Mathematics 110 (Calculus I) Laboratory Manual

Transcription

1 Mathematics 0 (Calculus I) Laboratory Manual Department of Mathematics & Statistics University of Regina rd edition by Douglas Farenick, Fotini Labropulu, Robert G. Petry Published by the University of Regina Department of Mathematics and Statistics

2 rd Edition Copyright 05 Douglas Farenick, Fotini Labropulu, Robert G. Petry. nd Edition Copyright 04 Douglas Farenick, Fotini Labropulu, Robert G. Petry. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version. or any later version published by the Free Software Foundation; with no Invariant Sections, no Front-Cover Tets, and no Back-Cover Tets. A copy of the license is included in the section entitled GNU Free Documentation License. Permission is granted to retain (if desired) the original title of this document on modified copies. History nd Edition produced in 04 entitled Math 0 (Calculus I) Laboratory Manual written by principal authors Douglas Farenick, Fotini Labropulu, and Robert G. Petry. Typeset in part by Kyler Johnson. Published by the University of Regina Department of Mathematics and Statistics. rd Edition produced in 05 entitled Math 0 (Calculus I) Laboratory Manual written by principal authors Douglas Farenick, Fotini Labropulu, and Robert G. Petry. Typeset in part by Kyler Johnson. Published by the University of Regina Department of Mathematics and Statistics. The source of this document (i.e. a transparent copy) is available via

3 Contents Module. Equations and Functions. Solving Equations Properties of Functions: Domains, Intercepts, Symmetry Function Algebra Review Eercises Module. Limits 7. Secant and Tangent Lines Calculating Limits Using Limit Laws Trigonometric Limits One-sided Limits Infinite Limits and Vertical Asymptotes Continuity Review Eercises Module. Differentiation. Average and Instantaneous Rate of Change Definition of Derivative Derivative Function Differentiation Rules: Power, Sums, Constant Multiple Rates of Change Differentiation Rules: Products and Quotients Parallel, Perpendicular, and Normal Lines Differentiation Rules: General Power Rule Differentiation Rules: Trigonometric Functions Differentiation Rules: Chain Rule Differentiation Rules: Implicit Differentiation Differentiation Using All Methods Higher Derivatives Related Rates Differentials Review Eercises Module 4. Derivative Applications 4. Relative and Absolute Etreme Values Rolle s Theorem and the Mean Value Theorem First Derivative Test Inflection Points Second Derivative Test Curve Sketching I Limits at Infinity and Horizontal Asymptotes Slant Asymptotes i

4 ii CONTENTS 4.9 Curve Sketching II Optimization Problems Review Eercises Module 5. Integration 5. Antiderivatives Series Area Under a Curve The Definite Integral Fundamental Theorem of Calculus Indefinite Integrals The Substitution Rule Integration Using All Methods Area Between Curves Distances and Net Change Review Eercises Answers 8 GNU Free Documentation License 65

5 Introduction One does not learn how to swim by reading a book about swimming, as surely everyone agrees. The same is true of mathematics. One does not learn mathematics by only reading a tetbook and listening to lectures. Rather, one learns mathematics by doing mathematics. This Laboratory Manual is a set of problems that are representative of the types of problems that students of Mathematics 0 (Calculus I) at the University of Regina are epected to be able to solve on quizzes, midterm eams, and final eams. In the weekly lab of your section of Math 0 you will work on selected problems from this manual under the guidance of the laboratory instructor, thereby giving you the opportunity to do mathematics with a coach close at hand. These problems are not homework and your work on these problems will not be graded. However, by working on these problems during the lab periods, and outside the lab periods if you wish, you will gain useful eperience in working with the central ideas of elementary calculus. The material in the Lab Manual does not replace the tetbook. There are no eplanations or short reviews of the topics under study. Thus, you should refer to the relevant sections of your tetbook and your class notes when using the Lab Manual. These problems are not sufficient practice to master calculus, and so you should solidify your understanding of the material by working through problems given to you by your professor or that you yourself find in the tetbook. To succeed in calculus it is imperative that you attend the lectures and labs, read the relevant sections of the tetbook carefully, and work on the problems in the tetbook and laboratory manual. Through practice you will learn, and by learning you will succeed in achieving your academic goals. We wish you good luck in your studies of calculus.

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7 Module Equations and Functions. Solving Equations -0: Solve the given equations = 0. 5 = = = = = = = = = 0 Page 8. Properties of Functions: Domains, Intercepts, Symmetry -6: Find the domain and the - and y-intercepts (if there are any) of the given functions.. f() =. h() = 6. f() = h() = f() = 0 5 Page 8 4. g() = f() = 6. h() = p() = 4 8. f() = g() = + 4. g() = + 4. p() = h() = 5. f() = f() = +

8 4 MODULE. EQUATIONS AND FUNCTIONS 7-: Determine whether each of the given functions is even, odd or neither. 7. f() = 8. g() = 9. h() = + 0. f(t) = 6t. g(u) = u u. f() = + 5. f() = 4 4. h() = f() = g(t) = 4t + t 7. h(z) = z f(t) = t g() = h(u) = + u u +. p() = 5 + s. u(s) = s Here is a graph of a function f: y (a) Sketch the graph of f(). (b) Sketch the graph of f( ). (c) Sketch the graph of f( + ). (d) Sketch the graph of f() +. (e) Is f even, odd, or neither? Page 40. Function Algebra. Suppose that f() = +. Determine: (a) f( + ) (b) f(f()) -5: Find the composite functions f g and g f and their domains for the given functions.. f() =, g() = +.

9 .. FUNCTION ALGEBRA 5. f() = , g() = 4. f(z) = z + 5, g(z) = z z + 5. f() = + 5 4, g() = + 6. Find two functions f() and g() such that f(g()) = +.

10 6 MODULE. EQUATIONS AND FUNCTIONS Module Review Eercises Page 40-5: Solve the given equations.. + = = 0. + = = = 0 6-9: Find the domain and the - and y-intercepts of the given functions. 6. h() = f() = g() = p() = : Determine whether each of the given functions is even, odd or neither. 0. f() = g(t) = t /. h(z) = z z +. g() = : Find the composite functions f g and g f and their domains. 4. f() = + 6, g() = / 5. f(t) = t + 5 t 4, g(t) = t + 6. f() =, g() = + 5 +

11 Module Limits. Secant and Tangent Lines. Consider the curve described by the function y = + +. (a) Show that the points (, 5) and (0, ) lie on the curve. (b) Determine the slope of the secant line passing through the points (0, ) and (, 5). (c) Let Q be the arbitrary point (, + + ) on the curve. Find the slope of the secant line passing through Q and (, 5). (d) Use your answer in (c) to determine the slope of the tangent line to the curve at the point (, 5). Page 4. Calculating Limits Using Limit Laws -5: Evaluate the following limits.. lim lim + 4. lim 0 4. lim 0 5. lim 4 6. lim t t 7. lim t t 8. lim + t 9. lim t 4 t 4 y 0. lim 5 y y y 6 ( + ) 9. lim lim u 5 u + 5 u + 5 t + t 6. lim t t 4. lim lim t t + 4t 5 t 7 Page 4

12 8 MODULE. LIMITS Page 4. Trigonometric Limits -: Evaluate the trigonometric limits. θ. lim θ 0 sin θ. lim θ 0 θ cot θ. lim 0 cos sin 7θ 4. lim θ 0 sin 5θ 7θ 5. lim θ 0 cos 5θ sin 6. lim 0 cos 7. lim π cos tan θ 8. lim θ π θ 4 sin t 9. lim t π cos t 0. lim 0 tan. lim θ π cos θ + sin θ Page 4.4 One-sided Limits. Find the one-sided limits of f at the values = 0, = and = 4. y Find the one-sided and two-sided limits of f at the values = 0, = and = 4. y 4 5

13 .5. INFINITE LIMITS AND VERTICAL ASYMPTOTES 9. Suppose f() = Find lim 4. Suppose f() = { 4 +4 if < + if. f(), lim f(), and lim f(), if they eist. + { + c if + 5c if >. Find all values for the constant c that make the two-sided limit eist at =..5 Infinite Limits and Vertical Asymptotes -4: Determine the following limits. For any limit that does not eist, identify if it has an infinite trend ( or ).. lim lim lim lim 0 sec Page 4 5-: Find the vertical asymptotes of the following functions. 5. f() = f() = g(t) = t + t 8. f() = + 9. f() = cos 0. y = f() = + +. F () = Continuity. Define precisely what is meant by the statement f is continuous at = a. -7: Use the continuity definition to determine if the function is continuous at the given value.. f() = + 5 +, at =. g(t) = t + t, at t = h(y) = y + 4y + 4, at y = y + Page 4 5. p(s) = s 4, at s =

14 0 MODULE. LIMITS { +, if 6. f() =, at =, if > 7. g(t) = + { t +, if t t 5t+6 t, if t >, at t = 8. Let c be a constant real number and f be the function f() = { + if < 0 + c if 0 (a) Eplain why, for c =, the function f is discontinuous at = 0. (b) Determine all real numbers c for which f is continuous at = Where is the function f() = + + continuous? 0. Using the Intermediate Value Theorem, show there is a real number c strictly between and such that c + c = 0.. Show that the equation + cos = 0 has a solution in the interval (0, ).

15 .6. CONTINUITY Module Review Eercises -5: Evaluate the limits. 5. lim + t. lim t 4 t t lim 4. lim t 0 + t t + 4t 4 8 t 5. lim 4 t t 4t 4 Page 4 6-9: Evaluate the trigonometric limits. 6. lim 0 sin(4) sin(5) 7. lim θ 0 sin(θ) tan(4θ) sin t 8. lim t π + cos t 9. lim 0 cos() + cos(4) 0-: Determine whether the functions are continuous at the given value. 0. f() = at =. h(t) = t + t t at t =. g() = {, if + 6, if > at =

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17 Module Differentiation. Average and Instantaneous Rate of Change. Consider the function f() = + +. (a) Find the slope of the secant line through points (, ) and (, 4) on the graph of f. (b) From your graph, estimate the slope of the tangent line to the curve at the point (, ). Net, numerically estimate the tangent slope by calculating the secant slope between the point (, ) and a point with value near. (c) What is the equation of the tangent line at (, )? (Use your estimate from (b) for the slope.) Page 4. After t seconds, a toy car moving along a straight track has position s(t) measured from a fied point of reference given by s(t) = t + t + cm. (a) How far is the car initially from the reference point? (b) How far from the reference point is the car after seconds? (c) What is the average velocity of the car during its first seconds of motion? (d) By calculator, estimate the instantaneous velocity of the car at time t = by computing the average velocities over small time intervals near t =.. One mole of an ideal gas at a fied temperature of 7 K has a volume V that is inversely proportional to the pressure P (Boyle s Law) given by V =.4 P, where V is in litres (L) and P is in atmospheres (atm). (a) What is the average rate of change of V with respect to P as pressure varies from atm to atm? (b) Use a calculator to estimate the instantaneous rate of change in the volume when the pressure is atm by computing the average rates of change over small intervals lying to the left and right of P = atm.

18 4 MODULE. DIFFERENTIATION. Definition of Derivative -6: Use the definition of the derivative to calculate f () for each of the following functions. Page 4. f() = +. f() =. f() = ( + ) 4. f() = + 5. f() = f() = 7. Prove that f() = ( ) is not differentiable at = by showing that the following left and right hand limits differ: f( + h) f() f( + h) f() lim lim. h 0 h h 0 + h. Derivative Function Page 44. Which of the following graphs represent functions that are differentiable at = 0? (Eplain why or why not). (a) y (b) y (c) y (d) y.4 Differentiation Rules: Power, Sums, Constant Multiple Page 44-6: Differentiate the following functions involving powers, sums and constant multiplication. (Any value that is not the function variable should be considered a constant.). f() = 4. f(u) = u 4 + u 4 ; Also find f ().. g() = 5. y = f() = sin(π/5) a 6. s(t) = g t + v 0 t + s 0 7. Calculate the instantaneous rates of change given in problems (b), (d), and (b) of Section. directly using the derivative. 8. Find the equation of the tangent line to the curve y = + at the point P (, ). 9. Find the value(s) of for which the curve y = has a horizontal tangent line.

19 .5. RATES OF CHANGE 5.5 Rates of Change -: The following problems consider the meaning of the derivative as a rate of change.. The concentration of carbon dioide in the Earth s atmosphere has been observed at Mauna Loa Observatory in Hawaii to be steadily increasing (neglecting seasonal oscillations) since 958. A best fit curve to the data measurements taken from 98 to 009 yields the following function for the concentration in parts per million (ppm) as a function of the year t : Page 44 C(t) = 0.04(t 98) +.8(t 98) + 4 (a) What was the level of CO in the air in the year 000? (b) At what rate was the CO level changing with respect to time in the year 000? (c) By what percentage did the CO level change between 000 and 005?. A conical tank has a height of 5 metres and radius at the top of metres. (a) Show that the volume of liquid in the tank when it is filled to a depth y is given by V = 4π 75 y (b) What is the rate of change of volume with respect to depth when the tank is filled to 4 metres?.6 Differentiation Rules: Products and Quotients -8: Differentiate the following functions involving products and quotients. (Any value that is not the function variable should be considered a constant.). f() = ( 4 + ) ( ) 5. f(θ) = θ + θ 4. h() = ( + π + ) ( θ 7 + ) 6. g() = + ; Also find g (4).. y = ( ) ( + ) ( + ) 4. f() = 4 6 (v + )(v + 4/v) 7. f(v) = v + v 8. h() = c + ( + ) ( + ) Page 45.7 Parallel, Perpendicular, and Normal Lines. Find the line through the point P (, ) that is parallel to the tangent to the curve y = + + at the point Q(, 6).. Find the normal line to the curve y = + at the point P (, ). Page 45. Find any points on the curve y = 4 with normal line having slope.

20 6 MODULE. DIFFERENTIATION.8 Differentiation Rules: General Power Rule Page 45-6: Differentiate the following functions requiring use of the General Power Rule. (Any value that is not the function variable should be considered a constant.). f() = ( + ) 9. g() =. f(t) = + 7 t + t y = ( ) h() = 5 n + 4c [ ( ) ] f() = Differentiation Rules: Trigonometric Functions Page 46. Find the derivative f () of f() = sin 4 using the definition of the derivative. -5: Differentiate the following functions involving trigonmetric functions.. f() = cos. f(t) = t sin t + tan t 4. H(θ) = csc θ cot θ ; Also find dh dθ 5. f() = (sin + cos )(sec cot ). θ=π/ 6. Calculate df dθ for the function f(θ) = sin θ + cos θ (a) Directly by using the rules of differentiation. (b) By first simplifying f with a trigonometric identity and then differentiating. 7-8: For each curve and point P, (a) Confirm the point P lies on the curve. (b) Find the equation of the tangent line at P. ( π 7. y = tan at point P 4, π ) 4 8. y = 5 + sin() cos() at point P (0, ).

21 .0. DIFFERENTIATION RULES: CHAIN RULE 7.0 Differentiation Rules: Chain Rule -: Differentiate the following functions requiring use of the Chain Rule. (Any value that is not the function variable should be considered a constant.). f() = ( ) 8. y = (csc + ) Page 46. g() = + ; Also find g ().. f(θ) = sin(θ ) 4. h(θ) = cot θ 5. f() = sec [( + ) ( + )] 6. y = 4 cos 7. f() = + sin 9. y = π tan θ + tan(πθ) ( 0. g() = + ) ( 4 ) 7. f() = ( ) +. A(t) = cos(ωt + φ) ; Also find da dt. f() = sin [ cos ( + )]. t=0 4. Find the value(s) of θ for which the curve f(θ) = cos θ sin θ has a horizontal tangent line.. Differentiation Rules: Implicit Differentiation -5: Calculate y for functions y = y() defined implicitly by the following equations. (Any value that is not or y should be considered a constant.). + y =. y y + = ; Also find dy d. (,y)=(,) Page 47. a + y b = 4. sin(y) = y 5. cos( + y) + = sin y 6. Consider the curve generated by the relation + y = 4. (a) Confirm that the point P (, ) lies on the curve. (b) Find the equation of the tangent line to the curve at the point P.

22 8 MODULE. DIFFERENTIATION Page 47. Differentiation Using All Methods -: Differentiate the following functions using any appropriate rules.. f() = y = y = f(θ) = cos θ + sin θ. g(t) = t + t t 4. f() = g() = ( + + ) ( + π) 6. h(y) = (y + 4) y g() = tan ( + ) cos () 0. y = 4 (sin t + 5). f() = y 4 + y = y + 6 Page 48. Higher Derivatives -4: Calculate the second derivative for each of the following functions.. f() = cot. y = sec. f() = ( ) 0 ; Also find f (). 4. y = 6 (Use implicit differentiation.) 5-6: When one uses derivatives, their simplification becomes important. 5. Show that f () = ( + 6 ) ( 6) for f() = Show that f () = ( ) 4 for f() = Page 48.4 Related Rates -7: Solve the following problems involving related rates.. An oil spill spreads in a circle whose area is increasing at a constant rate of 0 square kilometres per hour. How fast is the radius of the spill increasing when the area is 8 square kilometres?. A spherical balloon is being filled with water at a constant rate of cm /s. How fast is the diameter of the balloon changing when it is 5 cm in diameter?. An observer who is km from a launchpad watches a rocket that is rising vertically. At a certain point in time the observer measures the angle between the ground and her line of sight of the rocket to be π/ radians. If at that moment the angle is increasing at a rate of /8 radians per second, how fast is the rocket rising when she made the measurement?

23 .5. DIFFERENTIALS 9 4. A water reservoir in the shape of a cone has height 0 metres and radius 6 metres at the top. Water flows into the tank at a rate of 5 m / min, how fast is the level of the water increasing when the water is 0 m deep? 5. At 8 a.m., a car is 50 km west of a truck. The car is traveling south at 50 km/h and the truck is traveling east at 40 km/h. How fast is the distance between the car and the truck changing at noon? 6. A boy is walking away from a 5-metres high building at a rate of m/sec. When the boy is 0 metres from the building, what is the rate of change of his distance from the top of the building? 7. The hypotenuse of a right angle triangle has a constant length of cm. The vertical leg of the triangle increases at the rate of cm/sec. What is the rate of change of the horizontal leg, when the vertical leg is 5 cm long?.5 Differentials -: Find the volume V and the absolute error V of the following objects.. A cubical cardboard bo with side length measurement of l = 5.0 ± 0. cm.. A spherical cannonball with measured radius of r = 6.0 ± 0.5 cm. Page 48-4: Find the linear approimation (linearization) L() of the function at the given value of.. f() = 7 at -value a =. 4. f() = tan() at -value a = π/4.

24 0 MODULE. DIFFERENTIATION Module Review Eercises Page 49 -: For each function calculate f () using the definition of the derivative.. f() = +. f() = 4 +. f() = + 4-: Differentiate the functions. 4. y = π 5. g() = 6. h(y) = ( 4 + ) ( + sin ) y + 5 y + 7. f(θ) = cos θ + 4 cos ( θ ) 8. g() = sec ( + 4 ) cos() 9. f() = y + 4y = y + sin y. Find the equation of the tangent line to the curve y = sin cos() at the point P ( π, ).. Find the value(s) of for which the curve y = + + has a horizontal tangent line.. Find the value(s) of θ for which the curve f(θ) = cos(θ) cos θ has a horizontal tangent line. 4. The height h and radius r of a circular cone are increasing at the rate of cm/sec. How fast is the volume of the cone increasing when h = 8 cm and r = cm? 5. A right triangle has a constant height of 0 cm. If the base of the right triangle is increasing at the rate of 6 cm/sec, how fast is the angle between the hypotenuse and the base is changing when the base is 0 cm? 6. If the area of an equilateral triangle is increasing at a rate of 5 cm / sec, find the rate at which the length of a side is changing when the area of the triangle is 00 cm.

25 Module 4 Derivative Applications 4. Relative and Absolute Etreme Values. Identify the relative maima, relative minima, absolute maima, and absolute minima on each of the following graphs. (a) y (b) y (c) y 6 4 Page : Find the critical numbers of the given function.. f() = h() = + 4. f(s) = s s f() = g(t) = 4 t4 + t 5t + 6 on the interval [0, 0]. 7. H() = f(t) = t 4 9. g() = 5 0. F (θ) = sin(θ) θ -5: Find the absolute maimum and absolute minimum values and their locations for the given function on the closed interval.. f() = 4 + on [, ]

26 MODULE 4. DERIVATIVE APPLICATIONS. f() = on [, 0].. g(t) = t (t ) on [0, ]. 4. H() = on [, 8]. 5. f() = sin cos on [0, π] 4. Rolle s Theorem and the Mean Value Theorem Page 50. Using Rolle s Theorem, show that the graph of f() = 7 has a horizontal tangent line at a point with -coordinate between and 4. Net find the value = c guaranteed by the theorem at which this occurs.. By Rolle s Theorem it follows that if f is a function defined on [0, ] with the following properties: (a) f is continuous on [0, ] (b) f is differentiable on (0, ) (c) f(0) = f() then there eists a least one value c in (0, ) with f (c) = 0. Show that each condition is required for the conclusion to follow by giving a countereample in the case that (a), (b), or (c) is not required.. Suppose that f is continuous on [, 4], differentiable on (, 4), and that f( ) = 5 and f(4) =. Show there is a c in (, 4) with f (c) =. 4. Verify the Mean Value Theorem for the function f() = + on the interval [, ].

27 4.. FIRST DERIVATIVE TEST 4. First Derivative Test -5: Find the open intervals upon which the following functions are increasing or decreasing. Also find any relative minima and maima and their locations.. f() = + +. f() = +. f() = 4 cos on [0, π] 4. f() = if 0 5. f() = if = 0 Page 5 6-7: Each graph below is a graph of the derivative f of a function f. In each case use the graph of f to sketch a possible graph of f. 6. y 7. y Inflection Points -: Show that the following functions have no inflection points:. f() = 4 Page 5. f() =. Find the relative etrema (and their locations), the intervals of concavity, and the inflection points of the function f() =

28 4 MODULE 4. DERIVATIVE APPLICATIONS 4.5 Second Derivative Test Page 5 -: Determine the relative maima and minima of the following functions and their locations. Use the Second Derivative Test.. f() = +. f() = cos() 4 sin() on the interval ( π, π). Can you use the Second Derivative Test to categorize the critical number = 0 of the function f() = sin 4? Eplain why or why not. 4.6 Curve Sketching I Page 5 -: Find the domain, intercepts, intervals of increase and decrease, relative maima and minima, intervals of concave upward and downward, and inflection points of the given functions and then sketch their graphs.. f() = 6. g(t) = t t. F () = Limits at Infinity and Horizontal Asymptotes Page 5 -: Determine the following limits. For any limit that does not eist, identify if it has an infinite trend ( or ). 8. lim 5 +. lim +. lim + 4. lim 5. lim 6 + ( ) ] [ sec lim lim lim lim 0. lim lim (. lim ) (. lim + ) 6 + 5

29 4.8. SLANT ASYMPTOTES 5 4-: Find the horizontal asymptotes of the following functions. 4. f() = f() = g(t) = t + t 7. f() = + 8. f() = cos 9. y = f() = + +. F () = Slant Asymptotes -: Find any slant asymptotes of the graphs of the following functions.. f() = 4 +. g() = y = + + Page Curve Sketching II -: Apply calculus techniques to identify all important features of the graph of each function and then sketch it.. f() = Page 54. y =. f() = + +

30 6 MODULE 4. DERIVATIVE APPLICATIONS Page Optimization Problems -8: Solve the following optimization problems following the steps outlined in the tet.. The entrance to a tent is in the shape of an isosceles triangle as shown below. Zippers run vertically along the middle of the triangle and horizontally along the bottom of it. If the designers of the tent want to have a total zipper length of 5 metres, find the dimensions of the tent that will maimize the area of the entrance. Also find this maimum area Find the point on the line y = + closest to the point (, ) by (a) Using optimization. (b) Finding the intersection of the original line and a line perpendicular to it that goes through (, ).. The product of two positive numbers is 50. Find the two numbers so that the sum of the first number and two times the second number is as small as possible. 4. A metal cylindrical can is to be constructed to hold 0 cm of liquid. What is the height and the radius of the can that minimize the amount of material needed? 5. A pair of campers wish to travel from their campsite along the river (location A) to visit friends 0 km downstream staying in a cabin that is km from the river (location B) as shown in the following diagram: B A km 0 km (a) If the pair can travel at 8 km/h in the river downstream by canoe and km/h carrying their canoe by land, at what distance (see diagram) should they depart from the river to minimize the total time t it takes for their trip? (b) On the way back from the cabin they can only travel at 4 km/h in their canoe because they are travelling upstream. What distance will minimize their travel time in this direction? (c) Using the symbolic constants a for the downstream distance, b for the perpendicular land distance, w for the water speed and v for the land speed, find a general epression for optimal distance. Verify your results for parts (a) and (b) of this problem by substituting the appropriate constant values.

31 4.0. OPTIMIZATION PROBLEMS 7 (d) Does your general result from part (c) depend at all on the downstream distance a? Discuss. 6. A construction company desires to build an apartment building in the shape of a rectangular parallelpiped (shown) with fied volume of 000 m. The building is to have a square base. In order to minimise heat loss, the total above ground surface area (the area of the four sides and the roof) is to be minimised. Find the optimal dimensions (base length and height) of the building. h 7. A rectangular field is to be enclosed and then divided into three equal parts using metres of fencing. What are the dimensions of the field that maimize the total area? 8. Two power transmission lines travel in a parallel direction (north-south) 0 km apart. Each produces electromagnetic interference (EMI) with the one to the west producing twice the EMI of that of the one to the east due to the greater current the former carries. An amateur radio astronomer wishes to set up his telescope between the two power lines in such a way that the total electromagnetic interference at the location of the telescope is minimized. If the intensity of the interference from each line falls off as /distance, how far should the telescope be positioned from the stronger transmission line?

32 8 MODULE 4. DERIVATIVE APPLICATIONS Module 4 Review Eercises Page 56 -: Find the critical numbers of the given functions.. f() = +. g(t) = t t. F (θ) = cos(θ) + sin θ 4-5: Find the absolute maimum and absolute minimum values and their locations for the given function on the closed interval. 4. f() = on the closed interval [, ] g(t) = t 8 t on the closed interval [0, ]. 6-7: Find the domain, intercepts, asymptotes, relative maima and minima, intervals of increase and decrease, intervals of concave upward and downward, and inflections points. Then sketch the graph of the given functions. 6. f() = g() = 8-: Evaluate the given limits lim lim lim. lim + ( ) -: Find the horizontal and vertical asymptotes of the given functions.. y = g(t) = 9 + 4t t +

33 4.0. OPTIMIZATION PROBLEMS 9 4. A store with a rectangular floorplan is to sit in the middle of one side of a larger rectangular lot with parking on three sides as shown. 0 m 0 m 8 m If the narrow strips of parking on the side are to be 0 m wide while the parking in the front is to be 8 m wide, find the optimal dimensions of the store that will minimize the total lot area if the store itself must have an area of 000 m. What is the total lot area in this case? 5. A farmer wants to enclose a rectangular garden on one side by a brick wall costing $0/m and on the other three sides by a metal fence costing $5/m. Find the dimensions of the garden that minimize the cost if the area of the garden is 50 m. 6. A window shaped like a Roman arch consists of a rectangle surmounted by a semicircle. Find the dimensions of the window that will allow the maimum amount of light if the perimeter of the window is 0 m. 7-0: Find the antiderivative of the given functions. 7. f() = g() = f(θ) = sin θ + 5 cos θ + θ + 0. g(θ) = tan θ sec θ cos θ + cos θ -: Find function f satisfying the given conditions.. f () = + 6. f (t) = 0 5, f() =, f () =. f (θ) = 5 sin θ 4 cos θ + 0, f(0) =, f (0) =

34 0

35 Module 5 Integration 5. Antiderivatives. Why are F () = 4 4 and F () = 4 (4 + ) both antiderivatives of f() =? -6: Find the antiderivative of the given functions.. f() = Page 58. f() = g(t) = t + t 5. h() = π 6. f(θ) = cos θ sin θ + sec θ 7-9: Find the function(s) f satisfying the following. 7. f () = f (t) = t +6t, f() =, f () = 9. f (θ) = sin θ + cos θ + 5, f(0) =, f (0) = 0. Suppose f is a function with f () = 0 for all. Show that f has no points of inflection.

36 MODULE 5. INTEGRATION 5. Series Page 58-4: Evaluate the following series. (Any value that is not an inde being summed over should be treated as a postive integer constant.). 5 i= i + i. n i= i + n. 4 6k k= 4. n i(i ) i= 5. Area Under a Curve Page 59. Let A be the area of the region R bounded by the -ais, the lines = and = 4, and the curve f() = + shown below: y R (a) Write a formula for the n th sum S n approimating A using right endpoints of the approimating rectangles. (b) Use A = lim n S n and your answer from (a) to calculate A. 5.4 The Definite Integral Page 59-4: Use the interpretation of the definite integral as the net signed area between the function and the -ais to compute the following integrals d 6 d ( 4) d ( ) d

37 5.5. FUNDAMENTAL THEOREM OF CALCULUS 0 5. Find r d by interpreting the integral as an area. Here r > 0 is a positive constant. r 6. Simplify the following to a single definite integral using the properties of the definite integral. 7 f() d + f() d f() d 7-8: Use Riemann sums with right endpoint evaluation to evaluate the following definite integrals b 0 ( + ) d d where b > 0 is constant 5.5 Fundamental Theorem of Calculus -4: Compute the derivatives of the following functions using the Fundamental Theorem of Calculus.. F () =. h() = t + t + dt t + t + dt. g() = 4. H() = [ cos ( t )] dt t + dt Page 59 5-: Compute the following definite integrals using the Fundamental Theorem of Calculus π 4 ( + ) d d sec θ dθ π π (sin + ) d d π d. d. The error function, erf(), is defined by erf() = π eponential function. If f() = erf ( ) find f (). e t dt where e u is the natural

38 4 MODULE 5. INTEGRATION. The left-side of a symmetrical glacial river valley has an approimate parabolic shape described by the curve y = 4 + (in km) as shown. y A y = 4 + s P (, y) Q B Using calculus techniques the arc length s from point A at the top of the valley to the point P (, y) can be shown to equal s = t dt. Engineers wish to build a road connecting points A and B with a bridge spanning the valley at the point P to the corresponding point Q on the opposite side of the valley. If the cost to build the bridge is 5% more per kilometre than the cost of building the road (i.e. it is 5/4 times as much per km), at what point P (, y) should they start the bridge to minimize the total cost? (Hint: Due to the symmetry of the situation just minimize the cost to build from point A to the middle of the bridge.) 5.6 Indefinite Integrals Page 60. Eplain why we use the indefinite integral symbol, f() d, to represent the general form of the antiderivative of the function f(). -5: Evaluate the following indefinite integrals. (. 4 6 ) d 4. csc θ cot θ dθ. + d 5. (tan + ) d 6. Find the general form of the function y = f() such that the equation y = + 9 is satisfied.

39 5.7. THE SUBSTITUTION RULE The Substitution Rule -6: Evaluate the following indefinite integrals using the Substitution Rule. +. ( + + 4) 5 d 4. cos(θ) sin θ dθ. (5 + ) 5 + d d. ( cos t t + t ) dt 6. sec ( π ) d Page 6 7-: Evaluate the following definite integrals using the Substitution Rule d 0. ( + ) d π tan 4 θ sec θ dθ [ + + ( ) 5 ] d.. π π 4 T 0 sin t cos t dt ( πt cos T ) dt (T > 0 is constant) 5.8 Integration Using All Methods -: Evaluate the given integrals using any method.. ( + ) + 5 ( ) 4 d t t dt Page ( + ) d ( + ) d (cos θ + sec θ ) dθ ( + 4 ) 5 d π 4 0 ( + ) 4 d ( + ) d sin (θ) dθ 6. cos θ (sin θ + ) 0 dθ. sin(tan ) d

40 6 MODULE 5. INTEGRATION 5.9 Area Between Curves Page 6-7: Find the area of the region bounded by the given curves.. y = + 8 and y = 4 over the closed interval [, ].. y = 5 and y = 6 over the closed interval [, 6].. y = + 6, y = + 4. = y, = y y =, y = 6. y = 0, y =, + y = 7. y = 8, y =, y = and lying in the first quadrant ( > 0 and y > 0). 5.0 Distances and Net Change Page 6. A particle oscillates in a straight line with velocity v(t) = sin (πt) centimetres per second. Compute the particle s displacement over the following time intervals. (a) t = 0 to t = seconds. (b) t = to t = seconds. (c) t = 0 to t = seconds.. The flow rate at a particular location for a large river over the month of May was approimately f(t) = (t 5) + 00 in gigalitres per day. Here the time t is measured in days from the beginning of the month. How much water flowed past that location between the times t = 0 and t = 0 days?

41 5.0. DISTANCES AND NET CHANGE 7 Module 5 Review Eercises -6: Evaluate the given integrals.. 4 ( 5 + ) 8 d. sec (θ) [tan(θ) + ] 5 dθ Page ( ) 5 t 4 π/4 0 π 0 5 t ( ) 6 d dt cos 4 () sin() d cos (θ) dθ 7-9: Find the area of the region bounded by the given curves. 7. y = + and y = y =, y = 4, and + y =. 9. = 0 y and = + y.

42 Answers. Eercises (page ). =. = or =. Written as a solution set it is {, /}. = ± 7, so no real solution = 5. {,, 4} 6. = { } 7. 0, ± { ±, ± } { }, ± {, 0, ± 4. Eercises (page ). D = R = (, ); -int=0; y-int=0 }. D = [6, ); -int= 6; No y-int. D = R = (, ); -int=-; y-int=4 4. D = R = (, ); -int=-5, 0,; y-int=0 5. D = R {} = (, ) (, ); No -int; y-int= 6. D = R { } = (, ) (, ); No -int; y-int= 4 7. D = [, ]; -int=-, ; y-int= 8

43 ANSWERS 9 8. D = R = (, ); -int=, ; y-int= 9. D = R = (, ); -int= ; y-int= 4 0. D = R { 7} = (, 7) ( 7, ); -int= 5; y-int= 5 7. D = R { /, } = (, /) ( /, ) (, ); No -int; y-int= 0. D = { R 4} = [ 4, ); -int= 4; y-int=. D = (, 0 ] [ 0, ) ; -int= ± 0; No y-int 4. D = (, 6] (, ); -int= 6; No y-int 5. D = (, 0 ] [ 0, ) ; -int= ± 0; No y-int 6. D = (, 0) (0, ); -int=; No y-int 7. Even. Neither 5. Even 9. Odd 8. Odd. Odd 6. Odd 0. Neither 9. Neither. Even 7. Neither. Even 0. Odd 4. Even 8. Even. Even. (a) y (c) y - (b) y (d) y (e) f( ) = f(), and so f is odd.

44 40 ANSWERS. Eercises (page 4). (a) f( + ) = + 4 (b) f(f()) = f g() = f(g()) = ( + ) with D = R, g f() = g(f()) = with D = R. f g() = f(g()) = ( ) + 6( ) + 4 = with D = R, g f() = g(f()) = ( ) = with D = R ( ) z 6z 4. f g(z) = f(g(z)) = + 5 = + 0z + 5 z + (z + ) with D = R { }, g f(z) = z g(f(z)) = + 5 z with D = R 5. f g() = f(g()) = ( + ) + 5 ( + ) 4 = + with D = R {, }, g f() = g(f()) = ( ) = ( 4) with D = R {4} 6. f() =, g() = +. Module Review Eercises (page 6). = /, =. Written as a solution set: {/, }. { 5, 4}. {,, } 4. {±, ±} 5. {,, } { 6. D = R 4 }, -int= 5, y-int= 4 7. D = (, ] [, ), -int=,, y-int does not eist 8. D = (, 5) [ ),, -int=, y-int= D = [ 8, ) (, ), -int= 8, y-int= 0. Even. Even. Odd. Odd 4. f g() = f(g()) = + 6 with D = R, g f() = g(f()) = ( + 6) with D = R 5. f g(t) = f(g(t)) = t + t with D = R {, }, g f(t) = g(f(t)) = 4t + 0t + 5 (t 4) + with D = R {4}

45 ANSWERS 4 6. f g() = f(g()) = D = [, ) + with D = (, ), g f() = g(f()) = with. Eercises (page 7). (a) ( ) + ( ) ( ) + = 5, (0) + (0) (0) + = 5 (b) 0 ( ) = (c) = ( ) ( ) ( (d) ) = lim. Eercises (page 7) Eercises (page 8) π 8. 4 π Eercises (page 8). lim f = 4, lim f = 4, approaches infinity), lim f = 4 + lim f =, lim f =, + lim f = (limit does not eist but 4. lim f =, lim f =, lim f =, lim f =, lim f =, lim f =, lim f =, lim f does not eist, lim f = 4. lim f() =, lim f() = 5, lim f() does not eist as the left and right-handed limits are + not equal at =. 4. c = 7

46 4 ANSWERS.5 Eercises (page 9) Vertical asymptote: = 6. No vertical asymptotes 7. Vertical asymptote: t = 8. Vertical asymptotes: =, = 9. Vertical asymptote: = 0 0. Vertical asymptotes: = 4, = 4. Vertical asymptote: = 0. No vertical asymptotes.6 Eercises (page 9). f is continuous at = a if (a) a is in the domain of f (b) lim a f() eists (c) lim a f() = f(a).. Continuous. Continuous 4. Not continuous 5. Continuous 6. Not continuous 7. Not continuous 8. (a) lim f() = and lim f() = 4. Therefore lim f() does not eist (b) c = ± 9. R {, }. Note the limit actually eists at = but the function is not defined there. 0. Let f() = +. Then notice that f() = and f() =. Then since f() < 0 < f() and f is continuous, there eists a c (, ) such that f(c) = 0 by the IVT. Since f(c) = c + c the result follows.. f() = + cos is continuous on [0, ], f(0) = < 0, f().58 > 0, so by the Intermediate Value Theorem there is a c in (0, ) with f(c) = 0. i.e. c + cos c = 0 and c is therefore a solution to the equation.

47 ANSWERS 4 Module Review Eercises (page ) Continuous. Not continuous. Not continuous. Eercises (page ). (a) 8 (b) m (c) Point-slope form: y = ( )+, Slope-intercept form: y = 4. (a) cm (b) 7 cm (c) 8 cm/s (d) v 0 cm/s. (a) V P. Eercises (page 4) 7.47 L/atm (b).49 L/atm. f ( + h) + ( + ) () = lim = h 0 h. f () = lim h 0 (+h) h =. f ( + + h) ( + ) () = lim = + 4 h 0 h + h f () = lim = h 0 h + (+h)+ 5. f +h+ + + () = lim h 0 h 6. f () = lim h 0 +h h = ( + ) = = 7. lim h 0 lim h 0 + ( + h ) h ( + h ) h = lim h 0 = lim h 0 + h h h h = lim h h 0 h = lim h h 0 h = = lim h 0 + h h = lim h 0 + h h =

48 44 ANSWERS. Eercises (page 4). (a) Differentiable (b) Differentiable (c) Not Differentiable. Discontinuous at = 0. (d) Not differentiable. At = 0 the left and right hand limits for the derivative are not equal: f( + h) f() f( + h) f() lim lim, h 0 h h 0 + h therefore the limit itself (the derivative) does not eist. Geometrically no tangent line may be drawn at the point so there can be no derivative as that is the tangent slope..4 Eercises (page 4). f () = =. g () = 5 7 = f (u) = 4u 5 + 4u ; f () = 0 5. f () = a sin(π/5) a dy. d = ds 5 6. dt = gt + v 0 dy 7. (b): d = [ + = = = (d): v() = ds dt = [ t + 4t t= = 0 cm/s t= dv (b): dp = (.4) P P = =.49 L/atm P = 8. y = + 9. = 0, 4.5 Eercises (page 5). (a) C(000) = ppm (b) C (000) = ppm/year (c) C(005) C(000) C(000) 00 =.5%.. (a) The volume of a cone is V = πr h. Use similar triangles to show that the radius of the surface of the liquid is r = 5 y. (b) dv dy = 64π m 8.04 y=4 m 5 m

49 ANSWERS 45.6 Eercises (page 5). f () = ( 4 6 ) ( ) + ( 4 + ) ( ). f () = ( + π ) ( + ) + ( + π + ) ( 4) dy. d = () ( + ) ( + ) + ( ) ( ) ( + ) + ( ) ( + ) ( 4 + ) 4. df d = ( 6) 5. f (θ) = (θ + ) ( θ 7 ) ( θ + θ 4 ) (θ) (θ 7) = θ + 6θ + (θ 7) 6. g () = + ; g (4) = f (v) = [()(v+4/v)+(v+)( 4/v )](v +v) (v+)(v+4/v)(v+) (v +v) 8. h () = c + ( + ) + ( + ) (4 + ).7 Eercises (page 5). Point-slope form: y = 8( ) +, Slope-intercept form: y = 8 5. Point-slope form: y = 5 ( ) +, Slope-intercept form: y = (, ), (, 4).8 Eercises (page 6). f () = 8 ( + ) 8. dg d = + ( + ) = + ( + ) + 4. f 7(4t + ) (t) = (t + t + 4) 4. y = 7 ( ) ( + ) 5n n 5. h () = (5 n + 4c) [ ( 6. f ) ] [ 4 4 () = ( + ) ( + ) ] +

50 46 ANSWERS.9 Eercises (page 6). Use the sine addition identity followed by the fundamental limits involving sine and cosine to get f () = 4 cos 4.. f () = cos sin. f (t) = t (sin t + tan t) t ( cos t + sec t ) (sin t + tan t) 4. dh dθ = csc θ cot θ csc θ ; H (π/) = 0 5. f () = (cos sin )(sec cot ) + (sin + cos )(sec tan + csc ) 6. df dθ = 0 7. (a) Since tan(π/4) =, = π/4 and y = π/4 indeed satisfy the equation. (b) Point-slope form: y = ( + π/)( π/4) + π/4, Slope-intercept form: y = ( + π/) π /8 8. (a) = 5(0) + (0) () (b) y =.0 Eercises (page 7). f () = ( ) ( 8 7 ). g () =. f (θ) = θ cos(θ ) ( + ) (6) = 4. h (θ) = cot θ csc θ + ; g () = f () = sec [( + ) ( + )] tan [( + ) ( + )] [ ( ) ( + ) + ( + ) ( 6. y = ( 4 sin ) ( ) = 4 sin 7. f () = ( ) ( + sin ) ( sin cos ) = sin cos ( + sin ) 8. dy d = 5 (csc + )4 csc cot y = π sec θ + π sec (πθ) 0. g () = ( + ) = ( + ) ( 4 ) ( + ) ( 4 ) ( ) (7) ( 4 ) 6 ( 4 ) + 8 ( 4 ) 6 + )] +

51 ANSWERS 47. f () =. da dt ( ) + = ω sin(ωt + φ) ; da dt 4 ) = ( ( + ) ( + ) 4 = ω sin(φ) t=0. f () = ( cos [ cos( + ) ]) ( sin ( + )) ( + ) { } 7π { } π { π } 4. + nπ n an integer + nπ n an integer nπ n an integer. Eercises (page 7). y = y. y = 6 y y y ; dy d = 5 (,y)=(,). y = b a y 4. y = y cos(y) cos(y) 5. y = sin( + y) cos y + sin( + y) 6. (a) Noting that = ( ) it follows that the values = and y = = ( ) simultaneously satisfy the equation. (b) Point-slope form: y = ( + ) +, Slope-intercept form: y = + 4. Eercises (page 8). f () = dy d = g (t) = t + 4t t 4. f () = 5 ( ) 5. g () = + ( + π) + ( + + ) 6. h (y) = (y + 4) ( + 0)(y + 5) (y + 4) ( + 0) (y + 5) = (y + 4) (y + ) (y + 5) 7. y = 4 ( ) 4 ( + )

52 48 ANSWERS 8. f (θ) = sin(θ) + sin θ cos θ 9. g () = sec ( + ) cos() tan ( + ) sin() 0.. dy dt = 4 (sin t + 5) 4 cos t df d = ( 4 ) ( ) ( ) ( ) () ( ) = ( 4 ) ( ). y = y y 4 4 y + y. Eercises (page 8). d f d = csc cot. f () = 90( ) 8 ; f () = 90. y = 6 sec + 6 sec tan + sec tan + sec 4. y = y y y = y y.4 Eercises (page 8) π km/h 6 5π cm/sec km/sec 5 π m/min km/h 4 5 m/sec cm/sec.5 Eercises (page 9). V ± V = 5 ± 5 cm. V ± V = 88π ± 7π cm

53 ANSWERS 49. L() = + 4( ) 4. L() = + ( π/4) Module Review Eercises (page 0). f () =. f () =. f () = 4. y = + 5. g () = 6. h (y) = ( + ) + / + 5 / ) ( 4 y 0 y + 5 (y + ) 7. f (θ) = cos θ sin θ 8θ sin ( θ ) ( ) ( + sin ) ( + cos ) 8. g () = sec ( + 4 ) tan ( + 4 ) cos() sec ( + 4 ) sin() ( ) 4/ f () = (4 + 5) 0. dy d = y 4 y 4 y + 8y cos y. y = 6 + π +. =, =. θ = π + nπ, θ = 5π + nπ (n an integer) 4. 57π cm /sec 5. cm/sec 6. cm/sec () /4 4. Eercises (page ). Note: It is assumed that the functions etend beyond the graph of the plot with the same trend unless they are eplicitly terminated with a point. The values below are approimate. (a) Relative minima: f(.4) = 0.8 and f(.9) = 0.5; Relative maima: f(.0) =.0, f(0.4) =.0, and f(.) =.6. No absolute minimum nor absolute maimum.

54 50 ANSWERS (b) Absolute maimum value of f(0) = 0.5. This is also a relative maimum. (c) Relative minimum: f( ) = ; Relative maimum: f( ) = 6, f(0) =. Absolute maimum: f( ) = 6, No absolute minimum.. =, 4. = 0 4. s = 6 5. =, 6. t = 7. No critical numbers 8. t =, 9. = 5, 0, 5 0. θ in { π + nπ n an integer } { 5π } + nπ n an integer. Absolute maimum: f( ) = f() =, Absolute minimum: f. Absolute maimum: f( ) =, Absolute minimum: f ( /) = 4 7. Absolute maimum: g(0) = 0, Absolute minimum: g (/) = 4 4. Absolute maimum: H(8) =, Absolute minimum: H ( ) = 4 ( ± ) = 4 5. Absolute maimum: f(π/4) = f(5π/4) =, Absolute minimum: f(π/4) = f(7π/4) = 4. Eercises (page ). f is continuous on [, 4], differentiable on (, 4), and f( ) = f(4) = so Rolle s Theorem applies. Solving f (c) = 0 shows c = 7/ or c =, however only c = 7/ is in the open interval (, 4).

55 ANSWERS 5. Each of the following graphs of f give one possible countereample. There is no point in the interval [0, ] where the function has a horizontal tangent and hence the conclusion to Rolle s theorem is invalid. (a) Continuity fails: y (b) Differentiability fails: y (c) f(0) f(): y Use the Mean Value Theorem. 4. c = in (, ) has f (c) = 4. Eercises (page ) 0 ( 5) ( ) = 5.. Decreasing on: (, ); Increasing on: (, ); No relative maima; Relative minimum: f( ) = 0. Increasing on: (, ) (, ); No relative maima or minima (. Decreasing on: 0, 7π ) ( ) ( π 7π 6 6, π ; Increasing on: 6, π ) ; ( ) 6 π Relative maimum: f = π ( ) 7π ; Relative minimum: f = 7π Decreasing on: (, 0); Increasing on: (0, ); Relative minimum: f(0) = 0 5. Decreasing on: (, 0); Increasing on: (0, ); Relative maimum: f(0) =. Note that the First Derivative Test cannot be applied here because f() is discontinuous at = 0. One must return to the definition of relative maimum to evaluate the critical number = y 7. y Note that your graphs should be equivalent up to vertical shift by a constant.

56 5 ANSWERS 4.4 Eercises (page ). Notice that f () = 4 and 4 is positive for all values of ecept 0. The only potential inflection point would therefore occur at = 0 but the concavity is positive on both sides of = 0 and hence does not change at that value.. Notice that f () = which is negative for < 0 and positive for > 0. Therefore the concavity does, in fact, change at = 0. However the function is not defined at 0 so there is no point on the curve there (it is a vertical asymptote) and hence no inflection point eists.. Notice that f () = 5 ( )(+), f () = 0 ( ) ( + ) (. Relative maimum: f( ) = 6; Relative minimum: f() = 6; Inflection points:, ) +, ( 5 (0, ),, ) ( + ; Concave upward on: / ) (, 0 / ), ; Concave downward on:, / ) ( ( 5 0, / ) ; 4.5 Eercises (page 4). Relative maimum: f( ) = 7; Relative mininum: f() = 5 (. Relative maimum at f π ) ( π ) = ; Relative minimum: f = 5;. No since f () = cos sin 4 sin 4 vanishes at = 0 so the test is inconclusive. The first derivative test shows that = 0 is a the location of a local minimum of f. 4.6 Eercises (page 4). D = R = (, ); -int= 0, 6; y-int= 0; Increasing on: (, 0) (4, ); Decreasing on: (0, 4); Relative maimum: f(0) = 0; Relative minimum: f(4) = ; Concave upward on: (, ); Concave downward on: (, ); Inflection point: (, 6); Graph: y f() = 6 (0, 0) 6 (, 6) (4, )

57 ANSWERS 5. D = [0, ); t-int= 0, ; y-int= 0; Increasing on: (, ); Decreasing on: (0, ); No relative maima; Relative minimum: g() = ; Concave upward on: (0, ); Not concave downward anywhere; No inflection points; Graph: y g(t) = t t t (, ). D = R = (, ); No -int; y-int= ; Increasing on: (0, ); Decreasing on: (, 0); No relative maima; Relative minimum: F (0) = ; Concave upward on: (, ); Not concave downward anywhere; No inflection points; Graph: y F () = + 9 (0, ) 4.7 Eercises (page 4) Horizontal asymptote: y = 5. No horizontal asymptotes 6. Horizontal asymptotes: y =, y =

58 54 ANSWERS 7. Horizontal asymptote: y = 8. Horizontal asymptote: y = 0 (Hint: The Squeeze Theorem, generalized to a limit at infinity, can be used here to evaluate the limits.) 9. Horizontal asymptote: y = 5 0. Horizontal asymptote: y =. Horizontal asymptotes: y =, y = 4.8 Eercises (page 5). Slant asymptote: y = 0. No slant asymptotes. Slant asymptote: y = 4.9 Eercises (page 5). Notice f () = = ( + )( ) and f () = 6. D = R = (, ); -int=, ; y-int= ; No asymptotes; No symmetry; Increasing on: (, ) (, ); Decreasing on: (, ); Relative maima: f( ) = 0; Relative minimum: f() = 4; Concave upward on: (0, ); Concave downward on (, 0); Inflection point: (0, ); Graph: y f() = (, 0) (, 4)

59 ANSWERS 55. Notice that the first and second derivatives simplify to y () = 6 ( ) and y = 6( + ) ( ). D = R {, } = (, ) (, ) (, ); -int= 0; y-int= 0; Horizontal asymptote: y = ; Vertical asymptotes: =, = ; Symmetric about the y-ais; Increasing on: (, ) (, 0); Decreasing on: (0, ) (, ); Relative maima: f(0) = 0; No relative minimum; Concave upward on: (, ) (, ); Concave downward on (, ); No inflection points; Graph: y y = f() = (0, 0) = =. Notice that the derivatives simplify to f () = ( + ) and f () = 4 ( + ) 4. D = R { } = (, ) (, ); -int= 0; y-int= 0; Horizontal asymptote: y = ; Vertical asymptote: = ; No symmetry; Increasing on: (, ) (0, ); Decreasing on: (, 0); No relative maima; Relative minimum: f(0) = 0; Concave upward on: (, ) (, /); Concave downward on (/, ); Inflection point: (/, /9); Graph: y f() = ++ y = = (/, /9) 4.0 Eercises (page 6). Base length= 5 m, Height= 5 m, Area= 5 8 m. (/5, 6/5). For part (a) note it is easier to minimize the distance-squared than the distance. For part (b) the line perpendicular is y = ( )+ = +. To find the intersection of this and the original line we solve the two equations simultaneously since the point of interest must lie on both lines.. First number=0, Second number= Radius= π.7 cm, Height=.4 cm π

60 56 ANSWERS 5. (a) = 0.5 km 7 (b) = 0.56 km 5 b (c) = ( w ) v (d) Although the total time taken t does depend on a the value for does not. As epected, it does not matter how far upstream you start, you would still turn off at the same location. That said, a does play a role in the solution because valid values of must lie in the interval [0, a]. If a had been smaller than the solution for, say in part (a) had the starting distance been 0. km, then the critical number would no longer be in the interval and that solution would be invalid. One would consider the endpoints of the interval to see which was optimal. 6. = 40 m, h = 0 m 7. 8 m 4 m ( 8. 0 ) 5.9 km Module 4 Review Eercises (page 8). =, =. t = 0, t =. θ = π 6 + nπ, θ = 5π 6 + nπ (n an integer) 4. Absolute maimum: f() = 7, Absolute mininimum: f( ) = 7 5. Absolute maimum: g() = 7, Absolute minimum: g(0) = 0

61 ANSWERS D = R = (, ); -int=0, -8; y-int=0; No asymptotes; Increasing on: (, ); Decreasing on: (, ); No relative maima; Relative mininimum: f( ) = (; Concave ) upward on (, 0) (4, ); Concave downward on (0, 4); Inflection points: (0, 0), 4, (4) 4 ; Graph: y ( ) 4 4, (4) f() = (, ) 7. D = R {} = (, ) (, ); -int=0; y-int=0; Vertical asymptote: = ; Increasing on: (, 0) (4, ); Decreasing on (0, ) (, ); Relative maimum: f(0) = 0; Relative minimum: f(4) = 8,, Concave upward on (, ); Concave downward on (, ); No inflection points; Graph: y f() = ( ) (4, 8) = Vertical asymptote: =, Horizontal asymptote: y =. Vertical asymptote: t =, Horizontal asymptotes: y =, y = 4. = 0 m, y = m 5 m m, lot area=560 m 6. Maimal area occurs when Width=Circle diameter= π the width). m and Height= π m (i.e. half

62 58 ANSWERS 7. F () = 5 8 8/5 + 8 / C 8. G() = / C 9. F (θ) = cos θ + 5 sin θ + 4 θ4 + θ + C 0. G(θ) = sec θ sin θ + tan θ + C. f() = 4 5 5/ C + D. f(t) = 9 t8/ t 5 t t + 6. f(θ) = 5 sin θ + 4 cos θ + 5θ + 7θ 7 5. Eercises (page ). Both F and F differentiate to. As this problem suggests, any two antiderivatives of a function differ at most by a constant.. F () = C. F () = 4 + C 4. G(t) = t + 4t + C 5. H() = π + C 6. F (θ) = sin θ + cos θ + tan θ + C 7. f() = C + D 8. f(t) = 4 5 t 5 + t 5 t f(θ) = sin θ cos θ + 5 θ + θ If f () = 0, then f () = C where C is a real constant. If C is non-zero then f has the same concavity everywhere, while if C = 0 then f () = 0 implies f() = D + E so f is linear and hence has no point of inflection. 5. Eercises (page ) n ( n + n + n 6 + n ) = n + n + 7 6n

63 ANSWERS n + n + n 6 n + n 5. Eercises (page ). (a) S n = n n f i= (b) A = lim n S n = Eercises (page ) ( + i ) = n = n n 4n n i= ( 9 n + 7i ) n + 54i n = n + 9 n Notice that the area can be viewed as the quarter of a circle: y 6. r 0 Then the integral is 4 9 f() d 7. S n = n and so ( πr ). (Since the area is above the -ais the integral is positive.) ( 7n n + 7n ) 4n + n = 9n + 6n + 8 4n ( + ) d = lim S n = 9 n S n = b ( + n + 6n 5.5 Eercises (page ) ) and so b 0 d = lim n S n = b.. df d = + +. h () = g () = [ cos ( )] 4. Noting that H() = 0 t + dt + 0 gets H () = t + dt = 0 t + dt + 0 t + dt, one

64 60 ANSWERS The Fundamental Theorem of Calculus is not applicable here because the integrand is discontinuous on [, ]. The area under the curve (i.e. the integral) in fact diverges to +. To show that requires a consideration of improper integrals.. f () = π e ( ) ( ) = 6 π e 6 Calculus.). (, y) = ( km, 7 6 km) 5.6 Eercises (page 4) (Use the Chain Rule and the Fundamental Theorem of. The Fundamental Theorem of Calculus shows that the definite integral b a f() d is, assuming the conditions of the theorem are met, intimately connected to the antiderivative of f by the relation b a f() d = F (b) F (a) where F is an antiderivative of f. Thus in many cases finding a definite integral is a two-step process where first one finds an antiderivative of f and then secondly takes the difference of that function evaluated at the limits of integration. It is natural, therefore, to generally write the answer to the first step, namely the antiderivative of f, symbolically as f() d. (The notation is further convenient because it embeds the function we are antidifferentiating directly in the symbol in the same way we write df d abstractly for the derivative of f.) C C 4. csc θ + C (tan 5. + ) d = sec d = tan + C 6. The equation states that the derivative of y is + 9 so y must be the general form of the antiderivative of + 9 which is just its indefinite integral: ( y = + 9 ) d = C The general solution for any differential equation of the form y = f() is similarly y = f() d.

65 ANSWERS Eercises (page 5).. ( ) 4 + C ( 5 + ) + C. sin( t) + 4 t4 + C 4. ( sin θ) + C (4 + ) 5 4 (4 + ) + C tan ( π ) + C 7. Using u = 4 + 9, integral is = 8. Using u = tan θ, integral is = 5 u 9 0 du 4 = 49 u 4 du = 5 9. Breaking the integral into three separate integrals (one per term) and using u = on the last one that integral equals = 0 u5 ( du) = 6. Combining this with the definite integral of the first two terms gives the final answer 5. Alternatively, one can find the indefinite integral of the third term (i.e. substitute back to ) to get the antiderivative of the entire integrand as + 6 ( )6 + C and evaluate that at the original limits of. 0. Using u = +, integral is =. Using u = cos t, integral is = 9 u u du 4 = 4 57 ( ( du) = 6 ). Using u = πt π, integral is = cos(u) T du T T 0 π = 4π 5.8 Eercises (page 5) C C C 4. sin θ + tan θ + C 5. ( + 4 ) 6 + C 6

66 6 ANSWERS (sin θ + ) + C 6 ( ) 7 t C Note that since sine and tangent are odd, the integrand itself is odd. Since the limits are ±a, the integral vanishes. 5.9 Eercises (page 6). 9 units. units. 4. unit unit 5. 6 unit 6. unit 7. 6 units 5.0 Eercises (page 6). (a). (b) 0 v(t) dt = π cm v(t) dt = π cm (c) 0 cm (The particle is back where it started after seconds.) gigalitres Module 5 Review Eercises (page 7)... ( 5 + ) 9 + C 45 [tan(θ) + ]6 + C 5 ( 4/ t /5 4) + C

67 ANSWERS ( ) π

68 64

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