Chalk Dust Company

  

General Information
Home Page
Our Company
Your Instructor
How to Use DVDs
Quick Sketches - Homeschool Students
Quick Sketches - College Students
Frequently Asked Q's
Customer Comments
Internet Warning

Product Information
Prices, Ordering & Evaluations
Textbooks & Solutions Guides
Course Outlines, Reviews & Demos

Existing Customers
Quick Start Guide
Assignments
Read Me First
Technical Support
Replacement DVD/CD
Tests and Measurement

Exhibit Schedule
Education Links


Chalk Dust Company
PMB 256
16107 Kensington Dr
Sugar Land, TX 77479-4401

800.588.7564 (USA)
281.265.2495
281.265.3197 fax

sales@chalkdust.com

Precalculus 5th Edition
Corrections and Clarifications (Program from Nov 2009)
Section 1.5, problem 59.
The problem in the solutions guide does not match the problem in the textbook. Either omit the problem or change the problem in the textbook to:

59. f(x) = sqrt(x+6), g(x) = x^2-5

Section 2.6, problems 19 and 21.
The textbook questions request (a) domain, (b) continuous, (c) asymptotes but the solutions guide gives (a) domain, (b) asymptotes, (c) graph.

Section 2.6, problem 23.
In the solutions guide, g(6) should be 10 rather than 20.

Section 2.8, on the DVD.
On the video at about the 6-minute mark, in the chart, the amount corresponding with the year 1991 should be 16.9 rather than 19.9.

Chapter 3 Review, #23.
The function in the textbook is not the same as the function in the solutions guide. Either omit the problem or use the function in the solutions guide.

Section 4.4, problem 39.
In the solutions guide, the illustration is labeled incorrectly. The negative angle should be -225 degrees and the reference angle should be 45 degrees.

Precalculus
Corrections and Clarifications (Program from Sept 2003)
Section P.3, problem 94.
In the solutions guide, the sentence..."If p = 65, x = 45 units." Should be "If p = 655, x = 45 units."

Section P.4, problems 65 and 66.
A factorization is not reasonably possible so skip the algebraic solution and solve the equation by graphing.

Section P.4, problem 74.
In the solutions guide, the technique used in factoring is not one that is known at this time. Please omit the problem.

Section P.4, problem 80.
The method for solving the system of equations algebraically is presented later in the textbook, so please omit that part of the problem.

Section P.5, problem 54(b).
Please omit this problem. To this point, students have not learned a method that can be used to factor the resulting polynomial and a method that can be used is not presented until later in the textbook.

Chapter P Review Exercises, #100.
The answers are correct but the factorization in the third step in the solutions guide is incorrect. Here are all the steps.

216x4 - x = 0
x(216x3 - 1) = 0
x(6x - 1)(36x2 + 6x + 1) = 0
x = 0,1/6

Section 1.4, #86.
In the solutions guide, replace the equal sign under the radical with a plus sign.

Chapter 1 Test, #7.
Omit this problem from the test because it contains a trick that I would not expect students to catch. The trick has to do with window settings necessary to see the curves in the graph. Here are the window settings that work.

Xmin = -1
Xmax = 1
and here's the real kicker ...
Ymin = -.1
Ymax = .1

Chapter 1 Test, #14.
Omit the problem because there is a mistake in the solutions guide and because there were no examples containing this degree of sophistication.

Chapter 1 Test, #21.
The domain should be 0 < x ≤ 25 because when x is greater than 25 it will no longer be the shortest side of the rectangle.

Section 2.4, #77.
There is a mistake beginning with the third term in the sequence. The steps below describe how to generate the third term.
Each term is generated using the idea...
(previous term)2 + (previous term)

The previous term is -3/16 + (1/4)i which I write as -3/16 + i/4 to make the calculations easier. The next term is...

(-3/16 + i/4)2 + (-3/16 + i/4)
9/256 + 2(-3/16)(i/4) + i2/16 - 3/16 + i/4
9/256 + 2(-3/16)(i/4) - 1/16 - 3/16 + i/4
Put like terms next to each other to see the collections easier...
9/256 - 1/16 - 3/16 - 3i/32 + i/4
9/256 - 4/16 - 3i/32 + i/4
9/256 - 64/256 - 3i/32 + 8i/32
-55/256 + 5i/32

Section 2.5, #40.
In the solutions guide, the exponent on the first binomial should be understood 1 rather than 2.

Chapter 2 Review Exercises, problem 5.
In the solutions guide, the sign in the binomial squared should be plus rather than minus and the first-coordinate of the vertex should be -5/2 rather than 5/2.

Section 3.2, #72.
In the solutions guide, the answer should be 21,351 ft-lb rather than 21,351 ft-16.

Section 3.5, #67.
Change the coefficient of the given formula from -1.0 to -2.5 so that the problem in the text corresponds with the solution in the solutions guide.

Chapter 3 Test, problem 1.
The problem may be counted correct if the student describes a transformation graphically in which the graph of 2^x is reflected in the y-axis, moved one unit to the left, and moved down 3 units. Those transformations seem to be indicated by the function.

However, that is not the correct interpretation. We did not study the combination of reflection in the y-axis and horizontal shift in the same problem so an incorrect interpretation is likely.

A correct interpretation is possible by manipulating the exponent from -x + 1 to -(x-1). In that form the thought process for transformations is
1. Consider 2^x.
2. 2^(x-1) moves the graph one unit to the right.
3. 2^[-(x-1)] reflects the y-axis.
4. 2^[-(x-1)] - 3 moves the graph down 3.

Cumulative Test for Chapters 1-3, #22.
The function listed should be f(x) = 2x4 + 5x3 + 5x2 + 20x - 20.

Section 4.4, #124.
The problem does not appear in the solutions manual.
Draw a right triangle and label one of the acute angles theta.
Since cos θ = 2/9, label the side adjacent to theta 2 and the hypotenuse 9. Use the Pythagorean Theorem to solve for the third side. It turns out to be the square root of 77. Label the side opposite theta the square root of 77.

sin θ = sqrt 77 / 9
tan θ = sqrt 77 / 2
cot θ = 2 / sqrt 77
sec θ = 9 / 2
csc θ = 9 / sqrt 77

Section 4.5, problem 83.
In the solutions guide, the equation listed is incorrect. The regression equation is of the form

y = a sin(bx + c) + d
where
a = 2.445783614
b = 0.4786424479
c = 1.579779881
d = 3.364812504

Section 4.8, problem 25.
In the solutions guide, the problem is solved incorrectly. The length of the hypotenuse (denominator) should be 13,800 miles rather than 4,150 miles. The angle theta should be 16.85 degrees and the angle alpha should be 73.15 degrees.

Section 5.4, problem 65.
The answer in the solutions guide is not the same as the suggested answer in the text. The answer in the solutions guide should contain t in the first fraction and 2x in the second fraction.

Section 6.3, problem 2.
In the solutions guide, the coordinates of the terminal point should be used first in the calculations of u and v.

Chapter 6, Practice Test in the Solutions Guide (page 590), #8.
The answer should read 109.442 mi. rather than 190.442 mi.

Chapter 4-6 Cumulative Test in the Solutions Guide (page 639).
The problem listed as #47 should be #46. There is no problem number 47.

Section 7.2, problem 65.
In the Solutions Guide, the second equation should begin with a factor of 3 rather than 6.

Section 7.2, problem 66.
There are actually two mistakes in the solutions guide, one mistake in each equation. In equation one, notice that both variable terms become negative in the second group of equations, but only the x-term should be negative.

The second equation is set up incorrectly. Since the airplanes are traveling in opposite directions, it is the sum of their distances which will be 3200 miles after 2 hours. Therefore, the second equation should be 2x + 1.5y = 3200.

Section 7.2, even-numbered problems from #70-78.
All of these problems are set up incorrectly in the solutions guide. Just a quick glance at the first equation in each problem reveals they are set up as differences when, in fact, they should all be sums. Please select the odd-numbered problems only in this group.

Section 7.2, problems from #90-93.
Page 598 Textbook, Advanced Applications. These problems are optional. These system of equations are from a course in differential equations, which is Calculus IV in college. Problems 90 and 92 are not supported on pages 988 and 989 in the solutions guide. But problems 91 and 93 are supported in the solutions guide on pages 348-349.

Section 7.2, problems from #94-108.
Page 508 Textbook, Review. The solutions for these textbook problems are reflected in the solutions guide as a problem number that is two less than the number in the textbook. For example, Precalculus 7.2, problem 96 in the textbook is supported as problem 94 in the solutions guide etc. This continues through textbook section 7.2, where problem 108 is supported as problem 106 in the solutions guide.

Section 7.4, problem 11.
Notice that once A and B are replaced with expressions, the two fractions are the same no matter how you decide to work it. The order of the fractions may be different, but the overall problem is the same so it doesn't matter whether A is over x, or over 2x + 1.

Section 10.6, problem 78.
Change the word "some" in the problem to "any" and in the solutions guide the answer should be true.

Cumulative Test, Chapters 7-10, problems 27 and 39.
>>problem 27. In the solutions guide, the values of A and C are listed incorrectly. The correct values are A=1 and C=2.
>>problem 39. The second vertex should be (-2, 0) rather than (2, pi).

Chapter 11 Test, problem 16.
Please omit the problem unless you have a calculator capable of 3-dimensional graphing.

Section 11.2, problems 56,58,60,62.
In the solutions guide, equal signs have been replaced with plus signs.

Section 11.3, problems 2,4,6,8.
In the solutions guide, equal signs have been replaced with plus signs.

Section 12.3, problem 39.
In the solutions guide, the answers to parts (b) and (c) should be in dollars and not billions of dollars.

All chapters.
You may use the calculator anytime for most of the problems. However, there will be times when answers will need to be in "exact form", which means that radicals should not be estimated using a calculator.

Precalculus
Corrections and Clarifications (old program through August 2003)
Chapter P (Prerequisites) Chapter Test, page 81, problem 3; difficulty with graph.
If you did not get the graph in the solutions guide then you entered the equation incorrectly into your calculator. The difficulty may be in entering the absolute value of x. The absolute value button is in the MATH menu, then press the right arrow button to highlight NUM at the top of the screen and absolute value (ABS) is item #1 in that menu. When you enter the equation in the calculator it should look like this: Y1 = 4 - (3/4) abs (x ).
To find x and y-intercepts, you can either TRACE to those points or make the calculations algebraically. Algebraically, to find x-intercepts let y = 0 and solve for x. If y = 0 the equations becomes:
0 = 4 - (3/4) [absolute value of x] You could use .75 instead of 3/4 .
-4 = - (3/4) [absolute value of x] Now divide on both sides by - (3/4).
-4 / -(3/4) = [absolute value of x] The left side is a complex fraction and - signs cancel.
16/3 = [absolute value of x] For that to be true, x can either be 16/3 or -16/3 so the coordinates of the x- intercepts are (+ or - 16/6, 0).
To find the y-intercept, let x=0 and solve for y.

Chapter P (Prerequisites) Chapter Test, problem 19
In the solutions guide the trinomial is factored incorrectly leading to an incorrect answer for "a". The correct factorization is (a-20)(a-80) and the answers are a=80 and b=20.

Section 1.3, problem 51a. The graph in the solutions guide is incorrect. The graph should be a straight line extending from the lower left to the upper right of the coordinate plane and going through the origin.

Section 2.3, problems 35 - 38. The author of the solutions guide made several mistakes which can be seen by using a degenerate form of the same idea. There are two ideas at work here and the author got them mixed up.
Here are the ideas. Consider 7/2.
1. When you divide 2 into 7 the quotient is 3 with a remainder of 1 so 7/2 = 3 1/2 which is also 3 + 1/2. Note the form is "quotient plus remainder-over-divisor."
The equivalent idea in #35 is that
(x^3 - x^2 - 14x + 11) / x - 4) = (x^2 + 3x - 2) + 3 / (x - 4)
2. The number 7, when divided by 2, is 3 with a remainder of 1 so 7 = 2(3) + 1. Note the form is "divisor times quotient plus remainder."
The equivalent idea in #35 is that
(x^3 - x^2 - 14x + 11) = (x - 4)(x^2 + 3x - 2) + 3
Notice that the left side is f(x) so the function is written as divisor times quotient plus remainder, NOT as divisor times quotient plus remainder-over-divisor.
In #1 above, you start with a fraction and rewrite the fraction as a quotient plus a remainder over the divisor.
In #2 above, you start with a number (or function) and rewrite as a product plus a remainder.

Section 2.7, page 230, problem 78. (a) To derive the equation, note that the entire round trip takes a total of 4 hours (an average of 50 miles per hour for 200 miles). Structure an overall equation with this concept:
Time (one way) + Time (other way) = Total Time
dist / rate + dist / rate = 4
100/x + 100/y = 4
100y + 100x = 4xy
25y + 25x = xy
25y - xy = -25x
y(25-x) = -25x
y = -25x / (25 - x)
y = 25x / (x - 25)
(b) The vertical asymptote is the zero of the denominator, x = 25. Since the degree of the numerator and denominator are the same, the horizontal asymptote is found by using the leading coefficients in numberator and denominator, y = 25/1 = 25.
(c) You should observe that as x gets bigger, y gets smaller.
(e) It would be impossible because 20 miles per hour one way requires more than 4 hours for that part of the trip, so no speed would allow for an average of 50 miles per hour over both parts of the trip.

Chapter 2 Review, page 234, problem 75. When the (b) part asks that you verify the equation for area, you notice that the area equation is written in terms of x only. Understanding the configuration of the triangle and that the formula for finding the area of a triangle is A = (1/2)bh, you quickly substitute x and y for b and h (because of the configuration of the triangle) to get A = (1/2)xy. Now the trick is to write Y in terms of x.
To do that you use the coordinates of pairs of points in the slope formula. Using (2,3) and the upper point (0,y) you get m = (y - 3) / (0 - 2 ). The author left out the 0 for x in the denominator. That wouldn't be so bad if it weren't for the fact that the 0 for y was included in the next fraction. Using (2,3) and the lower point (x,0) you get m = (0 - 3) / (x - 2). Next you set those two fractions equal to one another and solve for Y. Leaving out the 0's and setting the fractions equal to one another it looks like this:
(y - 3) / -2 = -3 / (x - 2) Now solve for y by first multiplying on both sides of - 2.
y - 3 = 6 / (x - 2)
y = 6 / (x - 2) + 3 now you need a common denominator
y = 6 / (x - 2) + 3 (x - 2) / (x - 2)
y = 6 / (x - 2) + (3x - 6) / (x - 2) Now bring together the numerators
y = (6 + 3x - 6) / (x - 2)
y = 3x / (x - 2 )
Now go back to the area formula and substitute:
A = (1/2) xy And replace y with the expression above
A = (1/2) x [ 3x / (x - 2)] Symplify
A = 3x^2 / [ 2 (x - 2) ]

Section 3.1, problem 11. The third funtion, h(x), should read h(x) = 1/9 (3^x), and NOT 1/9 (3x).

Section 3.3, problem 47. In the solutions guide the logarithm should have base 3, and not the natural logarithmic base e(ln).

Section 4.6, problem 15.. In the solutions guide, the graph is incorrect. The correct graph is a reflection in the x-axis of the graph given.

Section 4.8, problem 25. In the solutions guide the problem is solved incorrectly. The length of the hypotenuse should be 13,800 miles rather than 4,150 miles. The correct answer is 73.15 degrees.

Section 5.3, problem #15. When you take the square root on both sides you should get sec x = +/- (2/root3), which means that x is a reference angle of pi/6 (or 30 degrees) in ALL FOUR QUADRANTS. In the first quadrant the angle is pi/6, in the second quadrant the angle is 5pi/6, in the third quadrant the angle is 7pi/6, and in the fourth quadrant the angle is 11pi/6, along with 2pi multiples of all of those. But there is a technique to avoid the 2pi multiples because angles in opposite quadrants are pi units (or 180 degrees) from one another. The third quadrant angle of 7pi/6 can be written in terms of the first quadrant angle because it is pi/6 plus pi, and generally, all of the first and third quadrant angles can be expressed as pi/6 + n(pi). In a similar way the second and fourth quadrant angles can be expressed as 5pi/6 + n(pi). Using a general answer involving "plus 2pi" would omit the third and fourth quadrant angles. Problem 21 is similar.

Section 5.3, problem 49. The graph and the equation above the graph do not match the given equation. The correct graph is a horizontal flip of the one given.

Section 5.4, problem 39. In the solutions guide, in the first step on the right side of the equal sign, the letters u and v in the first term should be exchanged so that the expression reads, sin v cos u - sin u cos v. Note in the second term that the two factors are not written in the order corresponding with the form on page 435. However, the order of factors is not important due to the commutative property of multiplication.

Section 5.4, problem 55. The text problem should use pi/2 instead of pi/4.

Section 6.5, page 509, problems 27-30. 5 + 2i is in complex form with a = 5 and b = 2. To put the number into trig form, first calculate r and theta, with r = root (a^2 + b^2) and theta = inv -tan (b/a), from page 502. You find r = root (29) = 5.39. Theta = inv -tan (2/5) = 0.38. The answers in the text use radian values for theta so be sure to have your calculator set to radian mode before you start.

Section 6.5, page 509, problems 31-44. Problems are in trig form with r and theta easily identifiable. For each of these, be careful about the angle mode on the calculator--- set to degrees or radians for each problem.
On 31, r = 2 and theta = 150 degrees (set calculator to degree mode). >From info on page 502, calculate a = r [cos (theta)] = 2 cos (150) and b = r [ sin (theta)] = 2sin (150).
Its easy to find exact values with a reference triangle of 30 degrees in the second quadrant. Cos (150) is - (root 3) / 2 so r [cos(150)] is 2 (-root2/2) = - root3 and since sin(150) is 1/2, r [sin(150)] is 1. With a = -root3 and b = 1, the complex form of a + bi is -root3 + li or -root3 + i.
One way to draw the graphs is to draw a line segment. First, make window settings appropriate for your graph or like the ones in the solutions manual . You could use Xmin = -2.5, Xmax = 1, Ymin + -1, Ymax = 2. Press GRAPH, MORE, select DRAW, LINE. Move the cursor to the approximate coordinates for (a,b), press enter; move the cursor to the origin (as close as you can get), press enter.

Section 7.2, Problem 65. In the Solutions Guide, the second equation should begin with a factor of 3 rather than 6.

Section 10.4, page 789, problem 15. Consider the problem from the point of tan 2theta = -24/7. Using the calculator, inverse tan of -24/7 gives -73. 7397; so now the equation becomes 2theta = -73.7397. Notice that the angle is in the 4th quadrant. To get an equivalent reference angle that is positive, add 180 degrees. You get 2theta = 106.2602047 degrees. You can verify the equivalence of the angles by noticing that tan(-73.73) and tan 106.26 are both equal to -3.428, which is -24/7. Now divide on both sides of the equation by two to get theta = 53.13.
The solutions guide is using one of the "Power-Reducing Formulas" that are listed on the inside back cover of the text. Start with sin^2 u = (1 - cos2u) / 2, then take the square root on both sides and you get sin u = root [(1 - cos2u) / 2]. There is a typo in the solutions guide in that it says cos theta when it should read 2theta and they use the value of cos 2theta in the next step. Notice that the next line down when figuring cos theta, the angle is listed as 2theta. Precalculus Corrections and Clarifications continued. Once the substitution of - 7/25 is made for cos 2theta and after the sign change, under the radical you have (1 + 7/25) / 2
(25/25 + 7/25) / 2
[(32/25)/2] which means 32/25 divided by 2, or 32/25 times which becomes 16/25. The square root of 16/25 gives the 4/5.

Section 10.4 Solutions Guide, page 451, problem 15. In the fifth line of the copy, under the radical and in the numberator it ready "1 - cos theta" and it should read "1 - cos 2 theta". The next line of copy contains a similar expression in which the 2 theta is used correctly.

Chapter 10 Practice Test, Solutions Guide, page 481. In polar form, p is the distance between focus and directrix (page 817 in the textbook).

Chapter 10 Practice Test, Solutions Guide, page 544, problem 12. This is the solution to a problem given on page 481. The correction begins at the third line and continues to the end of the problem. tan theta = -1/square root of 3
theta = 5pie/6 or theta = 11pie/6
Polar Coordinates: (12, 5pie/6) or (2, 11pie/6).

Chapter 12 Review Exercises, problems 59 and 65.
Problem 59: The approximate areas for 4,8,20, and 50 rectangles should be 7.5, 6.375, 5.74, and 5.4944 respectively.
Problem 65: In the solutions guide, the second term in the second binomial of the first two steps should be i/2 rather than i/3.

All chapters. You can use the calculator anytime for most problems. However, there will be times when answers will need to be in "exact form", which means that radicals should not be estimated using a calculator.

Section 12.3, Problem 33. In the solutions guide, the second to last line contains an error. The factor listed as x subscript 3 should be (x-3).

TOP OF THE PAGE

Copyright © 1996-2011 Chalk Dust Company
Website design and hosting by Mediawest Online.