Connecting Position, Velocity, and Acceleration of Functions Using Integrals - AP Calculus AB Study Guide

The Core Idea: Connecting Position, Velocity, and Acceleration of Functions Using Integrals

This topic explores the powerful relationship between motion and calculus, viewed through the lens of integration. While differentiation allows us to determine the rate of change (finding velocity from position), integration allows us to reverse the process. The core idea is that the definite integral of a rate of change function over an interval gives the net accumulation, or net change, of the original quantity over that interval.

Specifically for rectilinear motion (motion along a straight line), integrating a particle's velocity function over a time interval reveals the particle's net change in position, known as displacement. By extending this concept, we can also determine a particle's final position given its initial position and find the total distance it has traveled, which may be different from its displacement. This connection is a direct application of the Fundamental Theorem of Calculus and is essential for solving problems involving motion.

Key Formulas

The following formulas are derived directly from the relationship between a function and its rate of change, as defined by the definite integral. In the context of motion, let $v(t)$ be the velocity of a particle and $x(t)$ be its position at time $t$.

• Net Change (Displacement): The net change in position, or displacement, of a particle from time $t=a$ to $t=b$ is found by integrating the velocity function.

  $$\text{Displacement}={\int }_a^{b}v(t)dt=x(b)-x(a)$$

  Displacement can be positive, negative, or zero, indicating the net direction of movement.

• Total Distance Traveled: The total distance traveled by a particle from time $t=a$ to $t=b$ is found by integrating the absolute value of the velocity function (speed).

  $$\text{Total Distance}={\int }_a^b|v(t)|dt$$

  Total distance is always non-negative as it accumulates all motion regardless of direction.

• Final Position from Initial Position: The position of a particle at a specific time $t=b$ can be found by adding its displacement over the interval $[a,b]$ to its initial position at $t=a$.

  $$x(b)=x(a)+{\int }_a^{b}v(t)dt$$

  This is often referred to as an "initial value" problem setup.

Understanding Displacement vs. Total Distance

The most critical conceptual nuance in this topic is the distinction between displacement and total distance. While both are calculated using definite integrals of motion, they answer two different questions.

• Displacement (${\int }_a^{b}v(t)dt$) measures the net change in position. It is the straight-line distance between the particle's starting point and ending point. Imagine a particle starts at $x=2$, moves to $x=10$, and then returns to $x=5$. Its final position is $5$ and its initial position is $2$, so its displacement is $5-2=3$. The integral of velocity calculates this net result automatically; areas where $v(t)$ is positive (moving right) are added, and areas where $v(t)$ is negative (moving left) are subtracted.

• Total Distance (${\int }_a^b|v(t)|dt$) measures the entire path length traveled by the particle. In the same scenario, the particle first traveled $8$ units (from 2 to 10) and then $5$ units (from 10 to 5), for a total distance of $8+5=13$. By integrating the absolute value of velocity (which is speed), we ensure that all motion, regardless of direction, is accumulated positively. Any area between the $v(t)$ curve and the t-axis is treated as positive.

If a particle never changes direction on the interval $[a,b]$ (i.e., $v(t)$ is always non-negative or always non-positive), then its displacement and total distance traveled will be equal in magnitude.

Core Concepts & Rules

• The definite integral of a particle's velocity function over a time interval $[a,b]$ yields the particle's displacement (net change in position) on that interval.

• The definite integral of a particle's speed, $|v(t)|$, over a time interval $[a,b]$ yields the total distance the particle traveled on that interval.

• A particle's final position is the sum of its initial position and its displacement over the time interval.

• The Fundamental Theorem of Calculus is the underlying principle: integrating a rate of change $F^{\prime }(x)$ over $[a,b]$ gives the net change in the original function, $F(b)-F(a)$.

Step-by-Step Example 1: Calculating Displacement and Total Distance

A particle moves along the x-axis with a velocity given by $v(t)=2t-6$ for $t\ge 0$.

Part A: Find the displacement of the particle on the time interval $[1,5]$.

Step 1: Set up the definite integral for displacement.

Displacement is the integral of the velocity function over the given interval.

$$\text{Displacement}={\int }_1^5(2t-6)dt$$

Step 2: Find the antiderivative of $v(t)$.

The antiderivative of $2t-6$ is $t^2-6t$.

Step 3: Evaluate the definite integral using the Fundamental Theorem of Calculus.

$${\int }_1^5(2t-6)dt=[t^2-6t{]}_1^5$$

$$=(5^2-6(5))-(1^2-6(1))$$

$$=(25-30)-(1-6)$$

$$=(-5)-(-5)=0$$

The displacement of the particle is 0. This means the particle ended at the same position where it started.

Part B: Find the total distance traveled by the particle on the time interval $[1,5]$.

Step 1: Determine where the particle changes direction by finding where $v(t)=0$.

$$2t-6=0⟹2t=6⟹t=3$$

The particle changes direction at $t=3$. For $1\le t<3$, $v(t)$ is negative (moves left). For $3<t\le 5$, $v(t)$ is positive (moves right).

Step 2: Set up the integral for total distance, splitting it at $t=3$.

We must integrate the absolute value of $v(t)$. This means we make the integrand positive on any interval where it is negative.

$$\text{Total Distance}={\int }_1^5|2t-6|dt={\int }_1^3-(2t-6)dt+{\int }_3^5(2t-6)dt$$

$$={\int }_1^3(6-2t)dt+{\int }_3^5(2t-6)dt$$

Step 3: Evaluate both integrals.

$$[6t-t^2{]}_1^3+[t^2-6t{]}_3^5$$

$$=((6(3)-3^2)-(6(1)-1^2))+((5^2-6(5))-(3^2-6(3)))$$

$$=((18-9)-(6-1))+((25-30)-(9-18))$$

$$=(9-5)+(-5-(-9))$$

$$=4+4=8$$

The total distance traveled by the particle is 8 units.

Step-by-Step Example 2: Exam-Style Application

A particle's velocity is given by $v(t)=\cos (e^t)$ for $0\le t\le 2$. At time $t=0$, the particle is at position $x=-3$.

Part A: What is the position of the particle at time $t=2$?

Step 1: Recall the formula for final position.

The final position is the initial position plus the displacement.

$$x(2)=x(0)+{\int }_0^{2}v(t)dt$$

Step 2: Substitute the given values into the formula.

$$x(2)=-3+{\int }_0^2\cos (e^t)dt$$

Step 3: Use a calculator to evaluate the definite integral.

This integral cannot be evaluated by hand using AB Calculus techniques.

$${\int }_0^2\cos (e^t)dt\approx -0.451$$

Step 4: Calculate the final position.

$$x(2)\approx -3+(-0.451)=-3.451$$

The position of the particle at $t=2$ is approximately $-3.451$.

Part B: What is the total distance traveled by the particle from $t=0$ to $t=2$?

Step 1: Recall the formula for total distance.

Total distance is the integral of the absolute value of velocity (speed).

$$\text{Total Distance}={\int }_0^2|v(t)|dt$$

Step 2: Substitute the velocity function into the formula.

$$\text{Total Distance}={\int }_0^2|\cos (e^t)|dt$$

Step 3: Use a calculator to evaluate the integral.

$$\text{Total Distance}\approx {\int }_0^2|\cos (e^t)|dt\approx 1.358$$

The total distance traveled by the particle is approximately $1.358$.

Using Your Calculator

A graphing calculator is essential for solving motion problems with complex velocity functions that are difficult or impossible to integrate by hand.

To find displacement or final position:

Use the numerical integration feature (often fnInt or $\text{intf}(x)dx$).

• Problem: Find the position at $t=b$ given $v(t)$ and the position $x(a)$.

• Calculator Setup:x(a) + fnInt(Y1, X, a, b)

• Steps:

  1. Enter the velocity function v(t)` into `Y1`. 2. On the home screen, type the initial position `x(a)`. 3. Add the definite integral of `Y1` from the initial time $a to the final time $b$.

  2. The result is the final position $x(b)$.

To find total distance:

Use the numerical integration feature combined with the absolute value function (often $abs$).

• Problem: Find the total distance traveled on $[a,b]$ for a particle with velocity $v(t)$.

• Calculator Setup:fnInt(abs(Y1), X, a, b) * **Steps:** 1. Enter the velocity function $v(t)intoY1`.

  2. On the home screen, select the numerical integration function.

  3. For the function to integrate, select the absolute value function and input Y1.

  4. Enter the bounds of integration $a$ and $b$.

  5. The result is the total distance traveled.

AP Exam Quick Hit

Common Question Types

• Finding Final Position: Given a velocity function $v(t)$ (often one that requires a calculator to integrate) and an initial position $x(a)$, you will be asked to find the position at a later time $x(b)$. This requires setting up the expression $x(b)=x(a)+{\int }_a^{b}v(t)dt$.

• Calculating Total Distance: Given $v(t)$, find the total distance traveled on an interval $[a,b]$. This is almost always a calculator question and requires the setup ${\int }_a^b|v(t)|dt$.

• Interpreting Motion from a Graph of Velocity: Given the graph of $v(t)$, you might be asked to find the displacement (net area between the curve and the t-axis) or total distance (total area, treating all regions as positive).

Common Mistakes

• Confusing Displacement and Total Distance: The most common error is calculating displacement ($\int v(t)dt$) when asked for total distance ($\int |v(t)|dt$). Remember, total distance must be non-negative.

• Forgetting the Initial Condition: When asked for a particle's position at time $t=b$, students often calculate only the displacement (${\int }_a^{b}v(t)dt$) and forget to add the initial position $x(a)$.

• Incorrectly Splitting the Integral for Total Distance: When calculating total distance by hand, you must split the integral into separate parts for every t-value where $v(t)$ changes sign. A common mistake is to not find these points or to not negate the integrand on the intervals where $v(t)<0$.

• Omitting the Absolute Value on the Calculator: When finding total distance with a calculator, students may forget to wrap the function inside the absolute value function ($abs()$), which results in calculating displacement instead.