Chapter 12: Problem 23
Sketch the curve traced out by the vectorvalued function. Indicate the direction in which the curve is traced out. $$ \mathbf{F}(t)=\cos t \mathbf{i}+t \mathbf{j}-\sin t \mathbf{k} \text { for } 0 \leq t \leq 2 \pi $$
Short Answer
Expert verified
The curve is a helical path rising linearly in the y-direction with oscillating x and z components, traced as \( t \) increases from 0 to \( 2\pi \).
Step by step solution
01
Understand the Components
The given vector-valued function \( \mathbf{F}(t) = \cos t \mathbf{i} + t \mathbf{j} - \sin t \mathbf{k} \) represents a curve in 3-dimensional space. The functions for \( x(t) = \cos t \), \( y(t) = t \), and \( z(t) = -\sin t \) give the coordinates in the \( i \), \( j \), and \( k \) directions, respectively.
02
Identify the Parametric Expressions
We can identify the parametric expressions for the curve: \( x(t) = \cos t \), \( y(t) = t \), and \( z(t) = -\sin t \). These will help us to sketch the curve by understanding how each coordinate changes with \( t \).
03
Analyze the Behavior of Each Component
The \( x(t) = \cos t \) oscillates between -1 and 1 as \( t \) goes from 0 to \( 2\pi \). The \( y(t) = t \) increases linearly, indicating a continuous rise along the y-axis. The \( z(t) = -\sin t \) fluctuates from 0 to -1 and back to 0, creating an oscillating pattern in the z-direction.
04
Plot Key Points and Sketch the Curve
Consider key points at \( t = 0, \pi/2, \pi, 3\pi/2, 2\pi \). At these points,\( \cos t \) takes the values 1, 0, -1, 0, 1 respectively; \( t \) takes 0, \pi/2, \pi, 3\pi/2, 2\pi; and \( -\sin t \) takes 0, -1, 0, 1, 0. Plot these points in space to help sketch the curve.
05
Determine the Direction of the Curve
As \( t \) increases from 0 to \( 2\pi \), the direction is along the positive y-axis with oscillations back and forth in x and z. The direction of the curve follows the sequence of key points from Step 4.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parametric Equations
Parametric equations are a way to describe a curve by expressing the coordinates that define the curve as functions of a single variable, usually denoted as \( t \).
In the context of the exercise, the vector-valued function \( \mathbf{F}(t) = \cos t \mathbf{i} + t \mathbf{j} - \sin t \mathbf{k} \) consists of three parametric equations:
This allows us to understand the movement of a point along the curve by observing how it progresses in each dimension independently.
In the context of the exercise, the vector-valued function \( \mathbf{F}(t) = \cos t \mathbf{i} + t \mathbf{j} - \sin t \mathbf{k} \) consists of three parametric equations:
- \( x(t) = \cos t \)
- \( y(t) = t \)
- \( z(t) = -\sin t \)
This allows us to understand the movement of a point along the curve by observing how it progresses in each dimension independently.
3-Dimensional Space
In 3-dimensional space, we work with three axes: the x-axis, y-axis, and z-axis.
Each axis represents a different direction in space, and together they form a coordinate system that can describe any point within that space.
A vector-valued function, like the one given in the exercise, represents a curve by associating each point on the curve with a particular value of \( t \).
In the exercise \( \mathbf{F}(t) = \cos t \mathbf{i} + t \mathbf{j} - \sin t \mathbf{k} \), we see:
Each axis represents a different direction in space, and together they form a coordinate system that can describe any point within that space.
A vector-valued function, like the one given in the exercise, represents a curve by associating each point on the curve with a particular value of \( t \).
In the exercise \( \mathbf{F}(t) = \cos t \mathbf{i} + t \mathbf{j} - \sin t \mathbf{k} \), we see:
- \( \cos t \) indicating movement along the x-axis
- \( t \) guiding motion along the y-axis
- \(-\sin t \) driving changes along the z-axis
Curve Sketching
Curve sketching using parametric equations involves plotting points at various values of \( t \) and connecting them smoothly.
For the given function, we seek key points that highlight the path of the curve.
For example, by choosing \( t = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, \text{and} 2\pi \), we determine specific values for each coordinate:
For the given function, we seek key points that highlight the path of the curve.
For example, by choosing \( t = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, \text{and} 2\pi \), we determine specific values for each coordinate:
- At \( t = 0 \), we have: \( (x, y, z) = (1, 0, 0) \)
- \( t = \frac{\pi}{2} \): \( (x, y, z) = (0, \frac{\pi}{2}, -1) \)
- \( t = \pi \): \( (x, y, z) = (-1, \pi, 0) \)
- \( t = \frac{3\pi}{2} \): \( (x, y, z) = (0, \frac{3\pi}{2}, 1) \)
- \( t = 2\pi \): \( (x, y, z) = (1, 2\pi, 0) \)
Behavior Analysis
Analyzing the behavior of a vector-valued function helps us understand the characteristics and movement pattern of the curve.
In the vector function \( \mathbf{F}(t) = \cos t \mathbf{i} + t \mathbf{j} - \sin t \mathbf{k} \):
Understanding this behavior allows for better prediction of the curve's path and its directional qualities.
In the vector function \( \mathbf{F}(t) = \cos t \mathbf{i} + t \mathbf{j} - \sin t \mathbf{k} \):
- \( x(t) = \cos t \) oscillates between -1 and 1 as \( t \) progresses, creating a wave-like motion along the x-axis.
- \( y(t) = t \) increases linearly, suggesting a consistent upward or forward direction along the y-axis. This linearity implies the curve never turns back in the y-direction.
- \( z(t) = -\sin t \) moves from 0 to -1 and back to 0 within each cycle of \( \pi \), adding vertical oscillation along the z-axis.
Understanding this behavior allows for better prediction of the curve's path and its directional qualities.