Reading a Vector-Valued Function
A vector-valued function assigns a vector to each permitted input. In this unit, the input is usually a real parameter \(t\), and the output is an ordered pair or triple of real numbers. The key reading skill is to see each coordinate formula as its own scalar-valued function while remembering that the complete output is one vector.
This component form makes a vector-valued function concrete: its domain is controlled by the component formulas, and evaluating it means evaluating every component at the same parameter value. If the function describes position, the resulting vector records the coordinates of the point at that parameter.
For a function taking values in \(\mathbb{R}^2\), there are two components; for one taking values in \(\mathbb{R}^3\), there are three. The components are ordered: in the plane, the first is the \(x\)-coordinate and the second is the \(y\)-coordinate; in space, the third component gives the \(z\)-coordinate.
Angle brackets emphasize that these entries form a vector. The same function can also be written using parentheses, such as \((x(t),y(t))\), when the context makes clear that the ordered pair is the vector output. Do not confuse a component formula, such as \(x(t)\), with the whole vector \(\mathbf{r}(t)\).
Domain: Where Every Component Is Defined
The domain is the set of allowed parameter values. Sometimes the problem states it directly, for example \(0\leq t\leq 4\). When no domain is stated, the natural domain consists of the real inputs for which every component formula is defined. A parameter must work in all components at once, not just one of them.
For example, a square root requires its radicand to be nonnegative, and a denominator must be nonzero. A logarithm requires a positive argument. When several such restrictions occur, list them separately and then keep only the values satisfying all of them.
The formulas alone do not always tell the entire story. A problem may restrict a formula to a stated interval even though the components are defined beyond it. That restriction is part of the function’s domain and must be respected when evaluating the function.
Evaluating the Vector Output
To evaluate \(\mathbf{r}(a)\), substitute \(a\) for the parameter in every component formula. Keep the results in their original order and write them as one vector. The input \(a\) is a scalar; \(\mathbf{r}(a)\) is the vector output.
Confirm that \(a\) belongs to the function’s stated or natural domain.
Use the same parameter value in every coordinate formula.
Place the evaluated component values in their original coordinate positions.
Worked Example: Evaluate a Two-Component Function
Let \(\mathbf{r}(t)=\langle t^2-1,\,2t+3\rangle\), with domain \(\mathbb{R}\). Find \(\mathbf{r}(2)\), and identify its coordinates.
Step 1: Check the input. Both components are polynomials, which are defined for every real number. Thus \(2\) is in the domain.
Step 2: Evaluate the first component. Substitute \(t=2\): \(2^2-1=4-1=3\).
Step 3: Evaluate the second component. Substitute \(t=2\): \(2(2)+3=4+3=7\).
Check: The first coordinate \(3\) is the value of \(t^2-1\) at \(2\), and the second coordinate \(7\) is the value of \(2t+3\) at \(2\). The ordered output is therefore \(\langle 3,7\rangle\), not a single scalar.
Finding a Common Domain
When the natural domain is not stated, find the restrictions imposed by each component, then combine them. A restriction from one component cannot be ignored just because another component is defined at that input.
Worked Example: Combine a Logarithm and a Square Root
Consider \(\mathbf{r}(t)=\langle \ln(t-1),\,\sqrt{5-t}\rangle\). Find its natural domain and evaluate \(\mathbf{r}(2)\).
Step 1: Find the logarithm restriction. The logarithm requires a positive argument: \(t-1>0\), so \(t>1\).
Step 2: Find the square-root restriction. The radicand must be nonnegative: \(5-t\geq 0\), so \(t\leq 5\).
Step 3: Intersect the restrictions. Both conditions must hold, giving the natural domain \((1,5]\).
Step 4: Check and evaluate at \(t=2\). Since \(1<2\leq5\), the input is allowed. The first component is \(\ln(2-1)=\ln 1=0\). The second is \(\sqrt{5-2}=\sqrt3\).
Check: At \(t=2\), the logarithm’s argument is \(1>0\), and the square-root radicand is \(3\geq0\). Both component evaluations are valid, so the ordered vector is defined.
Position Vectors and Coordinates
In a motion problem, a vector-valued function often gives the position of a moving point as a function of time. If \(\mathbf{r}(t)=\langle x(t),y(t)\rangle\), then the point’s position at time \(t\) has coordinates \((x(t),y(t))\). In three dimensions, \(\mathbf{r}(t)=\langle x(t),y(t),z(t)\rangle\) gives coordinates \((x(t),y(t),z(t))\). The parameter values are restricted to the times specified by the problem or by the natural domain of the component formulas.
Evaluating a position function at a particular time tells where the point is at that time. It does not, by itself, describe how the point moves between times; questions about rates or changes in position belong to later work with vector functions.
Worked Example: Read a Position Vector in Three Dimensions
A particle has position function \(\mathbf{r}(t)=\langle 2t,\,t^2-1,\,4-t\rangle\) for \(0\leq t\leq3\). Find its position at \(t=3\).
Step 1: Check the stated domain. Since \(0\leq3\leq3\), the time \(t=3\) is allowed.
Step 2: Evaluate the first component. \(2(3)=6\).
Step 3: Evaluate the second component. \(3^2-1=9-1=8\).
Step 4: Evaluate the third component. \(4-3=1\).
Check: Substituting \(t=3\) into the original formula gives \(\langle 2(3),3^2-1,4-3\rangle=\langle6,8,1\rangle\). Thus the particle’s position at that time is the point with coordinates \((6,8,1)\).
Why Component Form Works
A vector in \(\mathbb{R}^n\) is determined by its \(n\) coordinates. This fact explains why a vector-valued function can be read and evaluated component by component: its output is precisely the ordered list of the component values at the same input.
To make the reasoning explicit, suppose \(\mathbf{r}(t)=\langle f_1(t),\ldots,f_n(t)\rangle\). At any parameter \(a\) in the domain, the definition gives the output vector \(\langle f_1(a),\ldots,f_n(a)\rangle\). Each coordinate is obtained by applying its scalar component function to \(a\). Because two vectors in \(\mathbb{R}^n\) are equal exactly when their corresponding coordinates are equal, the component formulas uniquely determine the vector output.
Common Mistake: Using Different Inputs or Ignoring a Restriction
A frequent error is to substitute the parameter into only one component, or to treat the component values as if they could come from different parameter values. Another is to declare an input valid because one component is defined there, even though another component is not.
Worked Example: Correct a Domain and Evaluation Error
Let \(\mathbf{q}(t)=\langle \sqrt{t+1},\,\frac{1}{t-2}\rangle\). A student claims that \(t=2\) is in the domain because the square-root component is defined there, and reports \(\mathbf{q}(2)=\langle\sqrt3,0\rangle\).
Identify the domain conditions. The square root requires \(t+1\geq0\), so \(t\geq-1\). The denominator requires \(t-2\ne0\), so \(t\ne2\). Therefore the natural domain is \([-1,\infty)\setminus\{2\}\).
Correct the input check. Although \(2\geq-1\), the second condition excludes \(2\). The vector \(\mathbf{q}(2)\) is not defined.
Show why the proposed calculation fails. The first component would give \(\sqrt{2+1}=\sqrt3\), but the second would require \(\frac{1}{2-2}=\frac10\), which is undefined. It cannot be replaced by \(0\).
Check a valid nearby input instead. At \(t=1\), the input satisfies \(1\geq-1\) and \(1\ne2\). Then \(\sqrt{1+1}=\sqrt2\) and \(\frac{1}{1-2}=-1\), so \(\mathbf{q}(1)=\langle\sqrt2,-1\rangle\). Both components are defined at this input.
Summary
To read a vector-valued function, identify its ordered component formulas and determine the parameter values for which every component is defined. If the domain is stated, use that restriction; otherwise, intersect the components’ natural domains. To evaluate the function, check the input and substitute it into every component, preserving the coordinate order.
Check Your Understanding
For each question, keep the domain conditions separate from the component evaluations.
- For \(\mathbf{r}(t)=\langle t+2,\,3t\rangle\), find \(\mathbf{r}(-1)\).
- Find the natural domain of \(\mathbf{v}(t)=\langle\sqrt{t-2},\,\frac{1}{t+1}\rangle\).
- If \(\mathbf{p}(t)=\langle t^2,\,2-t,\,\ln t\rangle\), what is the natural domain?
- A function is defined by \(\mathbf{s}(t)=\langle t,\,t^2\rangle\) for \(-2\leq t<1\). Is \(t=1\) an allowed input even though both formulas are defined there?
- For a position function \(\mathbf{r}(t)=\langle 1-t,\,t+4\rangle\), find the coordinates of the point at \(t=2\).
