From Velocity Back to Position
A velocity vector describes how each component of position changes with time. If a particle's position is represented by \(\mathbf r(t)\), then its velocity is \(\mathbf v(t)=\mathbf r'(t)\). Recovering position from velocity reverses this relationship: integrate the velocity components, then use a known position to determine the constants of integration.
The initial position matters. Integrating a velocity function alone gives a family of possible position functions, because adding a constant vector does not change the derivative. A condition such as \(\mathbf r(t_0)=\mathbf r_0\) selects the particular position function that fits the given motion at time \(t_0\).
The Componentwise Method
Suppose the motion is in three dimensions and the velocity is given in component form:
Write the unknown position as \(\mathbf r(t)=\langle x(t),y(t),z(t)\rangle\). Because vector differentiation is performed component by component, the equation \(\mathbf r'(t)=\mathbf v(t)\) means \(x'(t)=v_x(t)\), \(y'(t)=v_y(t)\), and \(z'(t)=v_z(t)\). Integrate each equation, retaining a separate constant for each component. Then substitute the specified time and position to determine those constants.
The same method applies in two dimensions, with two components and two constants. More generally, each position component needs its own constant. The constants can differ; they are collected into one constant vector after they have been found.
The variable \(u\) in the integral is a dummy variable: it records the values of velocity being accumulated between \(t_0\) and \(t\). The upper limit \(t\) is the variable in the resulting position function.
Why the Formula Works
Here is a direct justification. Define a candidate position by adding the initial position to the accumulated vector integral. By the Fundamental Theorem of Calculus, the derivative of the integral of each continuous component is that component evaluated at \(t\). Thus the derivative of the candidate is exactly the given velocity. At \(t=t_0\), the integral has equal upper and lower limits, so it is the zero vector, leaving the required initial position.
To see why the answer is unique, suppose \(\mathbf r_1(t)\) and \(\mathbf r_2(t)\) both have derivative \(\mathbf v(t)\) and both equal \(\mathbf r_0\) at \(t_0\). Their difference has derivative \(\mathbf 0\) in every component. A differentiable scalar function whose derivative is zero throughout an interval is constant there, so each component of \(\mathbf r_1-\mathbf r_2\) is constant. At \(t_0\), that difference is \(\mathbf 0\); therefore it is \(\mathbf 0\) throughout the interval. The two position functions agree.
Worked Example: A Polynomial Velocity and an Initial Position
Worked Example: A Polynomial Velocity and an Initial Position
A particle moves in the plane with velocity \(\mathbf v(t)=\langle 3t^2-2,4t+1\rangle\), measured in meters per second. Its position at \(t=0\) is \(\mathbf r(0)=\langle 5,-2\rangle\) meters. Find its position function and verify it.
Write \(\mathbf r(t)=\langle x(t),y(t)\rangle\). Integrate the first velocity component:
Integrate the second component separately:
Use \(x(0)=5\): \(0^3-2(0)+C_1=5\), so \(C_1=5\). Use \(y(0)=-2\): \(2(0)^2+0+C_2=-2\), so \(C_2=-2\). Therefore,
Differentiate to check the velocity:
Check the initial position as well:
Both the velocity condition and the initial-position condition are satisfied.
Worked Example: The Initial Time Is Not Zero
Worked Example: The Initial Time Is Not Zero
Suppose \(\mathbf v(t)=\langle 2t,\cos t,3\rangle\) and \(\mathbf r(2)=\langle 1,0,-4\rangle\). Find \(\mathbf r(t)\) and verify both conditions.
Integrate each component, using its own constant:
Now substitute \(t=2\) into each component. For \(x\), \(2^2+C_1=1\), so \(C_1=-3\). For \(y\), \(\sin 2+C_2=0\), so \(C_2=-\sin 2\). For \(z\), \(3(2)+C_3=-4\), so \(C_3=-10\). The position function is
Differentiate component by component:
At \(t=2\), the components are \(2^2-3=1\), \(\sin 2-\sin 2=0\), and \(3(2)-10=-4\). Thus \(\mathbf r(2)=\langle 1,0,-4\rangle\), as required.
This example illustrates why the reference time in the initial condition must be used carefully. The constants are determined by evaluating the antiderivative at \(t=2\), not at \(t=0\).
Worked Example: Different Types of Components
Worked Example: Different Types of Components
For \(t>-1\), let \(\mathbf v(t)=\langle e^t,\frac{1}{t+1},2t-3\rangle\), and suppose \(\mathbf r(0)=\langle 0,2,1\rangle\). Find and check the position function.
Integrating the components gives
The domain condition \(t>-1\) ensures that \(\ln(t+1)\) is defined. Apply the initial condition one component at a time. Since \(e^0+C_1=0\), \(C_1=-1\). Since \(\ln(1)+C_2=2\) and \(\ln(1)=0\), \(C_2=2\). Since \(0^2-3(0)+C_3=1\), \(C_3=1\). Hence
Differentiate:
Evaluate at \(t=0\): \(\mathbf r(0)=\langle 1-1,\ln(1)+2,0-0+1\rangle=\langle 0,2,1\rangle\). The derivative and initial-value checks both succeed.
The Equivalent Antiderivative-and-Constant Form
The definite-integral formula and componentwise antiderivatives express the same method. If \(\mathbf R(t)\) is any antiderivative of \(\mathbf v(t)\), so that \(\mathbf R'(t)=\mathbf v(t)\), then every position with that velocity has the form \(\mathbf r(t)=\mathbf R(t)+\mathbf C\), where \(\mathbf C\) is a constant vector. Substitution of \(\mathbf r(t_0)=\mathbf r_0\) gives \(\mathbf C=\mathbf r_0-\mathbf R(t_0)\). Thus
This is also what the definite-integral formula says: the accumulated change from \(t_0\) to \(t\) is added to the starting position. The antiderivative form is often convenient when the component integrals are straightforward; the definite-integral form makes the role of the initial position especially clear.
A Common Mistake: Ignoring the Initial Position
Suppose \(\mathbf v(t)=\langle 2t,3\rangle\) and \(\mathbf r(1)=\langle 4,-1\rangle\). A tempting but incorrect answer is \(\langle t^2,3t\rangle\), obtained by integrating and setting both constants to zero. Its derivative is indeed \(\langle 2t,3\rangle\), but its value at \(t=1\) is \(\langle 1,3\rangle\), not \(\langle 4,-1\rangle\). Matching the derivative alone is not enough.
Correct the calculation by retaining both constants: \(\mathbf r(t)=\langle t^2+C_1,3t+C_2\rangle\). The initial condition gives \(1+C_1=4\), so \(C_1=3\), and \(3+C_2=-1\), so \(C_2=-4\). Therefore,
Indeed, \(\mathbf r'(t)=\langle 2t,3\rangle\), and \(\mathbf r(1)=\langle 1+3,3-4\rangle=\langle 4,-1\rangle\). Always check both requirements: the derivative must equal the given velocity, and the position must match the specified value at the specified time.
A Reliable Procedure
For a position-recovery problem, keep the two conditions visible throughout the calculation.
Set \(\mathbf r(t)=\langle x(t),y(t),\ldots\rangle\) and equate each derivative to the corresponding velocity component.
Find an antiderivative for each component and include a separate constant for each one.
Substitute the given time into every component and solve for its constant.
Differentiate the proposed position and check that it equals the given velocity. Substitute the initial time once more to confirm the stated position.
Why the Initial Condition Is Essential
Velocity specifies the derivative of position, not the position itself. If \(\mathbf r(t)\) has derivative \(\mathbf v(t)\), then \(\mathbf r(t)+\mathbf C\) has the same derivative for every constant vector \(\mathbf C\). Consequently, velocity alone cannot determine the location of the particle. One known position fixes the otherwise-undetermined constants and selects a unique position function on the interval where the velocity is continuous.
This distinction is useful whenever a problem gives a rate of change and asks for the underlying quantity. The integration produces possible functions; the given value determines which one answers the problem. In vector motion, this reasoning applies to every coordinate at once, while each coordinate still has its own constant to determine.
Check Your Understanding
For each question, remember to check both the derivative and the specified position.
- If \(\mathbf v(t)=\langle 4t,2\rangle\) and \(\mathbf r(0)=\langle 3,-1\rangle\), what is \(\mathbf r(t)\)?
- If \(\mathbf v(t)=\langle 1,\cos t\rangle\) and \(\mathbf r(\pi)=\langle 0,2\rangle\), how do you determine the two constants?
- Why does integrating each component require a separate constant?
- What does the formula \(\mathbf r(t)=\mathbf r_0+\int_{t_0}^{t}\mathbf v(u)\,du\) give when \(t=t_0\)?
- How can differentiation and substitution verify that a proposed position function is correct?
