Tutorials › AP Calculus › Interpreting a Position Vector and Its Derivatives

Home  /  Tutorials  /  Vector-Valued Function Calculus

Vector-Valued Function Calculus · Tutorial 788 of 999

Interpreting a Position Vector and Its Derivatives

Connect a particle’s position vector and its derivatives to its location, velocity, and acceleration, including the meaning and units of their components.

Advanced 9 min read

What You'll Learn

  • Interpret a position vector as the particle’s location relative to a chosen origin.
  • Explain how velocity and acceleration describe rates of change of position and velocity.
  • Read the coordinate components of position, velocity, and acceleration.
  • Track how units change when differentiating with respect to time.
  • Distinguish a negative component from a negative magnitude or a reversal of the entire motion.

Position and Its Derivatives Describe Motion

A vector-valued function of time can describe the motion of a particle. At each time \(t\), its position vector points from a chosen origin to the particle’s location. The first derivative describes how that position changes, and the second derivative describes how the first derivative changes.

These vectors carry both geometric and physical information. Their components tell us what happens along each coordinate axis, while their units tell us whether we are describing location, change in location per unit time, or change in velocity per unit time.

Key idea: If \(\mathbf{r}(t)\) is a particle’s position vector, then \(\mathbf{r}'(t)\) is its velocity vector and \(\mathbf{r}''(t)\) is its acceleration vector, when these derivatives exist.

Position, Velocity, and Acceleration

In two dimensions, write the position vector as \(\mathbf{r}(t)=\langle x(t),y(t)\rangle\). The particle is at the point with coordinates \(x(t)\) and \(y(t)\) at time \(t\). The vector starts at the origin and ends at that point; it is not the path itself.

For a particle moving in three dimensions, the same idea uses three coordinates: \(\mathbf{r}(t)=\langle x(t),y(t),z(t)\rangle\). In either setting, velocity and acceleration are defined by differentiation with respect to time.

$$ \mathbf{v}(t)=\mathbf{r}'(t) \qquad\text{and}\qquad \mathbf{a}(t)=\mathbf{v}'(t)=\mathbf{r}''(t). $$

For two-dimensional motion, this means the velocity components are the rates of change of the position coordinates, and the acceleration components are the rates of change of those velocity components:

$$ \mathbf{r}(t)=\langle x(t),y(t)\rangle,\qquad \mathbf{v}(t)=\langle x'(t),y'(t)\rangle,\qquad \mathbf{a}(t)=\langle x''(t),y''(t)\rangle. $$

The same component-by-component interpretation applies in three dimensions. For example, if \(\mathbf{r}(t)=\langle x(t),y(t),z(t)\rangle\), then \(\mathbf{a}(t)=\langle x''(t),y''(t),z''(t)\rangle\).

Why the Derivatives Have These Interpretations

The derivative of position is velocity because a derivative measures an instantaneous rate of change. Over a short time interval of length \(h\), the average change in position per unit time is \(\bigl(\mathbf{r}(t+h)-\mathbf{r}(t)\bigr)/h\). Taking the limit as the interval shrinks gives the instantaneous velocity.

This also explains why the derivatives act component by component. For \(\mathbf{r}(t)=\langle x(t),y(t)\rangle\), the difference quotient is

$$ \frac{\mathbf{r}(t+h)-\mathbf{r}(t)}{h} = \left\langle \frac{x(t+h)-x(t)}{h}, \frac{y(t+h)-y(t)}{h} \right\rangle. $$

If the coordinate derivatives exist, the limit of each component exists. Taking the limit of the vector difference quotient therefore gives \(\mathbf{r}'(t)=\langle x'(t),y'(t)\rangle\). Applying the same reasoning to velocity shows that acceleration has components \(x''(t)\) and \(y''(t)\). Thus each coordinate’s rate of change contributes the corresponding component of the motion vector.

Units: If position is measured in meters and time in seconds, then position has units of meters, velocity has units of meters per second, and acceleration has units of meters per second squared. Differentiation with respect to time divides the units by one factor of time.

Reading Components and Signs

Each component describes change along one coordinate axis. For instance, a positive \(x\)-component of velocity means the \(x\)-coordinate is increasing at that instant; a negative \(x\)-component means it is decreasing. The same interpretation applies separately to the \(y\)- and \(z\)-components.

A negative component does not mean that the whole vector is negative. It indicates a direction along one axis, relative to the chosen coordinate system. Similarly, a negative component of acceleration means the corresponding velocity component is decreasing. It does not by itself tell us whether the particle’s overall speed is increasing or decreasing.

Position components are coordinates relative to the selected origin. Moving the origin changes the coordinate values used to describe position, but it does not change the meaning of velocity or acceleration as rates of change.

Worked Examples

Worked Example: Reading Two-Dimensional Motion

A particle’s position, in meters, is \(\mathbf{r}(t)=\langle t^2,6t-t^2\rangle\), where \(t\) is measured in seconds. Find and interpret its position, velocity, and acceleration at \(t=2\).

Differentiate the position components to obtain velocity:

$$ \mathbf{v}(t) =\left\langle \frac{d}{dt}(t^2),\frac{d}{dt}(6t-t^2)\right\rangle =\langle 2t,6-2t\rangle\ \text{m/s}. $$

Differentiate the velocity components to obtain acceleration:

$$ \mathbf{a}(t) =\langle 2,-2\rangle\ \text{m/s}^2. $$

Evaluate each vector at \(t=2\):

$$ \mathbf{r}(2)=\langle 2^2,6(2)-2^2\rangle =\langle 4,8\rangle\ \text{m}, $$ $$ \mathbf{v}(2)=\langle 2(2),6-2(2)\rangle =\langle 4,2\rangle\ \text{m/s}, \qquad \mathbf{a}(2)=\langle 2,-2\rangle\ \text{m/s}^2. $$

At two seconds, the particle is at coordinates \((4,8)\) meters. Its \(x\)-coordinate is increasing at \(4\) meters per second, and its \(y\)-coordinate is increasing at \(2\) meters per second. Its \(x\)-velocity component is increasing at \(2\) meters per second squared, while its \(y\)-velocity component is decreasing at \(2\) meters per second squared.

Check: Differentiating \(\langle t^2,6t-t^2\rangle\) gives \(\langle2t,6-2t\rangle\), and differentiating that result gives \(\langle2,-2\rangle\), matching the velocity and acceleration used above.

Worked Example: Interpreting Three-Dimensional Components

A particle’s position, in feet, is \(\mathbf{r}(t)=\langle3+2t,1-t^2,4t\rangle\), with time \(t\) measured in seconds. Find and interpret all three vectors at \(t=1\).

Differentiate each position coordinate to find the velocity components:

$$ \mathbf{v}(t) =\left\langle \frac{d}{dt}(3+2t), \frac{d}{dt}(1-t^2), \frac{d}{dt}(4t) \right\rangle =\langle2,-2t,4\rangle\ \text{ft/s}. $$

Differentiate again:

$$ \mathbf{a}(t)=\langle0,-2,0\rangle\ \text{ft/s}^2. $$

At \(t=1\), the position is

$$ \mathbf{r}(1)=\langle3+2(1),1-1^2,4(1)\rangle =\langle5,0,4\rangle\ \text{ft}. $$

The velocity and acceleration at that time are

$$ \mathbf{v}(1)=\langle2,-2(1),4\rangle =\langle2,-2,4\rangle\ \text{ft/s}, \qquad \mathbf{a}(1)=\langle0,-2,0\rangle\ \text{ft/s}^2. $$

The particle is at \((5,0,4)\) feet. Its \(x\)- and \(z\)-coordinates are each increasing at \(2\) and \(4\) feet per second, respectively, while its \(y\)-coordinate is decreasing at \(2\) feet per second. Only the \(y\)-component of velocity is changing; it is decreasing at \(2\) feet per second squared.

Check: The derivatives of \(3+2t\), \(1-t^2\), and \(4t\) are \(2\), \(-2t\), and \(4\). Differentiating \(\langle2,-2t,4\rangle\) gives \(\langle0,-2,0\rangle\), as stated.

Worked Example: Tracking Units and a Negative Component

A particle moves in a plane with position \(\mathbf{r}(t)=\langle12-3t,t^2+1\rangle\) kilometers, where \(t\) is measured in hours. Determine the position, velocity, and acceleration at \(t=2\), and explain the signs.

Differentiate the coordinates to get velocity and then acceleration:

$$ \mathbf{v}(t) =\left\langle \frac{d}{dt}(12-3t), \frac{d}{dt}(t^2+1) \right\rangle =\langle-3,2t\rangle\ \text{km/h}, $$ $$ \mathbf{a}(t) =\langle0,2\rangle\ \text{km/h}^2. $$

Evaluate at \(t=2\):

$$ \mathbf{r}(2)=\langle12-3(2),2^2+1\rangle =\langle6,5\rangle\ \text{km}, $$ $$ \mathbf{v}(2)=\langle-3,2(2)\rangle =\langle-3,4\rangle\ \text{km/h}, \qquad \mathbf{a}(2)=\langle0,2\rangle\ \text{km/h}^2. $$

The negative \(x\)-component of velocity means the \(x\)-coordinate is decreasing at \(3\) kilometers per hour. The positive \(y\)-component means the \(y\)-coordinate is increasing at \(4\) kilometers per hour. The \(y\)-component of velocity is itself increasing at \(2\) kilometers per hour squared, while the \(x\)-component of velocity is constant.

Check: Differentiating \(\langle12-3t,t^2+1\rangle\) gives \(\langle-3,2t\rangle\); differentiating once more gives \(\langle0,2\rangle\). Substitution of \(t=2\) gives the vectors displayed above, with units changing from kilometers to kilometers per hour and kilometers per hour squared.

A Common Interpretation Error

Suppose \(\mathbf{r}(t)=\langle t^2,4t\rangle\) meters. At \(t=2\), the position is \(\mathbf{r}(2)=\langle4,8\rangle\) meters. A common mistake is to call \(\langle4,8\rangle\) the velocity at that instant.

That is incorrect: these entries describe the particle’s coordinates, and their units are meters. Velocity is the derivative of position. Here, \(\mathbf{v}(t)=\langle2t,4\rangle\) meters per second, so \(\mathbf{v}(2)=\langle4,4\rangle\) meters per second. Differentiating again gives \(\mathbf{a}(t)=\langle2,0\rangle\) meters per second squared. The different units are a useful check that the three vectors are not being confused.

Takeaway: To interpret a vector correctly, identify what it represents, read each component along its coordinate axis, and attach the appropriate units. Position is not velocity, and velocity is not acceleration.

Check Your Understanding

Use the definitions and component interpretations above to answer the following questions.

  1. If \(\mathbf{r}(t)=\langle x(t),y(t)\rangle\), what do the components of \(\mathbf{r}'(t)\) represent?
  2. A position coordinate is measured in meters and time in seconds. What are the units of its first and second derivatives with respect to time?
  3. If the \(x\)-component of velocity is negative at an instant, what does that tell you about the \(x\)-coordinate?
  4. For \(\mathbf{r}(t)=\langle t^2,3t\rangle\) meters, find \(\mathbf{r}(1)\), \(\mathbf{v}(1)\), and \(\mathbf{a}(1)\), including units.
  5. Why does a negative acceleration component not, by itself, establish that the particle’s overall speed is decreasing?

Test yourself: AP Calculus practice exams

Ready to check what you know? Work an AP-style practice exam, then open the full solutions.

AB Multiple Choice AB Free Response BC Multiple Choice BC Free Response

Want all 1,114 practice problems?

This tutorial covers one topic. Calc Pass covers all of AP Calculus AB & BC: full review, every practice question, and step-by-step solutions to each one.

QR code to download the Calc Pass app

Get the Calc Pass App

Scan to download and practice on the go, with instant scoring.

Calc Pass Volume 1: 1,114 Practice Questions book cover

Vol. 1: 1,114 Questions

The complete practice question bank, organized by AB & BC topic.

View on Amazon
Calc Pass Volume 2: Solutions Manual book cover

Vol. 2: Solutions Manual

Fully worked solutions to every question in Volume 1.

View on Amazon

Previous

Next

Back to Tutorials