Problem 41 - Calculus Early Transcendentals

Question

Find a parametric description of the line segment from the point \(P\) to the point \(Q\). Solutions are not unique. $$P(0,0), Q(2,8)$$

Step-by-Step Solution

Answer

Question: Find a parametric description of the line segment connecting points P(0, 0) and Q(2, 8) in 2D space. Answer: One possible parametric description of the line segment from point P to point Q is given by the vector \(\vec{v}(t) = \begin{pmatrix} 2t \\ 8t \end{pmatrix}\) with \(0 \leq t \leq 1\).

Step 1: Find the Position Vectors of Points P and Q

We are given two points in 2D space. Let's find their position vectors. $$P(0, 0) \implies \vec{P} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$ $$Q(2, 8) \implies \vec{Q} = \begin{pmatrix} 2 \\ 8 \end{pmatrix}$$

Step 2: Find the Direction Vector of Line Segment PQ

To find the direction vector, we'll subtract the position vector of point P from the position vector of point Q: $$\vec{d} = \vec{Q} - \vec{P} = \begin{pmatrix} 2 \\ 8 \end{pmatrix} - \begin{pmatrix} 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 8 \end{pmatrix}$$

Step 3: Find the Parametric Description of Line Segment PQ

Now that we have the position vector of point P and the direction vector, we can find the parametric description of the line segment PQ using: $$\vec{v}(t) = \vec{P} + t\vec{d}$$ Substitute \(\vec{P}\) and \(\vec{d}\): $$\vec{v}(t) = \begin{pmatrix} 0 \\ 0 \end{pmatrix} + t\begin{pmatrix} 2 \\ 8 \end{pmatrix} = \begin{pmatrix} 0+2t \\ 0+8t \end{pmatrix}$$ This parametric description includes the line segment from P to Q, for \(t\) in the range \([0, 1]\). So one possible parametric description of the line segment from point P to point Q is given by the vector: $$\vec{v}(t) = \begin{pmatrix} 2t \\ 8t \end{pmatrix}$$ with \(0 \leq t \leq 1\).

Key Concepts

Position Vectors

To understand parametric equations, it's crucial to start with position vectors. A position vector is simply a vector that represents the location of a point in space relative to an origin. Imagine you have a map, and you want to describe where certain places are located. The position vector would be like giving directions to these places starting from a central point, often called the origin.

In this problem, we have the points \( P(0,0) \) and \( Q(2,8) \). The position vectors of these points demonstrate their coordinates in space. For point \( P \), the position vector is \( \vec{P} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \). This means that \( P \) is located exactly at the origin. On the other hand, point \( Q \) has a position vector of \( \vec{Q} = \begin{pmatrix} 2 \ 8 \end{pmatrix} \).

These position vectors are fundamental in determining other properties of the line segment created by these two points.

Direction Vector

Once you know the position vectors of points \( P \) and \( Q \), you can find the direction vector of the line segment that connects them. The direction vector is crucial because it shows the direction and the distance from one point to another.

The direction vector \( \vec{d} \) is calculated by subtracting the position vector of \( P \) from the position vector of \( Q \):

• First, note the position of \( Q \) as \( \vec{Q} = \begin{pmatrix} 2 \ 8 \end{pmatrix} \).
• Next, subtract the position of \( P \), \( \vec{P} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \).

The resulting direction vector is \( \vec{d} = \begin{pmatrix} 2 \ 8 \end{pmatrix} \). This tells us that as you move from \( P \) to \( Q \), you'd travel 2 units in the x-direction and 8 units in the y-direction.

The direction vector can be seen as the guide or trajectory of the line segment from \( P \) to \( Q \).

Line Segment

The line segment between two points, \( P \) and \( Q \), is essentially the path that connects them. Parametric equations are a powerful tool to express this line segment mathematically. By having a starting point (like \( P \)) and clearly defined direction (our direction vector), we can create a parametric equation that describes the line segment.

Parametric equations use a parameter, commonly denoted as \( t \), to express each point along the line segment. Here’s how it works for our points \( P \) and \( Q \):

• Start at the point \( P \), which is expressed as \( \begin{pmatrix} 0 \ 0 \end{pmatrix} \).
• Add \( t \) times the direction vector \( \begin{pmatrix} 2 \ 8 \end{pmatrix} \): \( \vec{v}(t) = \begin{pmatrix} 0 \ 0 \end{pmatrix} + t \begin{pmatrix} 2 \ 8 \end{pmatrix} \).

This results in the equation \( \vec{v}(t) = \begin{pmatrix} 2t \ 8t \end{pmatrix} \). For \( t \) in the range \([0, 1]\), \( \vec{v}(t) \) gives all the points on the line segment from \( P \) to \( Q \).

In essence, the parametric form simplifies understanding of how point transformations happen along the segment, conveying movement from point \( P \) to \( Q \) in a clear, mathematical way.