APPENDIXES. B Coordinate Geometry and Lines C. D Trigonometry E F. G The Logarithm Defined as an Integral H Complex Numbers I

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1 APPENDIXES A Numbers, Inequalities, and Absolute Values B Coordinate Geometr and Lines C Graphs of Second-Degree Equations D Trigonometr E F Sigma Notation Proofs of Theorems G The Logarithm Defined as an Integral H Comple Numbers I Answers to Odd-Numbered Eercises A

2 A APPENDIX A NUMBERS, INEQUALITIES, AND ABSOLUTE VALUES A NUMBERS, INEQUALITIES, AND ABSOLUTE VALUES Calculus is based on the real number sstem. We start with the integers:...,,,,,,,,,... Then we construct the rational numbers, which are ratios of integers. Thus an rational number r can be epressed as r m n where m and n are integers and n Eamples are (Recall that division b is alwas ruled out, so epressions like and are undefined.) Some real numbers, such as s, can t be epressed as a ratio of integers and are therefore called irrational numbers. It can be shown, with varing degrees of difficult, that the following are also irrational numbers: s s5 s sin log The set of all real numbers is usuall denoted b the smbol. When we use the word number without qualification, we mean real number. Ever number has a decimal representation. If the number is rational, then the corresponding decimal is repeating. For eample, (The bar indicates that the sequence of digits repeats forever.) On the other hand, if the number is irrational, the decimal is nonrepeating: s If we stop the decimal epansion of an number at a certain place, we get an approimation to the number. For instance, we can write.5965 where the smbol is read is approimatel equal to. The more decimal places we retain, the better the approimation we get. The real numbers can be represented b points on a line as in Figure. The positive direction (to the right) is indicated b an arrow. We choose an arbitrar reference point O, called the origin, which corresponds to the real number. Given an convenient unit of measurement, each positive number is represented b the point on the line a distance of units to the right of the origin, and each negative number is represented b the point units to the left of the origin. Thus ever real number is represented b a point on the line, and ever point P on the line corresponds to eactl one real number. The number associated with the point P is called the coordinate of P and the line is then called a coor-

3 APPENDIX A NUMBERS, INEQUALITIES, AND ABSOLUTE VALUES A dinate line, or a real number line, or simpl a real line. Often we identif the point with its coordinate and think of a number as being a point on the real line. _.6 _ 7 œ π FIGURE _ The real numbers are ordered. We sa a is less than b and write a b if b a is a positive number. Geometricall this means that a lies to the left of b on the number line. (Equivalentl, we sa b is greater than a and write b a.) The smbol a b (or b a) means that either a b or a b and is read a is less than or equal to b. For instance, the following are true inequalities: s s In what follows we need to use set notation. A set is a collection of objects, and these objects are called the elements of the set. If S is a set, the notation a S means that a is an element of S, and a S means that a is not an element of S. For eample, if Z represents the set of integers, then Z but. If S and T are sets, then their union S T is the set consisting of all elements that are in S or T (or in both S and T ). The intersection of S and T is the set S T consisting of all elements that are in both S and T. In other words, S T is the common part of S and T. The empt set, denoted b, is the set that contains no element. Some sets can be described b listing their elements between braces. For instance, the set A consisting of all positive integers less than 7 can be written as Z A,,,, 5, 6 We could also write A in set-builder notation as A is an integer and 7 which is read A is the set of such that is an integer and 7. a FIGURE Open interval (a, b) a FIGURE Closed interval [a, b] b b INTERVALS Certain sets of real numbers, called intervals, occur frequentl in calculus and correspond geometricall to line segments. For eample, if a b, the open interval from a to b consists of all numbers between a and b and is denoted b the smbol a, b. Using set-builder notation, we can write a, b a b Notice that the endpoints of the interval namel, a and b are ecluded. This is indicated b the round brackets and b the open dots in Figure. The closed interval from a to b is the set a, b a b Here the endpoints of the interval are included. This is indicated b the square brackets and b the solid dots in Figure. It is also possible to include onl one endpoint in an interval, as shown in Table.

4 A APPENDIX A NUMBERS, INEQUALITIES, AND ABSOLUTE VALUES TABLE OF INTERVALS N Table lists the nine possible tpes of intervals. When these intervals are discussed, it is alwas assumed that a b. Notation Set description Picture a, b a, b a, b a, b a, a,, b, b, a b a b a b a b a a b b (set of all real numbers) a a a a a a b b b b b b We also need to consider infinite intervals such as a, a This does not mean that ( infinit ) is a number. The notation a, stands for the set of all numbers that are greater than a, so the smbol simpl indicates that the interval etends indefinitel far in the positive direction. INEQUALITIES When working with inequalities, note the following rules. RULES FOR INEQUALITIES. If a b, then a c b c.. If a b and c d, then a c b d.. If a b and c, then ac bc.. If a b and c, then ac bc. 5. If a b, then a b. Rule sas that we can add an number to both sides of an inequalit, and Rule sas that two inequalities can be added. However, we have to be careful with multiplication. Rule sas that we can multipl both sides of an inequalit b a positive number, but Rule sas that if we multipl both sides of an inequalit b a negative number, then we reverse the direction of the inequalit. For eample, if we take the inequalit 5 and multipl b, we get 6, but if we multipl b, we get 6. Finall, Rule 5 sas that if we take reciprocals, then we reverse the direction of an inequalit (provided the numbers are positive). EXAMPLE Solve the inequalit 7 5. SOLUTION The given inequalit is satisfied b some values of but not b others. To solve an inequalit means to determine the set of numbers for which the inequalit is true. This is called the solution set.

5 APPENDIX A NUMBERS, INEQUALITIES, AND ABSOLUTE VALUES A5 First we subtract from each side of the inequalit (using Rule with c ): 7 Then we subtract 7 from both sides (Rule with c 7): 6 Now we divide both sides b 6 (Rule with c 6 ): These steps can all be reversed, so the solution set consists of all numbers greater than. In other words, the solution of the inequalit is the interval (, ). M 6 EXAMPLE Solve the inequalities. SOLUTION Here the solution set consists of all values of that satisf both inequalities. Using the rules given in (), we see that the following inequalities are equivalent: (add ) (divide b ) Therefore the solution set is, 5. M EXAMPLE Solve the inequalit 5 6. SOLUTION First we factor the left side: We know that the corresponding equation has the solutions and. The numbers and divide the real line into three intervals:,,, N A visual method for solving Eample is to use a graphing device to graph the parabola 5 6 (as in Figure ) and observe that the curve lies on or below the -ais when. = -5+6 On each of these intervals we determine the signs of the factors. For instance,, Then we record these signs in the following chart: Interval?? Another method for obtaining the information in the chart is to use test values. For instance, if we use the test value for the interval,, then substitution in 5 6 gives FIGURE 5 6

6 A6 APPENDIX A NUMBERS, INEQUALITIES, AND ABSOLUTE VALUES FIGURE The polnomial 5 6 doesn t change sign inside an of the three intervals, so we conclude that it is positive on,. Then we read from the chart that is negative when. Thus the solution of the inequalit is Notice that we have included the endpoints and because we are looking for values of such that the product is either negative or ero. The solution is illustrated in Figure 5. M EXAMPLE Solve., SOLUTION First we take all nonero terms to one side of the inequalit sign and factor the resulting epression: or As in Eample we solve the corresponding equation and use the solutions,, and to divide the real line into four intervals,,,,,, and,. On each interval the product keeps a constant sign as shown in the following chart: Interval Then we read from the chart that the solution set is _ or,, FIGURE 6 The solution is illustrated in Figure 6. M ABSOLUTE VALUE The absolute value of a number a, denoted b, is the distance from a to on the real number line. Distances are alwas positive or, so we have a a for ever number a For eample, s s In general, we have N Remember that if a is negative, then a is positive. a a a if a a if a

7 APPENDIX A NUMBERS, INEQUALITIES, AND ABSOLUTE VALUES A7 EXAMPLE 5 Epress SOLUTION without using the absolute-value smbol. if if if if M Recall that the smbol s means the positive square root of. Thus sr s means s r and s. Therefore, the equation sa a is not alwas true. It is true onl when a. If a, then a, so we have sa a. In view of (), we then have the equation sa a which is true for all values of a. Hints for the proofs of the following properties are given in the eercises. 5 PROPERTIES OF ABSOLUTE VALUES Suppose a and b are an real numbers and n is an integer. Then.. a b. a n a n b a ab a b b For solving equations or inequalities involving absolute values, it s often ver helpful to use the following statements. _a FIGURE 7 b a a a a a-b a-b FIGURE 8 Length of a line segment= a-b a b 6 Suppose a. Then. a if and onl if 5. if and onl if a a 6. if and onl if a or a For instance, the inequalit sas that the distance from to the origin is less than a, and ou can see from Figure 7 that this is true if and onl if lies between a and a. If a and b are an real numbers, then the distance between a and b is the absolute value of the difference, namel,, which is also equal to. (See Figure 8.) EXAMPLE 6 Solve. SOLUTION B Propert of (6), a b a a a a 5 5 is equivalent to b a 5 or 5 So 8 or. Thus or. M

8 A8 APPENDIX A NUMBERS, INEQUALITIES, AND ABSOLUTE VALUES EXAMPLE 7 Solve. SOLUTION B Propert 5 of (6), 5 5 is equivalent to 5 Therefore, adding 5 to each side, we have 5 7 FIGURE 9 7 and the solution set is the open interval, 7. SOLUTION Geometricall the solution set consists of all numbers whose distance from 5 is less than. From Figure 9 we see that this is the interval, 7. M EXAMPLE 8 Solve. SOLUTION B Properties and 6 of (6), is equivalent to or In the first case, which gives. In the second case 6, which gives. So the solution set is { or }, [, ) M Another important propert of absolute value, called the Triangle Inequalit, is used frequentl not onl in calculus but throughout mathematics in general. 7 THE TRIANGLE INEQUALITY If a and b are an real numbers, then a b a b Observe that if the numbers a and b are both positive or both negative, then the two sides in the Triangle Inequalit are actuall equal. But if a and b have opposite signs, the left side involves a subtraction and the right side does not. This makes the Triangle Inequalit seem reasonable, but we can prove it as follows. Notice that is alwas true because a equals either or. The corresponding statement for b is Adding these inequalities, we get If we now appl Properties and 5 (with replaced b a b and a b ), we obtain which is what we wanted to show. a a a a a b b b ( a b ) a b a b a b a b a b

9 APPENDIX A NUMBERS, INEQUALITIES, AND ABSOLUTE VALUES A9. 7. EXAMPLE 9 If and, use the Triangle Inequalit to estimate. SOLUTION In order to use the given information, we use the Triangle Inequalit with a and b 7: Thus. M A EXERCISES Rewrite the epression without using the absolute value smbol s if 8. if Solve the inequalit in terms of intervals and illustrate the solution set on the real number line The relationship between the Celsius and Fahrenheit temperature scales is given b C 5 9 F, where C is the temper- ature in degrees Celsius and F is the temperature in degrees Fahrenheit. What interval on the Celsius scale corresponds to the temperature range 5 F 95?. Use the relationship between C and F given in Eercise 9 to find the interval on the Fahrenheit scale corresponding to the temperature range C.. As dr air moves upward, it epands and in so doing cools at a rate of about C for each -m rise, up to about km. (a) If the ground temperature is C, write a formula for the temperature at height h. (b) What range of temperature can be epected if a plane takes off and reaches a maimum height of 5 km?. If a ball is thrown upward from the top of a building 8 ft high with an initial velocit of 6 fts, then the height h above the ground t seconds later will be During what time interval will the ball be at least ft above the ground? 6 Solve the equation for Solve the inequalit. h 8 6t 6t

10 A APPENDIX B COORDINATE GEOMETRY AND LINES Solve for, assuming a, b, and c are positive constants. 57. ab c bc 58. a b c a 59 6 Solve for, assuming a, b, and c are negative constants. 59. a b c Suppose that and. Use the Triangle Inequalit to show that. 6. Show that if, then. 6. Show that if a b, then a a b b. 6. Use Rule to prove Rule 5 of (). a b c b. 5.5 ab a b 65. Prove that. [Hint: Use Equation.] 66. Prove that a. b a b 67. Show that if a b, then a b. 68. Prove that. [Hint: Use the Triangle Inequalit with a and b.] 69. Show that the sum, difference, and product of rational numbers are rational numbers. 7. (a) Is the sum of two irrational numbers alwas an irrational number? (b) Is the product of two irrational numbers alwas an irrational number? B COORDINATE GEOMETRY AND LINES Just as the points on a line can be identified with real numbers b assigning them coordinates, as described in Appendi A, so the points in a plane can be identified with ordered pairs of real numbers. We start b drawing two perpendicular coordinate lines that intersect at the origin O on each line. Usuall one line is horiontal with positive direction to the right and is called the -ais; the other line is vertical with positive direction upward and is called the -ais. An point P in the plane can be located b a unique ordered pair of numbers as follows. Draw lines through P perpendicular to the - and -aes. These lines intersect the aes in points with coordinates a and b as shown in Figure. Then the point P is assigned the ordered pair a, b. The first number a is called the -coordinate of P; the second number b is called the -coordinate of P. We sa that P is the point with coordinates a, b, and we denote the point b the smbol Pa, b. Several points are labeled with their coordinates in Figure. II b _ O _ III _ I 5 IV a P(a, b) (_, ) _ 5 _ (_, _) _ (, ) (, _) (5, ) FIGURE FIGURE B reversing the preceding process we can start with an ordered pair a, b and arrive at the corresponding point P. Often we identif the point P with the ordered pair a, b and refer to the point a, b. [Although the notation used for an open interval a, b is the

11 APPENDIX B COORDINATE GEOMETRY AND LINES A same as the notation used for a point a, b, ou will be able to tell from the contet which meaning is intended.] This coordinate sstem is called the rectangular coordinate sstem or the Cartesian coordinate sstem in honor of the French mathematician René Descartes (596 65), even though another Frenchman, Pierre Fermat (6 665), invented the principles of analtic geometr at about the same time as Descartes. The plane supplied with this coordinate sstem is called the coordinate plane or the Cartesian plane and is denoted b. The - and -aes are called the coordinate aes and divide the Cartesian plane into four quadrants, which are labeled I, II, III, and IV in Figure. Notice that the first quadrant consists of those points whose - and -coordinates are both positive. EXAMPLE Describe and sketch the regions given b the following sets. (a) (b), (c ) {, } SOLUTION (a) The points whose -coordinates are or positive lie on the -ais or to the right of it as indicated b the shaded region in Figure (a). = = =_ FIGURE (a) (b) = (c) < fi P (, fi) P (, ) - fi- P (, ) (b) The set of all points with -coordinate is a horiontal line one unit above the -ais [see Figure (b)]. (c) Recall from Appendi A that if and onl if The given region consists of those points in the plane whose -coordinates lie between and. Thus the region consists of all points that lie between (but not on) the horiontal lines and. [These lines are shown as dashed lines in Figure (c) to indicate that the points on these lines don t lie in the set.] Recall from Appendi A that the distance between points a and b on a number line is a b b a. Thus the distance between points P, and P, on a horiontal line must be and the distance between and on a vertical line must be P, P,. (See Figure.) To find the distance P P between an two points P, and P,, we note that triangle P P P in Figure is a right triangle, and so b the Pthagorean Theorem we have P P s P P P P s M FIGURE s

12 A APPENDIX B COORDINATE GEOMETRY AND LINES DISTANCE FORMULA The distance between the points P, and P, is P P s EXAMPLE The distance between, and 5, is s5 s 5 s M LINES We want to find an equation of a given line L; such an equation is satisfied b the coordinates of the points on L and b no other point. To find the equation of L we use its slope, which is a measure of the steepness of the line. P (, ) P (, ) L Î=fi- =rise DEFINITION The slope of a nonvertical line that passes through the points P, and P, is m FIGURE 5 Î= - =run m=5 m= m= m= m= The slope of a vertical line is not defined. Thus the slope of a line is the ratio of the change in,, to the change in,. (See Figure 5.) The slope is therefore the rate of change of with respect to. The fact that the line is straight means that the rate of change is constant. Figure 6 shows several lines labeled with their slopes. Notice that lines with positive slope slant upward to the right, whereas lines with negative slope slant downward to the right. Notice also that the steepest lines are the ones for which the absolute value of the slope is largest, and a horiontal line has slope. Now let s find an equation of the line that passes through a given point P, and has slope m. A point P, with lies on this line if and onl if the slope of the line through P and P is equal to m; that is, FIGURE 6 m=_ m=_ m=_5 m=_ This equation can be rewritten in the form m m and we observe that this equation is also satisfied when and. Therefore it is an equation of the given line. POINT-SLOPE FORM OF THE EQUATION OF A LINE An equation of the line passing through the point P, and having slope m is m

13 APPENDIX B COORDINATE GEOMETRY AND LINES A EXAMPLE Find an equation of the line through, 7 with slope. SOLUTION Using with m,, and 7, we obtain an equation of the line as 7 which we can rewrite as or M EXAMPLE Find an equation of the line through the points, and,. SOLUTION B Definition the slope of the line is m Using the point-slope form with and, we obtain which simplifies to M b =m+b Suppose a nonvertical line has slope m and -intercept b. (See Figure 7.) This means it intersects the -ais at the point, b, so the point-slope form of the equation of the line, with and b, becomes b m This simplifies as follows. FIGURE 7 SLOPE-INTERCEPT FORM OF THE EQUATION OF A LINE with slope m and -intercept b is m b An equation of the line b =a =b In particular, if a line is horiontal, its slope is m, so its equation is b, where b is the -intercept (see Figure 8). A vertical line does not have a slope, but we can write its equation as a, where a is the -intercept, because the -coordinate of ever point on the line is a. Observe that the equation of ever line can be written in the form a 5 A B C FIGURE 8 because a vertical line has the equation a or a ( A, B, C a) and a nonvertical line has the equation m b or m b ( A m, B, C b). Conversel, if we start with a general first-degree equation, that is, an equation of the form (5), where A, B, and C are constants and A and B are not both, then we can show that it is the equation of a line. If B, the equation becomes A C or CA, which represents a vertical line with -intercept CA. If B, the equation

14 A APPENDIX B COORDINATE GEOMETRY AND LINES can be rewritten b solving for : A B C B (5, ) (, _) FIGURE 9-5=5 and we recognie this as being the slope-intercept form of the equation of a line ( m AB, b CB ). Therefore an equation of the form (5) is called a linear equation or the general equation of a line. For brevit, we often refer to the line A B C instead of the line whose equation is A B C. EXAMPLE 5 Sketch the graph of the equation 5 5. SOLUTION Since the equation is linear, its graph is a line. To draw the graph, we can simpl find two points on the line. It s easiest to find the intercepts. Substituting (the equation of the -ais) in the given equation, we get 5, so 5 is the -intercept. Substituting in the equation, we see that the -intercept is. This allows us to sketch the graph as in Figure 9. M EXAMPLE 6 Graph the inequalit 5. SOLUTION We are asked to sketch the graph of the set solving the inequalit for :, 5 and we do so b.5 =_ + FIGURE Compare this inequalit with the equation, which represents a line with 5 slope 5 and -intercept. We see that the given graph consists of points whose -coordinates are larger than those on the line 5. Thus the graph is the region that lies above the line, as illustrated in Figure. M PARALLEL AND PERPENDICULAR LINES Slopes can be used to show that lines are parallel or perpendicular. The following facts are proved, for instance, in Precalculus: Mathematics for Calculus, Fifth Edition b Stewart, Redlin, and Watson (Thomson BrooksCole, Belmont, CA, 6). 6 PARALLEL AND PERPENDICULAR LINES. Two nonvertical lines are parallel if and onl if the have the same slope.. Two lines with slopes m and m are perpendicular if and onl if m m ; that is, their slopes are negative reciprocals: m m EXAMPLE 7 Find an equation of the line through the point 5, that is parallel to the line 6 5. SOLUTION The given line can be written in the form 5 6

15 APPENDIX B COORDINATE GEOMETRY AND LINES A5 which is in slope-intercept form with m. Parallel lines have the same slope, so the required line has slope and its equation in point-slope form is 5 We can write this equation as 6. M EXAMPLE 8 Show that the lines and 6 are perpendicular. SOLUTION The equations can be written as from which we see that the slopes are m and and m Since m m, the lines are perpendicular. M B EXERCISES 6 Find the distance between the points..,,, 5.,,. 6,,,., 6, 5., 5,, 7 6. a, b, 7 Find the slope of the line through P and Q. 7. P, 5, Q, 8. P, 6, 9. P,, Q, 6. P,,. Show that the triangle with vertices A,, B,, and C, is isosceles.. (a) Show that the triangle with vertices A6, 7, B,, and C, is a right triangle using the converse of the Pthagorean Theorem. (b) Use slopes to show that ABC is a right triangle. (c) Find the area of the triangle.. Show that the points, 9,, 6,,, and 5, are the vertices of a square.. (a) Show that the points A,, B,, and C5, 5 are collinear (lie on the same line) b showing that AB BC AC. (b) Use slopes to show that A, B, and C are collinear. 5. Show that A,, B7,, C5,, and D, 7 are vertices of a parallelogram. 6. Show that A,, B,, C, 8, and D, 6 are vertices of a rectangle. 7 Sketch the graph of the equation , 7, b, a Q, Q6, Find an equation of the line that satisfies the given conditions.. Through,, slope 6. Through,, slope. Through, 7, slope. Through, 5, slope 5. Through, and, 6 6. Through, and, 7. Slope, -intercept 5 8. Slope, -intercept 9. -intercept, -intercept. -intercept 8, -intercept 6. Through, 5, parallel to the -ais. Through, 5, parallel to the -ais. Through, 6, parallel to the line 6. -intercept 6, parallel to the line 5. Through,, perpendicular to the line 5 8 (, ) 6. Through, perpendicular to the line 8 7 Find the slope and -intercept of the line and draw its graph

16 A6 APPENDIX C GRAPHS OF SECOND-DEGREE EQUATIONS ,, 5 Sketch the region in the -plane..,. 5., {, } {, and }, and,, {, } and 5. Find a point on the -ais that is equidistant from 5, 5 and,. 5. Show that the midpoint of the line segment from P, to P, is, 55. Find the midpoint of the line segment joining the given points. (a), and 7, 5 (b), 6 and 8, 56. Find the lengths of the medians of the triangle with vertices A,, B, 6, and C8,. (A median is a line segment from a verte to the midpoint of the opposite side.) 57. Show that the lines and 6 are not parallel and find their point of intersection. 58. Show that the lines 5 9 and 6 5 are perpendicular and find their point of intersection. 59. Find an equation of the perpendicular bisector of the line segment joining the points A, and B7,. 6. (a) Find equations for the sides of the triangle with vertices P,, Q,, and R, 6. (b) Find equations for the medians of this triangle. Where do the intersect? 6. (a) Show that if the - and -intercepts of a line are nonero numbers a and b, then the equation of the line can be put in the form a b This equation is called the two-intercept form of an equation of a line. (b) Use part (a) to find an equation of the line whose -intercept is 6 and whose -intercept is A car leaves Detroit at : PM, traveling at a constant speed west along I-96. It passes Ann Arbor, mi from Detroit, at :5 PM. (a) Epress the distance traveled in terms of the time elapsed. (b) Draw the graph of the equation in part (a). (c) What is the slope of this line? What does it represent? C GRAPHS OF SECOND-DEGREE EQUATIONS In Appendi B we saw that a first-degree, or linear, equation A B C represents a line. In this section we discuss second-degree equations such as 9 which represent a circle, a parabola, an ellipse, and a hperbola, respectivel. The graph of such an equation in and is the set of all points, that satisf the equation; it gives a visual representation of the equation. Conversel, given a curve in the -plane, we ma have to find an equation that represents it, that is, an equation satisfied b the coordinates of the points on the curve and b no other point. This is the other half of the basic principle of analtic geometr as formulated b Descartes and Fermat. The idea is that if a geometric curve can be represented b an algebraic equation, then the rules of algebra can be used to anale the geometric problem. CIRCLES As an eample of this tpe of problem, let s find an equation of the circle with radius r and center h, k. B definition, the circle is the set of all points P, whose distance from

17 APPENDIX C GRAPHS OF SECOND-DEGREE EQUATIONS A7 r P(, ) the center Ch, k is r. (See Figure.) Thus P is on the circle if and onl if. From the distance formula, we have s h k r PC r C(h, k) or equivalentl, squaring both sides, we get h k r FIGURE This is the desired equation. EQUATION OF A CIRCLE An equation of the circle with center h, k and radius r is h k r In particular, if the center is the origin,, the equation is r EXAMPLE Find an equation of the circle with radius and center, 5. SOLUTION From Equation with r, h, and k 5, we obtain 5 9 M EXAMPLE Sketch the graph of the equation 6 7 b first showing that it represents a circle and then finding its center and radius. SOLUTION We first group the -terms and -terms as follows: 6 7 Then we complete the square within each grouping, adding the appropriate constants to both sides of the equation: or Comparing this equation with the standard equation of a circle (), we see that h, k, and r s, so the given equation represents a circle with center, and radius s. It is sketched in Figure. (_, ) FIGURE = M

18 A8 APPENDIX C GRAPHS OF SECOND-DEGREE EQUATIONS PARABOLAS The geometric properties of parabolas are reviewed in Section.5. Here we regard a parabola as a graph of an equation of the form a b c. EXAMPLE Draw the graph of the parabola. SOLUTION We set up a table of values, plot points, and join them b a smooth curve to obtain the graph in Figure. 9 = FIGURE. M = = = Figure shows the graphs of several parabolas with equations of the form a for various values of the number a. In each case the verte, the point where the parabola changes direction, is the origin. We see that the parabola a opens upward if a and downward if a (as in Figure 5). (_, ) (, ) =_ =_ =_ (a) =a, a> (b) =a, a< FIGURE FIGURE 5 Notice that if, satisfies a, then so does,. This corresponds to the geometric fact that if the right half of the graph is reflected about the -ais, then the left half of the graph is obtained. We sa that the graph is smmetric with respect to the -ais. The graph of an equation is smmetric with respect to the -ais if the equation is unchanged when is replaced b. If we interchange and in the equation a, the result is a, which also represents a parabola. (Interchanging and amounts to reflecting about the diagonal line.) The parabola a opens to the right if a and to the left if a. (See

19 APPENDIX C GRAPHS OF SECOND-DEGREE EQUATIONS A9 Figure 6.) This time the parabola is smmetric with respect to the -ais because if, satisfies a, then so does,. FIGURE 6 (a) =a, a> (b) =a, a< The graph of an equation is smmetric with respect to the -ais if the equation is unchanged when is replaced b. = =- (, ) EXAMPLE Sketch the region bounded b the parabola and the line. SOLUTION First we find the points of intersection b solving the two equations. Substituting into the equation, we get, which gives FIGURE 7 (, _) so or. Thus the points of intersection are, and,, and we draw the line passing through these points. We then sketch the parabola b referring to Figure 6(a) and having the parabola pass through, and,. The region bounded b and means the finite region whose boundaries are these curves. It is sketched in Figure 7. M ELLIPSES The curve with equation a b (, b) where a and b are positive numbers, is called an ellipse in standard position. (Geometric properties of ellipses are discussed in Section.5.) Observe that Equation is unchanged if is replaced b or is replaced b, so the ellipse is smmetric with respect to both aes. As a further aid to sketching the ellipse, we find its intercepts. (_a, ) FIGURE 8 + = a@ b@ (, _b) (a, ) The -intercepts of a graph are the -coordinates of the points where the graph intersects the -ais. The are found b setting in the equation of the graph. The -intercepts are the -coordinates of the points where the graph intersects the -ais. The are found b setting in its equation. If we set in Equation, we get a and so the -intercepts are a. Setting, we get b, so the -intercepts are b. Using this information, together with smmetr, we sketch the ellipse in Figure 8. If a b, the ellipse is a circle with radius a.

20 A APPENDIX C GRAPHS OF SECOND-DEGREE EQUATIONS EXAMPLE 5 Sketch the graph of 9 6. SOLUTION We divide both sides of the equation b : 6 9 The equation is now in the standard form for an ellipse (), so we have a 6, b 9, a, and b. The -intercepts are ; the -intercepts are. The graph is sketched in Figure 9. (, ) (_, ) (, ) FIGURE = (, _) M HYPERBOLAS The curve with equation b =_ a b = a a b (_a, ) (a, ) is called a hperbola in standard position. Again, Equation is unchanged when is replaced b or is replaced b, so the hperbola is smmetric with respect to both aes. To find the -intercepts we set and obtain a and a. However, if we put in Equation, we get b, which is impossible, so there is no -inter- cept. In fact, from Equation we obtain FIGURE The hperbola - = a@ b@ =_ a b FIGURE The hperbola - = a@ b@ (, a) (, _a) = a b a b s a which shows that a and so. Therefore we have a or a. This means that the hperbola consists of two parts, called its branches. It is sketched in Figure. In drawing a hperbola it is useful to draw first its asmptotes, which are the lines ba and ba shown in Figure. Both branches of the hperbola approach the asmptotes; that is, the come arbitraril close to the asmptotes. This involves the idea of a limit, which is discussed in Chapter. (See also Eercise 55 in Section.5.) B interchanging the roles of and we get an equation of the form a b which also represents a hperbola and is sketched in Figure.

21 APPENDIX C GRAPHS OF SECOND-DEGREE EQUATIONS A EXAMPLE 6 Sketch the curve 9 6. SOLUTION Dividing both sides b 6, we obtain 9 which is the standard form of the equation of a hperbola (Equation ). Since a, the -intercepts are. Since b 9, we have b and the asmptotes are ( ). The hperbola is sketched in Figure. =_ = (_, ) (, ) FIGURE The hperbola 9 - =6 M If b a, a hperbola has the equation a (or a ) and is called an equilateral hperbola [see Figure (a)]. Its asmptotes are, which are perpendicular. If an equilateral hperbola is rotated b 5, the asmptotes become the - and -aes, and it can be shown that the new equation of the hperbola is k, where k is a constant [see Figure (b)]. =_ = FIGURE Equilateral hperbolas (a) - =a@ (b) =k (k>) SHIFTED CONICS Recall that an equation of the circle with center the origin and radius r is r, but if the center is the point h, k, then the equation of the circle becomes h k r Similarl, if we take the ellipse with equation a b

22 A APPENDIX C GRAPHS OF SECOND-DEGREE EQUATIONS and translate it (shift it) so that its center is the point h, k, then its equation becomes 5 h a k b (See Figure.) b (h, k) (-h)@ a@ (-k)@ + = b@ a@ + b@ = a (, ) b k FIGURE (, ) a (-h, -k) h Notice that in shifting the ellipse, we replaced b h and b k in Equation to obtain Equation 5. We use the same procedure to shift the parabola a so that its verte (the origin) becomes the point h, k as in Figure 5. Replacing b h and b k, we see that the new equation is k a h or a h k =a (h, k) =a(-h)@+k FIGURE 5 EXAMPLE 7 Sketch the graph of the equation. SOLUTION First we complete the square: In this form we see that the equation represents the parabola obtained b shifting so that its verte is at the point,. The graph is sketched in Figure 6. FIGURE 6 = -+ (, _) M

23 APPENDIX C GRAPHS OF SECOND-DEGREE EQUATIONS A EXAMPLE 8 Sketch the curve. SOLUTION This time we start with the parabola (as in Figure 6 with a ) and shift one unit to the right to get the graph of. (See Figure 7.) FIGURE 7 (a) =_ (b) =- M C EXERCISES Find an equation of a circle that satisfies the given conditions.. Center,, radius 5. Center, 8, radius. Center at the origin, passes through, 7. Center, 5, passes through, Show that the equation represents a circle and find the center and radius Under what condition on the coefficients a, b, and c does the equation a b c represent a circle? When that condition is satisfied, find the center and radius of the circle. Identif the tpe of curve and sketch the graph. Do not plot points. Just use the standard graphs given in Figures 5, 6, 8,, and and shift if necessar Sketch the region bounded b the curves..,., 5. Find an equation of the parabola with verte, that passes through the points, and,. 6. Find an equation of the ellipse with center at the origin that passes through the points (, s ) and (, 5s5 ). 7 Sketch the graph of the set. 7., 8., 9.,.,

24 A APPENDIX D TRIGONOMETRY D TRIGONOMETRY ANGLES Angles can be measured in degrees or in radians (abbreviated as rad). The angle given b a complete revolution contains 6, which is the same as rad. Therefore rad 8 and rad rad.7 rad EXAMPLE (a) Find the radian measure of 6. (b) Epress 5 rad in degrees. SOLUTION (a) From Equation or we see that to convert from degrees to radians we multipl b 8. Therefore rad (b) To convert from radians to degrees we multipl b 8. Thus 5 5 rad 8 5 M In calculus we use radians to measure angles ecept when otherwise indicated. The following table gives the correspondence between degree and radian measures of some common angles. Degrees Radians r FIGURE r a Figure shows a sector of a circle with central angle and radius r subtending an arc with length a. Since the length of the arc is proportional to the sie of the angle, and since the entire circle has circumference r and central angle, we have Solving this equation for and for a, we obtain a r a r a r Remember that Equations are valid onl when is measured in radians.

25 APPENDIX D TRIGONOMETRY A5 r rad r FIGURE r In particular, putting a r in Equation, we see that an angle of rad is the angle subtended at the center of a circle b an arc equal in length to the radius of the circle (see Figure ). EXAMPLE (a) If the radius of a circle is 5 cm, what angle is subtended b an arc of 6 cm? (b) If a circle has radius cm, what is the length of an arc subtended b a central angle of 8 rad? SOLUTION (a) Using Equation with a 6 and r 5, we see that the angle is 6 5. rad 8 (b) With r cm and rad, the arc length is a r cm M The standard position of an angle occurs when we place its verte at the origin of a coordinate sstem and its initial side on the positive -ais as in Figure. A positive angle is obtained b rotating the initial side counterclockwise until it coincides with the terminal side. Likewise, negative angles are obtained b clockwise rotation as in Figure. initial side terminal side initial side terminal side FIGURE FIGURE < Figure 5 shows several eamples of angles in standard position. Notice that different angles can have the same terminal side. For instance, the angles, 5, and have the same initial and terminal sides because 5 and rad represents a complete revolution. FIGURE 5 Angles in standard position = =_ π = π =_ 5π = π

26 A6 APPENDIX D TRIGONOMETRY THE TRIGONOMETRIC FUNCTIONS For an acute angle the si trigonometric functions are defined as ratios of lengths of sides of a right triangle as follows (see Figure 6). hpotenuse opposite sin opp hp csc hp opp adjacent FIGURE 6 cos tan adj hp opp adj sec cot hp adj adj opp P(, ) This definition doesn t appl to obtuse or negative angles, so for a general angle in standard position we let P, be an point on the terminal side of and we let r be the distance as in Figure 7. Then we define 5 OP sin r csc r r cos r sec r O FIGURE 7 tan cot Since division b is not defined, tan and sec are undefined when and csc and cot are undefined when. Notice that the definitions in () and (5) are consistent when is an acute angle. If is a number, the convention is that sin means the sine of the angle whose radian measure is. For eample, the epression sin implies that we are dealing with an angle of rad. When finding a calculator approimation to this number, we must remember to set our calculator in radian mode, and then we obtain sin. œ π FIGURE 8 π π π 6 œ If we want to know the sine of the angle we would write sin and, with our calculator in degree mode, we find that sin.5 The eact trigonometric ratios for certain angles can be read from the triangles in Figure 8. For instance, sin cos s s sin cos 6 6 s sin cos s tan tan 6 s tan s

27 APPENDIX D TRIGONOMETRY A7 sin > all ratios> The signs of the trigonometric functions for angles in each of the four quadrants can be remembered b means of the rule All Students Take Calculus shown in Figure 9. tan > FIGURE 9 cos > EXAMPLE Find the eact trigonometric ratios for. SOLUTION From Figure we see that a point on the terminal line for P(, s ). Therefore, taking s in the definitions of the trigonometric ratios, we have r is P {_, œ } sin s cos tan s œ FIGURE π π csc sec cot s s The following table gives some values of sin and cos found b the method of Eample. M sin cos s s s s s s s s 5 FIGURE 6 =œ EXAMPLE If cos 5 and, find the other five trigonometric functions of. SOLUTION Since cos 5, we can label the hpotenuse as having length 5 and the adjacent side as having length in Figure. If the opposite side has length, then the Pthagorean Theorem gives 5 and so, s. We can now use the diagram to write the other five trigonometric functions: csc 5 s sin s 5 EXAMPLE 5 Use a calculator to approimate the value of in Figure. SOLUTION From the diagram we see that sec 5 tan 6 tan s cot s M FIGURE Therefore tan M

28 A8 APPENDIX D TRIGONOMETRY TRIGONOMETRIC IDENTITIES A trigonometric identit is a relationship among the trigonometric functions. The most elementar are the following, which are immediate consequences of the definitions of the trigonometric functions. 6 csc sin sec cos cot tan tan sin cos cot cos sin For the net identit we refer back to Figure 7. The distance formula (or, equivalentl, the Pthagorean Theorem) tells us that r. Therefore sin cos r r r r r We have therefore proved one of the most useful of all trigonometric identities: 7 sin cos If we now divide both sides of Equation 7 b cos and use Equations 6, we get 8 tan sec Similarl, if we divide both sides of Equation 7 b sin, we get 9 cot csc The identities a b sin sin cos cos N Odd functions and even functions are discussed in Section.. show that sin is an odd function and cos is an even function. The are easil proved b drawing a diagram showing and in standard position (see Eercise 9). Since the angles and have the same terminal side, we have sin sin cos cos These identities show that the sine and cosine functions are periodic with period. The remaining trigonometric identities are all consequences of two basic identities called the addition formulas:

29 APPENDIX D TRIGONOMETRY A9 a b sin sin cos cos sin cos cos cos sin sin The proofs of these addition formulas are outlined in Eercises 85, 86, and 87. B substituting for in Equations a and b and using Equations a and b, we obtain the following subtraction formulas: a b sin sin cos cos sin cos cos cos sin sin Then, b dividing the formulas in Equations or Equations, we obtain the corresponding formulas for tan : a b tan tan tan tan tan tan tan tan tan tan If we put in the addition formulas (), we get the double-angle formulas: 5a 5b sin sin cos cos cos sin Then, b using the identit sin cos, we obtain the following alternate forms of the double-angle formulas for cos : 6a 6b cos cos cos sin If we now solve these equations for cos and sin, we get the following half-angle formulas, which are useful in integral calculus: 7a 7b cos sin cos cos Finall, we state the product formulas, which can be deduced from Equations and :

30 A APPENDIX D TRIGONOMETRY 8a 8b 8c sin cos sin sin cos cos cos cos sin sin cos cos There are man other trigonometric identities, but those we have stated are the ones used most often in calculus. If ou forget an of them, remember that the can all be deduced from Equations a and b. EXAMPLE 6 Find all values of in the interval, such that sin sin. SOLUTION Using the double-angle formula (5a), we rewrite the given equation as sin sin cos Therefore, there are two possibilities: or sin cos sin or cos,, or cos or, 5 The given equation has five solutions:,,, 5, and. M GRAPHS OF THE TRIGONOMETRIC FUNCTIONS The graph of the function f sin, shown in Figure (a), is obtained b plotting points for and then using the periodic nature of the function (from Equation ) to complete the graph. Notice that the eros of the sine function occur at the _ π _π π π π 5π π _ π (a) ƒ=sin _π _ π _ π π π π 5π π FIGURE (b) =cos

31 APPENDIX D TRIGONOMETRY A integer multiples of, that is, sin whenever n, n an integer Because of the identit cos sin (which can be verified using Equation a), the graph of cosine is obtained b shifting the graph of sine b an amount to the left [see Figure (b)]. Note that for both the sine and cosine functions the domain is, and the range is the closed interval,. Thus, for all values of, we have sin cos The graphs of the remaining four trigonometric functions are shown in Figure and their domains are indicated there. Notice that tangent and cotangent have range,, whereas cosecant and secant have range,,. All four functions are periodic: tangent and cotangent have period, whereas cosecant and secant have period. _π _ π _ π π π _π _ π π π π (a) =tan (b) =cot =sin =cos _ π _ π π π _π _ π _ π π π FIGURE (c) =csc (d) =sec

32 A APPENDIX D TRIGONOMETRY D EXERCISES 6 Convert from degrees to radians Convert from radians to degrees Find the length of a circular arc subtended b an angle of rad if the radius of the circle is 6 cm.. If a circle has radius cm, find the length of the arc subtended b a central angle of A circle has radius.5 m. What angle is subtended at the center of the circle b an arc m long? 6. Find the radius of a circular sector with angle and arc length 6 cm. 7 Draw, in standard position, the angle whose measure is given rad 7. rad. rad. rad 8 Find the eact trigonometric ratios for the angle whose radian measure is given Find the remaining trigonometric ratios. 9. sin, 5. tan,. sec.5,. cos,. cot, csc, 5 8 Find, correct to five decimal places, the length of the side labeled Prove each equation. 9. (a) Equation a (b) Equation b. (a) Equation a (b) Equation b. (a) Equation 8a (b) Equation 8b (c) Equation 8c 58 Prove the identit.. cos sin. sin. cos 5. sin cot cos 6. sin cos sin sec cos tan sin tan sin tan sin cot sec tan csc csc t sec t csc t π 5 8 cm tan tan tan sin cm 5 sec sin cm 5. sin sin cos cos cos 5. sin sin sin sin 55. sin csc cot cos 5 cm π 8 sin sin

33 APPENDIX D TRIGONOMETRY A tan tan sin cos cos sin sin sin cos cos cos cos 59 6 If sin and sec 5, where and lie between and, evaluate the epression. 59. sin 6. cos 6. cos 6. sin 6. sin 6. cos position as in the figure. Epress and in terms of use the distance formula to compute c.] and then 8. In order to find the distance across a small inlet, a point C is located as in the figure and the following measurements were recorded: C AC 8 m 9 m Use the Law of Cosines from Eercise 8 to find the required distance. A AB BC 65 7 Find all values of in the interval, that satisf the equation. 65. cos 66. cot 67. sin 68. tan 69. sin cos 7. cos sin 7. sin tan 7. cos cos 7 76 Find all values of in the interval, that satisf the inequalit. 7. sin 7. cos 75. tan 76. sin cos 77 8 Graph the function b starting with the graphs in Figures and and appling the transformations of Section. where appropriate. sin 77. cos 78. tan Prove the Law of Cosines: If a triangle has sides with lengths a, b, and c, and is the angle between the sides with lengths a and b, then c a b ab cos sec tan b P(, ) [Hint: Introduce a coordinate sstem so that c sin (a, ) is in standard 85. Use the figure to prove the subtraction formula [Hint: Compute in two was (using the Law of Cosines from Eercise 8 and also using the distance formula) and compare the two epressions.] 86. Use the formula in Eercise 85 to prove the addition formula for cosine (b). 87. Use the addition formula for cosine and the identities cos sin sin to prove the subtraction formula for the sine function. 88. Show that the area of a triangle with sides of lengths a and b and with included angle is 89. Find the area of triangle ABC, correct to five decimal places, if AB cos cos c C å A ab sin cos sin sin A(cos å, sin å) c B(cos, sin ) cos cm BC cm ABC 7 B

34 A APPENDIX E SIGMA NOTATION E SIGMA NOTATION A convenient wa of writing sums uses the Greek letter our letter S) and is called sigma notation. (capital sigma, corresponding to This tells us to end with i=n. This tells us to add. This tells us to start with i=m. μ a i n im DEFINITION If a m, a m,..., a n are real numbers and m and n are integers such that m n, then n a i a m a m a m a n a n im With function notation, Definition can be written as n f i f m f m f m fn f n im n im Thus the smbol indicates a summation in which the letter i (called the inde of summation) takes on consecutive integer values beginning with m and ending with n, that is, m, m,..., n. Other letters can also be used as the inde of summation. EXAMPLE (a) (b) (c) (d) (e) (f) i i n i 5 n n i 5 j 5 6 j n k i k n i i i M EXAMPLE Write the sum n in sigma notation. SOLUTION There is no unique wa of writing a sum in sigma notation. We could write n n i i or or n n j n n k The following theorem gives three simple rules for working with sigma notation. j k M

35 APPENDIX E SIGMA NOTATION A5 (a) (c) THEOREM n ca i c n a i im im If c is an constant (that is, it does not depend on i), then n a i b i n a i n b i im im im (b) n a i b i n a i n b i im im im PROOF To see wh these rules are true, all we have to do is write both sides in epanded form. Rule (a) is just the distributive propert of real numbers: ca m ca m ca n ca m a m a n Rule (b) follows from the associative and commutative properties: a m b m a m b m a n b n a m a m a n b m b m b n Rule (c) is proved similarl. EXAMPLE Find SOLUTION n. i n n i n terms M M EXAMPLE Prove the formula for the sum of the first n positive integers: SOLUTION This formula can be proved b mathematical induction (see page 77) or b the following method used b the German mathematician Karl Friedrich Gauss ( ) when he was ten ears old. Write the sum S twice, once in the usual order and once in reverse order: Adding all columns verticall, we get n nn i n i S n n S n n n S n n n n n On the right side there are n terms, each of which is n, so S nn or S nn M EXAMPLE 5 Prove the formula for the sum of the squares of the first n positive integers: n nn n i n i 6

36 A6 APPENDIX E SIGMA NOTATION SOLUTION Let S be the desired sum. We start with the telescoping sum (or collapsing sum): Most terms cancel in pairs. n i i n n i n n n n On the other hand, using Theorem and Eamples and, we have n i i n i i n i n i n i i i i i S nn n S n 5 n Thus we have n n n S n 5 n Solving this equation for S, we obtain S n n n or S n n n 6 nn n 6 N PRINCIPLE OF MATHEMATICAL INDUCTION Let S n be a statement involving the positive integer n. Suppose that. S is true.. If is true, then is true. S k S k Then S n is true for all positive integers n. N See pages 55 and 58 for a more thorough discussion of mathematical induction. SOLUTION Let S. is true because be the given formula.. Assume that is true; that is, Then k k k So S k S n S k is true. 6 k k k k 7k 6 6 kk k 6 kk k 6 kk 6k 6 k k k 6 k k k k 6 B the Principle of Mathematical Induction, S n is true for all n. M

37 APPENDIX E SIGMA NOTATION A7 We list the results of Eamples,, and 5 together with a similar result for cubes (see Eercises 7 ) as Theorem. These formulas are needed for finding areas and evaluating integrals in Chapter 5. (a) (c) (e) THEOREM Let c be a constant and n a positive integer. Then n n i n nn i i n i i nn (b) (d) n c nc i n nn n i i 6 N The tpe of calculation in Eample 7 arises in Chapter 5 when we compute areas. EXAMPLE 6 Evaluate ii. SOLUTION Using Theorems and, we have EXAMPLE 7 Find lim. SOLUTION n i n ii n i i n i n i i i i i n l n i n i n lim i nl n i n n lim nl n nn n n nn nn lim n l n i n i n n i lim nl nn n n 6 n n n nl n n lim nl n n n lim nn i n i n nn n n M M

38 A8 APPENDIX E SIGMA NOTATION E EXERCISES Write the sum in epanded form.. si.. i. i i 5. k 6. k j. Write the sum in sigma notation i k n i i n j s s s5 s6 s n n n. n n 5 Find the value of the sum. 8 i. i. 6 j j.. 5. n 6. n i 7. i i 8. i i i n i 9. i. n i. i i. n i n i n i.. n i i ii i 6 6 i i n j jn 6 i i i 8 k k5 n f i i i ii 8 cos k k 5i i 5. n i i i 6. Find the number n such that i Prove formula (b) of Theorem. 8. Prove formula (e) of Theorem using mathematical induction. 9. Prove formula (e) of Theorem using a method similar to that of Eample 5, Solution [start with i i.. Prove formula (e) of Theorem using the following method published b Abu Bekr Mohammed ibn Alhusain Alkarchi in about AD. The figure shows a square ABCD in which sides AB and AD have been divided into segments of lengths,,,..., n. Thus the side of the square has length nn so the area is nn. But the area is also the sum of the areas of the n gnomons G, G,..., G n shown in the figure. Show that the area of G is i i and conclude that formula (e) is true.. Evaluate each telescoping sum. (a) (c) n i i i i 99 i i (b) (d). Prove the generalied triangle inequalit: n 6 Find the limit.. lim. lim n l n i D i n n 5. lim i n l n i n n. 5 G A G 5... n B n G n a i i n i G 5 n i... 5 i 5 i i n a i a i i i a i n l n i G n C i n n

39 6. lim n l n i i n n n i 7. Prove the formula for the sum of a finite geometric series with first term a and common ratio r : n ar i a ar ar ar n ar n i r 8. Evaluate. n i 9. Evaluate i i. 5. Evaluate i j. APPENDIX F PROOFS OF THEOREMS A9 n i i i m n j F PROOFS OF THEOREMS In this appendi we present proofs of several theorems that are stated in the main bod of the tet. The sections in which the occur are indicated in the margin. SECTION. LIMIT LAWS Suppose that c is a constant and the limits and eist. Then. lim f t L M. l a. lim cf cl. l a lim f L l a f 5. lim if M l a t L M lim t M l a lim f t L M l a lim f t LM la PROOF OF LAW Let be given. We want to find such that In order to get terms that contain and, we add and subtract Lt as follows: f t LM f t Lt Lt LM f Lt Lt M f Lt Lt M f L t L t M We want to make each of these terms less than. Since lim l a t M, there is a number such that if if a a f L then t M Also, there is a number such that if, then then f t LM t M a (Triangle Inequalit) t M and therefore t t M M t M M M ( L )

40 A APPENDIX F PROOFS OF THEOREMS Since lim l a f L, there is a number such that if Let min,,. If a, then we have a,, and, so we can combine the inequalities to obtain a a a then f L ( M ) f t LM f L t L t M ( M ) ( M ) L ( L ) This shows that lim l a f t LM. M PROOF OF LAW If we take t c in Law, we get lim cf lim tf lim t lim f l a l a l a l a lim l a c lim l a f c lim l a f (b Law 7) M PROOF OF LAW Using Law and Law with c, we have lim f t lim f t lim f lim t l a l a l a l a lim l a f lim l a t lim l a f lim l a t M PROOF OF LAW 5 First let us show that To do this we must show that, given, there eists such that Observe that We know that we can make the numerator small. But we also need to know that the denominator is not small when is near a. Since lim l a t M, there is a number such that, whenever, we have and therefore if lim l a a t M M t Mt a t M then t M t M M M M t t M t t M t

41 APPENDIX F PROOFS OF THEOREMS A This shows that and so, for these values of, Also, there eists such that if if a a Let min,. Then, for, we have It follows that lim l a t M. Finall, using Law, we obtain f lim lim lim f lim l a l a t L M L l a t f l a t M then a then t M M t Mt M M t M Mt M t M M M t M M M THEOREM If f t for all in an open interval that contains a (ecept possibl at a) and then L M. lim f L l a and lim t M l a PROOF We use the method of proof b contradiction. Suppose, if possible, that L M. Law of limits sas that lim t f M L l a Therefore, for an, there eists such that if In particular, taking L M (noting that L M b hpothesis), we have a number such that a then t f M L Since if a a a then for an number a, we have t f M L L M if a then t f M L L M which simplifies to if a then t f But this contradicts f t. Thus the inequalit L M must be false. Therefore L M. M

42 A APPENDIX F PROOFS OF THEOREMS THE SQUEEZE THEOREM If f t h for all in an open interval that contains a (ecept possibl at a) and lim f lim h L l a l a then lim t L l a PROOF Let be given. Since lim l a f L, there is a number such that that is, if if a a then then f L L f L Since lim l a h L, there is a number such that that is, if if a a then then h L L h L Let min,. If a, then a and a, so L f t h L In particular, L t L t L and so. Therefore lim l a t L. M SECTION.5 THEOREM If f is a one-to-one continuous function defined on an interval a, b, then its inverse function f is also continuous. PROOF First we show that if f is both one-to-one and continuous on a, b, then it must be either increasing or decreasing on a, b. If it were neither increasing nor decreasing, then there would eist numbers,, and in a, b with such that f does not lie between f and f. There are two possibilities: either () f lies between f and f or () f lies between f and f. (Draw a picture.) In case () we appl the Intermediate Value Theorem to the continuous function f to get a number c between and such that f c f. In case () the Intermediate Value Theorem gives a number c between and such that f c f. In either case we have contradicted the fact that f is one-to-one. Let us assume, for the sake of definiteness, that f is increasing on a, b. We take an number in the domain of f and we let f ; that is, is the number in a, b such that f. To show that f is continuous at we take an such that the interval, is contained in the interval a, b. Since f is increasing, it maps the numbers in the interval, onto the numbers in the interval f and f, f reverses the correspondence. If we let denote the smaller of the numbers f and f, then the interval, is contained in the interval f, f and so is mapped into the interval b f,. (See the arrow diagram in Figure.) We have

43 APPENDIX F PROOFS OF THEOREMS A therefore found a number if such that then f f f( - ) { f f! f( + ) } f FIGURE { { } } a - + b This shows that lim and so f l f f is continuous at an number in its domain. M 8 THEOREM If f is continuous at b and lim l a t b, then lim f t f b l a PROOF Let be given. We want to find a number such that Since f if a is continuous at b, we have then lim f f b l b f t f b and so there eists such that if b then Since lim l a t b, there eists such that f f b if a Combining these two statements, we see that whenever we have t b, which implies that f t f b a. Therefore we have proved that lim l a f t f b. M sin SECTION. The proof of the following result was promised when we proved that lim. then t b l tan THEOREM If, then. PROOF Figure shows a sector of a circle with center O, central angle, and radius. Then AD OA tan tan We approimate the arc AB b an inscribed polgon consisting of n equal line segments

44 A APPENDIX F PROOFS OF THEOREMS D and we look at a tpical segment PQ. We etend the lines OP and OQ to meet AD in the points R and S. Then we draw RT PQ as in Figure. Observe that B O FIGURE Q T P A S R and so RTS 9. Therefore we have If we add n such inequalities, we get where L n RTO PQO 9 PQ RT RS L n AD tan is the length of the inscribed polgon. Thus, b Theorem.., we have lim n l L n tan But the arc length is defined in Equation 8.. as the limit of the lengths of inscribed polgons, so lim n l L n tan M SECTION. CONCAVITY TEST (a) If f for all in I, then the graph of f is concave upward on I. (b) If f for all in I, then the graph of f is concave downward on I. =ƒ PROOF OF (a) Let a be an number in I. We must show that the curve f lies above the tangent line at the point a, f a. The equation of this tangent is f a f a a ƒ f(a)+f ª(a)(-a) So we must show that f f a f a a FIGURE a whenever I a. (See Figure.) First let us take the case where a. Appling the Mean Value Theorem to f on the interval a,, we get a number c, with a c, such that f f a f c a Since f on I, we know from the Increasing/Decreasing Test that f is increasing on I. Thus, since a c, we have f a f c and so, multipling this inequalit b the positive number a, we get f a a f c a

45 APPENDIX F PROOFS OF THEOREMS A5 Now we add f a to both sides of this inequalit: f a f a a f a f c a But from Equation we have f f a f c a. So this inequalit becomes f f a f a a which is what we wanted to prove. For the case where a we have f c f a, but multiplication b the negative number a reverses the inequalit, so we get () and () as before. M SECTION. In order to give the promised proof of l Hospital s Rule, we first need a generaliation of the Mean Value Theorem. The following theorem is named after another French mathematician, Augustin-Louis Cauch ( ). N See the biographical sketch of Cauch on page. CAUCHY S MEAN VALUE THEOREM Suppose that the functions f and t are continuous on a, b and differentiable on a, b, and t for all in a, b. Then there is a number c in a, b such that f c tc f b f a tb ta Notice that if we take the special case in which t, then tc and Theorem is just the ordinar Mean Value Theorem. Furthermore, Theorem can be proved in a similar manner. You can verif that all we have to do is change the function h given b Equation.. to the function h f f a and appl Rolle s Theorem as before. f b f a tb ta t ta L HOSPITAL S RULE Suppose f and t are differentiable and t on an open interval I that contains a (ecept possibl at a). Suppose that lim f l a and lim t l a or that lim f l a and lim t l a (In other words, we have an indeterminate form of tpe or.) Then lim l a f t lim l a f t if the limit on the right side eists (or is or ).

46 A6 APPENDIX F PROOFS OF THEOREMS PROOF OF L HOSPITAL S RULE We are assuming that lim l a f and lim l a t. Let f L lim l a t We must show that lim l a f t L. Define F f Then F is continuous on I since f is continuous on I a and Likewise, G is continuous on I. Let I and a. Then F and G are continuous on a, and differentiable on a, and G there (since F f and G t). Therefore, b Cauch s Mean Value Theorem, there is a number such that a and Here we have used the fact that, b definition, Fa and Ga. Now, if we let l a, then l a (since a ), so lim la f t F F Fa F G G Ga G lim la if a if a F F lim lim G la G la A similar argument shows that the left-hand limit is also L. Therefore lim l a f t L G t lim F lim f Fa l a l a if a if a f t L This proves l Hospital s Rule for the case where a is finite. If a is infinite, we let t. Then t l as l, so we have lim l f t f t lim t l tt f tt lim t l ttt f t lim lim t l tt l f t (b l Hospital s Rule for finite a) M SECTION.8 In order to prove Theorem.8., we first need the following results. THEOREM. If a power series c n n converges when b (where b ), then it converges whenever b.. If a power series c n n diverges when d (where d ), then it diverges whenever. d

47 APPENDIX F PROOFS OF THEOREMS A7 PROOF OF Suppose that c n b n converges. Then, b Theorem..6, we have lim n l c n b n. According to Definition.. with, there is a positive integer N such that whenever n N. Thus, for n N, we have b c nb n b c n n c nb n n b n b n c nb n n b n b If, then, so is a convergent geometric series. Therefore, b the Comparison Test, the series nn c n n is convergent. Thus the series c n n is absolutel convergent and therefore convergent. d d PROOF OF Suppose that c n d n diverges. If is an number such that, then c cannot converge because, b part, the convergence of c n n n n would impl the convergence of c. Therefore c n n n d n diverges whenever. M M THEOREM For a power series c n n there are onl three possibilities:. The series converges onl when.. The series converges for all.. There is a positive number R such that the series converges if and diverges if. R R PROOF Suppose that neither case nor case is true. Then there are nonero numbers b and d such that converges for and diverges for. Therefore the set S is not empt. B the preceding theorem, the series diverges if c n n b d d c n n converges, so d for all S. This sas that d is an upper bound for the set S. Thus, b the Completeness Aiom (see Section.), S has a least upper bound R. If R, then S, so c diverges. If, then is not an upper bound for S and so there eists b S such that b. Since b S, c n b n converges, so b the preceding theorem R n n c n n converges. M THEOREM For a power series c n a n there are onl three possibilities:. The series converges onl when a.. The series converges for all.. There is a positive number R such that the series converges if and diverges if. a R a R PROOF If we make the change of variable u a, then the power series becomes c n u n and we can appl the preceding theorem to this series. In case we have convergence for u R and divergence for u R. Thus we have convergence for and divergence for. M a R a R

48 A8 APPENDIX F PROOFS OF THEOREMS SECTION. CLAIRAUT S THEOREM Suppose f is defined on a disk D that contains the point a, b. If the functions and are both continuous on D, then f a, b f a, b. f f PROOF For small values of h, h, consider the difference Notice that if we let t f, b h f, b, then B the Mean Value Theorem, there is a number c between a and a h such that Appling the Mean Value Theorem again, this time to f, we get a number d between b and b h such that Combining these equations, we obtain If h l, then c, d l a, b, so the continuit of at a, b gives Similarl, b writing h f a h, b h f a h, b f a, b h f a, b lim h l h ta h ta ta h ta tch h f c, b h f c, b f c, b h f c, b f c, dh h h f c, d h h lim c, d l a, b f c, d f a, b h f a h, b h f a, b h f a h, b f a, b and using the Mean Value Theorem twice and the continuit of at a, b, we obtain f f lim h l h h f a, b It follows that f a, b f a, b. M SECTION. 8 THEOREM If the partial derivatives f and f eist near a, b and are continuous at a, b, then f is differentiable at a, b. PROOF Let f a, b f a, b According to (..7), to prove that f is differentiable at a, b we have to show that we can write in the form f a, b f a, b where and l as, l,.

49 APPENDIX F PROOFS OF THEOREMS A9 Referring to Figure, we write f a, b f a, b f a, b f a, b (a, b+î) (a+î, b+î) (u, b+î) (a, b) (a, ) R FIGURE Observe that the function of a single variable t f, b is defined on the interval a, a and t f, b. If we appl the Mean Value Theorem to t, we get ta ta tu where u is some number between a and a. In terms of f, this equation becomes f a, b f a, b f u, b This gives us an epression for the first part of the right side of Equation. For the second part we let h f a,. Then h is a function of a single variable defined on the interval b, b and h f a,. A second application of the Mean Value Theorem then gives hb hb hv where v is some number between b and b. In terms of f, this becomes f a, b f a, b f a, v We now substitute these epressions into Equation and obtain f u, b f a, v f a, b f u, b f a, b f a, b f a, v f a, b f a, b f a, b where f u, b f a, b f a, v f a, b Since u, b l a, b and a, v l a, b as, l, and since f and f are continuous at a, b, we see that l and l as, l,. Therefore f is differentiable at a, b. M

50 A5 APPENDIX G THE LOGARITHM DEFINED AS AN INTEGRAL G THE LOGARITHM DEFINED AS AN INTEGRAL Our treatment of eponential and logarithmic functions until now has relied on our intuition, which is based on numerical and visual evidence. (See Sections.5,.6, and..) Here we use the Fundamental Theorem of Calculus to give an alternative treatment that provides a surer footing for these functions. Instead of starting with a and defining log a as its inverse, this time we start b defining ln as an integral and then define the eponential function as its inverse. You should bear in mind that we do not use an of our previous definitions and results concerning eponential and logarithmic functions. THE NATURAL LOGARITHM We first define ln as an integral. DEFINITION The natural logarithmic function is the function defined b ln t dt = t area=ln t FIGURE The eistence of this function depends on the fact that the integral of a continuous function alwas eists. If, then ln can be interpreted geometricall as the area under the hperbola t from t to t. (See Figure.) For, we have For, ln ln and so ln is the negative of the area shown in Figure. t t dt dt t dt FIGURE = t A area=_ln = t t V EXAMPLE (a) B comparing areas, show that ln. (b) Use the Midpoint Rule with n to estimate the value of ln. SOLUTION (a) We can interpret ln as the area under the curve t from to. From Figure we see that this area is larger than the area of rectangle BCDE and smaller than the area of trapeoid ABCD. Thus we have ln ( ) ln (b) If we use the Midpoint Rule with f t t, n, and t., we get E D B C t FIGURE ln dt. f.5 f.5 f.95 t M

51 APPENDIX G THE LOGARITHM DEFINED AS AN INTEGRAL A5 Notice that the integral that defines ln is eactl the tpe of integral discussed in Part of the Fundamental Theorem of Calculus (see Section 5.). In fact, using that theorem, we have d d t dt and so d d ln We now use this differentiation rule to prove the following properties of the logarithm function. LAWS OF LOGARITHMS number, then If and are positive numbers and r is a rational. ln ln ln. ln ln ln. ln r r ln PROOF. Let f lna, where a is a positive constant. Then, using Equation and the Chain Rule, we have Therefore f and ln have the same derivative and so the must differ b a constant: Putting in this equation, we get ln a ln C C C. Thus If we now replace the constant a b an number, we have. Using Law with, we have and so ln Using Law again, we have f a ln ln ln ln lna ln C lna ln ln a ln ln ln d d a a a ln ln ln ln ln ln ln The proof of Law is left as an eercise. M

52 A5 APPENDIX G THE LOGARITHM DEFINED AS AN INTEGRAL In order to graph ln, we first determine its limits: (a) lim ln l (b) lim ln l PROOF (a) Using Law with and r n (where n is an positive integer), we have ln n n ln. Now ln, so this shows that ln n l as n l. But ln is an increasing function since its derivative. Therefore ln l as l. (b) If we let t, then t l as l. Thus, using (a), we have =ln lim ln lim ln lim ln t l t l t t l M If ln,, then d d and d d FIGURE which shows that ln is increasing and concave downward on,. Putting this information together with (), we draw the graph of ln in Figure. Since ln and ln is an increasing continuous function that takes on arbitraril large values, the Intermediate Value Theorem shows that there is a number where ln takes on the value. (See Figure 5.) This important number is denoted b e. e =ln FIGURE 5 5 DEFINITION e is the number such that ln e. We will show (in Theorem 9) that this definition is consistent with our previous definition of e. THE NATURAL EXPONENTIAL FUNCTION Since ln is an increasing function, it is one-to-one and therefore has an inverse function, which we denote b ep. Thus, according to the definition of an inverse function, f &? f 6 ep &? ln and the cancellation equations are f f f f 7 In particular, we have epln and lnep ep since ln ep e since ln e We obtain the graph of ep b reflecting the graph of ln about the line.

53 APPENDIX G THE LOGARITHM DEFINED AS AN INTEGRAL A5 =ep = =ln (See Figure 6.) The domain of ep is the range of ln, that is,, ; the range of ep is the domain of ln, that is,,. If r is an rational number, then the third law of logarithms gives lne r r ln e r Therefore, b (6), epr e r Thus ep e whenever is a rational number. This leads us to define e, even for irrational values of, b the equation FIGURE 6 e ep In other words, for the reasons given, we define e to be the inverse of the function ln. In this notation (6) becomes 8 e &? ln and the cancellation equations (7) become 9 e ln lne for all = The natural eponential function f e is one of the most frequentl occurring functions in calculus and its applications, so it is important to be familiar with its graph (Figure 7) and its properties (which follow from the fact that it is the inverse of the natural logarithmic function). FIGURE 7 The natural eponential function PROPERTIES OF THE EXPONENTIAL FUNCTION The eponential function f e is an increasing continuous function with domain and range,. Thus e for all. Also lim e l lim e l So the -ais is a horiontal asmptote of f e. We now verif that f has the other properties epected of an eponential function. LAWS OF EXPONENTS If and are real numbers and r is rational, then.. e e e e e. e r e r e

54 A5 APPENDIX G THE LOGARITHM DEFINED AS AN INTEGRAL PROOF OF LAW Using the first law of logarithms and Equation, we have lne e lne lne lne Since ln is a one-to-one function, it follows that e e e. Laws and are proved similarl (see Eercises 6 and 7). As we will soon see, Law actuall holds when r is an real number. M We now prove the differentiation formula for. e d d e e PROOF The function e is differentiable because it is the inverse function of ln, which we know is differentiable with nonero derivative. To find its derivative, we use the inverse function method. Let e. Then ln and, differentiating this latter equation implicitl with respect to, we get d d d e d M GENERAL EXPONENTIAL FUNCTIONS If a and r is an rational number, then b (9) and (), a r e ln a r e r ln a Therefore, even for irrational numbers, we define a e ln a Thus, for instance, s e s ln e.. The function f a is called the eponential function with base a. Notice that a is positive for all because e is positive for all. Definition allows us to etend one of the laws of logarithms. We alread know that lna r r ln a when r is rational. But if we now let r be an real number we have, from Definition, ln a r lne r ln a r ln a Thus ln a r r ln a for an real number r

55 APPENDIX G THE LOGARITHM DEFINED AS AN INTEGRAL A55 The general laws of eponents follow from Definition together with the laws of eponents for. e 5 LAWS OF EXPONENTS If and are real numbers and a, b, then. a a a. a a a. a a. ab a b PROOF. Using Definition and the laws of eponents for, we have. Using Equation we obtain e ln a e ln a a a a e lna ln a e e a e ln a ln a ln a e e ln a a The remaining proofs are left as eercises. M The differentiation formula for eponential functions is also a consequence of Definition : 6 d d a a ln a lim a =, lim a =` _` ` PROOF d d a d d e ln a e d ln a ln a d a ln a M FIGURE 8 =a, a> If a, then ln a, so dd a a ln a, which shows that a is increasing (see Figure 8). If a, then ln a and so a is decreasing (see Figure 9). GENERAL LOGARITHMIC FUNCTIONS If a and a, then f a is a one-to-one function. Its inverse function is called the logarithmic function with base a and is denoted b log a. Thus 7 log a &? a lim a =`, lim a = _` ` FIGURE 9 =a, <a< In particular, we see that log e ln

56 A56 APPENDIX G THE LOGARITHM DEFINED AS AN INTEGRAL The laws of logarithms are similar to those for the natural logarithm and can be deduced from the laws of eponents (see Eercise ). To differentiate log, we write the equation as a a. From Equation we have ln a ln,so log a ln ln a Since ln a is a constant, we can differentiate as follows: d d log a d d ln ln a ln a d ln d ln a 8 d d log a ln a THE NUMBER e EXPRESSED AS A LIMIT In this section we defined e as the number such that ln e. The net theorem shows that this is the same as the number e defined in Section. (see Equation.6.5). 9 e lim l PROOF Let f ln. Then f, so f. But, b the definition of derivative, f lim hl Because f, we have f h f h lim l ln ln lim lim ln lim ln l l l lim ln l f f Then, b Theorem.5.8 and the continuit of the eponential function, we have e e e lim l ln lim eln lim l l M G EXERCISES. (a) B comparing areas, show that ln.5 5 (b) Use the Midpoint Rule with n to estimate ln.5.. Refer to Eample. (a) Find the equation of the tangent line to the curve t that is parallel to the secant line AD. (b) Use part (a) to show that ln.66.. B comparing areas, show that n ln n n. (a) B comparing areas, show that ln ln. (b) Deduce that e.

57 APPENDIX H COMPLEX NUMBERS A57 5. Prove the third law of logarithms. [Hint: Start b showing that both sides of the equation have the same derivative.] 6. Prove the second law of eponents for e [see ()]. 7. Prove the third law of eponents for e [see ()]. 8. Prove the second law of eponents [see (5)]. 9. Prove the fourth law of eponents [see (5)].. Deduce the following laws of logarithms from (5): (a) log a log a log a (b) log a log a log a (c) log a log a H COMPLEX NUMBERS _+i _-i Im _i FIGURE Comple numbers as points in the Argand plane i +i -i Re A comple number can be represented b an epression of the form a bi, where a and b are real numbers and i is a smbol with the propert that i. The comple number a bi can also be represented b the ordered pair a, b and plotted as a point in a plane (called the Argand plane) as in Figure. Thus the comple number i i is identified with the point,. The real part of the comple number a bi is the real number a and the imaginar part is the real number b. Thus the real part of i is and the imaginar part is. Two comple numbers a bi and c di are equal if a c and b d; that is, their real parts are equal and their imaginar parts are equal. In the Argand plane the horiontal ais is called the real ais and the vertical ais is called the imaginar ais. The sum and difference of two comple numbers are defined b adding or subtracting their real parts and their imaginar parts: a bi c di a c b di For instance, a bi c di a c b di i 7i 7i 5 6i The product of comple numbers is defined so that the usual commutative and distributive laws hold: a bic di ac di bic di ac adi bci bdi Since i, this becomes a bic di ac bd ad bci EXAMPLE i 5i 5i i 5i 5i 6i 5 i M Division of comple numbers is much like rationaliing the denominator of a rational epression. For the comple number a bi, we define its comple conjugate to be a bi. To find the quotient of two comple numbers we multipl numerator and denominator b the comple conjugate of the denominator. i EXAMPLE Epress the number in the form a bi. 5i

58 A58 APPENDIX H COMPLEX NUMBERS SOLUTION We multipl numerator and denominator b the comple conjugate of 5i, namel 5i, and we take advantage of the result of Eample : i 5i i 5i 5i 5i i i M Im i =a+bi The geometric interpretation of the comple conjugate is shown in Figure : is the reflection of in the real ais. We list some of the properties of the comple conjugate in the following bo. The proofs follow from the definition and are requested in Eercise 8. _i =a-bi Re PROPERTIES OF CONJUGATES w w w w n n FIGURE Im The modulus, or absolute value, of a comple number a bi is its distance from the origin. From Figure we see that if a bi, then bi œ = a@+b@ =a+bi b Notice that sa b a Re a bia bi a abi abi b i a b FIGURE and so This eplains wh the division procedure in Eample works in general: Since i, we can think of i as a square root of. But notice that we also have i i and so i is also a square root of. We sa that i is the principal square root of and write s i. In general, if c is an positive number, we write sc sc i With this convention, the usual derivation and formula for the roots of the quadratic equation a b c are valid even when b ac : b sb ac a EXAMPLE Find the roots of the equation. w w ww SOLUTION Using the quadratic formula, we have w w s s s i M

59 APPENDIX H COMPLEX NUMBERS A59 We observe that the solutions of the equation in Eample are comple conjugates of each other. In general, the solutions of an quadratic equation a b c with real coefficients a, b, and c are alwas comple conjugates. (If is real,, so is its own conjugate.) We have seen that if we allow comple numbers as solutions, then ever quadratic equation has a solution. More generall, it is true that ever polnomial equation a n n a n n a a of degree at least one has a solution among the comple numbers. This fact is known as the Fundamental Theorem of Algebra and was proved b Gauss. POLAR FORM Im r a a+bi b Re We know that an comple number a bi can be considered as a point a, b and that an such point can be represented b polar coordinates r, with r. In fact, a r cos b r sin as in Figure. Therefore we have FIGURE a bi r cos r sin i Thus we can write an comple number in the form rcos i sin where r sa b and tan b a Im œ FIGURE 5 π _ π 6 +i Re œ -i The angle is called the argument of and we write. Note that arg is not unique; an two arguments of differ b an integer multiple of. EXAMPLE Write the following numbers in polar form. (a) i (b) w s i SOLUTION (a) We have r s s and tan, so we can take. Therefore the polar form is s cos (b) Here we have r w s and tan s. Since w lies in the fourth quadrant, we take and w cos i sin The numbers and w are shown in Figure 5. M i sin arg

60 A6 APPENDIX H COMPLEX NUMBERS The polar form of comple numbers gives insight into multiplication and division. Let r cos i sin r cos i sin be two comple numbers written in polar form. Then r r cos i sin cos i sin r r cos cos sin sin isin cos cos sin Im Therefore, using the addition formulas for cosine and sine, we have r r cos i sin + Re This formula sas that to multipl two comple numbers we multipl the moduli and add the arguments. (See Figure 6.) A similar argument using the subtraction formulas for sine and cosine shows that to divide two comple numbers we divide the moduli and subtract the arguments. FIGURE 6 Im r r cos i sin r In particular, taking and (and therefore and ), we have the following, which is illustrated in Figure 7. _ r Re If rcos i sin, then r cos i sin. FIGURE 7 EXAMPLE 5 Find the product of the comple numbers i and s i in polar form. Im œ =+i œ π w Re w=œ -i SOLUTION From Eample we have and So, b Equation, i s cos s i cos i sin 6 i(s i) s cos s cos i sin i sin 6 i sin 6 6 FIGURE 8 This is illustrated in Figure 8. M

61 APPENDIX H COMPLEX NUMBERS A6 Repeated use of Formula shows how to compute powers of a comple number. If rcos i sin then and r cos i sin r cos i sin In general, we obtain the following result, which is named after the French mathematician Abraham De Moivre (667 75). DE MOIVRE S THEOREM If rcos i sin and n is a positive integer, then n rcos i sin n r n cos n i sin n This sas that to take the nth power of a comple number we take the nth power of the modulus and multipl the argument b n. EXAMPLE 6 Find. SOLUTION Since, it follows from Eample (a) that i i i has the polar form So b De Moivre s Theorem, ( i) i i s cos s cos i sin 5 cos 5 5 i sin i De Moivre s Theorem can also be used to find the nth roots of comple numbers. An nth root of the comple number is a comple number w such that w n Writing these two numbers in trigonometric form as i sin M w scos i sin and using De Moivre s Theorem, we get and rcos i sin s n cos n i sin n rcos i sin The equalit of these two comple numbers shows that s n r or s r n and cos n cos and sin n sin

62 A6 APPENDIX H COMPLEX NUMBERS From the fact that sine and cosine have period Thus n k it follows that ncos k k w r i sin n n or Since this epression gives a different value of w for k,,,..., n,we have the following. k n ROOTS OF A COMPLEX NUMBER Let rcos i sin and let n be a positive integer. Then has the n distinct nth roots where k,,,..., n. ncos k k w k r i sin n n n Notice that each of the nth roots of has modulus r. Thus all the nth roots of lie on the circle of radius r n in the comple plane. Also, since the argument of each successive nth root eceeds the argument of the previous root b n, we see that the nth roots of are equall spaced on this circle. EXAMPLE 7 Find the si sith roots of 8 and graph these roots in the comple plane. SOLUTION In trigonometric form, 8cos i sin. Appling Equation with n 6, we get w k 8 6cos i sin 6 6 k w k k w _œ œ w Im œ i _œ i w w FIGURE 9 The si sith roots of =_8 w w Re We get the si sith roots of 8 b taking k,,,,, 5 in this formula: w 8 6cos i sin s 6 6 s i w 8 6cos i sin s i 5 5 w 86cos i sin s s 6 6 i 7 7 w 86cos i sin s s 6 6 i w 86cos i sin s i w 5 86cos i sin s s 6 6 i All these points lie on the circle of radius s as shown in Figure 9. M

63 APPENDIX H COMPLEX NUMBERS A6 COMPLEX EXPONENTIALS We also need to give a meaning to the epression e when i is a comple number. The theor of infinite series as developed in Chapter can be etended to the case where the terms are comple numbers. Using the Talor series for e (..) as our guide, we define e n n n!!! and it turns out that this comple eponential function has the same properties as the real eponential function. In particular, it is true that 5 e e e If we put i, where is a real number, in Equation, and use the facts that i, i i i i, i, i 5 i,... we get e i i i! i cos i sin i!! i!! i 5!! 6 i! i5 5! 5! 6! i! 5 5! Here we have used the Talor series for cos and sin (Equations..6 and..5). The result is a famous formula called Euler s formula: 6 e i cos i sin Combining Euler s formula with Equation 5, we get 7 e i e e i e cos i sin EXAMPLE 8 Evaluate: (a) e i (b) e i N We could write the result of Eample 8(a) as e i This equation relates the five most famous numbers in all of mathematics:,, e, i, and. SOLUTION (a) From Euler s equation (6) we have (b) Using Equation 7 we get e i cos i sin i e i e cos i sin e i i e Finall, we note that Euler s equation provides us with an easier method of proving De Moivre s Theorem: rcos i sin n re i n r n e in r n cos n i sin n M

64 A6 APPENDIX H COMPLEX NUMBERS H EXERCISES i8 i Evaluate the epression and write our answer in the form a bi.. 5 6i i.. 5i i i( i) 7. i i 8. i i 9.. i i. i. i. s5. s s 5 7 Find the comple conjugate and the modulus of the number. 5. 5i 6. s i 7. i 8. Prove the following properties of comple numbers. (a) w w (b) w w (c) n n, where n is a positive integer [Hint: Write a bi, w c di.] 9 Find all solutions of the equation Write the number in polar form with argument between and. 5. i 6. s i 7. i 8. 8i 9 Find polar forms for w, w, and b first putting and w into polar form. 9. s i, w s i. s i, w 8i. s i, w i. (s i), w i ( i) (9 5 i) 6 Find the indicated power using De Moivre s Theorem.. i. ( si) 5 5. (s i) 5 6. i 8 7 Find the indicated roots. Sketch the roots in the comple plane. 7. The eighth roots of 8. The fifth roots of 9. The cube roots of i. The cube roots of i 6 Write the number in the form a bi... e i e i.. e i Use De Moivre s Theorem with n to epress cos and sin in terms of cos and sin. 8. Use Euler s formula to prove the following formulas for cos and sin : cos ei e i 9. If u f it is a comple-valued function of a real variable and the real and imaginar parts f and t are differentiable functions of, then the derivative of u is defined to be u f it. Use this together with Equation 7 to prove that if F e r, then F re r when r a bi is a comple number. 5. (a) If u is a comple-valued function of a real variable, its indefinite integral u d is an antiderivative of u. Evaluate e i d (b) B considering the real and imaginar parts of the integral in part (a), evaluate the real integrals e cos d and e i e i i e sin ei e i i e sin d (c) Compare with the method used in Eample in Section 7..

65 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES A65 I ANSWERS TO ODD-NUMBERED EXERCISES CHAPTER EXERCISES. N PAGE. (a) (b).8 (c), (d).5,. (e),,, (f ),. 85, 5 5. No 7. Yes,,,,, 9. Diet, eercise, or illness. T., , 9. 5, 6 _9,, t t 5. T.,., 5. midnight amount noon t (, ) (, ) _ 7. Height of grass price 5. f f 5, 5 f s if 6 if 5 5. AL L L, L 5. A s, 55. S 8, 57. V 6, (a) R (%) (b) $, $9 Wed. Wed. Wed. Wed. Wed. t 9. (a) N 6 (b) In millions: 9; t., 6, a a, a a, a 5a, 6a a, a a, a a, 9a 6a a a, a 6ah h a h. h 5. a 7. { } (, ) (, ) 9.,., 5, (c) 5 T (in dollars) 5 I (in dollars),,,,, I (in dollars) 6. f is odd, t is even 6. (a) 5, (b) 5, 65. Odd 67. Neither 69. Even EXERCISES. N PAGE. (a) Root (b) Algebraic (c) Polnomial (degree 9) (d) Rational (e) Trigonometric (f) Logarithmic. (a) h (b) f (c) t

66 A66 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES 5. (a) b, b= b= (c) [See graph in (b).] where b is the -intercept. b=_ (d) About.5 per population (e) About 6% (f) No. (a) (ft) =+b (b) m m, where m is the slope. See graph at right. (c) m=_ (, ) m= m= 896 (ear) Linear model is appropriate (b) (c) ft (d) No (ft) 7. Their graphs have slope. 9. f. (a) 8., change in mg for ever ear change (b) 8. mg 9. (a) F (b) 5, change in F for ever (, ) C change;, Fahrenheit temperature corresponding 9 F= C+ 5 to C c=_ c=_ -=m(-) c= c= c= ,8 7,7,77; 9 million EXERCISES. PAGE N. (a) f (b) f (c) f (d) f (e) f (f) f (g) f (h) f. (a) (b) (c) (d) 5 (e) 5. (a) (b) (c) 896 (ear) (d) (_, _) C 5. (a) T 6 N 7 6 (b) 6, change in F for ever chirp per minute change (c) 76F 7. (a) P.d 5 (b) 96 ft 9. (a) Cosine (b) Linear. (a) 5 Linear model is appropriate 7. s =_# _ =(+)@ 6, (b) (b). (c) 6, π =+ cos

67 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES A =sin(/) _ =_( +8) 5. Lt sin 65 t 8 7. (a) The portion of the graph of f to the right of the -ais is reflected about the -ais. (b) (c) π _8 = sin = sin π _ =_ =œ + = + =œ tt cos t, f t st h, t, f 9. h s, t sec, f 5. (a) (b) (c) (d) Does not eist; f 6 6 is not in the domain of t. (e) (f) 5. (a) rt 6t (b) A rt 6t ; the area of the circle as a function of time 55. (a) s sd 6 (b) d t (c) s s9t 6; the distance between the lighthouse and the ship as a function of the time elapsed since noon 57. (a) H (b) V (c) V Yes; m m 6. (a) f( 6 (b) t 6. (a) Even; even (b) Odd; even 65. Yes EXERCISES. N PAGE 5. (c). t 5 t Vt Ht Vt Ht 5 t 9. f t 5,, f t,, ft 5 6,, ft, { s}. (a) f t,, (b) t f,, (c) f f,, (d) t t,, (a) f t cos,, (b) t f cos,, (c) f f 9,, (d) t t coscos,, _ 5. (a) f t 6 5,, (b) t f, {, (c) f f, { (d) t t 5, {, } 5 7. f t h 9. f t h s 6. t, f. t s, f 9... _...5 _.5 _5

68 A68 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES. _π π π _ 5 π 5. 5 = =5 = All approach as l, all pass through,, and = all are increasing. The larger the base, the faster the rate of increase = = The functions with base 5 = = greater than are increasing and those with base less than are decreasing. The latter are reflections of the former about the -ais. _ 7. No ,.88. t (a) (b) (c) 9. ^œ _ œ œ $œ %œ _ Œ $œ _.5.5 %œ Œ (d) Graphs of even roots are similar to s, graphs of odd roots are similar to s. As n increases, the graph of s n becomes steeper near and flatter for _ = - =,. (a) e (b) e (c) e (d) e (e) e 5. (a), (b),, 7. f. At (a) (b) t (c),59 (d) t 6.9 h 6, =_ =- e _ =_ _ If c.5, the graph has three humps: two minimum points and a maimum point. These humps get flatter as c increases until at c.5 two of the humps disappear and there is onl one minimum point. This single hump then moves to the right and approaches the origin as c increases.. The hump gets larger and moves to the right.. If c, the loop is to the right of the origin; if c, the loop is to the left. The closer c is to, the larger the loop. EXERCISES.5 PAGE 58 N. (a) f a, a (b) (c), (d) See Figures (c), (b), and (a), respectivel. 7. ab t, where a and b.77676; 598 million; 77 million EXERCISES.6 N PAGE 7. (a) See Definition. (b) It must pass the Horiontal Line Test.. No 5. Yes 7. No 9. No. Yes. No F 9 5 C ; the Fahrenheit temperature as a function of the Celsius temperature; 7.5,. f,. f s ln

69 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES A69 5. e f s (a) It s defined as the inverse of the eponential function with base a, that is, log a &? a. (b), (c) (d) See Figure.. (a) (b) 5. (a) (b) 7. ln 5 9. ln ( )s sin. =log All graphs approach.5 =ln as l, all pass =log through,, and all are increasing. The =log 5 larger the base, the slower the rate of increase. 5 f. About,8,588 mi 5. (a) (b) _5 _ 7. (a) se (b) ln 5 9. (a) 5 log or 5 ln ln (b) ( s e) 5. (a) ln (b) e 5. (a) (, (b) f ln ] ln, [, s) f! 6 =log (+5) f! f =-ln π _ π π _ The second graph is the reflection of the first graph about the line. 7. (a), (b), 7. (a) t f c (b) h c f CHAPTER REVIEW N PAGE 7 True-False Qui. False. False 5. True 7. False 9. True. False. False Eercises. (a).7 (b)., 5.6 (c) 6, 6 (d), (e), (f) No; it fails the Horiontal Line Test. (g) Odd; its graph is smmetric about the origin.. a h 5. (, ) (, ),,, 7. 6,, 9. (a) Shift the graph 8 units upward. (b) Shift the graph 8 units to the left. (c) Stretch the graph verticall b a factor of, then shift it unit upward. (d) Shift the graph units to the right and units downward. (e) Reflect the graph about the -ais. (f) Reflect the graph about the line (assuming f is one-to-one)... =sin! =_sin π =sin π = (+ ) = The graph passes the Horiontal Line Test., where D ss7 6; two of the epressions are comple. 57. (a) f n ln lnn; the time elapsed when there are n bacteria (b) After about 6.9 hours 59. (a) (b) 6. (a) (b) 6. (a) (b) 67. s f () (s 6)(s D 7 s D 7 s ) 5. =_ = + 7. (a) Neither (b) Odd (c) Even (d) Neither 9. (a) f t ln 9,,, (b) t f ln 9,, (c) f f ln ln,, (d) t t 9 9,,..9.88; about 77.6 ears

70 A7 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES. 5. (a) 9 (b) (c) s (d) (b) lim 5 l f means that the values of f can be made arbitraril large negative b taking sufficientl close to through 7. (a). ears values larger than. 5. (a) (b) (c) Does not eist (d) (e) Does not eist 7. (a) (b) (c) Does not eist (d) (e) (f) Does not eist (g) (h) 9. (a) (b) (c) (d) (e) (f) 7,,, 6 P. (a) (b) (c) Does not eist (b) t ln ; the time required for the population. 5. 9P to reach a given number P. (c) ln 8. ears PRINCIPLES OF PROBLEM SOLVING PAGE 8 N. a sh 6h, where a is the length of the altitude and h is the length of the hpotenuse. 7, ; 5. (a).788 (b) [, s) ( s, ] 5. mih 9. f n n CHAPTER EXERCISES. N PAGE 87. (a)., 8.8, 7.8,., 6.6 (b). (c). (a) (i). (ii).658 (iii).556 (iv).55 (v). (vi).895 (vii).8756 (viii).9875 (b) (c) 5. (a) (i) fts (ii) 5.6 fts (iii).8 fts (iv).6 fts (b) fts 7. (a) (i).65 ms (ii) 5.6 ms (iii) 7.55 ms (iv) 7 ms (b) 6. ms 9. (a),.7,.87,.7,.,.87,,.65,.66, 5,.; no (c). EXERCISES. PAGE 96 N. Yes. (a) lim l f means that the values of f can be made arbitraril large (as large as we please) b taking sufficientl close to (but not equal to ). 7. (a).998,.6859,.588,.5868,.885,.898,.65; (b).57,.6,.97,.978,.99,.;. 9. No matter how man times we oom in toward the origin, the graph appears to consist of almost-vertical lines. This indicates more and more frequent oscillations as l...9,.; sin, EXERCISES. PAGE 6 N. (a) 6 (b) 8 (c) (d) 6 (e) Does not eist (f) Does not eist (a), (b) Does not eist 5. (a) (b) (i) (ii) (iii) Does not eist (iv) 7. (a) (i) (ii) (b) No (c) sin

71 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES A7 9. (a) (i) (ii) Does not eist (iii) 9. (b) (i) n (ii) n (c) a is not an integer ; EXERCISES. N PAGE 7. 7 (or an smaller positive number).. (or an smaller positive number) (or an smaller positive number) 7..,. (or smaller positive numbers) 9. (a). (b).. (a) (b) Within approimatel.5 cm (c) Radius; area; ; ; 5;.5. (a).5 (b).5 5. (a).9 (b), where B 6 8 s6 8. Within. EXERCISES.5 PAGE 8 N. lim l f f. (a) (removable), ( jump), ( jump), (infinite) (b), neither;, left;, right;, right (a) (b) Discontinuous at t,,, Cost (in dollars) s cm s B 6B , left 9., right;, left (, e) (, ) (, ) (, ) (, ) (, ) (, ).. (a) t (b) t 5. (b).86, (b) None 6. Yes EXERCISES.6 N PAGE. (a) As becomes large, f approaches 5. (b) As becomes large negative, f approaches.. (a) (b) (c) (d) (e) (f) 5.,,, 7. = f is not defined. 7. lim f does not eist. 9. lim f f. l Time (in hours) = l =, = 9. = = a b (a), (b) 9. ;. ;, f 9., 5., _π. [, ) 5., 7.,,

72 A7 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES 5. (a) (b) An infinite number of times 9.. 7; _.5. (a) 5; 5 6 (b) (a) (b) (a) v* (b)..7 s _ 6 6. N 5 6. N 6, N 65. (a) EXERCISES.7 N PAGE 5 f f f f. (a) (b) lim l. (a) (b) (c) a a a. f, a or f, a. 5. f, a 5 f cos, a or f cos, a 7. ms ; ms 9. Temperature Greater (in magnitude) 7 8 (in F) _ (a) 8a 6a (b), 8 9 (c). (a) Right: t and t 6; left: t ; standing still: t and t (b) v (m/s) _ 5 Time (in hours). (a) (i) percentear (ii) percentear (iii) 6 percentear (b).5 percentear (c) 5 percentear. (a) (i) $.5unit (ii) $.5unit (b) $unit 5. (a) The rate at which the cost is changing per ounce of gold produced; dollars per ounce (b) When the 8th ounce of gold is produced, the cost of production is $7o. (c) Decrease in the short term; increase in the long term 7. The rate at which the temperature is changing at : AM; Fh 9. (a) The rate at which the ogen solubilit changes with respect to the water temperature; mglc (b) S6.5; as the temperature increases past 6C, the ogen solubilit is decreasing at a rate of.5 mglc. 5. Does not eist EXERCISES.8 N PAGE 6 t (seconds). fts 5. a ms ; ms ; ; ms 7 ms 7. t,, t, t, t. (a).5 (b) (c) (d) (e) (f) (g).5 fª

73 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES A7. (a) II (b) IV (c) I (d) III fª fª 9. _ Differentiable at ; not differentiable at 9.. fª fª f fª 6 _ a f, b f, c f a acceleration, b velocit, c position f f, f. 96 to 97 =Mª(t) t _ f e 7. f, f f f 6, 6 f 6, 9. (a) 5. fª a 7 f fªªª if 6 if 6 f fª f, fª or f 6 6 _ 6 7. (a),,, (b),, (c) f 9. f,,. f t 5 8t,,. f,, 5. t s, [, ), (, ) 7. Gt,,,,,, t 9. f,,. (a) f. (a) The rate at which the unemploment rate is changing, in percent unemploed per ear (b) t Ut t Ut 5. corner; discontinuit 7. vertical tangent; corner (a) (b) All (c) CHAPTER REVIEW N PAGE 66 f True-False Qui. False. True 5. False 7. True 9. True. False. True 5. True 7. False 9. False Eercises. (a) (i) (ii) (iii) Does not eist (iv) (v) (vi) (vii) (viii) (b), (c), (d),,, ,. 9. (a) (i) (ii) (iii) Does not eist (iv) (v) (vi)

74 A7 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES (b) At and (c). 5. (a) 8 (b) (a) (i) ms (ii).75 ms (iii).65 ms (iv).55 ms (b).5 ms 9. (a) (b) 6 (c) 9. f t t Vr r 5. As 6s 6 7. G (s) e 9. F 5. a b. s (s) (s) H 9. u 5 t 5 t. A Be. 5. Tangent: ; normal: e (a) (c) (a) The rate at which the cost changes with respect to the interest rate; dollars(percent per ear) (b) As the interest rate increases past %, the cost is increasing at a rate of $(percent per ear). (c) Alwas positive. fª f 9 6, f 8 f 5, f (a) vt t, at 6t (b) (c) a 6 ms 5.,,, , , 6. P No ƒ ƒ 5. (a) f 5 5 (b) (, 5 ], (, 5 ) (c) 6 7. (discontinuit), (corner), (discontinuit), 5 (vertical tangent) 9. The rate at which the total value of US currenc in circulation is changing in billions of dollars per ear; $. billionear 5. PROBLEMS PLUS _ N PAGE a s5 9.. (b) Yes (c) Yes; no. (a) (b) (c) f CHAPTER fª _6 EXERCISES. N PAGE 8 f. (a) See Definition of the Number e (page 79). (b).99,.;.7 e.8. f 5. ft 7. f 69. (a) Not differentiable at or f (b) 9 (, ) ƒ if if a, b 75. m, b ; EXERCISES. N PAGE f e 5. e 7. t 5 9. V 6 6. F tt t 7 t t ƒ

75 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES A75 7. r e r 9. v sv 9. (a) sec cos (b) sec tan sin cos t.. f ACe B Ce (c) cos sin cot csc ft ( st) f c c 7. e ; 8 e EXERCISES. N PAGE 9. ;. cos e s (s).. ; 7. F 5. (a) (b) 9.. tt t F.5 t. sina 5. e k k (_,.5) 7. t cos sin e cos.5 5. F 7. (a) e 9. e, e 7. r 9. costan sec.. (a) 6 (b) 9 (c). sin ln cos. sec tan (a) (b) e e 5. sin 9. (a) t t (b) t tt e e (c) t t 7. cos cotsin csc sin 5. Two, ( s, ( s)) 9. ft sec e t e t e tan t sec t 5. $.67 billionear 55. (c) e. f t sine sin t cose sin t e sin t sin t cos t 57. f e, f e, f 6 6e, f 8 e. t r pln ara r n p a r, f 5 e ; f (n) n nn e 5. costan sec sinssintan ssintan EXERCISES. N PAGE h s, h. f 6 sin. f cos csc 9. e ; cos sin e sin cos tt t cos t t sin t 7. h csc cot e cot csc 55. (a) (b) 9.. f sec tan tan sec tan sec (, ). 5. cos sin f e csc cot _. s s. _.5 5. (a) (b) π 57. (a) f s 59. n,, n,, n an integer π (a) (b) 6, π 65. (a) (b) Does not eist (c) 67. (a) F e f e (b) G e f f cos π 77. vt 5 cost cms db 79. (a) 7 t cos 7. (a) sec tan dt 5 5. (b).6 9. ; cos 8. vt e.5t cos t.5 sin t. (a) f tan sec (b) f cos sin 5. n, n an integer 5. (a) vt 8 cos t, at 8 sin t s (b) s,, s; to the left 7. 5 ftrad 9... sin s 7 cos sin sin

76 A76 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES 8. dvdt is the rate of change of velocit with respect to time; 6. (s, ) 65.,,, 67. (b) 69. dvds is the rate of change of velocit with respect to displacement 85. (a) ab t where a. and b.56 EXERCISES.6 PAGE N (b) 67.6 A. The differentiation formula is simplest. 87. (b) The factored form cosln 9. (b) n cos n sinn. f 5. f ln EXERCISES.5 PAGE N 7. f 9. f sin cos ln5 5s 5 ln. (a) 6 (b),.. t Ft 6. (a) (b), t t 5. ln 5. f u u lnu sin ln log. tan tan cos. ln; ln s 5. e e ; s s 9. e sin cos ln f ; e cos cos ln. 6, e e, f ;,, cos. (a) 9 5 (b) (a) Eight;.,.58 5 _ (, ) 9. sin tan cot sec tan. ln sin. sin cos ln 5. cos tan ln cos 7. tan sec ln tan tan f n n n! n _ (b), (c) s 9. ( 5 s, 5 ). a b s s 9. G arccos s 5. ht 5. e s e 55. arcsin s EXERCISES.7 N PAGE. (a) t t 6 (b) 9 fts (c) t, 6 (d) t, t 6 (e) 96 ft (f) t 8, (g) 6t ; 6 ms s (h) t 6, s t, s a s t, s s 8 (i) Speeding up when t or t 6; slowing down when t or t 6 5

77 . (a) (b) 8s fts (c) t,, 8 sin t (d) t 8 (e) ft (f) (g) (h) t, s _ 6 cost; _ (i) Speeding up when t, t 6; slowing down when t, 6 t 8 5. (a) Speeding up when t or t ; slowing down when t (b) Speeding up when t or t ; slowing down when t or t 7. (a) t s (b) t.5 s; the velocit has an absolute minimum. 9. (a) 5. ms (b) s7 ms. (a) mm mm; the rate at which the area is increasing with respect to side length as reaches 5 mm (b) A. (a) (i) (ii) (iii) (b) (c) A r r 5. (a) (b) (c) The rate increases as the radius increases. 7. (a) 6 kgm (b) kgm (c) 8 kgm At the right end; at the left end 9. (a).75 A (b) 5 A; t s. (a) dvdp CP (b) At the beginning. t ln ; 685 bacteriah 5. (a) 6 millionear; 78.5 millionear (b) Pt at bt ct d, where a.97, b 7.6, c,8.979, d 7,7,77 (c) Pt at bt c (d).8 millionear; 75.9 millionear (smaller) (e) 8.6 millionear 7. (a).96 cms;.69 cms; (b) ; 9.6 cmscm; 85. cmscm (c) At the center; at the edge 9. (a) C..5 (b) $ard; the cost of producing the st ard (c) $.. (a) p p ; the average productivit increases as new workers are added...6 Kmin 5. (a) and (b) C (c),, 5, 5; it is possible for the species to coeist. s t =, s= a 5 8 ft ft.5 t 8, s t, s s fts 8 6 ft ft s. ft ft APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES A77 EXERCISES.8 N PAGE 9. About 5. (a). t (b) 79 (c),6 bacteriah (d) ln ln.. h 5. (a) 58 million, 87 million (b) 6 million (c) 97 million; wars in the first half of centur, increased life epectanc in second half 7. (a) Ce.5t (b) ln.9 s 9. (a) t mg (b) 9.9 mg (c) 99. ears. 5 ears. (a) 7F (b) 6 min 5. (a).c (b) 67.7 min 7. (a) 6.5 kpa (b) 9.9 kpa 9. (a) (i) $88.8 (ii) $8.5 (iii) $85.8 (iv) $85.6 (v) $85. (vi) $85.8 (b) dadt.5a, A EXERCISES.9 PAGE 5 N. dvdt ddt. 8 cm s 5. 5 mmin (a) The plane s altitude is mi and its speed is 5 mih. (b) The rate at which the distance from the plane to the station is increasing when the plane is mi from the station (c) (d) (e) 5s mih. (a) The height of the pole (5 ft), the height of the man (6 ft), and the speed of the man (5 fts) (b) The rate at which the tip of the man s shadow is moving when he is ft from the pole 5 (c) (d) 6 5 (e) fts mih 7. 87s fts cmmin. 55. kmh., 8, cm min 5. cmmin ftmin 9.. m s 7. 8 cm min. 8. s mmin 7. (a) 6 fts (b).96 rads 9. 9 kmmin. 7 65s 96 kmh. s ms EXERCISES. N PAGE 5. L 6. L 5. s ; s.9.95, =- s (, ) =œ - _ (, ) (a) d cos sin d (b) d t t dt _

78 A78 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES. (a) d (b) u du 5. (a) d e d (b).;. 7. (a) d sec d (b). 9..6, d.8.., d.5 = (a) 7 cm,., % (b) 6 cm,.6, 5. (a) ; 8. (b) ; (a) rh r (b) r h. (a).8, 5. (b) Too large.6% 8 7 cm cm EXERCISES. N PAGE 59. (a) (b). (a) (b) e e (a) (b). sech 5, sinh, csch, tanh 5, coth 5. (a) (b) (c) (d) (e) (f ) (g) (h) (i). f cosh. h tanh 5. e cosh sinh 7. f t e t sech e t tanhe t 9. sinh sech. G tanh cosh. s 5. sinh 7. s 5. (a).57 (b) (b) sinh cosh 55. (ln s), s) CHAPTER REVIEW N PAGE 6 d=î True-False Qui. True. True 5. False 7. False 9. True. True d 6r r dr =- d Î d=î Î d Eercises s s 7 5. s 7. sin cos e 9. t coss s sin s. t s. e sec tan tan 9. c e c sin. ln ln ln. 5. cos 7. cos ln 5 9. cot sin cos. 6 tan. 5 sec csc cos(tan s )(sec s ) s cos tansin sec sin s 8. cosh sinh 5. tanh 7. cosh ssinh 9. sin(e stan )e stan sec 5. 7 stan s s ; 6. (a) s5 (b) 7, 8 (c) (, ) (, ) _ 65. (, s), (5, s) 69. (a) (b) 7. t t 7. tt 75. te e 77. tt 79. ft t f f t 8. ftsin tsin cos 8., vt Ae ct c cost sint, at Ae ct c cost c sint 89. (a) vt t ; at 6t (b) t ; t (c) (d) (e) t ; t t 5 ƒ _ a v position

79 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES A79 9. kgm 9. (a). t (b),. (a) (b) (c) 5,9 bacteriah (d) ln 5ln.. h 95. (a) C e kt (b) h 97. cm min 99. fts. fth. (a) L ; s ; s.. (b).. _ cm PROBLEMS PLUS N PAGE 66. ( 9. (, 5 s, ) ). (a) ss rads (b) (cos s8 cos ) cm (c) 5. T,, T,, N (, 5 ), N ( 5, ) 7. (b) (i) 5 (or 7) (ii) 6 (or 7) 9. R approaches the midpoint of the radius AO.. sin a. se 7.,,, 9. s sin ( cos s8 cos ) cms CHAPTER EXERCISES. PAGE 77 N Abbreviations: abs., absolute; loc., local; ma., maimum; min., minimum. Absolute minimum: smallest function value on the entire domain of the function; local minimum at c: smallest function value when is near c. Abs. ma. at s, abs. min. at r, loc. ma. at c, loc. min. at b and r 5. Abs. ma. f 5, loc. ma. f 5 and f6, loc. min. f and f cm min 5. Abs. ma. f 5 7. None 9. Abs. min. f. Abs. ma. f 9, abs. and loc. min. f. Abs. ma. f ln 5. Abs. ma. f 7. Abs. ma. f 9. 5.,., ( s5 ) 5., 7., 9 9., 8 7, f 5, f 7 9. f 8, f 9 5. f 66, f 5. f, f 55. f (s), f s f6 s, f f se, f s 8 e 6. f ln, f ( ) ln 6. f a a b 65. (a).9,.8 (b), 5 s 5 s (a).,. (b) 6 s, C 7. Cheapest, t.855 (June 99); most epensive, t.68 (March 998) 7. (a) (b) v r r 7 kr (c) 7 kr# a a b b a b ab 6 n n an integer 6, 5 5. (a) (b) r r r (c) EXERCISES. N PAGE f is not differentiable on 7..8,.,., (a), (b) (c) s (, ) _.. 5. f is not continous at ln[ 6 ( e 6 )] No. No

80 A8 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES EXERCISES. N PAGE 95 Abbreviations: inc., increasing; dec., decreasing; CD, concave downward; CU, concave upward; HA, horiontal asmptote; VA, vertical asmptote; IP, inflection point(s). (a),,, 6 (b),,, (c), (d),,, 6 (e),. (a) I/ D Test (b) Concavit Test (c) Find points at which the concavit changes. 5. (a) Inc. on, 5; dec. on, and 5, 6 (b) Loc. ma. at 5, loc. min. at 7., 7 9. (a) Inc. on,,, ; dec. on, (b) Loc. ma. f 8; loc. min. f (c) CU on, ; CD on (, ) ; IP (, 7 ). (a) Inc. on,,, ; dec. on,,, (b) Loc. ma. f ; loc. min. f (c) CU on (, s), (s, ) ; CD on (s, s) ; IP (s, 9 ). (a) Inc. on,, 5, ; dec. on, 5 (b) Loc. ma. f s; loc. min. f5 s (c) CU on, 7; CD on,, 7, ; IP,, 7, 5. (a) Inc. on ( ; dec. on (, ln, ) ln ) (b) Loc. min. f ( ln ) (c) CU on, 7. (a) Inc. on, e ; dec. on e, (b) Loc. ma. f e e (c) CU on e 8, ; CD on, e 8 ; IP (e 8, 8 e ) 9. Loc. ma. f 7, loc. min. f. Loc. ma. f ( ) 5. (a) f has a local maimum at. (b) f has a horiontal tangent at (a) Inc. on (, ), (, 6), 8, ; dec. on (, ), (6, 8) (b) Loc. ma. at, 6; loc. min. at, 8 (c) CU on (, 6), 6, ; CD on (, ) (d) (e) See graph at right.. (a) Inc. on,,, ; dec. on, (b) Loc. ma. f 7; loc. min. f (c) CU on (, ) ; CD on (, ; IP (, ) ) (d) See graph at right. 5. (a) Inc. on,,, ; dec. on,,, (b) Loc. ma. f, f ; loc. min. f (c) CU on (s, s) ; CD on (, s), s, ) ; IP (s, 9 ) (d) See graph at right. 7. (a) Inc. on,,, ; dec. on, (b) Loc. ma. h 7; loc. min. h (c) CU on, ; CD on, ; IP, (d) See graph at right. 9. (a) Inc. on, ; dec. on, (b) Loc. min. A (c) CU on, (d) See graph at right. 6 8 _ (_, 7) _ (_, 7) _, œ 9 (_, ) (_, ) _ 7, _ _ (, _) (, _), œ 9 (, ) _ = _. (a) Inc. on, ; dec. on, (b) Loc. min. C (c) CU on,,, ; CD on, ; IPs,, (, 6s) (d) See graph at right.. (a) Inc. on, ; dec. on, (b) Loc. min. f (c) CU on, 5; CD on,, 5, ; IP (, 5 ), (5, 5 ) (d) See graph at right. π 5, 5π 5, π π _ (π, _) _ (_, _) {, 6 Œ }

81 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES A8 5. (a) HA, VA, 55. e 57. e e 6 (b) Inc. on,,, ; 6. se 65. e a = 8. (a) dec. on,,, (c) Loc. ma. f EXERCISES.5 N PAGE (d) CU on,,, ; CD on,. A. B. -int. ; -int. (e) See graph at right. =_ = C. About, D. None 7. (a) HA E. Inc. on, F. None G. CU on, ; CD on, ; (b) Dec. on, (c) None IP (, ) H. See graph at right. (d) CU on, (e) See graph at right. 9. (a) VA, e (b) Dec. on, e (c) None (d) CU on (, ); CD on, e; IP (, ) (e) See graph at right. 5. (a) HA, VA (b) Inc. on,,, (c) None (d) CU on,, (, ); CD on (, ) ; IP (, e ) (e) See graph at right. 5., 55. (a) Loc. and abs. ma. f s, no min. (b) ( s7) 57. (b) CU on.9,.57,.7, 5.5; CD on,.9,.57,.7, 5.5, ; IP.9,.,.57,.6,.7,.6, 5.5,. 59. CU on,.6,., ; CD on.6,. 6. (a) The rate of increase is initiall ver small, increases to a maimum at t 8 h, then decreases toward. (b) When t 8 (c) CU on, 8; CD on 8, 8 (d) 8, 5 6. K K; CD min, when the rate of increase of drug level in the bloodstream is greatest; 85.7 min, when rate of decrease is greatest 67. f 9 7 EXERCISES. N PAGE. (a) Indeterminate (b) (c) (d),, or does not eist (e) Indeterminate. (a) (b) Indeterminate (c) pq ln aa = =_ (, ) =e =. A. B. -int. ; -int., (7 s5) C. None D. None E. Inc. on (, 5); dec. on,, 5, F. Loc. min. f 5; loc. ma. f 5 7 G. CU on, ; CD on, ; IP, H. See graph at right. 5. A. B. -int. ; -int., C. None D. None E. Inc. on, ; dec. on, F. Loc. min. f 7 G. CU on,,, ; CD on, ; IP (, ),, 6 H. See graph at right. 7. A. B. -int. C. None D. None E. Inc. on,,, ; dec. on, F. Loc. ma. f ; loc. min. f G. CU on (s, ) ; CD on (, s ); IP (s, 9( s 6)) H. See graph at right. 9. A. B. -int. ; -int. C. None D. VA, HA E. Dec. on,,, F. None G. CU on, ; CD on, H. See graph at right. (, _5) (_, _7), (5, 7), _

82 A8 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES. A. B. -int.. A. B. -int. ; -int. 9 C. About -ais D. VA, HA C. About the origin = E. Inc. on,,, ; D. HA dec. on (, ),, E. Inc. on, F. None (, ) F. Loc. ma. f G. CU on, ; 9 =_ G. CU on,,, ; CD on, ; IP, CD on, H. See graph at right. H. See graph at right. 5. A. {, },, B. -int. C. About (, ) D. VA. A. B. -int. ; -int. E. Dec. on,,, C. About (, ) D. HA, 6 F. None E. Inc. on, ; G. CU on (, s), (, s) ; dec. on,,, CD on (s, ), (s, ) ; F. Loc. min. f ; loc. ma. f 6 IP (s, s) 6; H. See graph at right. G. CU on (s, ), (s, ) ; _, _ 6 CD on (, s), (, s) ; 7. A. B. -int. ; -int., s C. About the origin IP (, ), (s, s) D. None E. Inc. on,,, ; dec. on, H. See graph at right. F. Loc. ma. f ; _, loc. min. f 5. A.,, B. -int. G. CU on, ; CD on, ;, C. None D. HA ; VA,, 9 IP, E. Inc. on, ; _œ, dec. on,,, H. See graph at right., _ F. Loc. ma. f G. CU on, ; 9. A. B. -int. ; -int. CD on,,, ; IP (, C. About -ais D. None 9) H. See graph at right E. Inc. on, ; dec. on, F. Loc. min. f 7. A. B. -int., -int. = G. CU on, ; C. About -ais D. HA CD on,,, ; (_, ) (, ) E. Inc. on, ; dec. on, F. Loc. min. f IP, H. See graph at right. (, _) G. CU on, ; _,, CD on,,, ; IP (, ). A. B. -int. ; -int. ( n an integer) (, ) H. See graph at right C. About the origin, period D. None E. Inc. on ; π, 9. A., 5 B. -int. ; -int., 5 œ 5, dec. on 9 C. None D. None F. Loc. ma. f n ; E. Inc. on (, ; dec. on ( ), 5) loc. min. f n _π F. Loc. ma. f ( ) _ 9 s5 G. CU on n, n; _ G. CD on, 5 π CD on n, n ; IP n, _, _ H. See graph at right. H. See graph at right. n n n n, n, œ, π. A.,, B. -int., C. None D. None E. Inc. on (, ) ; dec. on (, ) F. None G. CD on,,, H. See graph at right. _. A., B. -int. ; -int. C. About -ais D. VA E. Inc. on, ; dec. on, F. Loc. min. f G. CU on, H. See graph at right. π π

83 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES A8 5. A., 5. C. None D. None A. B. -int. C. None E. Inc. on, 5, 7, ; D. HA, dec. on,, 5, 7 E. Dec. on F. None = F. Loc. min. f 6, f 7 76 ; loc. ma. f 5 56 s s G. CU on (ln, ) ; CD on, ln ) ; s IP (ln, 9) ln, 9 G. CU on,,, ; H. See graph at right. CD on, ; IP,,, π 5π 7π H. See graph at right. 7. A. All in n, n ( n an integer) π π π B. -int. n C. Period D. VA n E. Inc. on n, n; dec. on n, n 7. A. All reals ecept n ( n an integer) F. Loc. ma. f n G. CD on n, n B. -int. ; -int. H. C. About the origin, period D. VA n E. Inc. on n, n F. None _π _π _π _π π π π π G. CU on n, n ; CD on n, n; IP n, H. =_π =_π =π =π n 9. A. B. -int. C. Period D. None Answers for E G are for the interval,. E. Inc. on,,, ; dec. on, F. Loc. ma. f e; loc. min. f e G. CU on,,, where, ; CD on, ; IP when, H. e _π _π π. A. B. -int. C. None D. HA, E. Inc. on F. None G. CU on, ; CD on, ; IP (, ) H. See graph at right.. A., B. None C. None D. VA E. Inc. on, ; dec. on, F. Loc. min. f G. CU on, H. See graph at right. sin ( ( s5)) π π (, ) 9. A. B. -int. ; -int. C. About (, ) D. HA E. Inc. on (s, s) ; dec. on (, s), (s, ) F. Loc. min. f (s) se; loc. ma. f (s) se G. CU on (s, ), (s, ) ; CD on (, s), (, s) ; IP (s, se ),, H., 5. A. B. -int. C. None D. None E. Inc. on ( ; dec. on (, 5 ln 5 ln, ) ) F. Loc. min. f ( 5 ln ) ( ) 5 ( ) 5 G. CU on, H. See graph at right. 5. m 55. (, m ) =c œ œ e L/ local minimum (, ) L

84 A8 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES 6. A. B. -int. ; -int. (5 s7) C. None D. VA ; SA E. Dec. on (, ), (, ) F. None G. CU on (, ) ; CD on (, ) H. See graph at right 6. A. B. None C. About (, ) D. VA ; SA E. Inc. on,,, ; dec. on,,, F. Loc. ma. f ; loc. min. f G. CU on, ; CD on, H. See graph at right. 65. A. B. -int. ; -int. C. None D. SA E. Inc. on, F. None G. CU on (, s), (, s); CD on (s, ), (s, ) ; IP (s, s),, H. See graph at right VA, asmptotic to π EXERCISES.6 PAGE N. Inc. on.9,.5,.58, ; dec. on,.9,.5,.58; loc. ma. f.5 ; loc. min. f.9 5., f ; CU on,.6,.5, ; CD on.6,.5; IP.6,.,.5,.999 (, ) (, ) IP,,., 6,5,,.9,,8, π ƒ {_œ, _ œ +}. (, ) _ ƒ =_ (_, _) =+ = =_+ (, ) = {œ, œ +} 5.8, 8,5, _5 5. Inc. on,.7,.7,.,., ; dec. on,.6,.6, ; loc. ma. f ; CU on,.7,.56,.,.6, ; CD on.7,.56,.,.6; IP.56,.9 7. Inc. on.9,.7,.89, ; dec. on,.9,.7,.89; loc. ma. f ; loc. min. f , f ; CU on,.8,.8, ; CD on.8,.8; IP.8, 8.77,.8,.8 _ 9. Inc. on (8 s6, 8 s6) ; dec. on (, 8 s6), 8 s6,,, ; CU on ( s8, s8),, ; CD on (, s8), ( s8, ).95. (a) ƒ,, f _ f _,, 5 _5 5 _.5 6 f 6, _, 75 ƒ f ƒ..7 _6.96. Inc. on 5,., 8.9, ; dec. on, 5,., 8.9; loc. ma. f. 5,8; loc. min. f 5 9,7,, f 8.9,7,; CU on,.,,.9, 5.8, ; CD on.,,.9, 5.8;, _.5.75 _.5 (b) lim l f (c) Loc. min. f (se) e; CD on, e ; CU on e, œ e e

85 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES A85. Loc. ma. f 5.6.8, f.8 8.5, f 5..5; loc. min. f. Inc. on,,, ; CU on,.,,.;. CD on.,,., ; IP., ƒ _ ƒ _.. (a) 5. 8 _ 5. 5 f f CU on 5., 5.,,.5,.,,,,, ; CD on, 5., 5.,,.5,.; IP 5.,.5, 5.,.5,,,.5,.,., Inc. on,.; dec. on., ; loc. ma. f..; CU on.9, ; CD on,.9; IP.9, (b) lim, lim l l (c) Loc. ma. f e e e (d) IP at.58,.7 5. Ma. f.59, f.68, f.96 ; min. f , f.6.9, f.7.5; IP.6,.99998,.66,.99998,.7,.7,.75,.77,.8,.. f π. f. 5 _π π 9. Inc. on.9,.5,,.77,.9, 8.6,.79,., 7.8, ; dec. on.5,.,.77,., 8.6,.79,., 7.8; loc. ma. f.5.6, f.77.58, f 8.6.6, f..9; loc. min. f.79., f 7.8.9; CU on 9.6,.5, 5.8, 8.65; CD on.9,.,,.,.9, 9.6,.5, 5.8, 8.65, ; IPs at 9.6,.95,.5,.7, 5.8,.9, 8.65, For c, there is no IP and onl one etreme point, the origin. For c, there is a maimum point at the origin, two minimum points, and two IPs, which move downward and awa from the origin as c l.. f _.. _5 _.

86 A86 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES 9. There is no maimum or minimum, regardless of the value of c. For c, there is a vertical asmptote at, lim l f, and lim l f. c is a transitional value at which f for. For c, lim l f, lim l f, and there are two IPs, which move awa from the -ais as c l. c=.5 c= c= c=_ c=_ c=_.5 7. (a) Positive (b) _ c=_ c=_ c= c= c=.5 c=. c=. _6 6 c=. For c, the maimum and minimum values are alwas, but the etreme points and IPs move closer to the -ais as c increases. c is a transitional value: when c is replaced b c, the curve is reflected in the -ais.. For, the graph has local maimum and minimum values; for c it does not. The function increases for c and decreases for c. As c changes, the IPs move verticall but not horiontall. 5. c _5 5 For c, lim l f and lim l f. For c, lim l f and lim l f. As c increases, the maimum and minimum points and the IPs get closer to the origin. c= c= c=_ c=_ c=.5 c= c=_.5 EXERCISES.7 N PAGE 8. (a), (b).5,.5., 5. 5 m b 5 m 7. N 9. (a) 5,5 ft@ (b) 5 5,5 ft@ 9 ft@ A 75 5 (c) A (d) 5 75 (e) (f),6.5 ft. ft b 5 ft. cm 5. $ ( 8 7, 7 7 ) 9. (, s). Square, side sr. L, s L 5. Base sr, height r 7. r (s) 9. r ( s5). cm, 6 cm. (a) Use all of the wire for the square (b) s(9 s) m for the square 5. Height radius s V cm 7. V R (9s). E r. (a) S csc csc s cot (b) (c) 6s[h s(s)] cos (s) Row directl to B km east of the refiner 9. s ( s ) ft from the stronger source 5. a b 5. (b) (i) $,9; $unit; $9unit (ii) (iii) $unit 55. (a) p 9 (b) $ (a) p 55 (b) $75 (c) $ m in At a distance 5 s5 from A 7. L W 7. (a) About 5. km from B (b) C is close to B; C is close to D; WL s5, where BC (c).7; no such value (d) s.6 75

87 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES A87 EXERCISES.8 N PAGE 8. (a)., (b) No , , , ,.9975, , , (b) (a).97,.7,.5785 (b) , , % EXERCISES.9 N PAGE 5. F C. F 5 C 5. F C 7. F 5 7 C 9. F C. F 5 8 C if 5 8 C if Fu u 6u C G sin 5 cos C F 5e sinh C F ln C F 5 6. C D 5. 8 C D 7. e t Ct Dt E sin t tan t s 5. if ; 5 if sin cos cos 5. ln ln ln b 5. F fts 75. 6,5 kmh.8 ms 77. (a).95 mi (b).675 mi (c) min s (d) 55.5 mi CHAPTER REVIEW N PAGE 7 True-False Qui. False. False 5. True 7. False 9. True. True. False 5. True 7. True 9. True Eercises. Abs. ma. f 5, abs. and loc. min. f ; loc. min. f. Abs. ma. f, abs. and loc. min. f ( ) Abs. ma. f ; abs. min. f ; loc. ma. f ; loc. min. f s s A. B. -int. C. None D. None E. Dec. on, F. None G. CU on, ; CD on, ; IP, H. See graph at right. 9 = =_ _ (, ) (, ) (, ) _π π 57. st cos t sin t 59. st 6 t t t 6. st sin t cos t 6t 6. (a) st 5.9t (b) s s (c) 9.8s ms (d) About 9.9 s ft 69. $ s F. A. B. -int. ; -int., C. None D. None E. Inc. on (, ), dec. on (, ) F. Loc. min. f ( ) 7 56 G. CU on (, ),, ; CD on (, ) ; IP (, 6 ),, H. See graph at right.. A., B. None C. None D. HA ; VA, E. Inc. on, ; dec. on,,,,, F. Loc. min. f G. CU on,,, ; CD on, H. See graph at right.

88 A88 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES 5. A Inc. on.,,.6, ; dec. on,.,,.6; B. -int., -int. C. None loc. ma. f ; loc. min. f..96, f.6 9.; 8 D. VA 8; SA 8 CU on,.,., ; E. Inc. on, 6,, ; CD on.,.; IP.,.98,.,. 6, dec. on 6, 8, 8, F. Loc. ma. f 6 ; f loc. min. f f _. G. CU on 8, ; CD on, 8 H. See graph at right. 7. A., B. -int. ; -int., C. None D. None E. Inc. on (, ), dec. on (, ) F. Loc. min. f ( ) 9 s6 G. CU on, œ 6 _, _ 9 H. See graph at right. 9. A. B. -int. C. About -ais, period D. None E. Inc. on n, n, n an integer; dec. on n, n F. Loc. ma. f n ; loc. min. f n G. CU on ; CD on ; IP H. n, n n, n 5 _π _π { }. A. π B. None C. About (, ) D. HA _ E. Dec. on,,, π _ F. None G. CU on, ; CD on, H. See graph at right.. A. B. -int., -int. C. None D. HA E. Inc. on (, ), dec. on (, ) F. Loc. ma. G. CU on, ; CD on, ; IP, e H. _ π, e! {, π (n, ) f ( ) e 9..8,.; (s, e ) _5 _..96,.8,.;.57,.57;.6,.75,.6,.. For C, f is periodic with period and has local maima at, n an integer. For C, f has no graph. For C, f has vertical asmptotes. For C, f is continuous on. As C increases, f moves upward and its oscillations become less pronounced. 9. (a) (b) CU on 5. sr 55. s cm from D 57. L C 59. $ f sin sin C f C f t t cos t f ( st t tan t 75. (b).e cos.9 (c) 5 n _ 77. No 79. (b) About 8.5 in. b in. (c) s in., s in. 5 _.5 _.5 F. 5. Inc. on (s, ), (, s) ; dec. on (, s), (s, ) ; loc. ma. f (s) 9 s, loc. min. f (s) 9 s; CU on (s6, ), (s6, ) ; CD on (, s6), (, s6) ; IP (s6, 5, (s6, 5 6 s6) 6 s6).5 ƒ _5 5 _.5 PROBLEMS PLUS PAGE 5 N 5. 7.,,,..5 a.5. m, m 5. a e e 9. (a) T Dc, T h sec c D h tan c, T sh D c (c) c.85 kms, c 7.66 kms, h. km. (s ) h

89 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES A89 CHAPTER 5 EXERCISES 5. N PAGE 76 EXERCISES 5. N PAGE 6. (a), 5 (b)., 9. =ƒ 5 5 =ƒ. 6 The Riemann sum represents the sum of the areas of the two rectangles above the -ais minus the sum of the areas of the three rectangles below the -ais; that is, the net area of the rectangles with respect to the -ais. ƒ= (a).798, underestimate (b).85, overestimate ƒ=cos ƒ=cos..986 The Riemann sum represents the sum of the areas of the three rectangles above the -ais minus the area of the rectangle below the -ais. 6 5 ƒ= - π π π π (a) 8, (b) 5, 5.75 (c) 5.75, M 6 (d) 7..5,.7,.,.5;. 9. (a) Left:.8,.797,.79; right:.76,.777, ft,.8 ft. 6. L, 7 L ft lim n l i i i lim s n cos n l n n n 5in 5n i n. The region under the graph of tan from to 6 n n n n. (a) lim n i 5 (b) (c) n l n 6 i 5. sin b, π π π π (a) (b) 6 (c) 7. 75, ,.9869, n R n The values of R n appear to be approaching ln d 9. 8 s. d in lim n l. lim n i in 5 n n l isin 5i n n n 5. (a) (b) (c) (d) e 5 e 7. 5 f d m f d M b Comparison Propert tan d s s d e d e 69. d 7. EXERCISES 5. N PAGE 87. One process undoes what the other one does. See the Fundamental Theorem of Calculus, page 87.. (a),, 5, 7, (d) (b) (, ) (c) g _

90 A9 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES 5. (a), (b) =t@ 9. sin C 5 _ t t sin. F s sec. h arctan 5. stan stan sec ln e.. The function f is not continuous on the interval,, so FTC cannot be applied. 5. The function f sec tan is not continuous on the interval,, so FTC cannot be applied t sin s sin 57. s s 6. (a) sn, sn, n an integer (b),, (sn, sn ), and (sn, sn ), n an integer (c).7 6. (a) Loc. ma. at and 5; loc. min. at and 7 (b) (c) (, ),, 6, 8, 9 (d) See graph at right. _ = t f, a (b) Average ependiture over, t; minimie average ependiture EXERCISES 5. N PAGE C 8 C 9. t t t t C. s C. cos cosh C 5. csc C 7. tan C _. 8. e s ,.; The increase in the child s weight (in pounds) between the ages of 5 and 5. Number of gallons of oil leaked in the first hours 5. Increase in revenue when production is increased from to 5 units 55. Newton-meters (or joules) 57. (a) m (b) 59. (a) vt (b) 6 t t 5 ms m 6. 6 kg 6.. mi 65. $58, (b) At most %; EXERCISES 5.5 N PAGE 6. e C cos 9 C C 9. 6 cos C C. C. ln 5 C 5. ( cos t C 7. sa b C 9. ln C. sin st C. 7 sin 7 C e C C 9. e tan C. sin C. cot C ln 9.. ln sin C C C sec C tan ln C C 8 C 9. sin C _ e se (s )a s 5 lne All three areas are equal L cos t L 5 6 F f _ 8 _ CHAPTER 5 REVIEW N PAGE _.5 True-False Qui. True. True 5. False 7. True 9. True. False. False 5. False _6 ƒ F 6 m π

91 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES A9 Eercises. (a) 8 (b) 5.7. =ƒ =ƒ f is c, f is b, f t dt is a Does not eist 9. sin ( sin t ln C s C C 9. e s C. lncos C. ln C 5. ln sec C s sin C. 5. F 5. t cos 8 7. (e e s ) 9. s d s Number of barrels of oil consumed from Jan.,, through Jan., , c f e e =9 (6, 9) = =œ (, ) = = PROBLEMS PLUS N PAGE.. f e 9.,. (a) n n (b) bb b aa a 7. (s ) CHAPTER 6 = (, ) EXERCISES 6. N PAGE.. e e ln ln ln s ,.9; s ft. cm 5. (a) Car A (b) The distance b which A is ahead of B after minute (c) Car A (d) t. min 7. 5 s m ; m ln m EXERCISES 6. N PAGE. 9 = =- = =. 6. π _, ( s) = =œ π, = =+sec = (, ) = = =

92 A9 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES = = = (, ) e =e _ (, _) 7. 9 = (, ) = _ =_ 7. 6 =(-)@ 7 = -+7 (, ) (, ) tan d sin d s [5 (s ) ] d.88,.88; Solid obtained b rotating the region cos, about the -ais. Solid obtained b rotating the region above the -ais bounded b and about the -ais 5. cm 7. (a) 96 (b) r h 5. h (r h) 5. b h 55. cm (a) 8R r sr d (b) 65. (b) 5 r h r sr sr d EXERCISES 6. PAGE 6 N. Circumference, height ; s r r R ln d. sin d 5. ssin d Solid obtained b rotating the region, about the -ais. Solid obtained b rotating the region bounded b (i),, and, or (ii),, and about the line ln.. r 5. r h 8 EXERCISES 6. PAGE N J. 9 ft-lb 5. 8 J 7. ft-lb 5 9. (a). J (b).8 cm. W W 875. (a) 65 ft-lb (b) ft-lb 5. 65, ft-lb J 9. 5 J..6 6 J m 9. Gm m a b 5 ft-lb

93 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES A9 EXERCISES 6.5 N PAGE e (a) (b), (c) 5. (a) (b) (c)., cos sin cos C 9. ln C. t arctan t 8 ln 6t C. t tan t lnsec t C 5. ln ln C 7. e sin cos C 9.. e. 5. ln e 7. 6 ( 6 s) 9. sin ln sin C. 5 ln 6 5 ln 6 5. s sin s cos s C ln C 9. e C 7 ƒ F F 59F 9. 6 kgm. 5. L CHAPTER 6 REVIEW N PAGE 6 Eercises ( )(cos ah h ) d 5. (a) 5 (b) 6 (c) (a).8 (b) Solid obtained b rotating the region cos, about the -ais. Solid obtained b rotating the region, sin about the -ais s m 7.. J 9. (a) ft-lb (b). ft. f PROBLEMS PLUS N PAGE 8. (a) f t t (b) f s. 5. (b).6 (c).676 m (d) (i) 5. ins (ii) 7 s 6.5 min (a) V h f d (c) f skac Advantage: the markings on the container are equall spaced.. b a 5. B 6A CHAPTER 7 EXERCISES 7. N PAGE 57. ln 9 C. 5. r e r C 7 5 sin 5 5 cos 5 C.. (b) cos sin 8 6 sin C 5. (b), ln ln 6 ln 6 C e ,.87; e 6. ln 9 6. e t t t m 65. EXERCISES C N PAGE cos 5 cos C. 5. sin sin 5 sin 7 C ssin 5 8 sin 5 sin C ln sin cos ln cos C sin C. tan C. tan C tan 5 t tan t tan t C sec sec C. sec tan ln sec C. 6 tan 6 tan C 5. sec ln sec tan C 7. s 9.. ln csc cot C csc 5 csc 5 C sin 6 cos 6 cos C sin 6 C 7. sin C 9. tan 5 t C _ 5 F f _ 8 7 sin sin C

94 A9 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES 5. sin cos C s cos t EXERCISES 7. PAGE 7 N. s 99 C. 8s 9 C 5. s8 7. s5 5 C 9. ln(s 6 ) C. sin s C. 6 sec s 9 C 5. 6a ln. 9 s 7 C (s ) s C. sin s5 C 5. s ln(s ) C 7. s ln s C 9. sin s C. 6 (s8 sec 7) 7..8, ;.. rsr r r R arcsinrr. EXERCISES 7. N PAGE 8. (a) A B (b). (a) A B C (b) A 5. (a) A B C D At B Ct D (b) t t Et F t 7. 6 ln 9. ln 5 6 ln C C ln ln 9 5 ln (or 9 5 ln 8 ) _π a ln b C 6 ln 5 6 ln tan C ln ln C ln ln 9 tan C ln 5 tan ln 6 ln tan s ln 8 F. 5. f π _π D E A B C B D 5 6 ln C C ln (s ) tan (s ) C π 6 ln ln 6 sin 8 sin 9 C F _ ln ƒ (s 5 ) 5 Rr C C s 8 C s tan s ln s s C. ln C 5. s s 7. ln 6s e 6 ln s 6 C C e 9. ln tan t ln tan t C 5. ( ) ln s7 tan C s7 5. ln.55 tan 55. ln ln tan C C 6. ln 6. ln 65. t ln P 9 ln.9p 9 C, where C., (a) ,55 7,98 8,95 6,5 5 (b) ln 5 ln,9 6,5 ln 5 s9 The CAS omits the absolute value signs and the constant of integration. EXERCISES 7.5 N PAGE 88 sin sin C sin ln csc cot C.. 5. ln 9 7. e e 9.. ln 5 tan 5 ln 5 C. 8 cos 8 6 cos 6 C (or sin sin 6 8 sin 8 C) 5. s C 7. sin cos sin C (or sin 8 cos C) 9. e e C. arctan s s C ln 5 5 ln C 7. ln e C ln 7. sin s. sin C s C ln sec ln sec C. C. e C 5. e C 7. ln s ln s ln C s C C 5. m coshm m sinhm coshm C m tan ln sec 8 6 C 75,77 tan 6,5s9 6 8,55 ln 7 C

95 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES A ln s ln( s ) C 7. F sin cos 7 8 sin cos 7 6 sin cos c 7 c c C 8 sin cos 56 sin cos 56 ; 59. sinsin sin sin C 6. ( s )e s C ma. at, min. at ; IP at.7,, and.5 6. tan cos C 65. C. 67. s s ln( s) ln( s) ƒ e ln e C s arcsin C F 7. 8 ln 6 ln 8 tan C 75. s e ln s e π s e C 77. tan C 79. sin cos EXERCISES 7.7 PAGE 55 9 cos C 8. e C. (a) L 6, R, M 9.6 (b) L is an underestimate, R and M are overestimates. EXERCISES 7.6 N PAGE 9 (c) T 9 I (d) L n T n I M n R n. s7 s sin (ss7) C. (a) T (underestimate). sec tan ln sec tan C (b) M.9897 (overestimate) T I M tan ln cos C 5. (a) , E M.65 (b) , E S s 99 C. e 7. (a).79 (b).5 (c).. tan ln cos C 9. (a).6879 (b).79 (c) e arctane e C. (a).598 (b).599 (c) s sin. (a).568 (b).7856 (c).675 s7 5. (a).95 (b).5 (c).56 6 C 7. (a).675 (b).676 (c) sin lnsin C 9. (a) T 8.9, M s ln e s e C (b),.78 EM.9 s (c) n 7 for T n, n 5 for M n. tan sec 8 tan sec 8 ln sec tan C. (a) T.985, E T.676; 5. ln s ln ln[ln s ln ] C M.88, E M.88; 7. se cos e C S., E S ln 5 s C (b),. EM.99 ES.7 5. tan sec (c) n 59 for T n, n 6 for M n, n for S n tan C 7. s ln(s ) C. (a).8 (b) (c) (d) (e) The actual error is much smaller. C (f ).9 (g) (h).59. ln cos tan tan C (i) The actual error is smaller. (j) n 5 s. (a) ln ; C 5. n L n R n T n M n both have domain,, F ln ln ; ma. at, min. at ; IP at.7,, and f F n E L E R E T E M Observations are the same as after Eample.

96 A96 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES 7. Observations are the same as after Eample. 9. (a) 9.8 (b).6 (c).5. (a). (b) fts 5.,77 megawatt-hours EXERCISES 7.8 N PAGE 55 Abbreviations: C, convergent; D, divergent. (a) Infinite interval (b) Infinite discontinuit (c) Infinite discontinuit (d) Infinite interval. t ;.95,.9995, ; D 9. e. D. 5. D 7. D D D 9.. D. 5. D 7. 8 e 9. ln 8 9. e. e 5. Infinite area 7. (a) n n.5.5 t T n E T It appears that the integral is convergent. M n E M =sec@ t sin d ,.67957,.677 S n E S π.5 9 = +9 _7 7 (c). ƒ= 9. C 5. D 5. D p, p 59. p, p 65. sgmr 67. (a) = sin@ =F(t) 7 t (in hours) (b) The rate at which the fraction Ft increases as t increases (c) ; all bulbs burn out eventuall (a) Fs s, s (b) Fs s, s (c) Fs s, s 77. C ; ln 79. No CHAPTER 7 REVIEW N PAGE 58 True-False Qui. False. False 5. False 7. False 9. (a) True (b) False. False. False Eercises. 5 ln. ln cosln t C 9. 5 ln 5. s. e s (s s ) C ln ln C sec ln sec tan C 8 ln tan [ ( )] C. ln s. ln C s C 5. ln tan s tan (s) C sin C s 5 5. s s C 7. sin 8 cos C 9. 8 e. 6. D 5. ln ln arctan C s ln s C

97 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES A sin s sin ln(sin s sin ) C 9. s ln( s ). 6. No. 7(5s5 s) 5. a 6. (a).95 (b).995 (c) (a).8, n 68 (b).67, n mi. 69. (a).8 (b).7867,.66 (c) n 6[ln(s ) s] 7. C (a) 56 a (b) 5sa PROBLEMS PLUS PAGE 5 9. (a) b a b sin (sa b a) N sa b. About.85 inches from the center. 7. f. b b a a ba e (b) a ab sin (sb a b). sin sb (s5) a. b c f s f d. a CHAPTER 8 6 [ ln(s7 ) ln(s ) s7 s] r EXERCISES 8. PAGE 5 N. s5. s 5. s9 6 d sin d 6 7. (8s8 ) 9... ln(s ) 5. ln 7. s e s ln(s e ) ln(s ) 6 9. s ln( s) (a), (b) L, L 6., L 7.5 EXERCISES 8. PAGE 57 N. (a) 87.5 lbft (b) 875 lb (c) 56.5 lb. 6 lb N N 9.. lb. ah N 5. (a) N (b) 5 N 7. (a) 5.6 lb (b) 5.6 lb (c).88 lb (d). 5 lb N. ;. ; ; (, 7 ) 5., (, 9 ) 9 e, e s.. (, ) (s ), (s ) 5. 6; 6; ( 8, ) 7..78,.. (, ) 5. r h (c) s (d) d s5 ln( ( s5)) s ln( s) EXERCISES 8. N PAGE 55. $8,. $,866,9. 5. $ $, 9. 77; $7,75 kb k a k. (6s 8) $9.75 million. kb k a k 5..9 cm s Lmin Lmin. s 7 [ 9 s] 5. s (s ) m in... EXERCISES 8. N PAGE 57. (a) (b) s 6 6 d s 6 6 d. (a) d (b) d 5. 7(5s5 ) tan EXERCISES 8.5 PAGE 56 N. (a) The probabilit that a randoml chosen tire will have a lifetime between, and, miles (b) The probabilit that a randoml chosen tire will have a lifetime of at least 5, miles. (a) f for all and f d (b) 8 s.5 5. (a) (b) 7. (a) f for all and f d (b) 5. (a) e.5. (b) e.5.55 (c) If ou aren t served within minutes, ou get a free hamburger.. % 5. (a).668 (b) 5.%

98 A98 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES 9. (b) ; a (c) EXERCISES 9. PAGE 578 N. (a) (b) (iv),, (d) e (e) a (i) (ii) _ CHAPTER 8 REVIEW N PAGE 56 _ (iii) Eercises 5.. (a) 6 (b) lb. ( 8 5, ). (, ) $ (a) f for all and f d (b).55 (c) 5, es 5. III 5. IV (b) (a). (a) e 8. (b) (c) 8 ln 5.55 min e 5.9 PROBLEMS PLUS N PAGE 56 _ (c) _. s. (a) rr d (b).6 6 mi (d) mi.. 5. (a) P P t d (b) P thr the LH r e H sr d r 7. Height s b, volume 7 s6 )b 9.. m., ( 8 CHAPTER 9 5. EXERCISES 9. N PAGE 57. (a), 5. (d) 7. (a) It must be either or decreasing (c) (d) 9. (a) P (b) P (c) P, P. (a) At the beginning; stas positive, but decreases (c) P(t) M 7. c= c ;,, c= _ t c= P() t c=_

99 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES A99 9. (a) (i). (ii). (iii).6 (b) =.5 h=.. h=. h=..... Underestimates (c) (i).98 (ii).58 (iii).77 It appears that the error is also halved (approimatel)..,, 6.5, (a) (i) (ii).98 (iii).7 (iv).68 (c) (i).6 (ii).9 (iii). (iv). It appears that the error is also divided b (approimatel). 7. (a), (d) Q 6 (b) (c) Yes; Q (e).77 C.... t EXERCISES 9. N PAGE 586. K. Ks ln sec C ste t e t C 9. u Ae tt. s 9. cos sin e 5. u st tan t 5 7. a sin a s 9. e. Ke. (a) sin C (b) sin, s s (c) No =sin _œ π/ œπ/ 5. cos cos 5 7. (a), (c) 5 (b) s C _. Qt e t ; 5. Pt M Me kt ; M 7. (a) a (kt sa) (b) t ksa btan b b a b tan a b 9. (a) Ct C rke kt rk (b) rk; the concentration approaches rk regardless of the value of C. (a) 5e t kg (b) 5e.. kg. About.9% 5. tk kt 7. (a) dadt ksa M A (b), where C _5 9. C. C _ EXERCISES 9. sm sa sm sa N PAGE 598 and A A. (a) ;.5 (b) Where P is close to or ; on the line P 5; P ; P (c) P _ 5 P = P = P =8 P =6 5 P = P = 6 t _5 5 _ - =C =k At M CesM Ce sm kt.5.5 Solutions approach ; some increase and some decrease, some have an inflection point but others don t; solutions with P and P have inflection points at P 5 (d) P, P ; other solutions move awa from P and toward P. (a). 7 kg (b).55 ears

100 A APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES 5. (a) dpdt 65 P P, P in billions (b) 5.9 billion (c) In billions: 7.8, 7.7 (d) In billions: 5.8, 7.6,. 7. (a) ddt k (b) e kt (c) :6 PM. P Et t 9,;,658.5 P Lt 9,.75e.76t m kp. (a) Pt m (b) k P m ke kt (c) m kp, m kp (d) Declining 5. (a) Fish are caught at a rate of 5 per week. (b) See part (d) (c) P 5, P 75 (d) P P 5: P l ; P 5: P l 5; P 5: P l 75 8 P (in thousands), P L P E 5 9, t (ear) EXERCISES 9.5 N PAGE 66. Yes. No 5. e Ce 7. ln C 9. s C. sin d C. u t t C sin t 5. e 7. v t e t 5e t 9. cos. 5. e C C 5 7. (a) It e 5t (b) e.57 A 9.. Qt e t, It e t Pt M Ce kt P(t) M _ c= c=5 c=7 c=_5 c=_ c=_ 5 _5 c= c= c=5 c=7 c=_5 c=_ c=_ 8 t P() t t5 5 75ke (e) Pt ke t5 where k, 9. 5 t, t ;.75 kgl 5. (b) mtc (c) mtct mce ctm m tc EXERCISES 9.6 PAGE 6 N 7. (b) P P : P l ; P : P l ; P : P l 8 6 (c) Pt 6 8 t KmkK t mk P KP me KmkK t K P P me 9. (a) Pt P e krsinrt sin (b) Does not eist. (a) predators, pre; growth is restricted onl b predators, which feed onl on pre. (b) pre, predators; growth is restricted b carring capacit and b predators, which feed onl on pre.. (a) The rabbit population starts at about, increases to, then decreases back to. The fo population starts at, decreases to about, increases to about 5, decreases to, and the ccle starts again. (b) R F R F t t t t

101 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES A (a) Population stabilies at 5. (b) (i) W, R : Zero populations (ii) W, R 5: In the absence of wolves, the rabbit population is alwas 5. (iii) W 6, R : Both populations are stable. (c) The populations stabilie at rabbits and 6 wolves. (d) Species R Species t CHAPTER 9 REVIEW N PAGE 65 True-False Qui. True. False 5. True 7. True Eercises. (a) (b) c ; 6,, (iv) (iii) (ii) (i) t= t= t= t=, 5 W R t= t W (a) Pt ; 56 (b) t ln e.t 7. (a) Lt L (b) Lt 5 e.t L Le kt 9. 5 das.. (a) Stabilies at, (b) (i), : Zero populations (ii),, : In the absence of birds, the insect population is alwas,. (iii) 5,, 75: Both populations are stable. (c) The populations stabilie at 5, insects and 75 birds. (d) (insects) 5, 5, 5, 5, 5, 5. (a) k cosh k a k or k cosh k k cosh kb h (b) PROBLEMS PLUS N PAGE 68. f e (b) f L L ln C (c) No L L. (a) 9.8 h (b) ; 68 ft h (c) 5. h. 6 5 CHAPTER k ln h h RVt C insects birds,9, ft EXERCISES. PAGE 66 N.. t=5 {+œ 5, 5} (birds) t k sinh kb t=π {, π@}. (a)..8 t= (, ) t= (, ) 5 t= (, ) 5 _ (b) (c) and ; there is a local maimum or minimum 5. ( C)e sin 7. sln C 9. rt 5e tt.. C ln 5. (a) (b) (_5, ) t= (_8, _) t=_ (_, ) t= (, 5) t=

102 A APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES 7. (a) (b) 5, (_, 5) t= 9. (a) (b), (, ) t= 5, t= (, ) t= (, _) t= (7, ) t=_ (, _) t=. (b) 5t, 7 8t, t. (a) cos t, sin t, t (b) cos t, sin t, t 6 (c) cos t, sin t, t 7. The curve is generated in (a). In (b), onl the portion with is generated, and in (c) we get onl the portion with.. a cos, b sin ; a b, ellipse. a 5. (a) Two points of intersection O. (a),. (a), (b) (b) (, ) (, _) 5. (a) ln 7. (a), (b) (b) (, ) (b) One collision point at, when t (c) There are still two intersection points, but no collision point. 7. For c, there is a cusp; for c, there is a loop whose sie increases as c increases. _ Moves counterclockwise along the circle from, to,. Moves times clockwise around the ellipse 5, starting and ending at,. It is contained in the rectangle described b and (, ) t= t= _ 9. As n increases, the number of oscillations increases; a and b determine the width and height. EXERCISES. N PAGE 66. t t cos t sin t. 5. e _ (_, ) t= (, _) t=_ t= _ t, t, t e t, e t e t, t tan t, sec t, t 6, 6 7. Horiontal at, vertical at, 9. Horiontal at (s, ) (four points), vertical at,..6, ; (5 6 65, e 6 5 ) _

103 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES A , (c) π O 8.5 _, π 7. (a) d sin r d cos 9. 7, 9 9 ),,. ab. e 5. r d 7. s t dt s sin t cos t dt.67. s. s ln( s) s ln( s) 5. s e 8 ( 6. (a) (b) (c),,, 5 (, π), O π π, _ O (, s) _ π 7. e e (a) s, s t, (s, s) 5. (a) (i) (s, 7) (ii) (s, ) (b) (i), (ii), r= O O π π _, r= O =_ π = π 6 r=. = 7π 5 5 r= r= (b) t e t se t t t t dt (7s 6) 6. 5a (99s6 ) EXERCISES. PAGE 67 N. (a) π (b), O π 5 π, _ O _ π O. s 5. Circle, center O, radius 7. Circle, center (, ), radius 9. Horiontal line, unit above the -ais. r sec. r cot csc 5. r c cos 7. (a) (b) O = 5π π, =_ π 6 O, 7,,, 5,,

104 A APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES. 5. O O 69. Center b, a, radius sa b _..8 _ = π 6 5 = π _.6 _ = 5π 6 = π 6 O = π O = π 77. B counterclockwise rotation through angle 6,, or about the origin 79. (a) A rose with n loops if n is odd and n loops if n is even (b) Number of loops is alwas n 8. For a, the curve is an oval, which develops a dimple as a l. When a, the curve splits into two parts, one of which has a loop. 7 (, π) (, ) EXERCISES. N PAGE ,. 8 s (, ) (6, ) O O O (a) For c, the inner loop begins at and ends at ; for c, it begins at O and ends at. sin c sin /c sin c sin c. 5. = π s Horiontal at (s, ), (s, ) ; vertical at,,, 65. Horiontal at (, ),, [the pole], and (, 5) ; vertical at (, ), (, ), (, ) 67. Horiontal at,,, ; vertical at (, where sin ( s, ) s) ( s, ) s (, 6), (, 56) 9 s ( s s) 7., and the pole 9., where, 5,, 7 and, where,, 9, 7 s

105 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES A5 ( s, ), ( s, )9.,,,,.,, (, s),., and the pole. Intersection at (b) ( s) _ ,.5; area = _ = EXERCISES.5 N PAGE 66.,,,.,, (, 8, ( 8, ), 8 _ 6) 6 = 6.,,, ; (s5, ); (, _) (-œ 5, _) (, _) (+œ 5, _), _ 6 =_ 8 5.,,, 5, 7.,, 5,, (_, 5) (_5, _) = (_, _) = 9., focus (, ), directri.,,,.,, (, s) _ œ 5 œ 5 _ 5. 7., foci (, s5),, (, s5) 9 (, ),, (, ) 5. Parabola, 7. Ellipse, (s, ),, 9. Hperbola,,,, ; (, s5) ,76,6,75,96 5. (a),5,65,9,75 (b) 8 mi 55. (a) Ellipse (b) Hperbola (c) No curve b c where c a b a ab ln a b c EXERCISES.6 N PAGE r. r 7 sin cos 8 5. r 7. r sin cos 9. (a) (b) Parabola (c) (d) π, = O (,_)

106 A6 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES. (a) (b) Ellipse (c) r (d).9 cos π, AU km..6 8 km. (a) (b) Ellipse (c) 9 (d) 9, π (, π) O (, ) π 5, =_ = 9 π, 9 8, O CHAPTER REVIEW N PAGE 669 True-False Qui. False. False 5. True 7. False 9. True Eercises. 8. (, 6), t=_ (5, ), t= (, ), = π, 5. (a) (b) Hperbola (c) (d) -,, π O 8 5. t, st; t, t ; tan t, tan t, t 7. (a) π (b) (s, ),, (s, 7) O π 7. (a) =_ 8, (, s) 9.. π, = π 6 _ = - (, π) O (, ) π, (b) r sin _. 5. (, π) (, ) O π _, = π, O 9. The ellipse is nearl circular when e is close to and becomes more elongated as e l. At e, the curve becomes a parabola. _ e=. e=. e=.6 e=.8 7. r cos sin r= sin

107 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES A7.. CHAPTER 5. sin t cos t sin t, 7. ( 8, ) cos t cos t EXERCISES. PAGE 68 N 9. Vertical tangent at Abbreviations: C, convergent; D, divergent ( a, s a), a, ;. (a) A sequence is an ordered list of numbers. It can also be horiontal tangent at defined as a function whose domain is the set of positive integers. a,, ( a, sa) (b) The terms a (a, ) n approach 8 as n becomes large. (a, ) (c) The terms a n become large as n becomes large...8,.96,.99,.998, ,,, 8, a 5n 7., 5, 9, 7, 9. a n n.. a 5., 5, 7, 9,, 5 n ( ) n ; es; n D e. 8., 5.. ln. D 5. D (5s5 ) 5. D ln s 55. (a) 6,.6, 9., 6.8, 8. (b) D 57. r 59. Convergent b the Monotonic Sequence Theorem; 5 L 8. 7,95 6. Decreasing; es 6. Not monotonic; no. All curves have the vertical asmptote. For c, the 65. Decreasing; es ( s5) curve bulges to the right. At c, the curve is the line. 7. (b) ( s5) 7. (a) (b) 9, For c, it bulges to the left. At c there is a cusp at (, ). For c, there is a loop. EXERCISES. PAGE 69 N 5.,,, 7. ( 5, ),,. (a) A sequence is an ordered list of numbers whereas a series is œ the sum of a list of numbers. (, ) (_, ) (b) A series is convergent if the sequence of partial sums is a convergent sequence. A series is divergent if it is not convergent. s s r 5 6,8 cos 57. acot sin cos, a sin PROBLEMS PLUS N PAGE 67. ln. [ s, s], 5. (a) At (, ) and (, ) (b) Horiontal tangents at (, ) and (s, s ) ; vertical tangents at (, ) and (s, s ) (d) (g) œ s..,.9,.6,.9968,.6,.99987,.,.99999,.,.; convergent, sum ,.676,.778,.876,.9987,.888,., 9., 9.666, 9.6; divergent ,.65,.5,.5579,.5975,.6,.665,.66667,.6877,.6989; convergent, sum sa n d sa n d ss n d {s n} {a n} ss n d

108 A8 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES 9. (a) C (b) D. 9. D D. D. D D 9. D. D. ee e ; ; 5. All ; 5. cos 55. a, a n for n, sum nn 57. (a) S n D c n (b) c The series is divergent. nn 7. s n is bounded and increasing. 7. (a) n! 75. (a), 5 6,, 9 ; (c) n! EXERCISES. N PAGE 7. C, 9, 9,,, 7 9, 8 9, =. a a a a... (s ). D 5. C 7. C 9. D. C. D 5. C 7. C 9. C. D. C 5. C 7. p 9. p.,. (a).5977, error. (b).65, error.5 (c) n b e 7..,.66,.889,.79,.8,.75,.789,.75,.78,.755; error An underestimate. p is not a negative integer 5. b n is not decreasing EXERCISES.6 N PAGE 79 Abbreviations: AC, absolutel convergent; CC, conditionall convergent. (a) D (b) C (c) Ma converge or diverge. AC 5. CC 7. AC 9. D. AC. AC 5. AC 7. CC 9. AC. AC. D 5. AC 7. D 9. D. (a) and (d) 5. (a) , error.5 (b) n,.699 EXERCISES.7 N PAGE 7. C. D 5. C 7. D 9. C. C. C 5. C 7. D 9. C. C. D 5. C 7. C 9. C. D. C 5. C 7. C EXERCISES.8 PAGE 77 N. A series of the form n c n a n, where is a variable and a and the c n s are constants.,, 5.,, 7.,, 9.,,., (, ].,, 5.,, 7., [, ) 9.,,. b, a b, a b., { 5., [ }, ] 7.,, 9. (a) Yes (b) No. k k. No 5. (a), (b), (c) s s s _ ss n d sa n d EXERCISES. N PAGE 79. (a) Nothing (b) C. C 5. D 7. C 9. C. C. C 5. C 7. D 9. D. C. C 5. D 7. C 9. C. D..9, error , error. 5. Yes EXERCISES.5 N PAGE 7. (a) A series whose terms are alternatel positive and negative (b) b n b n and lim n l b n, where an (c) Rn bn. C 5. C 7. D 9. C. C. D 5. C 7. C 9. D _ s s s 7.,, f. EXERCISES.9 N PAGE n n,, n n,, 9 n,, n n _8 n J 8 n n n,, n

109 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES A9.. (a) (b) (c) 5. ln 5 7. n5, R 5 n 9.. n n n n,, 5. n n n n n, R n n n n n, R n nn n, R n n _ s f s n n n 6 n, R n s s s n n, R.5 _.5 n n n, R n s s f s s s s f s s e n n! n, R n n n! n, R n 5 n 9 n, R 9 n n n n! n n n n n! n n n n n! n, R n 5 n n, R n n! n n n n, R n n n n! n, R n, R n n n n! n, R 5 n n n, R n! n n n n n! n, R.5 T =T =T =T. 5. C n n n, R (b).9 7.,,,,, EXERCISES. N PAGE 76. b 8 f 8 58! C n n, R n n n n n! n, R 5 n n n t 8n 8n, R n 9.. n! n, R n n, R n n n, R n!., R. n (a) (b) n n! n, R n _.5 _ n T T T _.5 T T Tß 5 n n n! 5 n n n n! f 6 _6.5 Tˆ=T =T =T f T =T =Tß=T T T n T n T f T Tß

110 A APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES 7. C n 9. C n 5.. n n! n, R n n e s 67. e EXERCISES. PAGE 755 N. (a) T T, T T, T T 5, T n 6n n!, R 7. T =T T =T 6 9. T f.6.6 _.5 (b) (c) As n increases, T n is a good approimation to f on a larger and larger interval.. _π f 8 6 Tß T T T f _ T =T T T f π T T T 6 5 T T _ T f T T f _. T 5 π. (a) 6 (b) (a) 9 8 (b) (a) (b).5 9. (a) (b).6. (a) 6 (b) Four m, no 7. (c) The differ b about 8 9 km. π T 8 T 6 5 T T f f T CHAPTER REVIEW N PAGE 759 True-False Qui. False. True 5. False 7. False 9. False. True. True 5. False 7. True 9. True _. π f T π Eercises.. D e 9.. C. C 5. D 7. C 9. C. C. CC 5. AC e e 5..97

111 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES A 7.,error ,, 6, 9s,. s9, s85, 6,..5, [.5,.5). (a) 5. n n s n n! 6 n! (b) 6 (c) n n n, R n n, R. A plane parallel to the -plane and units to the left of it n 5. A half-space consisting of all points in front of the plane n n 7. All points on or between the horiontal planes and 6, R n n! 9. All points on or inside a sphere with radius s and center O 5 9 n n, R 6. All points on or inside a circular clinder of radius with ais n n! 6n the -ais 55. C ln n r R n n n! 7. (a) (,, ) (b) L 57. (a) 8 6 C (b) (c).6 P T f 6 PROBLEMS PLUS N PAGE 76. 5!5!,897,86,. (b) if, cot if k, k an integer 5. (a) s n n, l n n, p n n n (c) 5 s 9.,,. ln 5 5. (a) e n5 e n5 (b) CHAPTER , a plane perpendicular to AB EXERCISES. PAGE 777 N. (a) Scalar (b) Vector (c) Vector (d) Scalar. AB l DC l, DA l CB l, DE l EB l, EA l CE l 5. (a) u (b) (c) v v+w v w u+v (d) A _v B L u-v w w+v+u u u v EXERCISES. PAGE 769 N.,,. Q; R 5. A vertical plane that intersects the -plane in the line, (see graph at right) =- =-, = 7. a, 9. a, A(, ) A(_, ) B(_, ) a a B(, ) PQ 6 QR RP 6 7., s, ; isosceles triangle 9. (a) No (b) Yes. 5; 9, (a circle). 5. 8,,,5. a,,. 5, A(,, ) a B(,, _) k_, l k6, _l k5, l

112 A APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES 5.,,. (a) Scalar (b) Meaningless (c) Vector (d) Meaningless (e) Meaningless (f) Scalar 5. ; into the page 7. 5,,, 5,, k,, l 9. s6, s6, s6, s6, s6, s6 7 k,, _l (a) 6,, (b). (a),, 5 (b) s9 k,, _l sin 8.6 N m. 7 N. (b) s97 7., 8,,,, 9. (a) No (b) No (c) Yes 9. i j k, i j 9 k, s, s i 9 j 9 k s58 i 7 s58 j EXERCISES.5 PAGE 8 N. (a) True (b) False (c) True (d) False (e) False 5., s fts, 8.57 fts (f) True (g) False (h) True (i) True ( j) False 9. s7 6.6 N, 9. (k) True. s9. mih, N8W. r i. j.5 k t i j k;. T 96 i.9 j, T 96 i.9 j t,. t,.5 t 5. i js r i 6k ti j k; 9. (a), (b) (d) s 9 7, t 7 t, t, 6 t sa 7. 5t,, t;, a 5 c 9. t, t, t; b tb. t, t, t;. A sphere with radius, centered at,, EXERCISES. N PAGE 78. Yes 5. (a) 5 6. (b), (c), (d) are meaningful (b),,, (,, ),,, rt i j k ti 7j k, t. u v cos 9. cos 9 s7 cos, u w 9. Parallel. Skew 95 s , 5, 9 s (a) Neither (b) Orthogonal 9.. (c) Orthogonal (d) Parallel (,, ),, 5. Yes 7. i j ks [or i j ks] 9.,, ; 65, 56, 5 5s 5s s (, _, ). 7, 7, 6 7; 7, 65, 9 (,, ). s, s, s; 55, 55, 55 (,, ) 9 5., 9 5, 7., 7 9, 5 9, s, i j (5,, ) k.,, s or an vector of the form s, t, s s, s, t 5. J 7. cos 89 ft-lb 9. 5.,, 5 5.,, 7.,, 5. cos (s) Perpendicular 5. Neither, Parallel EXERCISES. N PAGE (a), t, t (b) cos s i 8 k. 5 i j k 5. i j k 57., 7. t i t j t k 9.. i j k a b c

113 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES A 6. t, t, t 65. P and P are parallel, P and P are identical s (s) s6 5. Hperboloid of two sheets EXERCISES.6 PAGE 8 N. (a) Parabola (b) Parabolic clinder with rulings parallel to the -ais (c) Parabolic clinder with rulings parallel to the -ais. Elliptic clinder 5. Parabolic clinder 7. Ellipsoid (,, ) (, 6, ) (,, ) 9. Hperbolic paraboloid 7. Clindrical surface. VII. II 5. VI 7. VIII Hperboloid of two sheets with ais the -ais (,, 6) 9. (a) k, k, hperbola k ; k, k, hperbola k ; k, k, circle (b) The hperboloid is rotated so that it has ais the -ais (c) The hperboloid is shifted one unit in the negative -direction. Elliptic paraboloid with ais the -ais. 6 Elliptic paraboloid with verte,, and ais the -ais (,, _6). Ellipsoid with center,, (,, ) (,, ). Elliptic cone with ais the -ais 5. Circular cone with verte,, and ais parallel to the -ais (, _, )

114 A APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES Hperboloid of two sheets 5. Ellipsoid (,, ) (,, ) (,, ) (,, ). = (,, _) 7. 6 =œ + PROBLEMS PLUS N PAGE , paraboloid 7. (a) (b) Circle (c) Ellipse 5.. (s.5) m. (a) c cc cc (b) t, t (c) CHAPTER EXERCISES. PAGE 8 N.,.,, 5. i j k π (,, ) CHAPTER REVIEW N PAGE 8 (π,, ) True-False Qui. True. True 5. True 7. True 9. True. False. False 5. False 7. True Eercises. (a) 69 (b) 68, (c) Center,,, radius 5. u v s; u v s; out of the page 5., 7. (a) (b) (c) (d) 9. cos ( ) 7. (a),, (b) s. 66 N, N 5. t, t, t 7. t, t, 5t 9.. (,,). Skew s6 9. Plane. Cone.. = 5. rt t, t, t, t ; t, t, t, t 7. rt t, t, 5t, t ; t, t, 5t, t 9. VI. IV. V 5. 7.,,,,,

115 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES A5 9. r h r (c) r lim ; T h l h r r. (a), (c) (b) rt, t (_, ). _ rª(_) r(_) 5. (a), (c) 7. (a), (c) œ, œ π rª π r r() (, ) rª(). _ 7. rt t i t j t k 9. cos t, sin t, cos t. Yes EXERCISES. N PAGE 88. (a) C R (b), (d) _ r(.5) r(.) r() r(.5)-r().5 r(.5) C r(.) r(.)-r(). R Q r(.5)-r() Q r(.)-r() P T() _ (b) rt cos t i sin t j (b) rt e t i e t j 9. rt t cos t sin t, t, cos t t sin t. rt e t k. rt te t i [ t k 5. rt b tc 7.,, 9. 5 j 5 k., t, t, s, s, s,,, 6t, 6t, 6t,. t, t, t 5. t, t, t 7. t, t, t 9. t, t, t. 66. i j 5 k 5. i j k 7. e t i t j t ln t tk C t i t j ( t ) k t cos t sin t cos t sin t EXERCISES. PAGE 86 N. s9. e e rts s9 s i s9 s j 5 s9 s k 5. sin,, cos 7. (a) (s9) cos t, 5s9, (s9) sin t, sin t,, cos t (b) 9 9. (a) e t e t, se t, se t e t set, e t,, (b) se t e t. t s / 9. 5s. ( ln, s) ; approaches. (a) P (b).,.7 5. =k() r() P 7. a is f, b is

116 A6 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES 9. k(t) t 6s cos t cos t. s i e t j e t k, e t j e t k, e t e t 7 cos t. e t cos t sin ti sin t cos tj t k, e t sin t i cos t j t k, e t st t 5. vt t i t j k, rt ( t i t j t k 7. (a) (b) integer multiples of rt ( t t) i t sin t j ( cos t) k π π 6π. (se t ). 5. 6, ( 5 ) 8, ( 5 ) t,,,,,,,, t. rt t i t j 5 t k,. (a) km (b). km (c) 5 ms 5. ms 7.., , (a) 6 m (b).6 upstream.6.. vt s5t 5 9.,, 57. t t Å m EXERCISES. N PAGE 86. (a).8i.8j.7k,.i.j.6k,.8i.8j.k,.8i.8j.k (b).i.8j.5k,.58. vt t, v() at, (_, ) vt st a(). 6t, 6 5., 7. e t e t, s 9..5 cms, 9. cms. t CHAPTER REVIEW N PAGE 85 True-False Qui. True. False 5. False 7. True 9. False. True Eercises. (a) 5. vt sin t i cos t j at cos t i sin t j vt s5 sin t (, ) π v π a, œ (, ) (,, ) (,, ) 7. vt i t j at j s t vt (,, ) 9. t, t, t,, 6t,, t s9t 8 a() v() (b) rt i sin t j cos t k, rt cos t j sin t k. rt cos t i sin t j 5 cos tk, t 5. i j k (a) t, t, st t (b) t, t, t tst 8 t 6 t 5t (c) st 8 t 6 t 5t t t vt ln t i j e t k 7., vt s ln t ln t e t, at ti e t k

117 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES A7 9. (a) About.8 ft above the ground, 6.8 ft from the athlete (b). ft (c) 6. ft from the athlete. (c) e t v d e t R 7.,, = PROBLEMS PLUS N PAGE 85. (a) v Rsin t i cos t j (c) a r. (a) 5. (a) 9, v t.9 ft to the right of the table s edge, 5 fts (b) 7.6 (c). ft to the right of the table s edge ,, _ CHAPTER EXERCISES. N PAGE 865., horiontal plane. (a) 7; a temperature of 5C with wind blowing at kmh feels equivalent to about 7C without wind. (b) When the temperature is C, what wind speed gives a wind chill of C? kmh (c) With a wind speed of kmh, what temperature gives a wind chill of 9C? 5C (d) A function of wind speed that gives wind-chill values when the temperature is 5C (e) A function of temperature that gives wind-chill values when the wind speed is 5 kmh. Yes 5. (a) 5; a -knot wind blowing in the open sea for 5 h will create waves about 5 ft high. (b) f, t is a function of t giving the wave heights produced b -knot winds blowing for t hours. (c) f v, is a function of v giving the wave heights produced b winds of speed v blowing for hours. 7. (a) (b) (c), 9. (a) e (b),, (c),.,. 5, plane 5., parabolic clinder (.5,, ) (,, ) (,, ) =_., 9 5.,, 9 + = s, elliptic paraboloid top half of cone (,, ). 56, 5. Steep; nearl flat

118 A8 APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES k. ln k. ke 5. k 7. _ 9 k = = 55. (a) C (b) II 57. (a) F (b) I 59. (a) B (b) VI 6. Famil of parallel planes 6. Famil of hperboloids of one or two sheets with ais the -ais 65. (a) Shift the graph of f upward units (b) Stretch the graph of f verticall b a factor of (c) Reflect the graph of f about the -plane (d) Reflect the graph of f about the -plane and then shift it upward units _5 5 5 f appears to have a maimum value of about 5. There are two local maimum points but no local minimum point. 69. _ 5 = _5 _ 9. = _ The function values approach as, become large; as, approaches the origin, f approaches or, depending on the direction of approach. 7. If c, the graph is a clindrical surface. For c, the level curves are ellipses. The graph curves upward as we leave the origin, and the steepness increases as c increases. For c, the level curves are hperbolas. The graph curves upward in the -direction and downward, approaching the -plane, in the -direction giving a saddle-shaped appearance near (,, ). 7. c,, 75. (b).75. _