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Motion Along Space Curves · Tutorial 160 of 999

Initial Conditions and Integration Constants

Use initial position and velocity data to determine each integration constant and verify the resulting vector motion.

Intermediate 10 min read

What You'll Learn

  • Distinguish the constants introduced when integrating velocity from those introduced when integrating acceleration.
  • Translate vector initial conditions into separate scalar equations for each component.
  • Solve for constants when the initial time is nonzero.
  • Use both initial velocity and initial position to reconstruct a position function from acceleration.
  • Check a reconstructed motion by differentiation and substitution.

Why Initial Conditions Matter

Integrating a velocity function determines position only up to an additive constant vector. Integrating acceleration introduces a constant vector for velocity, and integrating again introduces another for position. Initial conditions supply the information needed to determine those constants.

The initial time need not be zero. If an object’s position is known at \(t=t_0\), the position condition must be applied at that time: substitute \(t_0\) into the component equations and set the results equal to the corresponding position coordinates. The same applies to a specified initial velocity.

Definition: An initial condition is a specified value of a motion quantity, such as position or velocity, at a stated time \(t_0\). For example, \(\mathbf r(t_0)=\mathbf r_0\) specifies position and \(\mathbf v(t_0)=\mathbf v_0\) specifies velocity at that time.

Vector Constants Become Component Equations

Write the position, velocity, and acceleration of an object as \(\mathbf r(t)\), \(\mathbf v(t)\), and \(\mathbf a(t)\). By definition, \(\mathbf v(t)=\mathbf r'(t)\) and \(\mathbf a(t)=\mathbf v'(t)\). In three dimensions, write a vector function in components:

$$ \mathbf r(t)=\langle x(t),y(t),z(t)\rangle,\qquad \mathbf v(t)=\langle v_x(t),v_y(t),v_z(t)\rangle. $$

An antiderivative of a vector function is found by integrating its components. Each component has its own scalar constant of integration. Those three constants can be collected into one constant vector, but using separate symbols makes it easier to apply the initial condition correctly.

$$ \int \mathbf v(t)\,dt = \left\langle \int v_x(t)\,dt+C_x,\; \int v_y(t)\,dt+C_y,\; \int v_z(t)\,dt+C_z \right\rangle. $$

If \(\mathbf r(t_0)=\langle x_0,y_0,z_0\rangle\), then the constants are determined by the three equations \(x(t_0)=x_0\), \(y(t_0)=y_0\), and \(z(t_0)=z_0\). The time \(t_0\) must be substituted into the entire component expression, not just into its constant.

Why the Initial-Value Formula Works

The link between an initial condition and an integration constant follows from the fundamental theorem of calculus, applied to each component. If \(\mathbf r'(t)=\mathbf v(t)\), then, on an interval where \(\mathbf v\) is continuous,

$$ \mathbf r(t)-\mathbf r(t_0) = \int_{t_0}^{t}\mathbf v(s)\,ds. $$

Here \(s\) is a variable of integration. The equation holds component by component: for example, \(x(t)-x(t_0)=\int_{t_0}^{t}v_x(s)\,ds\). Therefore, once \(\mathbf r(t_0)=\mathbf r_0\) is specified, the position is determined by

$$ \mathbf r(t)=\mathbf r_0+\int_{t_0}^{t}\mathbf v(s)\,ds. $$

This formula also explains how the constant appears in an indefinite integral. If \(\mathbf F(t)\) is any componentwise antiderivative of \(\mathbf v(t)\), then \(\mathbf r(t)=\mathbf F(t)+\mathbf C\), where \(\mathbf C\) is a constant vector. Substituting \(t=t_0\) gives \(\mathbf r_0=\mathbf F(t_0)+\mathbf C\), so

$$ \mathbf C=\mathbf r_0-\mathbf F(t_0). $$

Thus an initial position does not generally equal the integration constant. It equals the position expression evaluated at the initial time, including both the antiderivative terms and the constant.

Worked Example: Finding Position from Velocity

Worked Example: Finding Position from Velocity

Suppose

$$ \mathbf v(t)=\langle 3t^2-2,\;4t+1,\;6\rangle, \qquad \mathbf r(2)=\langle 1,-3,5\rangle. $$

Find \(\mathbf r(t)\). Integrate each velocity component, keeping a separate constant:

$$ \mathbf r(t) = \langle t^3-2t+C_x,\;2t^2+t+C_y,\;6t+C_z\rangle. $$

Apply the position condition at \(t=2\). For the first component,

$$ x(2)=2^3-2(2)+C_x=8-4+C_x=4+C_x=1, \qquad C_x=-3. $$

For the second component,

$$ y(2)=2(2^2)+2+C_y=8+2+C_y=10+C_y=-3, \qquad C_y=-13. $$

For the third component,

$$ z(2)=6(2)+C_z=12+C_z=5, \qquad C_z=-7. $$

Therefore,

$$ \mathbf r(t)=\langle t^3-2t-3,\;2t^2+t-13,\;6t-7\rangle. $$

Check the result by differentiating:

$$ \mathbf r'(t)=\langle 3t^2-2,\;4t+1,\;6\rangle=\mathbf v(t). $$

Check the initial position as well:

$$ \mathbf r(2) = \langle 8-4-3,\;8+2-13,\;12-7\rangle = \langle 1,-3,5\rangle. $$

Both checks agree with the given data.

Recovering Position from Acceleration

When acceleration is given, there are two integrations and therefore two stages of constants. First integrate acceleration to obtain velocity, then use the initial velocity to determine the constants in that velocity expression. Next integrate velocity to obtain position, and use the initial position to determine the new constants. The position condition cannot replace the velocity condition: the two conditions apply to different functions.

Key takeaway: Integrating acceleration introduces constants in the velocity components. Integrating those velocity components introduces a second set of constants in the position components. Use initial velocity to find the first set and initial position to find the second.

Worked Example: Using Initial Velocity and Position

Worked Example: Using Initial Velocity and Position

Suppose the acceleration is

$$ \mathbf a(t)=\langle 6t,-4,2\rangle, $$

and at \(t=1\) the velocity and position are

$$ \mathbf v(1)=\langle 2,-1,3\rangle, \qquad \mathbf r(1)=\langle 4,0,-2\rangle. $$

First integrate acceleration:

$$ \mathbf v(t)=\langle 3t^2+C_x,\;-4t+C_y,\;2t+C_z\rangle. $$

Substitute \(t=1\) into each component and apply the initial velocity. The equations are

$$ 3+C_x=2,\qquad -4+C_y=-1,\qquad 2+C_z=3. $$

Solving gives \(C_x=-1\), \(C_y=3\), and \(C_z=1\), so

$$ \mathbf v(t)=\langle 3t^2-1,\;-4t+3,\;2t+1\rangle. $$

Now integrate this velocity to find position:

$$ \mathbf r(t) = \left\langle t^3-t+D_x,\;-2t^2+3t+D_y,\;t^2+t+D_z \right\rangle. $$

Use the position condition at \(t=1\). In the first component, \(1^3-1=0\), so \(D_x=4\). In the second, \(-2(1)^2+3(1)=1\), so \(1+D_y=0\) and \(D_y=-1\). In the third, \(1^2+1=2\), so \(2+D_z=-2\) and \(D_z=-4\). The resulting position is

$$ \mathbf r(t)=\langle t^3-t+4,\;-2t^2+3t-1,\;t^2+t-4\rangle. $$

Differentiate position to check velocity, and then differentiate velocity to check acceleration:

$$ \mathbf r'(t)=\langle 3t^2-1,\;-4t+3,\;2t+1\rangle=\mathbf v(t), $$
$$ \mathbf v'(t)=\langle 6t,-4,2\rangle=\mathbf a(t). $$

Finally, substitute \(t=1\):

$$ \mathbf v(1)=\langle 2,-1,3\rangle,\qquad \mathbf r(1)=\langle 4,0,-2\rangle. $$

The reconstructed functions satisfy the differential relationships and both initial conditions.

Worked Example: A Nonzero Initial Time

Worked Example: A Nonzero Initial Time

Let

$$ \mathbf v(t)=\langle 2t+1,\;3-t,\;t^2\rangle, \qquad \mathbf r(3)=\langle 5,-2,7\rangle. $$

Find the position function. Integrating each component gives

$$ \mathbf r(t)= \left\langle t^2+t+C_x,\;3t-\frac{t^2}{2}+C_y,\;\frac{t^3}{3}+C_z \right\rangle. $$

Now use \(t=3\), not \(t=0\). In the first component, \(3^2+3=12\), so \(12+C_x=5\) and \(C_x=-7\). In the second, \(3(3)-3^2/2=9-9/2=9/2\), so \(9/2+C_y=-2\) and \(C_y=-13/2\). In the third, \(3^3/3=9\), so \(9+C_z=7\) and \(C_z=-2\). Hence

$$ \mathbf r(t)= \left\langle t^2+t-7,\;3t-\frac{t^2}{2}-\frac{13}{2},\;\frac{t^3}{3}-2 \right\rangle. $$

Differentiate to verify the given velocity:

$$ \mathbf r'(t)=\langle 2t+1,\;3-t,\;t^2\rangle=\mathbf v(t). $$

Substitute \(t=3\) to check the position:

$$ \mathbf r(3) = \left\langle 9+3-7,\;9-\frac{9}{2}-\frac{13}{2},\;9-2 \right\rangle = \langle 5,-2,7\rangle. $$

The constants are determined by evaluating the complete antiderivative at the specified time.

A Common Mistake: Setting the Constant Equal to the Initial Position

In the first example, the first position component has the form \(x(t)=t^3-2t+C_x\), and the condition is \(x(2)=1\). A common error is to set \(C_x=1\), as though the constant were automatically the initial position. That gives \(x(t)=t^3-2t+1\), for which

$$ x(2)=2^3-2(2)+1=8-4+1=5\ne 1. $$

The correction is to substitute the initial time into the entire expression:

$$ x(2)=8-4+C_x=1, \qquad C_x=-3. $$

The constant equals the initial position only when the other terms in that component happen to evaluate to zero at the initial time. At a nonzero initial time, there is no reason to expect that to happen.

A Reliable Procedure

For motion reconstructed from velocity or acceleration, keep the order of the data and integrations explicit:

1
Integrate component by component.
Introduce a separate constant in each component of every antiderivative.
2
Apply the matching initial condition.
Substitute the stated time into the velocity expression to use initial velocity, or into the position expression to use initial position.
3
Solve each scalar equation.
Match the \(x\)-, \(y\)-, and \(z\)-components separately; a vector equality represents three component equations.
4
Check the result.
Differentiate to recover the specified velocity or acceleration, then substitute the initial time to verify the given data.

These steps are useful because they separate two different checks. Differentiation checks that the reconstructed functions have the correct rates of change. Substitution checks that the constants place the motion at the given initial state.

Check Your Understanding

For each question, keep the initial time and the component equations visible as you work.

  1. If \(x(t)=t^2+C\) and \(x(3)=4\), what is \(C\)?
  2. Why does integrating a three-component velocity function introduce three scalar constants?
  3. If acceleration is integrated to obtain velocity, which initial condition determines the constants in that velocity expression?
  4. For \(\mathbf r(t)=\langle t^2+C_x,\,2t+C_y,\,t^3+C_z\rangle\) and \(\mathbf r(2)=\langle 1,5,0\rangle\), find \(C_x,C_y,C_z\).
  5. What two kinds of checks help verify a position function reconstructed from acceleration and initial data?

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