Integrating vector functions and recovering position

Understand the idea

Integrate a vector-valued function one component at a time. Integrating velocity gives a family of position functions, and an initial position determines the separate constants. A definite integral of velocity gives displacement, which is a vector from the initial point to the final point.

Use the same input interval for every component. Each component can require a different integration technique and a different constant. When an initial time is not zero, a position formula using a definite integral from that time often makes the condition easiest to track.

The magnitude of a displacement vector is the direct endpoint distance. It usually differs from total distance traveled, because a curved or reversing path can be longer than the straight connection between its endpoints.

\mathbf r(t)=\mathbf r(t_0)+\int_{t_0}^{t}\mathbf v(s)\,ds,\qquad \int_a^b\mathbf v(t)\,dt=\mathbf r(b)-\mathbf r(a)

A method you can apply

Integrate each component of velocity and keep its constant.
Apply every component of the initial position at the stated initial time.
Differentiate the resulting position to check the velocity and verify the initial point.

Worked example

Given $\displaystyle \mathbf v(t)=\langle2t,3t^2\rangle$ and $\displaystyle \mathbf r(0)=\langle1,-2\rangle$ , find position and displacement from $\displaystyle 0$ to $\displaystyle 2$ .

Integrating gives $\displaystyle \mathbf r(t)=\langle t^2+C_1,t^3+C_2\rangle$ .
The initial position gives $\displaystyle C_1=1$ and $\displaystyle C_2=-2$ .
Thus $\displaystyle \mathbf r(2)=\langle5,6\rangle$ and subtracting $\displaystyle \mathbf r(0)$ gives displacement $\displaystyle \langle4,8\rangle$ .

Position is

\displaystyle \mathbf r(t)=\langle t^2+1,t^3-2\rangle

and the displacement is

\displaystyle \langle4,8\rangle

.

Try it yourself

Recover position from $\displaystyle \mathbf v(t)=\langle\cos t,2\rangle$ and $\displaystyle \mathbf r(0)=\langle0,1\rangle$ . Find position at $\displaystyle \pi/2$ .

Show solution and reasoning

Integrating and applying the condition gives $\displaystyle \mathbf r(t)=\langle\sin t,2t+1\rangle$ . Consequently $\displaystyle \mathbf r(\pi/2)=\langle1,\pi+1\rangle$ .