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

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

In the study of motion, differentiation allows us to move from a position function to a velocity function. Integration, as the inverse process of differentiation, allows us to reverse this. By integrating a particle's velocity function over a specific time interval, we can determine its change in position, known as displacement. This is a powerful application of the Fundamental Theorem of Calculus, where the definite integral of a rate of change (velocity) gives the net change in the original quantity (position).

This topic distinguishes between two crucial measurements of motion: displacement and total distance traveled. While displacement is the net change between the starting and ending positions, total distance accounts for the entire path taken by the particle, including any changes in direction. This distinction is captured by integrating the velocity function, $v(t)$, for displacement, versus integrating the speed function, $|v(t)|$, for total distance. Understanding how to set up and evaluate these specific definite integrals is fundamental to analyzing rectilinear motion.

Key Formulas

The following formulas are derived directly from the application of the definite integral to rectilinear motion. Let a particle move along a line with velocity $v(t)$ and position $x(t)$.

Displacement (Net Change in Position)

The displacement of a particle over the time interval $[t_1,t_2]$ is the net change in its position. It is calculated by the definite integral of the velocity function.

$$\text{Displacement}={\int }_{t_1}^{t_2}v(t)dt$$

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

Final Position

To find the position of a particle at a specific time $t_2$, you must know its position at an initial time $t_1$ and add the displacement that occurred during the interval $[t_1,t_2]$.

$$x(t_2)=x(t_1)+{\int }_{t_1}^{t_2}v(t)dt$$

This is often referred to as an "initial value problem" structure. The final position is the initial position plus the accumulated change.

Total Distance Traveled

The total distance traveled by a particle over the time interval $[t_1,t_2]$ is the sum of all distances moved, regardless of direction. It is calculated by the definite integral of the speed function, $|v(t)|$.

$$\text{Total Distance}={\int }_{t_1}^{t_2}|v(t)|dt$$

Since speed $|v(t)|$ is always non-negative, the total distance traveled will also always be non-negative.

Understanding Displacement vs. Total Distance

The distinction between displacement and total distance is a critical concept in calculus and physics. Both are calculated using definite integrals of motion, but the choice of integrand—velocity or speed—leads to fundamentally different results.

• Displacement: The integral ${\int }_{t_1}^{t_2}v(t)dt$ accumulates the signed values of the velocity. When the particle moves in the positive direction ($v(t)>0$), the integral adds positive area. When the particle moves in the negative direction ($v(t)<0$), the integral adds negative area (subtracts area). The final result is the net change, $x(t_2)-x(t_1)$. Imagine a person walking 10 meters forward and 3 meters back. Their displacement is $+7$ meters, but the total distance they walked is $13$ meters.

• Total Distance: The integral ${\int }_{t_1}^{t_2}|v(t)|dt$ accumulates the absolute value of the velocity (speed). By taking the absolute value, any movement in the negative direction is treated as a positive contribution to the total path length. This integral calculates the sum of all areas between the velocity curve and the t-axis, treating all areas as positive. This corresponds to the total odometer reading of a car, which only increases regardless of the direction of travel.

To calculate total distance by hand, one must find all times $t$ in the interval $[t_1,t_2]$ where $v(t)=0$. These are the points where the particle may change direction. The integral must then be split at these points, and any portion where $v(t)$ is negative must be negated to make its contribution positive.

Core Concepts & Rules

• The definite integral of a particle's velocity function, $v(t)$, over a time interval $[t_1,t_2]$ gives the particle's displacement (or net change in position) on that interval.

• To find a particle's final position at time $t_2$, you must add the displacement from $t_1$ to $t_2$ to the known initial position at time $t_1$.

• The definite integral of a particle's speed function, $|v(t)|$, over a time interval $[t_1,t_2]$ gives the total distance the particle traveled on that interval.

• Displacement can be positive, negative, or zero, as it considers the direction of motion.

• Total distance is always non-negative, as it measures the total path length covered.

Step-by-Step Example 1: Analytical Application

A particle moves along the x-axis with velocity given by $v(t)=t^2-3t+2$ for $t\ge 0$. The particle is at position $x=5$ at time $t=0$.

(a) Find the displacement of the particle during the first 3 seconds.

(b) Find the position of the particle at $t=3$.

(c) Find the total distance traveled by the particle during the first 3 seconds.

Solution

(a) Displacement

Displacement is the definite integral of the velocity function from $t=0$ to $t=3$.

$$\text{Displacement}={\int }_0^3(t^2-3t+2)dt$$

Step 1: Find the antiderivative of $v(t)$.

$$\int (t^2-3t+2)dt=\frac{t^3}{3}-\frac{3t^2}{2}+2t$$

Step 2: Apply the Fundamental Theorem of Calculus.

$${\left[\frac{t^3}{3}-\frac{3t^2}{2}+2t\right]}_0^3=\left(\frac{3^3}{3}-\frac{3(3^2)}{2}+2(3)\right)-(0)$$

$$=\left(9-\frac{27}{2}+6\right)=15-13.5=1.5$$

The displacement of the particle is $1.5$ units.

(b) Final Position

The final position is the initial position plus the displacement.

$$x(3)=x(0)+{\int }_0^{3}v(t)dt$$

Step 1: Use the initial position $x(0)=5$ and the displacement calculated in part (a).

$$x(3)=5+1.5=6.5$$

The position of the particle at $t=3$ is $x=6.5$.

(c) Total Distance

Total distance is the integral of the speed, ${\int }_0^3|v(t)|dt$.

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

$$t^2-3t+2=0$$

$$(t-1)(t-2)=0$$

The particle changes direction at $t=1$ and $t=2$.

Step 2: Determine the sign of $v(t)$ on the intervals $[0,1]$, $[1,2]$, and $[2,3]$.

• On $[0,1]$, let $t=0.5$: $v(0.5)=(0.5{)}^2-3(0.5)+2=0.25-1.5+2=0.75>0$.

• On $[1,2]$, let $t=1.5$: $v(1.5)=(1.5{)}^2-3(1.5)+2=2.25-4.5+2=-0.25<0$.

• On $[2,3]$, let $t=2.5$: $v(2.5)=(2.5{)}^2-3(2.5)+2=6.25-7.5+2=0.75>0$.

Step 3: Split the integral at $t=1$ and $t=2$. Negate the integral over the interval where $v(t)$ is negative.

$$\text{Total Distance}={\int }_0^1(t^2-3t+2)dt+{\int }_1^2-(t^2-3t+2)dt+{\int }_2^3(t^2-3t+2)dt$$

Step 4: Evaluate each integral.

$${\int }_0^{1}v(t)dt={\left[\frac{t^3}{3}-\frac{3t^2}{2}+2t\right]}_0^1=(\frac{1}{3}-\frac{3}{2}+2)-0=\frac{2-9+12}{6}=\frac{5}{6}$$

$${\int }_1^{2}v(t)dt={\left[\frac{t^3}{3}-\frac{3t^2}{2}+2t\right]}_1^2=(\frac{8}{3}-6+4)-(\frac{5}{6})=\frac{2}{3}-\frac{5}{6}=-\frac{1}{6}$$

$${\int }_2^{3}v(t)dt={\left[\frac{t^3}{3}-\frac{3t^2}{2}+2t\right]}_2^3=(1.5)-(\frac{8}{3}-6+4)=1.5-(\frac{2}{3})=\frac{3}{2}-\frac{2}{3}=\frac{5}{6}$$

Step 5: Sum the absolute values of the displacements for each interval.

$$\text{Total Distance}=\left|\frac{5}{6}\right|+\left|-\frac{1}{6}\right|+\left|\frac{5}{6}\right|=\frac{5}{6}+\frac{1}{6}+\frac{5}{6}=\frac{11}{6}$$

The total distance traveled is $11/6$ units.

Step-by-Step Example 2: Exam-Style Application (Calculator Active)

For $0\le t\le 6$, a particle moves along the x-axis. The velocity of the particle is given by $v(t)=2\sin (e^{t/4})$. The particle is at position $x=-1$ at time $t=0$.

(a) Find the position of the particle at time $t=6$.

(b) Find the total distance traveled by the particle from $t=0$ to $t=6$.

Solution

(a) Final Position

The position at $t=6$ is the initial position $x(0)$ plus the displacement from $t=0$ to $t=6$.

$$x(6)=x(0)+{\int }_0^{6}v(t)dt$$

Step 1: Set up the expression with the given values.

$$x(6)=-1+{\int }_0^{6}2\sin (e^{t/4})dt$$

Step 2: Use a graphing calculator to evaluate the definite integral. Ensure the calculator is in Radian Mode.

$${\int }_0^{6}2\sin (e^{t/4})dt\approx 3.9335$$

Step 3: Add the initial position to the displacement.

$$x(6)\approx -1+3.9335=2.9335$$

Rounding to three decimal places, the position of the particle at $t=6$ is $x=2.934$.

(b) Total Distance

The total distance traveled is the definite integral of the speed, $|v(t)|$.

$$\text{Total Distance}={\int }_0^6|v(t)|dt$$

Step 1: Set up the integral using the absolute value of the velocity function.

$$\text{Total Distance}={\int }_0^6|2\sin (e^{t/4})|dt$$

Step 2: Use a graphing calculator to evaluate this definite integral.

$${\int }_0^6|2\sin (e^{t/4})|dt\approx 6.5708$$

Rounding to three decimal places, the total distance traveled by the particle from $t=0$ to $t=6$ is $6.571$.

Using Your Calculator

For problems involving complex functions that are difficult or impossible to integrate by hand, the graphing calculator is an essential tool.

Calculating Displacement and Final Position

To solve a problem like Example 2(a), $x(6)=-1+{\int }_0^{6}2\sin (e^{t/4})dt$:

1. Set Mode: Ensure your calculator is in RADIAN mode.

2. Enter Expression: On the home screen, type the initial position followed by the integral.

   • On a TI-84 style calculator, this would be: -1 + fnInt(2*sin(e^(X/4)), X, 0, 6)

   • The fnInt command is typically found under the MATH menu (e.g., MATH -> 9: fnInt).

3. Execute: Press ENTER to get the result.

Calculating Total Distance

To solve a problem like Example 2(b), $TotalDistance={\int }_0^6|2\sin (e^{t/4})|dt$:

1. Set Mode: Ensure your calculator is in RADIAN mode.

2. Enter Expression: Use the fnInt command, but place the function inside the absolute value function $abs()$.

   • On a TI-84 style calculator: fnInt(abs(2*sin(e^(X/4))), X, 0, 6)

   • The $abs()$ command is typically found under the MATH menu, in the NUM sub-menu (e.g., MATH -> NUM -> 1: abs()).

3. Execute: Press ENTER to get the result.

Pro-Tip: For multi-part questions involving the same function, store $v(t)$ in the Y= editor (e.g., as Y1). Then, you can use Y1 in your integral commands: $fnInt(abs(Y1),X,0,6)$. This saves time and reduces typing errors.

AP Exam Quick Hit

Common Question Types

• Calculator-Active FRQ: You will be given a complex velocity function $v(t)$ and an initial position $x(t_1)$. You will be asked to find the position at a later time $t_2$ and the total distance traveled over $[t_1,t_2]$. This tests your ability to correctly set up the integrals for position and total distance and to use your calculator effectively.

• Non-Calculator Problem (MCQ or FRQ part): You will be given a simple, integrable velocity function (e.g., polynomial, basic trig). A common task is to find the total distance traveled, which requires you to find the roots of $v(t)$ within the interval, split the integral at those roots, and sum the absolute values of the results.

• Graph-Based FRQ: You will be given the graph of the velocity function $v(t)$. You will be asked to find displacement and total distance by interpreting the integral as the area under the curve. Displacement is the net signed area (areas above the t-axis minus areas below). Total distance is the sum of the absolute values of the areas of all regions between the graph and the t-axis.

Common Mistakes

• Confusing Displacement and Total Distance: The most common error is calculating displacement ($\int v(t)dt$) when asked for total distance ($\int |v(t)|dt$). Always integrate speed for total distance.

• Forgetting the Initial Position: When asked for a final position $x(t_2)$, students often calculate only the displacement ($\int v(t)dt$) and forget to add the initial position $x(t_1)$. Remember the formula: $Final=Initial+Change$.

• Incorrectly Evaluating $\int |v(t)|dt$ by Hand: Students sometimes calculate $\int v(t)dt$ and then simply take the absolute value of the final answer: $|\int v(t)dt|$. This is incorrect and only works if $v(t)$ never changes sign on the interval. You must split the integral where $v(t)$ is negative.

• Calculator Mode Error: Using Degree mode instead of Radian mode when trigonometric functions are involved in $v(t)$. AP Calculus problems almost exclusively use radians.

• Notation Error: Writing the integral for total distance as $|\int v(t)dt|$ instead of the correct $\int |v(t)|dt$ on an FRQ. While the final numerical answer might be the same in some cases, the setup is conceptually incorrect and may lose points.