Chapter 11: Problem 20
Sketch the curve whose vector equation is \(\mathbf{r}(t)=\) \(t \mathbf{i}+\frac{1}{2} t^{2} \mathbf{j}+\frac{1}{3} t^{3} \mathbf{k},-2 \leq t \leq 3\)
Short Answer
Expert verified
Sketch the curve by plotting points at various \( t \) values and smoothly connecting them in 3D space.
Step by step solution
01
Understand the Vector Equation
The vector equation is given by \( \mathbf{r}(t) = t\mathbf{i} + \frac{1}{2}t^2\mathbf{j} + \frac{1}{3}t^3\mathbf{k} \). This expresses a parametric curve in three-dimensional space, where each coordinate is expressed as a function of the parameter \( t \).
02
Identify the Range of \( t \)
The parameter \( t \) varies from -2 to 3. This means we will evaluate the vector equation for values of \( t \) within this interval to understand the shape and behavior of the curve.
03
Generate Key Points on the Curve
Evaluate the vector equation \( \mathbf{r}(t) \) at specific values of \( t \) (such as \( t = -2, -1, 0, 1, 2, 3 \)):- For \( t = -2 \), \( \mathbf{r}(-2) = (-2)\mathbf{i} + \frac{1}{2}(-2)^2\mathbf{j} + \frac{1}{3}(-2)^3\mathbf{k} = -2\mathbf{i} + 2\mathbf{j} -\frac{8}{3}\mathbf{k} \).- For \( t = -1 \), \( \mathbf{r}(-1) = -\mathbf{i} + \frac{1}{2}\mathbf{j} -\frac{1}{3}\mathbf{k} \).- Continue this process for \( t = 0, 1, 2, 3 \).
04
Plot the Points and Analyze Curve
Plot these points in a 3D coordinate system. Each point corresponds to a position vector originating from the origin to the calculated \( \mathbf{r}(t) \). Connect the points smoothly to sketch the curve, noting that the curve is polynomial and smooth.
05
Observe the Characteristics of the Curve
The curve is a polynomial space curve due to the powers of \( t \). While sketching, observe that: - As \( t \) moves from -2 to 3, the curve starts from the direction of \( -\mathbf{i} \), moves upwards due to increasing influence of \( \mathbf{j} \) and \( \mathbf{k} \) components.- The curve tends to swirl around the space due to the cubic term, going further away from the origin as \( t \) increases.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Calculus
Vector calculus is a mathematical field focused on differentiation and integration of vector fields. In this exercise, we use a vector equation to describe a curve in space. The vector \[ \mathbf{r}(t) = t \mathbf{i} + \frac{1}{2} t^2 \mathbf{j} + \frac{1}{3} t^3 \mathbf{k} \] represents the movement through three-dimensional space.
Each part of the equation corresponds to a direction, specified by the unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) along the x, y, and z axes, respectively.
Each part of the equation corresponds to a direction, specified by the unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) along the x, y, and z axes, respectively.
- \( t \mathbf{i} \) changes linearly, affecting movement along the x-axis proportionally to \( t \).
- \( \frac{1}{2} t^2 \mathbf{j} \) adds a quadratic influence in the y direction, showing acceleration.
- \( \frac{1}{3} t^3 \mathbf{k} \) introduces a cubic relationship along the z-axis, indicating an even more rapid change.
3D Coordinate Systems
Three-dimensional coordinate systems are essential for visualizing problems in vector calculus. In our exercise, the vector equation \[ \mathbf{r}(t) = t \mathbf{i} + \frac{1}{2} t^2 \mathbf{j} + \frac{1}{3} t^3 \mathbf{k} \] is graphed on a 3D coordinate plane, involving x, y, and z axes.
Each axis corresponds to a component of the vector:
Each axis corresponds to a component of the vector:
- \( x \) corresponds to the \( t \mathbf{i} \) portion, moving the point left or right.
- \( y \) corresponds to \( \frac{1}{2} t^2 \mathbf{j} \), moving the point up or down.
- \( z \) relates to \( \frac{1}{3} t^3 \mathbf{k} \), affecting depth.
Polynomial Spaces
Polynomial spaces are crucial for understanding the nature of curves presented by parametric equations. In the given problem \[ \mathbf{r}(t) = t \mathbf{i} + \frac{1}{2} t^2 \mathbf{j} + \frac{1}{3} t^3 \mathbf{k} \], a polynomial expression is built from linear, quadratic, and cubic terms of \( t \).
Each degree of the polynomial influences the curve's behavior and shape:
Each degree of the polynomial influences the curve's behavior and shape:
- The linear term \( t \mathbf{i} \) lays a straight path guiding initial direction.
- A quadratic term \( \frac{1}{2} t^2 \mathbf{j} \) introduces a parabolic curve.
- The cubic term \( \frac{1}{3} t^3 \mathbf{k} \) adds complexity, causing the path to twist and turn.