Integrate each vector component
A path from its velocity
A drone has the velocity \(\langle 2t,\ 3t^2\rangle\) meters per second at time \(t\) seconds. The first part is how fast the x-coordinate changes, and the second part is how fast the y-coordinate changes.
To find the path, undo each rate by itself. The x-coordinate changes at the rate \(2t\), so it is \(t^2\) plus a constant. The y-coordinate changes at the rate \(3t^2\), so it is \(t^3\) plus a constant.
Drag the time slider to move the drone. The orange arrow shows the velocity, and its two green legs show the two parts.
Move the start with the two sliders. The whole path moves, but each faint orange arrow keeps its size and direction, because a shift of the path does not change its velocity.
Many paths, one velocity
Move the shift sliders. The path moves without a change of shape, because every point moves the same distance to the right and the same distance up.
The dashed line from the origin to the start of the path shows these two distances, which are the two constants \(\cy{C_1}\) and \(\cy{C_2}\). They are separate numbers, so they can differ.
At the same time, the orange arrow is the same on the path and on the two faint copies. Many paths therefore have one velocity, and any two of them differ by a constant vector \(\cy{\vec C}\).
Notation
A velocity vector has two parts, and each part is a function of time.
\[\cy{\vec v}(t) = \langle f(t),\ g(t)\rangle\]Find an antiderivative of each part by itself. The x-part \(F\) has \(F' = f\), and the y-part \(G\) has \(G' = g\).
\[\cb{\vec R}(t) = \langle F(t),\ G(t)\rangle\]Add one constant to each part, because each part can shift by its own amount. Write the two constants as one constant vector.
\[\begin{aligned}\int \vec v\,dt &= \langle F(t) + C_1,\ G(t) + C_2\rangle \\ &= \vec R(t) + \vec C\end{aligned}\]The integral of a vector function is the vector of the integrals of its parts. Each part must be continuous, so that it has an antiderivative.
To check an answer, differentiate each part. The derivatives must equal \(f\) and \(g\).
Example 1: An exponential and a line
A drone has the velocity \(\vec v(t) = \langle e^{2t},\ 8t - 1\rangle\). Find every vector function whose derivative is \(\vec v\).
Integrate the x-part. The derivative of \(\frac12e^{2t}\) is \(e^{2t}\), because the chain rule gives a factor 2 that cancels the \(\frac12\).
\[\int e^{2t}\,dt = \tfrac12 e^{2t}\]Integrate the y-part with the power rule.
\[\int (8t - 1)\,dt = 4t^2 - t\]Add a constant to each part. Every choice of \(C_1\) and \(C_2\) gives one path, and the faint path is the choice \(C_1 = C_2 = 0\).
\[\vec R(t) = \langle \tfrac12 e^{2t} + C_1,\ 4t^2 - t + C_2\rangle\]
Example 2: A sine and a cubic
A particle has the velocity \(\vec v(t) = \langle 3\sin t,\ 4t^3 - 2\rangle\). Find every vector function whose derivative is \(\vec v\).
Integrate the x-part. The derivative of \(-3\cos t\) is \(3\sin t\).
\[\int 3\sin t\,dt = -3\cos t\]Integrate the y-part with the power rule.
\[\int (4t^3 - 2)\,dt = t^4 - 2t\]A tempting wrong move: The answer is \(\langle -3\cos t,\ t^4 - 2t\rangle\), because its derivative is \(\vec v\).
\[\vec R(t) = \langle -3\cos t,\ t^4 - 2t\rangle\]This function is only one antiderivative. The red path has the same orange arrow at the same time, so it is another one. A formula with no constants cannot describe it.
Add a constant to each part. The red path is the case \(C_1 = 2\), \(C_2 = -1\).
\[\begin{aligned}\vec R(t) &= \langle -3\cos t + C_1, \\ &\qquad t^4 - 2t + C_2\rangle\end{aligned}\]
Practice
A robot has the velocity \(\vec v(t) = \langle 3\sqrt{t},\ 6\cos 2t\rangle\) for \(t > 0\). Find every vector function whose derivative is \(\vec v\).
Integrate each part by itself. The two sliders move the path without a change of its orange arrow, so each part needs its own constant.
Solution
Write the root as a power. Then use the power rule on the x-part.
\[\int 3t^{1/2}\,dt = 2t^{3/2}\]Integrate the y-part. The derivative of \(3\sin 2t\) is \(6\cos 2t\), because the chain rule gives a factor 2.
\[\int 6\cos 2t\,dt = 3\sin 2t\]Add a constant to each part.
\[\vec R(t) = \langle 2t^{3/2} + C_1,\ 3\sin 2t + C_2\rangle\]
Questions
Questions 1 and 2 use three paths of a particle for \(0 \le t \le 2\). Path A is \(\langle 2t + 1,\ t^2 + 1\rangle\). Path B is faint. Path C is red. The orange arrow on each path shows the velocity at \(t = 1\).
Which path has the same velocity as path A at every time?
- Path B
- Path C
- Neither, because the start points differ
Answer
Path B
Path B comes from path A by adding a constant vector to every position. Which vector is it?
- \(\langle -4,\ 2\rangle\)
- \(\langle 4,\ -2\rangle\)
- \(\langle -3,\ 3\rangle\)
- \(\langle 2,\ -4\rangle\)
Answer
\(\langle -4,\ 2\rangle\)
Which vector function has the derivative \(\langle 6t^2,\ 1/t\rangle\) for \(t > 0\)?
- \(\langle 2t^3 + C_1,\ \ln t + C_2\rangle\)
- \(\langle 2t^3 + C_1,\ -1/t^2 + C_2\rangle\)
- \(\langle 2t^3 + C,\ \ln t + C\rangle\)
Answer
\(\langle 2t^3 + C_1,\ \ln t + C_2\rangle\)
Summary
You integrate a vector function one part at a time and add one constant to each part.
This works because each part has its own antiderivative, and a constant vector shifts the whole path without a change of its velocity.
You use it to find a path from its velocity, or a velocity from its acceleration.
This is one of four skills in topic 9.5 of Which & Why: AP Calculus.