Problem 23
Question
Eliminate the parameter \(t\), write the equation in Cartesian coordinates, then sketch the graphs of the vector-valued functions. $$ \mathbf{r}(t)=t^{3} \mathbf{i}+2 t \mathbf{j} $$
Step-by-Step Solution
Verified Answer
The Cartesian equation is \( y = 2x^{1/3} \). The graph is a cubic root curve, stretched vertically by a factor of 2.
1Step 1: Identify the Parametric Equations
The given vector-valued function is \( \mathbf{r}(t) = t^3 \mathbf{i} + 2t \mathbf{j} \). This can be separated into parametric equations: \( x(t) = t^3 \) and \( y(t) = 2t \). We will use these equations to eliminate the parameter \( t \).
2Step 2: Solve for t in terms of x or y
From the parametric equation \( x = t^3 \), solve for \( t \) in terms of \( x \). We have \( t = x^{1/3} \).
3Step 3: Substitute t into the y equation
Substitute \( t = x^{1/3} \) into the equation for \( y \): \( y = 2t \). This gives \( y = 2x^{1/3} \).
4Step 4: Write the Cartesian Equation
The equation in Cartesian coordinates, obtained by eliminating the parameter \( t \), is \( y = 2x^{1/3} \).
5Step 5: Graph the Cartesian Equation
Graph the equation \( y = 2x^{1/3} \). This represents the cube root function with a vertical stretch factor of 2. The graph passes through the origin and extends into both the positive and negative directions for \( x \) values. The curve increases gently as \( x \) becomes positive and goes to negative values for negative \( x \).
Key Concepts
Vector-Valued FunctionsCartesian CoordinatesGraphing Parametric Equations
Vector-Valued Functions
Vector-valued functions are mathematical expressions that use vectors instead of scalar variables to define a relationship. In essence, these functions map a parameter, commonly denoted as \( t \), to a vector. This vector can be composed of components along the \( x \), \( y \), and sometimes \( z \) axes. For instance, consider the vector-valued function \( \mathbf{r}(t) = t^3 \mathbf{i} + 2t \mathbf{j} \). This expression can be understood as a function providing us with
- an \( x \)-component: \( t^3 \)
- a \( y \)-component: \( 2t \)
Cartesian Coordinates
Cartesian coordinates are a system used to uniquely determine each point in a plane or space by a set of numerical coordinates. These coordinates represent points through a pair (or triplet in three dimensions) of perpendicular lines, called axes. Commonly, the horizontal axis is the \( x \)-axis, and the vertical is the \( y \)-axis. In this exercise, we will focus on converting a vector-valued function into a Cartesian equation.By manipulating vector components, such as \( x(t) = t^3 \) and \( y(t) = 2t \), we transform them to eliminate the parameter \( t \) and express the function purely in terms of \( x \) and \( y \). Such a transformation results in a Cartesian equation, "flattening" the path described by the vector into a more visually interpretable format on a two-dimensional plane, like \( y = 2x^{1/3} \). Transforming vector-valued functions into Cartesian coordinates simplifies graphing and visual analysis.
Graphing Parametric Equations
Graphing parametric equations involves plotting the relationship defined by parametric expressions. Once we eliminate the parameter \( t \) from the equations, we transform the function into a familiar \( y = f(x) \) format, which can be easier to graph on the Cartesian plane.For example, let's consider the contrived example of the equation \( y = 2x^{1/3} \), derived from the original parametric equations \( x(t) = t^3 \) and \( y(t) = 2t \).
- This equation represents a stretched cube root function.
- It passes through the origin, meaning that when \( x = 0 \), \( y \) is also zero.
- The curve stretches toward both positive and negative values of \( x \).
Other exercises in this chapter
Problem 22
Find the arc length of the curve on the given interval.The position function for a particle is \(\mathbf{r}(t)=a \cos (\omega t) \mathbf{i}+b \sin (\omega t) \m
View solution Problem 22
Eliminate the parameter \(t\), write the equation in Cartesian coordinates, then sketch the graphs of the vector-valued functions. \(\mathbf{r}(t)=2 t \mathbf{i
View solution Problem 24
Compute the first, second, and third derivatives of \(\mathbf{r}(t)=3 t \mathbf{i}+6 \ln (t) \mathbf{j}+5 e^{-3 t} \mathbf{k}\).
View solution Problem 24
Find the arc length of the curve on the given interval.Given \(\mathbf{r}(t)=\left\langle 2 e^{t}, e^{t} \cos t, e^{t} \sin t\right\rangle\), determine the tang
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