Recovering Position from Velocity

From Velocity Back to Position

A velocity function describes how position changes with time. If a particle’s position is \(\mathbf{r}(t)\), its velocity is \(\mathbf{v}(t)\), and the relationship between them is

\mathbf{r}'(t)=\mathbf{v}(t).

Recovering position from velocity reverses this relationship: integrate the velocity, then use a known position to determine the constants of integration. The velocity tells us the particle’s change in position, but by itself it does not tell us where the particle was at any particular time. That information comes from the specified position.

Key idea: Integrate every component of the velocity function. Then use the known position at time \(t_0\) to determine the integration constant in each coordinate.

Componentwise Integration and the Position Condition

Write the velocity and position of a particle in three dimensions as \(\mathbf{v}(t)=\langle v_1(t),v_2(t),v_3(t)\rangle\) and \(\mathbf{r}(t)=\langle x(t),y(t),z(t)\rangle\). Since differentiation of a vector-valued function is componentwise, the equation \(\mathbf{r}'(t)=\mathbf{v}(t)\) means \(x'(t)=v_1(t)\), \(y'(t)=v_2(t)\), and \(z'(t)=v_3(t)\).

Consequently, if the velocity components have antiderivatives \(F_1,F_2,F_3\), the position has the form \(\mathbf{r}(t)=\langle F_1(t)+C_1,F_2(t)+C_2,F_3(t)+C_3\rangle\). The constants \(C_1,C_2,C_3\) are determined using the supplied position condition \(\mathbf{r}(t_0)=\mathbf{r}_0\). Each coordinate equation is solved using the matching coordinate of \(\mathbf{r}_0\).

Definition: Given a continuous velocity function \(\mathbf{v}\) on an interval and a position \(\mathbf{r}_0\) at time \(t_0\), recovering position means finding \(\mathbf{r}(t)\) such that \(\mathbf{r}'(t)=\mathbf{v}(t)\) and \(\mathbf{r}(t_0)=\mathbf{r}_0\).

Equivalently, the position can be written using a definite vector integral:

\mathbf{r}(t)=\mathbf{r}_0+\int_{t_0}^{t}\mathbf{v}(u)\,du.

The integration variable \(u\) is a dummy variable: it distinguishes the variable being integrated from the endpoint \(t\). The vector integral is evaluated component by component. This formula works when the velocity is continuous between \(t_0\) and \(t\); more generally, the same relationship holds whenever the relevant definite integrals and the fundamental theorem of calculus apply.

Why the Formula Works

The formula follows from the fundamental theorem of calculus applied to each coordinate. Let \(V_i(t)=\int_{t_0}^{t}v_i(u)\,du\). If \(v_i\) is continuous, then \(V_i'(t)=v_i(t)\) and \(V_i(t_0)=0\). Thus the \(i\)-th coordinate of \(\mathbf{r}_0+\int_{t_0}^{t}\mathbf{v}(u)\,du\) has derivative \(v_i(t)\), and at \(t=t_0\) it equals the \(i\)-th coordinate of \(\mathbf{r}_0\). It satisfies both required conditions.

These conditions also determine position uniquely. If \(\mathbf{r}_1\) and \(\mathbf{r}_2\) both have derivative \(\mathbf{v}\) and the same position at \(t_0\), then each coordinate of \(\mathbf{r}_1-\mathbf{r}_2\) has derivative zero. By the mean value theorem, a differentiable scalar function with zero derivative on an interval is constant there. Since each coordinate difference is zero at \(t_0\), it is zero throughout the interval. Therefore \(\mathbf{r}_1(t)=\mathbf{r}_2(t)\).

This also explains why an integration constant cannot simply be omitted: integrating velocity determines the change in position, while the known position determines the particle’s location. The definite-integral formula incorporates both pieces of information directly.

Worked Examples

Worked Example: Polynomial Velocity with a Position at \(t=1\)

Suppose time is measured in seconds and position in metres. A particle has velocity \(\mathbf{v}(t)=\langle 2t,3,-4t\rangle\) and position \(\mathbf{r}(1)=\langle 1,-2,5\rangle\). Find its position function.

Integrate each component separately:

x(t)=\int 2t\,dt=t^2+C_1,\qquad y(t)=\int 3\,dt=3t+C_2,\qquad z(t)=\int -4t\,dt=-2t^2+C_3.

Now apply the position condition at \(t=1\). For the first coordinate, \(1^2+C_1=1\), so \(C_1=0\). For the second, \(3(1)+C_2=-2\), so \(C_2=-5\). For the third, \(-2(1)^2+C_3=5\), so \(C_3=7\). Therefore,

\mathbf{r}(t)=\langle t^2,3t-5,-2t^2+7\rangle.

Check by differentiating: \(\mathbf{r}'(t)=\langle 2t,3,-4t\rangle=\mathbf{v}(t)\). At \(t=1\), \(\mathbf{r}(1)=\langle 1,3-5,-2+7\rangle=\langle 1,-2,5\rangle\), as required.

Worked Example: Trigonometric and Exponential Components

Let \(\mathbf{v}(t)=\langle \cos t,2t,e^t\rangle\), with \(\mathbf{r}(0)=\langle 2,-1,3\rangle\). Find \(\mathbf{r}(t)\).

Integrating the three components gives

x(t)=\sin t+C_1,\qquad y(t)=t^2+C_2,\qquad z(t)=e^t+C_3.

Substitute \(t=0\) and use the specified coordinates. Since \(\sin 0=0\), \(0+C_1=2\), so \(C_1=2\). Since \(0^2+C_2=-1\), \(C_2=-1\). Since \(e^0+C_3=3\), \(1+C_3=3\), so \(C_3=2\). Hence

\mathbf{r}(t)=\langle \sin t+2,t^2-1,e^t+2\rangle.

Differentiate to check: \(\mathbf{r}'(t)=\langle \cos t,2t,e^t\rangle\), which is the given velocity. Evaluate the position condition: \(\mathbf{r}(0)=\langle 0+2,0-1,1+2\rangle=\langle 2,-1,3\rangle\). Both checks pass.

Worked Example: A Nonzero Initial Time and Mixed Components

Suppose \(\mathbf{v}(t)=\langle 3t^2-2,4\sin(2t),6e^{-t}\rangle\) and \(\mathbf{r}(0)=\langle -1,4,0\rangle\). Determine the position function.

Integrate each component. For the middle component, use that the derivative of \(\cos(2t)\) is \(-2\sin(2t)\):

x(t)=t^3-2t+C_1,\qquad y(t)=-2\cos(2t)+C_2,\qquad z(t)=-6e^{-t}+C_3.

Apply the position at \(t=0\). The first coordinate gives \(0-0+C_1=-1\), so \(C_1=-1\). Since \(\cos(0)=1\), the second gives \(-2+C_2=4\), so \(C_2=6\). Since \(e^0=1\), the third gives \(-6+C_3=0\), so \(C_3=6\). Thus

\mathbf{r}(t)=\langle t^3-2t-1,-2\cos(2t)+6,-6e^{-t}+6\rangle.

Differentiate each coordinate: \(\frac{d}{dt}(t^3-2t-1)=3t^2-2\), \(\frac{d}{dt}(-2\cos(2t)+6)=4\sin(2t)\), and \(\frac{d}{dt}(-6e^{-t}+6)=6e^{-t}\). The derivative is the given velocity. At \(t=0\), the coordinates are \(-1\), \(-2+6=4\), and \(-6+6=0\), confirming the specified position.

Common Pitfall: Losing a Sign or a Coordinate Constant

Consider the second coordinate in the preceding example. A common error is to write \(\int 4\sin(2t)\,dt=2\cos(2t)\). This has the wrong sign: differentiating \(2\cos(2t)\) gives \(-4\sin(2t)\), not \(4\sin(2t)\). The correct antiderivative is \(-2\cos(2t)\), because

\frac{d}{dt}\bigl(-2\cos(2t)\bigr) =(-2)\bigl(-2\sin(2t)\bigr) =4\sin(2t).

Another error is to use the same numerical constant in every coordinate without checking the position condition. The constants belong to separate scalar antiderivatives; each is determined by its corresponding coordinate. A reliable check is to differentiate the entire position vector and then substitute \(t_0\) into the entire vector. Checking only one coordinate does not verify the other two.

Using the Definite-Integral Form

The componentwise antiderivative method is useful when the velocity components have familiar antiderivatives. The definite-integral form gives the same result while building in the specified position from the start. For instance, in the first example,

\mathbf{r}(t) =\langle 1,-2,5\rangle+ \int_{1}^{t}\langle 2u,3,-4u\rangle\,du.

Evaluate each integral: \(\int_1^t 2u\,du=t^2-1\), \(\int_1^t 3\,du=3t-3\), and \(\int_1^t -4u\,du=-2t^2+2\). Adding these changes to the initial coordinates gives

\mathbf{r}(t) =\langle 1+t^2-1,-2+3t-3,5-2t^2+2\rangle =\langle t^2,3t-5,-2t^2+7\rangle.

This agrees with the position obtained from indefinite integration. The integral from the known time to the current time records displacement; adding that displacement to the known position recovers position. The order of the limits matters: reversing them changes the sign of the displacement.

Why Recovering Position Matters

A velocity function describes instantaneous change, whereas a position function describes location throughout the motion. Recovering position connects a rate of change to the path followed over time. It also makes the role of information clear: knowing velocity determines position up to a constant vector, and one specified position fixes that vector. If no position is supplied, the velocity alone cannot identify a unique position function.

For a practical calculation, use this sequence:

Integrate each coordinate.
Find an antiderivative for every component of the velocity and include a separate constant in each coordinate.

Apply the position condition.
Substitute the specified time into each coordinate and solve for its constant using the matching coordinate of the given position.

Verify both requirements.
Differentiate the result to recover the velocity, then evaluate it at the specified time to recover the given position.

Check Your Understanding

For each question, show the coordinate calculations and use both checks whenever you recover a position function.

If \(\mathbf{v}(t)=\langle 4t,2,-3\rangle\) and \(\mathbf{r}(0)=\langle 1,0,5\rangle\), what is \(\mathbf{r}(t)\)?
Why does integrating three velocity components produce three integration constants?
For \(\mathbf{v}(t)=\langle \sin t,2e^t,1\rangle\) and \(\mathbf{r}(0)=\langle 0,3,-2\rangle\), determine the position function.
What does the definite integral of velocity from \(t_0\) to \(t\) represent, and why must it be added to \(\mathbf{r}(t_0)\)?
A proposed second coordinate is \(y(t)=2\cos(2t)+C\) for velocity component \(4\sin(2t)\). Differentiate it and explain the correction needed.