Book cover for University Calculus: Early Transcendentals

University Calculus: Early Transcendentals

Joel Hass, Maurice D. Weir, George B. Thomas, Jr.

ISBN #9780321999580

3rd Edition

6,517 Questions

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42,715 Students Helped

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Summary
Learning Objectives
Key Concepts
Example Problems
Explanations
Common Mistakes

Summary

This chapter establishes the foundation for multivariable calculus by extending familiar two-dimensional coordinate systems into three dimensions. Key topics include the establishment of a right-handed Cartesian coordinate system, the definition of coordinate planes and octants, and the derivation of the distance formula in space. Additionally, it covers the standard equation of a sphere and introduces vectors as quantities with both magnitude and direction, represented in component form. Mastery of these concepts is crucial for solving advanced geometric and applied problems in higher mathematics and various fields of science and engineering.

Learning Objectives

1

Understand and describe the three-dimensional Cartesian coordinate system, including axes, planes, and octants.

2

Apply the distance formula in space to compute distances between two points.

3

Derive and interpret equations for spheres and planes in three-dimensional space.

4

Explain the definition and properties of vectors, including their representation in component form and standard position.

5

Analyze and solve geometric problems using coordinate equations and inequalities.

Key Concepts

CONCEPT

DEFINITION

Cartesian Coordinates

A system for locating points in space using an ordered triple (x, y, z) corresponding to distances along three mutually perpendicular axes. In a right-handed system, the thumb (z-axis) points out when fingers curl from the x-axis to the y-axis.

Coordinate Planes

The three planes determined by the coordinate axes: the xy-plane (z = 0), the yz-plane (x = 0), and the xz-plane (y = 0).

Octants

The eight regions of space determined by the coordinate planes. The first octant is where all coordinates are positive.

Distance Formula in Space

An extension of the Pythagorean Theorem used to calculate the distance between two points P1(x1, y1, z1) and P2(x2, y2, z2): d = √[(x2 - x1)² + (y2 - y1)² + (z2 - z1)²].

Sphere

The set of all points in space that are a fixed distance (radius) from a fixed point (center). Its standard equation is (x - x₀)² + (y - y₀)² + (z - z₀)² = a².

Vector

A quantity that has both magnitude and direction. It is represented by a directed line segment, and in component form is written as v = ⟨v₁, v₂, v₃⟩.

Standard Position (of a Vector)

A representation of a vector with its initial point at the origin, used to uniquely define the vector’s direction and magnitude.

Example Problems

Example 1

Give a geometric description of the set of points in space whose coordinates satisfy the given pairs of equations. $$x=2, \quad y=3$$

Example 2

Give a geometric description of the set of points in space whose coordinates satisfy the given pairs of equations. $$x=-1, \quad z=0$$

Example 3

Give a geometric description of the set of points in space whose coordinates satisfy the given pairs of equations. $$y=0, \quad z=0$$

Example 4

Give a geometric description of the set of points in space whose coordinates satisfy the given pairs of equations. $$x=1, \quad y=0$$

Example 5

Give a geometric description of the set of points in space whose coordinates satisfy the given pairs of equations. $$x^{2}+y^{2}=4, \quad z=0$$
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Step-by-Step Explanations

QUESTION

How do you find the distance between P1(2, 1, 5) and P2(-2, 3, 0)?

STEP-BY-STEP ANSWER:

Step 1: Write down the coordinates for P1 and P2. P1 = (2, 1, 5) and P2 = (-2, 3, 0).
Step 2: Substitute into the distance formula: d = √[(-2 - 2)² + (3 - 1)² + (0 - 5)²].
Step 3: Compute each term: (-2 - 2)² = (-4)² = 16, (3 - 1)² = (2)² = 4, (0 - 5)² = (-5)² = 25.
Step 4: Sum them: 16 + 4 + 25 = 45.
Step 5: Take the square root: d = √45 = 3√5 (approximately 6.708).
Final Answer: The distance between P1 and P2 is 3√5.

Distance Between Two Points

QUESTION

How can you find the center and radius of the sphere given by x² + y² + z² + 3x - 4z + 1 = 0?

STEP-BY-STEP ANSWER:

Step 1: Group the x, y, and z terms: (x² + 3x) + y² + (z² - 4z) = -1.
Step 2: Complete the square for the x-terms: x² + 3x becomes (x + 3/2)² - (3/2)².
Step 3: The y-term is already a perfect square (y² = (y - 0)²).
Step 4: Complete the square for the z-terms: z² - 4z becomes (z - 2)² - 2².
Step 5: Substitute these back into the equation: (x + 3/2)² - (9/4) + y² + (z - 2)² - 4 = -1.
Step 6: Combine constants: - (9/4) - 4 = - (9/4 + 16/4) = - (25/4). Move constant to right-hand side: (x + 3/2)² + y² + (z - 2)² = -1 + 25/4 = 21/4.
Step 7: Read the center and radius from the standard form: Center = (-3/2, 0, 2) and Radius = √(21/4) = (√21)/2.
Final Answer: The center is (-3/2, 0, 2) and the radius is (√21)/2.

Equation of a Sphere by Completing the Square

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Common Mistakes

  • Confusing right-handed and left-handed coordinate systems, which can lead to incorrect orientation of axes and angles.
  • Forgetting to complete the square properly when converting a sphere's equation into standard form.
  • Mixing up the roles of coordinate planes, especially when identifying equations like x = constant or z = constant.
  • Neglecting the importance of vector direction and mistakenly equating vectors that have the same magnitude but different directions.
  • Applying the distance formula without carefully subtracting corresponding coordinates, which can result in sign errors.