Integrating Vector-Valued Functions - AP Calculus BC Study Guide

The Core Idea: Integrating Vector-Valued Functions

The integration of vector-valued functions is the fundamental process for determining the net change in position (displacement) and the final position of a particle moving in a plane. Just as the integral of a single-variable rate function gives the net change in that quantity, the integral of a vector-valued velocity function, ${\vec{r}}^{\prime }(t)$, gives the net change in the position vector, $\vec{r}(t)$. This process is performed component-wise, meaning we integrate the horizontal component of velocity to find the horizontal displacement and the vertical component of velocity to find the vertical displacement.

By applying the Fundamental Theorem of Calculus to each component, we can establish a direct relationship between a particle's velocity over an interval and its change in position. The core application of this concept is the "position update" formula: the particle's position at a later time $t=b$ is found by taking its known position at an earlier time $t=a$ and adding the vector that represents the total displacement from $a$ to $b$. This allows us to predict the location of a particle if we know its velocity function and a single point in its history.

Key Formulas

The principles of integrating vector-valued functions are captured in three essential formulas derived from the Fundamental Theorem of Calculus.

1. The Definite Integral of a Vector-Valued Function

   The definite integral of a vector-valued function $\vec{r}(t)=⟨x(t),y(t)⟩$ is a new vector whose components are the definite integrals of the original component functions.

   $${\int }_a^b\vec{r}(t)dt=⟨{\int }_a^{b}x(t)dt,{\int }_a^{b}y(t)dt⟩$$

   This formula establishes that integration is a component-wise operation.

2. The Fundamental Theorem of Calculus for Vector-Valued Functions

   The definite integral of the derivative of a vector-valued function (the velocity vector, ${\vec{r}}^{\prime }(t)$) gives the net change in the original function (the displacement vector, $\vec{r}(b)-\vec{r}(a)$).

   $${\int }_a^b{\vec{r}}^{\prime }(t)dt=\vec{r}(b)-\vec{r}(a)$$

   This connects the rate of change of position (velocity) to the net change in position (displacement).

3. The Position Update Formula

   By rearranging the Fundamental Theorem of Calculus, we can solve for the final position, $\vec{r}(b)$. The position of a particle at time $t=b$ is its initial position at $t=a$ plus its displacement from $t=a$ to $t=b$.

   $$\vec{r}(b)=\vec{r}(a)+{\int }_a^b{\vec{r}}^{\prime }(t)dt$$

   This is the primary tool for solving problems involving a particle's motion in the plane. In component form, it is written as:

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

   $$y(b)=y(a)+{\int }_a^b{y}^{\prime }(t)dt$$

Understanding Displacement vs. Final Position

A critical nuance in this topic is the distinction between displacement and final position. The definite integral of a velocity vector, ${\int }_a^b{\vec{r}}^{\prime }(t)dt$, does not directly yield the particle's location at time $t=b$. Instead, it calculates the displacement vector, which represents the net change in position, or the straight-line vector from the starting point $\vec{r}(a)$ to the ending point $\vec{r}(b)$.

Consider the analogy of a person walking. If they start at a landmark (initial position) and we only know the directions and distances they walked (velocity integrated over time), we can determine their displacement—for example, "they ended up 3 miles east and 4 miles north of where they started." To find their final location on a map, we must know the location of the landmark they started from and then add the displacement vector.

The position update formula, $\vec{r}(b)=\vec{r}(a)+{\int }_a^b{\vec{r}}^{\prime }(t)dt$, formalizes this idea. The term $\vec{r}(a)$ is the initial position vector (the "landmark"), and the integral term is the displacement vector. Forgetting to add the initial position $\vec{r}(a)$ is a common conceptual error that leads to reporting the displacement as the final answer. Always remember that the integral of velocity gives the change in position, which must be added to a known position to find a new one.

Core Concepts & Rules

• Component-Wise Integration: To find the definite integral of a vector-valued function, you must integrate each component function separately over the given interval.

• Integral of Velocity is Displacement: The definite integral of a velocity vector $\vec{v}(t)={\vec{r}}^{\prime }(t)$ over an interval $[a,b]$ results in the displacement vector, which represents the net change in the particle's $x$ and $y$ coordinates from time $a$ to time $b$.

• The Fundamental Theorem of Calculus Governs Motion: The relationship ${\int }_a^b{\vec{r}}^{\prime }(t)dt=\vec{r}(b)-\vec{r}(a)$ is the vector extension of the FTC and is the theoretical basis for solving problems involving position and velocity.

• Initial Condition is Necessary for Final Position: To determine the specific position of a particle at a time $t=b$, you must know its position at some other time $t=a$ (the initial condition) in addition to its velocity function. The final position is the sum of the initial position vector and the displacement vector.

Step-by-Step Example 1: Finding Position from Velocity (Analytical)

Problem: A particle moves in the $xy$-plane with a velocity vector given by $\vec{v}(t)=⟨3t^2,\sin (\pi t)⟩$. At time $t=1$, the particle is at the point $(2,5)$. Find the position of the particle at time $t=2$.

Solution:

Step 1: Identify the Goal and Given Information

• We are given the velocity vector: ${\vec{r}}^{\prime }(t)=\vec{v}(t)=⟨3t^2,\sin (\pi t)⟩$.

• We are given an initial position: $\vec{r}(1)=⟨2,5⟩$.

• We need to find the final position: $\vec{r}(2)$.

Step 2: Set up the Position Update Formula

We use the formula $\vec{r}(b)=\vec{r}(a)+{\int }_a^b{\vec{r}}^{\prime }(t)dt$.

Here, $a=1$, $b=2$, $\vec{r}(a)=\vec{r}(1)=⟨2,5⟩$, and ${\vec{r}}^{\prime }(t)=⟨3t^2,\sin (\pi t)⟩$.

$$\vec{r}(2)=\vec{r}(1)+{\int }_1^2⟨3t^2,\sin (\pi t)⟩dt$$

Step 3: Calculate the Displacement Vector by Integrating Component-Wise

The displacement vector is ${\int }_1^2⟨3t^2,\sin (\pi t)⟩dt$. We find this by integrating each component.

• x-component of displacement:

  $${\int }_1^{2}3t^{2}dt={\left[t^3\right]}_1^2=(2{)}^3-(1{)}^3=8-1=7$$

• y-component of displacement:

  $${\int }_1^2\sin (\pi t)dt={\left[-\frac{1}{\pi }\cos (\pi t)\right]}_1^2=\left(-\frac{1}{\pi }\cos (2\pi )\right)-\left(-\frac{1}{\pi }\cos (\pi )\right)$$

  $$=\left(-\frac{1}{\pi }(1)\right)-\left(-\frac{1}{\pi }(-1)\right)=-\frac{1}{\pi }-\frac{1}{\pi }=-\frac{2}{\pi }$$

So, the displacement vector from $t=1$ to $t=2$ is $⟨7,-\frac{2}{\pi }⟩$.

Step 4: Add the Displacement Vector to the Initial Position Vector

Now we add the result from Step 3 to the initial position $\vec{r}(1)$.

$$\vec{r}(2)=\vec{r}(1)+\text{displacement}$$

$$\vec{r}(2)=⟨2,5⟩+⟨7,-\frac{2}{\pi }⟩$$

$$\vec{r}(2)=⟨2+7,5-\frac{2}{\pi }⟩=⟨9,5-\frac{2}{\pi }⟩$$

Final Answer: The position of the particle at time $t=2$ is $\left(9,5-\frac{2}{\pi }\right)$.

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

Problem: For $t\ge 0$, a particle moves in the $xy$-plane with velocity vector $\vec{v}(t)=⟨\ln (t^2+1),\cos (e^t)⟩$. At time $t=0$, the particle is at $(-3,4)$. Find the position of the particle at time $t=2$.

Solution:

Step 1: Identify the Goal and Given Information

• Velocity vector: $\vec{v}(t)={\vec{r}}^{\prime }(t)=⟨\ln (t^2+1),\cos (e^t)⟩$.

• Initial position: $\vec{r}(0)=⟨-3,4⟩$.

• We need to find the final position: $\vec{r}(2)$.

Step 2: Set up the Position Update Formula in Component Form

The functions in the velocity vector are difficult to integrate by hand, so this is a calculator-active problem. We set up the integrals for the $x$ and $y$ coordinates of the position at $t=2$.

• Final x-position: $x(2) = x(0) + \int_{0}^{2} x'(t) \,dt = -3 + \int_{0}^{2} \ln(t^2+1) \,dt$-Finaly-position:$$y(2)=y(0)+{\int }_0^2{y}^{\prime }(t)dt=4+{\int }_0^2\cos (e^t)dt$**Step 3: Use a Calculator to Evaluate the Definite Integrals** We use the numerical integration feature of a graphing calculator. - **For the x-position:** Calculate the value of the integral: Formula[12] Now, add the initial x-position: Formula[13] - **For the y-position:** Calculate the value of the integral: Formula[14] Now, add the initial y-position: Formula[15] **Step 4: State the Final Position Vector** Combine the calculated $x and $y$ coordinates into a final position vector, rounding to the appropriate number of decimal places (typically three or four on the AP Exam).

$$\vec{r}(2)\approx ⟨-1.444,3.509⟩$$

Final Answer: The position of the particle at time $t=2$ is approximately $(-1.444,3.509)$.

Using Your Calculator

For problems involving vector-valued functions where the components cannot be easily integrated by hand, a graphing calculator is essential. The primary function you will use is the numerical definite integral command (e.g., fnInt on TI-84 style calculators).

To find the final position $\vec{r}(b)=⟨x(b),y(b)⟩$ given $\vec{r}(a)=⟨x_a,y_a⟩$ and $\vec{v}(t)=⟨x^{\prime }(t),y^{\prime }(t)⟩$:

1. Define Velocity Components: Store the $x$-component of velocity, $x^{\prime }(t)$, in Y1 and they$-component, $y^{\prime }(t)$, in Y2`.

   • Y1 = x'(t)

   • Y2 = y'(t)

2. Calculate Final x-Position: On the home screen, type the expression for x(b).

   • x_a + fnInt(Y1, X, a, b)

   • For Example 2 above: -3 + fnInt(ln(X^2+1), X, 0, 2)

   • Press ENTER to get the value of $x(b)$.

3. Calculate Final y-Position: On the home screen, type the expression for $y(b)$.

   • y_a + fnInt(Y2, X, a, b)

   • For Example 2 above: 4 + fnInt(cos(e^X), X, 0, 2)

   • Press ENTER to get the value of $y(b)$.

4. Combine and Report: Combine the results into an ordered pair or vector $⟨x(b),y(b)⟩$, rounding as specified in the problem.

This method is efficient and minimizes transcription errors. The calculator's role is purely to evaluate the definite integrals that you have set up correctly using the Fundamental Theorem of Calculus.

AP Exam Quick Hit

Common Question Types

• Finding Final Position: This is the most common application. You are given a velocity vector $\vec{v}(t)$, an initial position $\vec{r}(a)$ at time $t=a$, and asked to find the position $\vec{r}(b)$ at time $t=b$.

  • Example: "A particle's velocity is $\vec{v}(t)=⟨\sqrt{t},2t-1⟩$. If the particle is at $(4,0)$ at $t=1$, what is its position at $t=4$`?"

• Finding Displacement: You are given a velocity vector $\vec{v}(t)$ and asked for the net change in position (displacement) over an interval $[a,b]$. This tests whether you know the integral alone gives displacement, without needing an initial condition.

  • Example: "Find the displacement of a particle from $t=0$ to $t=\pi$ if its velocity is $\vec{v}(t)=⟨\cos (t),1⟩$."

• Finding an Initial Position (Working Backwards): You are given a velocity vector $\vec{v}(t)$ and a position $\vec{r}(b)$ at time $t=b$. You are asked to find the position $\vec{r}(a)$ at an earlier time $t=a$. This requires rearranging the position update formula to $\vec{r}(a)=\vec{r}(b)-{\int }_a^b\vec{v}(t)dt$.

  • Example: "A particle has velocity $\vec{v}(t)=⟨e^{-t},2⟩$. At $t=3$, the particle is at $(1,8)$. Find the position of the particle at $t=0$."

Common Mistakes

• Forgetting the Initial Condition: Calculating the displacement ${\int }_a^b\vec{v}(t)dt$ and reporting it as the final position, instead of adding it to the initial position $\vec{r}(a)$.

• Confusing Displacement and Position: Stating that the definite integral of velocity gives the final position. It gives the net change in position.

• Antidifferentiation Errors: Making a mistake when finding the antiderivative of one or both components in an analytical problem (e.g., sign errors, incorrect application of the chain rule in reverse).

• Incorrect Bounds of Integration: Using the wrong start or end time ($a$ or $b$) in the definite integral setup.

• Calculator Setup Errors: When using a calculator, entering the wrong function, bounds, or initial condition. It is crucial to write the integral setup on paper first to guide your calculator work and to be eligible for partial credit.