Use an initial position or velocity
One radar reading picks one path
A drone has the velocity \(\langle 2t,\ 3\rangle\), so its path has the form \(\langle t^2 + C_1,\ 3t + C_2\rangle\). The faint curves show some of these paths, and the orange point marks the drone on the bright path at the chosen time.
A radar reading puts the drone at \((5, 3)\) when \(t = 2\), and the red ring marks that point.
Keep the time at 2, and move the C₁ and C₂ sliders until the orange point reaches the red ring. Only one path does this, because the reading fixes both constants.
At \(t = 2\), \(t^2 = 4\) and \(3t = 6\), so the reading gives \(4 + C_1 = 5\) and \(6 + C_2 = 3\). This means \(C_1 = 1\) and \(C_2 = -3\).
A known position at a known time
The red ring is a known position \((x_0, y_0)\), and the orange point is the path at the time \(t_0\) of the reading. Move the slider to change \(t_0\). The path moves, but its point at \(t_0\) stays on the ring.
At \(t_0\), the x-part of the path must equal \(x_0\), and the y-part must equal \(y_0\). Each condition fixes one constant.
You substitute the time at which the position is given, whether or not it is 0.
Notation
Start with the antiderivative. It has one constant in each part.
\[\vec r(t) = \langle F(t) + \cb{C_1},\ G(t) + \cb{C_2}\rangle\]A position \((x_0, y_0)\) at the time \(t_0\) gives one equation for each part. Substitute \(t_0\) into each part.
\[\begin{aligned}F(t_0) + C_1 &= x_0, \\ G(t_0) + C_2 &= y_0\end{aligned}\]Solve each equation for its constant. Then write the constants back into \(\vec r\).
\[\begin{aligned}C_1 &= x_0 - F(t_0), \\ C_2 &= y_0 - G(t_0)\end{aligned}\]A path from an acceleration needs two readings, one for each integration. The velocity at \(t_0\) fixes the constants of \(\vec v\), and the position at \(t_0\) fixes the constants of \(\vec r\).
Find \(\vec v\) first, because you integrate \(\vec v\) to find \(\vec r\).
Another form builds the reading into the integral. The integral from \(t_0\) to \(t_0\) is 0, so the formula gives \(\vec r(t_0)\) at \(t = t_0\).
\[\vec r(t) = \vec r(t_0) + \cb{\int_{t_0}^{t} \vec v(u)\,du}\]
Example 1: From acceleration to position
A particle has the acceleration \(\vec a(t) = \langle 2,\ 6t\rangle\). At \(t = 0\), its velocity is \(\langle 1, -2\rangle\) and its position is \((3, 5)\). Find \(\vec v(t)\) and \(\vec r(t)\).
Integrate the acceleration. At \(t = 0\), the terms with \(t\) are 0, so the velocity \(\langle 1, -2\rangle\) gives \(C_1 = 1\) and \(C_2 = -2\).
\[\vec v(t) = \langle 2t + 1,\ 3t^2 - 2\rangle\]Integrate the velocity, and call the new constants \(D_1\) and \(D_2\). The position \((3, 5)\) at \(t = 0\) gives \(D_1 = 3\) and \(D_2 = 5\).
\[\begin{aligned}\vec r(t) &= \langle t^2 + t + D_1, \\ &\qquad t^3 - 2t + D_2\rangle\end{aligned}\]Write \(D_1 = 3\) and \(D_2 = 5\) into \(\vec r\). They differ from \(C_1\) and \(C_2\) because they come from a different reading.
\[\vec r(t) = \langle t^2 + t + 3,\ t^3 - 2t + 5\rangle\]
Example 2: A reading at t = 2
A particle has the velocity \(\vec v(t) = \langle 3t^2,\ 2t + 1\rangle\). It is at the point \((10, 4)\) when \(t = 2\). Find \(\vec r(t)\).
Integrate each part.
\[\vec r(t) = \langle t^3 + C_1,\ t^2 + t + C_2\rangle\]A tempting wrong move: The reading is \((10, 4)\), so the constants are 10 and 4.
\[\vec r(t) = \langle t^3 + 10,\ t^2 + t + 4\rangle\]This treats \((10, 4)\) as the position at \(t = 0\), but the reading is for \(t = 2\). At \(t = 2\), this path is at \((18, 10)\), and the red dashed line shows how far that is from the ring.
Substitute \(t = 2\) into each part, and set each part equal to the reading.
\[\begin{aligned}2^3 + C_1 &= 10, \\ 2^2 + 2 + C_2 &= 4\end{aligned}\]Solve for the constants. The orange point at \(t = 2\) now sits on the ring.
\[\begin{aligned}C_1 = 2,&\quad C_2 = -2, \\ \vec r(t) &= \langle t^3 + 2,\ t^2 + t - 2\rangle\end{aligned}\]
Practice
A particle has the acceleration \(\vec a(t) = \langle 3\cos t,\ 2\rangle\). At \(t = 0\), its velocity is \(\langle 0, -1\rangle\) and its position is \((-2, 4)\). Find \(\vec r(t)\). Then find the position at \(t = \pi\).
Find the velocity first, with its two constants. Then integrate the velocity, and use the position at t = 0 for the next two constants. On the figure, use the sliders to move the path until it starts on the red ring.
Solution
Integrate the acceleration. Because \(\sin 0 = 0\), the velocity \(\langle 0, -1\rangle\) at \(t = 0\) gives \(C_1 = 0\) and \(C_2 = -1\).
\[\vec v(t) = \langle 3\sin t,\ 2t - 1\rangle\]Integrate the velocity, and call the new constants \(D_1\) and \(D_2\). Because \(\cos 0 = 1\), the position \((-2, 4)\) at \(t = 0\) gives \(-3 + D_1 = -2\) and \(D_2 = 4\).
\[\vec r(t) = \langle -3\cos t + 1,\ t^2 - t + 4\rangle\]Substitute \(t = \pi\) into each part.
\[\vec r(\pi) = \langle 4,\ \pi^2 - \pi + 4\rangle\]
Questions
A drone has the velocity \(\langle 4,\ 2t\rangle\). A radar reading puts it at \((6, 3)\) when \(t = 1\), and the red ring marks the reading. The figure shows path P and path Q, with a point on each at \(t = 1\).
Which path fits the radar reading?
- Path P
- Path Q
- Both, because both paths have the velocity \(\langle 4, 2t\rangle\)
Answer
Path P
The path has the form \(\langle 4t + C_1,\ t^2 + C_2\rangle\). Which equation fixes \(C_2\)?
- \(1 + C_2 = 3\)
- \(0 + C_2 = 3\)
- \(2 + C_2 = 3\)
- \(1 + C_2 = 6\)
Answer
\(1 + C_2 = 3\)
A particle has the velocity \(\langle 2t,\ 6t^2\rangle\). It is at \((4, 5)\) when \(t = 1\). Where is it when \(t = 0\)?
- \((3, 3)\)
- \((4, 5)\)
- \((5, 7)\)
- \((3, 5)\)
Answer
\((3, 3)\)
Summary
You substitute the time of the known position or velocity into each part, and each equation gives one constant.
This works because the reading is true at its own time, and each part of the path has one unknown constant.
You use it when a problem gives a rate and a position or a velocity at one time.
This is one of four skills in topic 9.5 of Which & Why: AP Calculus.