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Chapter 12: Problem 17

Sketch the line segment represented by each vector equation. (a) \(\mathbf{r}=(1-t) \mathbf{i}+t \mathbf{j} ; 0 \leq t \leq 1\) (b) \(\mathbf{r}=(1-t)(\mathbf{i}+\mathbf{j})+t(\mathbf{i}-\mathbf{j}) ; 0 \leq t \leq 1\)

Short Answer

Expert verified
(a) Line segment from (1,0) to (0,1); (b) Line segment from (1,1) to (1,-1).

Step by step solution

01

Understanding Vector Equation (a)

The vector equation \( \mathbf{r} = (1-t) \mathbf{i}+t \mathbf{j} \) represents a line segment. Here, \( \mathbf{i} = (1,0) \) and \( \mathbf{j} = (0,1) \). By plugging in the values of \( t \) from 0 to 1, we can find the coordinates of the points on the segment.
02

Compute Endpoints of Vector (a)

When \( t=0 \), \( \mathbf{r} = (1-0)\mathbf{i} + 0\mathbf{j} = \mathbf{i} = (1,0) \). When \( t=1 \), \( \mathbf{r} = (1-1)\mathbf{i} + 1\mathbf{j} = \mathbf{j} = (0,1) \). So the line segment runs from point (1,0) to (0,1).
03

Plot Line Segment for Vector (a)

To sketch the line segment for \( \mathbf{r} = (1-t)\mathbf{i}+t\mathbf{j} \), plot the points (1,0) and (0,1) on the cartesian plane and draw a straight line connecting them.
04

Understanding Vector Equation (b)

The equation \( \mathbf{r} = (1-t)(\mathbf{i}+\mathbf{j}) + t(\mathbf{i}-\mathbf{j}) \) requires evaluating both vectors for given \( t \) values. Simplify the expression to find points on the line.
05

Simplify and Find Endpoints of Vector (b)

The vector \( (1-t)(\mathbf{i}+\mathbf{j}) + t(\mathbf{i}-\mathbf{j}) \) can be expanded as \( (1-t)(1,1) + t(1,-1) \). When \( t=0 \), the expression becomes \( \mathbf{r} = (1,1) \); when \( t=1 \), it becomes \( \mathbf{r} = (1,-1) \). The line segment runs from (1,1) to (1,-1).
06

Plot Line Segment for Vector (b)

On the coordinate plane, plot the points (1,1) and (1,-1) and draw a straight line segment connecting them, forming a vertical line.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Line Segments
Line segments are part of geometry and vector analysis, representing the set of points between and including two endpoints. Unlike lines, line segments have a fixed length since they start and end at specific points.
Understanding this feature helps in sketching them accurately:
  • Endpoints: The line starts at one endpoint and ends at another.
  • Fixed length: Since it has defined start and endpoints, the length never changes.
  • Direction: Depending on how you draw from one endpoint to another, a line segment has directionality from start to end.
For example, in the vector equation from the exercise, each line segment can be determined by evaluating the endpoints when different values of the parameter, \( t \), are inserted. Identifying these bounds, like when \( t = 0 \) and \( t = 1 \), helps find the coordinates of these crucial endpoints, enabling the line segment's sketch.
Cartesian Plane
The Cartesian plane is a two-dimensional plane defined by two perpendicular axes, usually labeled \( x \)-axis (horizontal) and \( y \)-axis (vertical). This plane forms the backbone for plotting points, lines, and shapes in a visual format.
Here are common features:
  • Coordinates: Any point is represented by a pair \((x, y)\) describing its exact location.
  • Origin: The point \((0, 0)\) where both axes intersect.
  • Quadrants: They divide the plane into four parts, each with positive or negative values for \( x \) and \( y \).
When plotting line segments like those from the vector equations in the exercise, using the Cartesian plane simplifies the process. By identifying and plotting the endpoints, you gain a visual understanding of the relationship between these points and the resulting line that connects them.
Vectors in Mathematics
Vectors are mathematical objects used to represent quantities having both magnitude and direction. In mathematics, vectors often denote positions, velocities, and forces, translating across many fields such as physics and engineering.
Important aspects include:
  • Direction and magnitude: Highlighting how far and in which direction from a point.
  • Components: Vectors can be broken into parts like \( \mathbf{i} \) and \( \mathbf{j} \), which are unit vectors pointing in the horizontal and vertical directions, respectively.
  • Vector equations: Express how vectors relate or transform; equations can represent operations such as translations or rotations in space.
In our context involving these vector equations, vectors convey the linear path between points on the plane. By altering \( t \), you change the vector's point along the segment, directly echoing the line segment's characteristics on an analytical plane.

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Most popular questions from this chapter

Prove: If \(\mathbf{r}(t)\) is a smoothly parametrized function, then the angles between \(\mathbf{r}^{\prime}(t)\) and the vectors \(\mathbf{i}, \mathbf{j}\), and \(\mathbf{k}\) are continuous functions of \(t\).

Consider the various forces that a passenger in a car would sense while traveling over the crest of a hill or around a curve. Relate these sensations to the tangential and normal vector components of the acceleration vector for the car's motion. Discuss how speeding up or slowing down (e.g., doubling or halving the car's speed) affects these components.

Find the arc length of the graph of \(\mathbf{r}(t)\). $$ \mathbf{r}(t)=t^{3} \mathbf{i}+t \mathbf{j}+\frac{1}{2} \sqrt{6} t^{2} \mathbf{k} ; \quad 1 \leq t \leq 3 $$

Suppose that the position function of a particle moving along a circle in the \(x y\) -plane is \(\mathbf{r}=5 \cos 2 \pi t \mathbf{i}+5 \sin 2 \pi t \mathbf{j}\) (a) Sketch some typical displacement vectors over the time interval from \(t=0\) to \(t=1\). (b) What is the distance traveled by the particle during the time interval?

If an automobile of mass \(m\) rounds a curve, then its inward vector component of acceleration \(a_{N} \mathbf{N}\) is caused by the frictional force \(\mathbf{F}\) of the road. Thus, it follows from the vector form of Newton's second law [Equation (19)] that the frictional force and the normal scalar component of acceleration are related by the equation \(\mathbf{F}=m a_{N} \mathbf{N}\). Thus, $$ \|\mathbf{F}\|=m \kappa\left(\frac{d s}{d t}\right)^{2} $$ Use this result to find the magnitude of the frictional force in newtons exerted by the road on a \(500 \mathrm{~kg}\) go-cart driven at a speed of \(10 \mathrm{~km} / \mathrm{h}\) around a circular track of radius \(15 \mathrm{~m}\). \(\left[\right.\) Note: \(\left.1 \mathrm{~N}=1 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}^{2} .\right]\)
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