Question

Create a vector-valued function whose graph matches the given description. A line through (2,3) with a slope of 5 .

Short answer

The vector-valued function is \( \mathbf{r}(t) = \langle t, 5t - 7 \rangle \).

Step-by-step solution

Understand the Slope-Intercept Form

The slope-intercept form of a line is given by the equation \( y = mx + b \), where \( m \) is the slope, and \( b \) is the y-intercept. A vector-valued function for a line can be expressed using a point and the slope.

Identify Known Values

From the problem, the line passes through the point \( (2,3) \) and has a slope of \( 5 \). This means the line equation in slope-intercept form would be \( y = 5x + b \). We need to find the value of \( b \) using the point \( (2, 3) \).

Find the Y-Intercept

Substitute \( x = 2 \) and \( y = 3 \) into \( y = 5x + b \) to find \( b \). This gives us:\[ 3 = 5(2) + b \]\[ 3 = 10 + b \]Solving for \( b \), we find:\[ b = 3 - 10 = -7 \]

Write the Equation of the Line

Now that we have both the slope and y-intercept, the equation of the line is \( y = 5x - 7 \).

Express the Line as a Vector-Valued Function

A vector-valued function for the line can be expressed as \( \mathbf{r}(t) = \langle x, y \rangle \). Since lines can be parameterized in terms of \( x \), we can express \( x \) as \( t \) which evolves over time. The function becomes:\[ \mathbf{r}(t) = \langle t, 5t - 7 \rangle \]Here, \( t \) represents any real number, which parametrizes the movement along the line.

Confirm the Function's Validity

Verify that \( \mathbf{r}(t) \) passes through the point \( (2,3) \) when \( t = 2 \):\[ \mathbf{r}(2) = \langle 2, 5(2) - 7 \rangle = \langle 2, 3 \rangle \]This confirms that the vector function correctly represents the line through the specified point with the given slope.

Key concepts

Slope-Intercept Form

The slope-intercept form of a line is one of the most commonly used equations in mathematics. It is represented as \( y = mx + b \). Here, \( m \) stands for the slope, which indicates how steep the line is. The variable \( b \) represents the y-intercept, which is the point where the line crosses the y-axis. This form is incredibly useful because it quickly shows both the slope and the y-intercept, making it easy to understand the line's behavior at a glance.
Whenever you have a line equation in this form, you can quickly identify how much the line ascends or descends by its slope value and exactly where it will touch the vertical y-axis through the y-intercept value.

Y-Intercept

The y-intercept is a fundamental component in the slope-intercept equation \( y = mx + b \). It's the point where the line crosses the y-axis, specified by the value of \( b \).
To find the y-intercept, you can substitute the point where the line crosses the y-axis (when \( x = 0 \)) into the equation. For instance, in this exercise, you take the point \( (2, 3) \) and the slope \( m = 5 \) to solve \( 3 = 5 \times 2 + b \). Solving this equation gives us \( b = -7 \), confirming the line crosses the y-axis at \(-7\).
This intercept tells you exactly where to find the line in relation to the y-axis, an essential step in sketching or understanding the line.

Parametrization

Parametrization is a way to describe a line using a parameter, like \( t \), that evolves over time.
This approach is especially useful in vector calculus and when dealing with vector-valued functions, where describing a line in terms of a parameter simplifies the process of modeling lines and curves. In this exercise, we transform the line equation \( y = 5x - 7 \) into a vector-valued function \( \mathbf{r}(t) = \langle t, 5t - 7 \rangle \). The parameter \( t \) acts as a controlling variable, moving easily along the line's path.
This method helps in various applications, such as computer graphics and physics, allowing for smooth transitions along the described path.

Line Equation

The line equation combines several concepts, providing a unified way to describe a line in a plane.
It encompasses elements such as slope, y-intercept, and the relationship between them, resulting in equations like the familiar slope-intercept form \( y = mx + b \). For this exercise, with known values like slope \( m = 5 \) and point \( (2,3) \), finding the line equation involves solving for the y-intercept \( b \,(-7)\) to finalize the line as \( y = 5x - 7 \).
This line equation is then expressed as a vector-valued function \( \mathbf{r}(t) = \langle t, 5t - 7 \rangle \), highlighting how a line can be transformed into different representations. Understanding these transformations and how they relate is crucial for applying lines to practical scenarios.