part. The velocity is constant, so the area is a rectangle with height 3 and width \(\pi\).
\[\int_0^{\pi} 3\,dt = 3\pi\]Write the two changes as one vector, which is the teal arrow from the start to the end.
\[\Delta\vec r = \langle 4,\ 3\pi\rangle\]
Example 2: An interval that does not start at 0
A particle has the velocity \(\vec v(t) = \langle 2t,\ 3\rangle\). Find its displacement from \(t = 1\) to \(t = 3\).
Find an antiderivative of each part.
\[\vec R(t) = \langle t^2,\ 3t\rangle\]A tempting wrong move: The displacement is the antiderivative at the end time.
\[\vec R(3) = \langle 9,\ 9\rangle\]The arrow and the shaded areas start at \(t = 0\), so they include the part from 0 to 1, which is outside the interval. They must start at \(t = 1\).
Subtract the value at the start time in each part, which removes the area from 0 to 1.
\[\begin{aligned} \Delta\vec r &= \vec R(3) - \vec R(1) \\ &= \langle 9 - 1,\ 9 - 3\rangle = \langle 8,\ 6\rangle \end{aligned}\]The teal arrow has length 10, because \(\sqrt{8^2 + 6^2} = 10\).
Practice
A particle is at the point \((4, 1)\) when \(t = 0\). Its velocity is \(\vec v(t) = \langle t^2,\ 2t - 3\rangle\). Find the displacement from \(t = 0\) to \(t = 3\). Then find the position at \(t = 3\).
Each shaded area is the change in one coordinate. Find the two areas, count the red area as negative, and add the result to the start point.
Solution
Integrate the x-part from 0 to 3.
\[\int_0^3 t^2\,dt = 9\]Integrate the y-part. The red area from 0 to 1.5 cancels the green area from 1.5 to 3, so the integral is 0.
\[\int_0^3 (2t - 3)\,dt = \left[t^2 - 3t\right]_0^3 = 0\]The displacement is \(\langle 9, 0\rangle\). Add it to the start point to find the position at \(t = 3\).
\[\vec r(3) = (4, 1) + \langle 9,\ 0\rangle = (13,\ 1)\]
Questions
A boat has the velocity \(\langle 3,\ 4 - 2t\rangle\) for \(0 \le t \le 5\), and it starts at \((2, 10)\). The graphs show the two parts of the velocity. On the second graph, the green area is 4 and the red area is 9.
What is the y-part of the displacement from \(t = 0\) to \(t = 5\)?
- \(-5\)
- \(13\)
- \(-9\)
- \(4\)
Answer
\(-5\)
Where is the boat at \(t = 5\)?
- \((17, 5)\)
- \((15, -5)\)
- \((17, 14)\)
Answer
\((17, 5)\)
A particle has the velocity \(\langle 6t^2 - 6,\ 4\rangle\). What is its displacement from \(t = 1\) to \(t = 3\)?
- \(\langle 40,\ 8\rangle\)
- \(\langle 36,\ 12\rangle\)
- \(\langle 48,\ 4\rangle\)
Answer
\(\langle 40,\ 8\rangle\)
Summary
Integrating each part of the velocity from the start time to the end time gives the displacement.
This works because each coordinate changes by the area under its rate, so only the two end positions decide the displacement.
It finds a change in position, and with a known start point it also gives the final position.
This is one of four skills in topic 9.5 of Which & Why: AP Calculus.