Book cover for Thomas Calculus

Thomas Calculus

George B. Thomas, Jr.

ISBN #9780321878960

13th Edition

6,812 Questions

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127,035 Students Helped

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Summary
Learning Objectives
Key Concepts
Example Problems
Explanations
Common Mistakes

Summary

This section introduces the calculus of vector-valued functions, showing how they describe curves and motions in space. Key concepts include the representation of a particle’s path by the position vector, and the interpretation of derivatives as tangent vectors that yield velocity and acceleration. Techniques such as differentiating component-wise, applying rules (product, dot, cross, and chain rules) and computing limits provide the mathematical foundation for analyzing motions in both the plane and space. These tools are essential in fields like physics, engineering, and computer graphics.

Learning Objectives

1

Explain the calculus of vector-valued functions, including how to compute derivatives and integrals component-wise.

2

Describe how to represent curves in space using position vectors and their parametric equations.

3

Demonstrate how to find tangent vectors, velocity, speed, and acceleration for particles moving along curves.

4

Apply differentiation rules such as the product, dot product, cross product, and chain rules to vector functions.

5

Analyze motion in space and the significance of continuity, limits, and constant-length vector functions.

Key Concepts

CONCEPT

DEFINITION

Vector-Valued Function

A function whose inputs are real numbers (or points in space) and whose outputs are vectors, often used to represent curves in space.

Position Vector (r(t))

A vector function r(t)=Æ’(t)i + g(t)j + h(t)k which defines the location of a point in space at time t.

Component Functions

The scalar functions Æ’(t), g(t), and h(t) that comprise each corresponding coordinate in the position vector.

Tangent Vector

The derivative r′(t) of the position vector, representing both the direction of motion and the instantaneous rate of change of position.

Continuity of r(t)

A vector function is continuous at t=t0 if each of its component functions is continuous at t0; i.e., lim(t→t0) r(t) = r(t0).

Velocity and Speed

Velocity is the derivative of the position vector (r′(t)); speed is the magnitude of the velocity vector, |r′(t)|.

Acceleration

The derivative of the velocity vector, given by r″(t), which indicates how the velocity changes over time.

Differentiation Rules

Rules (constant, scalar multiple, sum/difference, product, dot product, cross product, and chain rule) that allow the differentiation of vector functions component-wise.

Example Problems

Example 1

In Exercises $1-4, \mathbf{r}(t)$ is the position of a particle in the $x y$ -plane at time $t .$ Find an equation in $x$ and $y$ whose graph is the path of the particle. Then find the particle's velocity and acceleration vectors at the given value of $t .$ \begin{equation} \mathbf{r}(t)=(t+1) \mathbf{i}+\left(t^{2}-1\right) \mathbf{j}, \quad t=1 \end{equation}

Example 2

${r}(t)$ is the position of a particle in the $x y$ -plane at time $t .$ Find an equation in $x$ and $y$ whose graph is the path of the particle. Then find the particle's velocity and acceleration vectors at the given value of $t .$ \begin{equation} \mathbf{r}(t)=\frac{t}{t+1} \mathbf{i}+\frac{1}{t} \mathbf{j}, \quad t=-\frac{1}{2} \end{equation}

Example 3

${r}(t)$ is the position of a particle in the $x y$ -plane at time $t .$ Find an equation in $x$ and $y$ whose graph is the path of the particle. Then find the particle's velocity and acceleration vectors at the given value of $t .$ \begin{equation} \mathbf{r}(t)=e^{t} \mathbf{i}+\frac{2}{9} e^{2 t} \mathbf{j}, \quad t=\ln 3 \end{equation}

Example 4

${r}(t)$ is the position of a particle in the $x y$ -plane at time $t .$ Find an equation in $x$ and $y$ whose graph is the path of the particle. Then find the particle's velocity and acceleration vectors at the given value of $t .$ \begin{equation} \mathbf{r}(t)=(\cos 2 t) \mathbf{i}+(3 \sin 2 t) \mathbf{j}, \quad t=0 \end{equation}

Example 5

Exercises $5-8$ give the position vectors of particles moving along various curves in the $x y$ -plane. In each case, find the particle's velocity and acceleration vectors at the stated times and sketch them as vectors on the curve. \begin{equation} \mathbf{r}(t)=(\sin t) \mathbf{i}+(\cos t) \mathbf{j} ; \quad t=\pi / 4 \text { and } \pi / 2 \end{equation}
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Step-by-Step Explanations

QUESTION

Find the derivative of r(t) = (cos t)i + (sin t)j + t k.

STEP-BY-STEP ANSWER:

Step 1: Differentiate each component separately. The derivative of cos t is -sin t.
Step 2: The derivative of sin t is cos t.
Step 3: The derivative of t is 1.
Step 4: Combine the derivatives to form the derivative vector: r′(t) = (-sin t)i + (cos t)j + 1 k.
Final Answer: r′(t) = (-sin t)i + (cos t)j + k.

Derivative of a Vector-Valued Function

QUESTION

Explain how to compute lim(t → t0) r(t) for r(t) = ƒ(t)i + g(t)j + h(t)k.

STEP-BY-STEP ANSWER:

Step 1: Find the limit of each component: lim(t → t0) ƒ(t), lim(t → t0) g(t), and lim(t → t0) h(t).
Step 2: Express the overall limit as the vector of these limits: lim(t → t0) r(t) = (lim ƒ(t))i + (lim g(t))j + (lim h(t))k.
Final Answer: Compute each component's limit and then reassemble the vector.

Limit of a Vector Function

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Common Mistakes

  • Failing to differentiate each component of the vector function separately.
  • Confusing the dot product and cross product rules during differentiation.
  • Overlooking the importance of the derivative being tangent to the curve, particularly in determining the direction of motion.
  • Misapplying the chain rule by not correctly adjusting for the inner function’s derivative.
  • Assuming that a constant-length vector automatically implies zero velocity; in reality, the velocity vector is orthogonal to the position vector when the length is constant.