Vector-Valued Functions - AP PreCalculus Study Guide

The Core Idea: Vector-Valued Functions

Vector-valued functions extend the concept of a traditional function by producing a vector as an output instead of a single number. In AP Precalculus, we focus on functions that take a single real number as an input, often denoted by the parameter $t$, and map it to a two-dimensional vector. This creates a powerful way to describe motion and paths in a plane.

The core idea is to define the position of a point in the $xy$-plane using a single variable, $t$. The vector-valued function, $r(t)$, gives the position vector—a vector from the origin to the point's location—at any given value of $t$. The horizontal and vertical components of this position vector are themselves functions of $t$, denoted $x(t)$ and $y(t)$. As the parameter $t$ changes, the $x$ and $y$ components change accordingly, causing the tip of the position vector to trace out a curve.

Key Representations

A vector-valued function can be expressed in two primary forms. Both representations describe the same mathematical object: a function that maps a real number $t$ to a vector in the plane. The functions $x(t)$ and $y(t)$ are called the component functions.

1. Component Form:

   This form explicitly shows the horizontal and vertical component functions inside angled brackets.

   $$r(t)=⟨x(t),y(t)⟩$$

   • $r(t)$: The vector-valued function, which outputs a vector for each input t.

   • $t$: The parameter, a real number from the function's domain.

   • $x(t)$: The horizontal component function. It determines the x-coordinate of the output vector.

   • $y(t)$: The vertical component function. It determines the y-coordinate of the output vector.

2. Standard Unit Vector Form:

   This form represents the vector as a linear combination of the standard unit vectors $i$ and $j$. In this context, $i$ represents the vector $⟨1,0⟩$ and $j$ represents the vector $⟨0,1⟩$.

   $$r(t)=x(t)i+y(t)j$$

   • $x(t)i$: The horizontal vector component. The scalar function $x(t)$ scales the unit vector $i$.

   • $y(t)j$: The vertical vector component. The scalar function $y(t)$ scales the unit vector $j$.

Understanding The Domain and Range

A critical aspect of vector-valued functions is understanding their domain and range, as defined by the Essential Knowledge.

• Domain: The domain of a vector-valued function $r(t)$ is a subset of the real numbers. This means the input to the function, the parameter $t$, is always a scalar (a single number). The domain of $r(t)=⟨x(t),y(t)⟩$ is the intersection of the domains of the component functions $x(t)$ and $y(t)$. For the function to be defined, both component functions must be defined.

• Range: The range of a vector-valued function $r(t)$ is a set of vectors. For every valid real number $t$ you input, the function outputs a corresponding two-dimensional vector. For example, if $r(3)=⟨5,-2⟩$, the input is the real number $3$ and the output is the vector $⟨5,-2⟩$.

The parameter $t$ acts as an independent variable that "drives" the function. As $t$ varies over its domain, the output vectors change, and their endpoints trace a path in the $xy$-plane. This is why vector-valued functions are fundamental to describing parametric motion.

Core Concepts & Rules

• Function Definition: A vector-valued function is a function whose input is a real number (parameter $t$) and whose output is a vector.

• Output is a Vector: The result of evaluating a vector-valued function at a specific $t$ is always a vector, not a scalar or a coordinate point.

• Component Functions: The behavior of a vector-valued function $r(t)$ is entirely determined by its component functions, $x(t)$ and $y(t)$.

• Component Form Notation: The function can be written as $r(t)=⟨x(t),y(t)⟩$.

• Unit Vector Notation: The function can also be written as $r(t)=x(t)i+y(t)j$. These two forms are equivalent.

• Domain: The domain of $r(t)$ consists of all values of $t$ for which both $x(t)$ and $y(t)$ are defined.

Step-by-Step Example 1: Evaluating a Vector-Valued Function

Problem:

A particle's position is given by the vector-valued function $r(t)=⟨3t-4,\sqrt{t+1}⟩$. Find the position vector of the particle at $t=8$.

Solution:

Step 1: Identify the Component Functions

From the given function $r(t)=⟨3t-4,\sqrt{t+1}⟩$, we can identify the horizontal and vertical component functions:

• Horizontal component: $x(t)=3t-4$

• Vertical component: $y(t)=\sqrt{t+1}$

Step 2: Substitute the Parameter Value into Each Component Function

We are asked to find the position vector at $t=8$. We substitute this value into both $x(t)$ and $y(t)$.

• For $x(t)$:

  $$x(8)=3(8)-4$$

• For $y(t)$:

  $$y(8)=\sqrt{8+1}$$

Step 3: Calculate the Value of Each Component

• Calculate $x(8)$:

  $$x(8)=24-4=20$$

• Calculate $y(8)$:

  $$y(8)=\sqrt{9}=3$$

Step 4: Assemble the Final Position Vector

Combine the calculated components back into the vector form. The position vector at $t=8$ is:

$$r(8)=⟨20,3⟩$$

This can also be written in standard unit vector form as $r(8)=20i+3j$.

Step-by-Step Example 2: Interpreting from Unit Vector Form

Problem:

The motion of an object is described by the vector-valued function $r(t)=(t^2+2t)i+\left(\frac{t}{t-1}\right)j$.

(a) Find the object's position vector at $t=3$.

(b) Determine the domain of the function $r(t)$.

Solution:

(a) Find the position vector at $t=3$

Step 1: Identify the Component Functions

The function is given in standard unit vector form. The coefficient of $i$ is $x(t)$ and the coefficient of $j$ is $y(t)$.

• $x(t)=t^2+2t$

• $y(t)=\frac{t}{t-1}$

Step 2: Evaluate Each Component Function at $t=3$

• Evaluate $x(3)$:

  $$x(3)=(3{)}^2+2(3)=9+6=15$$

• Evaluate $y(3)$:

  $$y(3)=\frac{3}{3-1}=\frac{3}{2}$$

Step 3: Write the Resulting Vector

Combine the components to form the position vector $r(3)$.

• In component form: $r(3)=⟨15,\frac{3}{2}⟩$

• In standard unit vector form: $r(3)=15i+\frac{3}{2}j$

(b) Determine the domain of the function $r(t)$

Step 1: Find the Domain of the Horizontal Component $x(t)$

The horizontal component is $x(t)=t^2+2t$. This is a polynomial function, and its domain is all real numbers, $(-\infty ,\infty )$.

Step 2: Find the Domain of the Vertical Component $y(t)$

The vertical component is $y(t)=\frac{t}{t-1}$. This is a rational function. Its domain consists of all real numbers except those that make the denominator zero.

• Set the denominator to zero: $t-1=0⟹t=1$.

• The domain of $y(t)$ is all real numbers except $t=1$. In interval notation, this is $(-\infty ,1)\cup (1,\infty )$.

Step 3: Find the Intersection of the Component Domains

The domain of the vector-valued function $r(t)$ is the set of all $t$ values for which both component functions are defined. We need to find the intersection of the domains from Step 1 and Step 2.

• Domain of $x(t)$: $(-\infty ,\infty )$

• Domain of $y(t)$: $(-\infty ,1)\cup (1,\infty )$

• The intersection of these two sets is $(-\infty ,1)\cup (1,\infty )$.

Therefore, the domain of $r(t)$ is all real numbers $t$ such that $t\ne 1$.

Using Your Calculator

While the definition and evaluation of vector-valued functions are analytical, a graphing calculator is an excellent tool for visualizing the curve traced by the function's output vectors. This is done using the calculator's parametric mode.

Problem: Visualize the path traced by $r(t)=⟨4\cos (t),2\sin (t)⟩$ for $0\le t\le 2\pi$.

Steps (for a TI-84 style calculator):

1. Set to Parametric Mode:

   • Press the [MODE] button.

   • Navigate down to the line that reads FUNCTION PARAMETRIC POLAR SEQ.

   • Highlight PARAMETRIC (or PAR) and press [ENTER].

2. Enter Component Functions:

   • Press the [Y=] button. The editor will now show pairs of equations like $X_{1}T$ and $Y_{1}T$.

   • In $X_{1}T$, enter the horizontal component function: $4cos(T)$. (The $[X,T,\theta ,n]$ button will produce a $T$ in this mode).

   • In $Y_{1}T$, enter the vertical component function: $2sin(T)$.

3. Set the Parameter Window ($T$ values):

   • Press the [WINDOW] button.

   • Set $Tmin=0$. This is the starting value for the parameter $t$.

   • Set $Tmax=2\pi$. (You can type $2$ then [2nd]${[}^{]}$ to get $\pi$). This is the ending value for $t$.

   • Set $Tstep$ to a small value for a smooth curve, for example, $0.1$ or $\pi /24$. This determines how often the calculator plots a point.

4. Set the Viewing Window ($X$ and $Y$ values):

   • Based on the functions $x(t)=4\cos (t)$ and $y(t)=2\sin (t)$, we know the maximum $x$ value is 4 and the maximum $y$ value is 2.

   • Set $Xmin=-5$, $Xmax=5$, and $Xscl=1$.

   • Set $Ymin=-3$, $Ymax=3$, and $Yscl=1$. This gives some space around the graph.

5. Graph the Curve:

   • Press the [GRAPH] button. The calculator will draw an ellipse, which is the path traced by the tip of the vector $r(t)$ as $t$ goes from $0$ to $2\pi$.

AP Exam Quick Hit

Common Question Types

• Evaluating at a Point: You will be given a vector-valued function $r(t)$ and asked to find its value at a specific $t=a$.

  • Example: "If the position of a particle is given by $r(t)=⟨\ln (t),t^3-2⟩$, what is the position vector at $t=1$?"

  • Answer:$r(1)=⟨\ln (1),1^3-2⟩=⟨0,-1⟩$.

• Identifying Component Functions: You will be given a function in either component or unit vector form and asked to identify $x(t)$ or $y(t)$.

  • Example: "A curve is described by the vector-valued function $r(t)=(5t+3)i+\cos (t)j$. Which of the following is the vertical component function?"

  • Answer:$y(t)=\cos (t)$.

• Determining the Domain: You will be given a vector-valued function and asked to find its domain based on the domains of its component functions.

  • Example: "What is the domain of the function $r(t)=⟨\frac{1}{t-2},\sqrt{t}⟩$?"

  • Answer: The domain of $\frac{1}{t-2}$ is $t\ne 2$. The domain of $\sqrt{t}$ is $t\ge 0$. The intersection is $[0,2)\cup (2,\infty )$.

Common Mistakes

• Incorrect Output Type: Stating the answer as a coordinate point $(a,b)$ or a scalar value instead of a vector $⟨a,b⟩$. The output of a vector-valued function is always a vector.

• Swapping Components: Accidentally placing the $y(t)$ result in the horizontal component of the answer vector and the $x(t)$ result in the vertical component.

• Parameter vs. Component Confusion: For a function like $r(t)=⟨t^2,2t⟩$, a common mistake is to evaluate $r(3)$ as $⟨3,3⟩$ instead of correctly calculating $⟨(3{)}^2,2(3)⟩=⟨9,6⟩$. The parameter $t$ must be substituted into the component functions.

• Domain Calculation Errors: Finding the domain of only one component function and ignoring the other. The domain of $r(t)$ is the intersection of the domains of both$x(t)$ and $y(t)$.

• Misinterpreting $i$ and $j$: Treating the standard unit vectors $i$ and $j$ as variables to be solved for, rather than as fixed vector constants representing direction.