Problem 22
Question
Eliminate the parameter \(t\), write the equation in Cartesian coordinates, then sketch the graphs of the vector-valued functions. \(\mathbf{r}(t)=2 t \mathbf{i}+t^{2} \mathbf{j}\) (Hint: Let \(x=2 t\) and \(y=t^{2} .\) Solve the first equation for \(x\) in terms of \(t\) and substitute this result into the second equation.)
Step-by-Step Solution
Verified Answer
The Cartesian equation is \(y = \frac{x^2}{4}\), a parabola opening upwards.
1Step 1: Express Parameter in Terms of One Variable
We start by using the information given in the hint. We have that \(x = 2t\). To express \(t\) in terms of \(x\), we solve for \(t\) by dividing both sides by 2: \(t = \frac{x}{2}\).
2Step 2: Substitute to Find the Cartesian Equation
Now that we have \(t\) in terms of \(x\), substitute \(t = \frac{x}{2}\) into the equation for \(y\): \(y = t^2\). Thus, \(y = \left(\frac{x}{2}\right)^2\).
3Step 3: Simplify the Cartesian Equation
Simplify the expression for \(y\). \(y = \frac{x^2}{4}\). So the Cartesian equation is \(y = \frac{x^2}{4}\).
4Step 4: Sketch the Graph
The Cartesian equation \(y = \frac{x^2}{4}\) represents a parabola that opens upward. This parabola has its vertex at the origin \((0,0)\), and it is vertically compressed by a factor of 4 compared to the standard parabola \(y=x^2\).
Key Concepts
Cartesian coordinatesVector-valued functionsParabola graphing
Cartesian coordinates
The Cartesian coordinate system is a method for visually representing mathematical concepts in a plane. It uses two perpendicular axes: the x-axis (horizontal) and the y-axis (vertical). Each point in this system is defined by an ordered pair \(x, y\), where both coordinates have specific numerical values.
To convert parametric equations, like the one given in our problem, into Cartesian coordinates, involves expressing both parametric variables in terms of x and y. This often results in a two-variable equation that lies entirely within the familiar Cartesain plane. This conversion makes it easier to visualize and analyze the functions.
In the exercise, we've translated the vector-valued function into a Cartesian equation: \(y = \frac{x^2}{4}\). This step is crucial because it transforms the function into a form that's widely recognizable and interpretable in the Cartesian plane.
To convert parametric equations, like the one given in our problem, into Cartesian coordinates, involves expressing both parametric variables in terms of x and y. This often results in a two-variable equation that lies entirely within the familiar Cartesain plane. This conversion makes it easier to visualize and analyze the functions.
In the exercise, we've translated the vector-valued function into a Cartesian equation: \(y = \frac{x^2}{4}\). This step is crucial because it transforms the function into a form that's widely recognizable and interpretable in the Cartesian plane.
Vector-valued functions
Vector-valued functions are an elegant way to describe a range of mathematical concepts, including curves in two and three dimensions. Unlike simple numerical functions, vector-valued functions map a parameter, often \(t\), directly to vector quantities. These vectors typically describe position, and in our exercise, are written in terms of unit vectors, usually \(\mathbf{i}\) and \(\mathbf{j}\), which represent the x and y components, respectively.
For example, the vector-valued function \(\mathbf{r}(t) = 2t \mathbf{i} + t^2 \mathbf{j}\) maps every input of \(t\) to a vector located at \(2t\) on the x-axis and \(t^2\) on the y-axis. This means for each value of \(t\), you get a unique position specified by two numbers, an x-value and a y-value, forming the path the point will trace as \(t\) changes.
This functional representation is vital in physics and engineering to describe not only paths but also velocities and accelerations of moving objects.
For example, the vector-valued function \(\mathbf{r}(t) = 2t \mathbf{i} + t^2 \mathbf{j}\) maps every input of \(t\) to a vector located at \(2t\) on the x-axis and \(t^2\) on the y-axis. This means for each value of \(t\), you get a unique position specified by two numbers, an x-value and a y-value, forming the path the point will trace as \(t\) changes.
This functional representation is vital in physics and engineering to describe not only paths but also velocities and accelerations of moving objects.
Parabola graphing
Graphing a parabola involves depicting a curve that is symmetrical about a vertical line, often shown as \(y = ax^2 + bx + c\). The graph forms a 'U' or 'n' shape and is one of the simplest and most recognizably curved figures in mathematics.
In our converted Cartesian equation \(y = \frac{x^2}{4}\), we observe a standard parabola form's distinctive feature centered at the origin, \( (0,0) \). The equation tells us that the parabola opens upwards and that it's vertically compressed compared to the basic parabola having \(y = x^2\). The compression ensures that as x moves away from the origin, y increases at a slower rate than it would in the case of \(y = x^2\).
When sketching, it's crucial to note the vertex position, direction of opening, and any coefficient impacts on the shape like vertical compression or stretching.
In our converted Cartesian equation \(y = \frac{x^2}{4}\), we observe a standard parabola form's distinctive feature centered at the origin, \( (0,0) \). The equation tells us that the parabola opens upwards and that it's vertically compressed compared to the basic parabola having \(y = x^2\). The compression ensures that as x moves away from the origin, y increases at a slower rate than it would in the case of \(y = x^2\).
When sketching, it's crucial to note the vertex position, direction of opening, and any coefficient impacts on the shape like vertical compression or stretching.
Other exercises in this chapter
Problem 22
Find the following.\(\frac{d}{d t}[r(t) \cdot s(t)]\)
View solution 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 23
Eliminate the parameter \(t\), write the equation in Cartesian coordinates, then sketch the graphs of the vector-valued functions. $$ \mathbf{r}(t)=t^{3} \mathb
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