Problem 43 - Calculus Early Transcendentals

Question

Find a parametric description of the line segment from the point \(P\) to the point \(Q\). Solutions are not unique. $$P(-1,-3), Q(6,-16)$$

Step-by-Step Solution

Answer

Question: Find the parametric representation of the line segment that connects points P(-1, -3) and Q(6, -16) using the given parameter t, where 0 ≤ t ≤ 1. Answer: The parametric representation of the line segment is R(t) = (-1 + 7t, -3 - 13t).

Step 1: Parameter and Vector from P to Q

Let's denote \(t\) as the parameter (where \(0 \le t \le 1\)) and \(\overrightarrow{PQ}\) as the line segment's vector from P to Q. First, we need to find the vector between the two given points \(P(-1, -3)\) and \(Q(6, -16)\): $$\overrightarrow{PQ} = Q - P = (6 - (-1), -16 - (-3)) = (7, -13)$$

Step 2: Parametric Equation of the Line Segment

Now we will use the parameter \(t\) and the vector \(\overrightarrow{PQ}\) to create the parametric equation of the line segment that connects points P and Q. Let \(R(t)\) represent the points on the line segment, where \(0 \le t \le 1\): $$R(t) = P + t\overrightarrow{PQ} = (-1, -3) + t(7, -13)$$ So the line segment can be described by the parametric equations: $$R(t) = (-1 + 7t, -3 - 13t)$$ For each \(t\) value between 0 and 1, this represents a point on the line segment connecting point P to point Q.

Key Concepts

Vector Analysis

Vector analysis is a mathematical tool that helps us understand quantities that have both magnitude and direction. In this exercise, we're dealing with the vector \(\overrightarrow{PQ}\), which connects two points \(P(-1, -3)\) and \(Q(6, -16)\).

To find this vector, we subtract the coordinates of point \(P\) from those of point \(Q\):

• The difference in the x-coordinates is \(6 - (-1) = 7\).
• The difference in the y-coordinates is \(-16 - (-3) = -13\).

So, the vector \(\overrightarrow{PQ}\) is \((7, -13)\).

These components tell us how far and in what direction we should move from \(P\) to reach \(Q\). Understanding vectors is crucial for navigating spaces, physics models, and many other applications in science and engineering.

Line Segments

A line segment is simply a portion of a line with two endpoints. In our exercise, we are looking at the line segment from point \(P\) to point \(Q\). This can be visualized as a straight path between these two points.

The parametric equation for a line segment relies on these endpoints and a parameter \(t\). This parameter \(t\) usually ranges from 0 to 1, representing the start point and endpoint of the segment, respectively.

The parametric equation given is \(R(t) = (-1 + 7t, -3 - 13t)\). This equation:

• Begins at \(P\) when \(t = 0\), which gives the starting point coordinates.
• Ends at \(Q\) when \(t = 1\), which will equal the ending point coordinates.
• At any \(t\) value between 0 and 1, the equation gives the coordinates of a point on the segment.

By adjusting \(t\), we traverse along the line segment from \(P\) to \(Q\), illustrating how parameters can map a path in geometric space.

Coordinate Geometry

Coordinate geometry, or analytic geometry, is a bridge between algebra and geometry through graphing figures using coordinate points. In this problem, we're using coordinate geometry to describe a line segment between two points in the Cartesian plane.

Each point \(P(-1, -3)\) and \(Q(6, -16)\) can be plotted on a Cartesian plane using their respective (x, y) coordinates. By plotting these points and connecting them, we form a line segment.

The parametric equation \(R(t) = (-1 + 7t, -3 - 13t)\) is a way of translating the algebraic manipulation of these coordinates into geometric form.

• Algebra helps us find the vector and the equation representing all points between \(P\) and \(Q\).
• Geometry allows us to visualize this relationship as a segment on a graph.

Coordinate geometry thus provides us with tools to both analyze the algebraic aspects and understand the spatial relationships of geometric figures.