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Calc2_IRB_ExplM_09-12.qxd 11/17/06 Name: 3:53 PM Page 129 Group Members: Exploration 9-3a: Integration by Parts Practice Date: Objective: Integrate by parts to find antiderivatives quickly. Find the integral. Show the steps you take. 1. x 2 2. x 5 3. e sin 3x dx ln 4x dx 5x 4. x(ln x) 5. sin x cos x dx (Be clever!) 6. sin x dx, in terms of 10 10 3 dx sin 8 x dx cos 6x dx 7. What did you learn as a result of doing this Exploration that you did not know before? Calculus: Concepts and Applications Instructor’s Resource Book ©2005 Key Curriculum Press Exploration Masters / 129 Calc2_IRB_ExplM_09-12.qxd Name: 11/17/06 3:53 PM Page 130 Group Members: Exploration 9-4a: Reduction Formulas Date: Objective: Integrate by parts to find a reduction formula, and use that formula in a volume problem. 1. Integrate by parts to find a formula for sin10 x dx in terms of sin8 x dx. 4. Problem 3 involves sin6 x dx. Use the reduction formula to evaluate this indefinite integral. Use the result to find the volume of the solid algebraically, using the fundamental theorem. Does the result agree with the answer to Problem 3? 2. Based on the pattern you observe in the answer to Problem 1, write a formula for sinn x dx in terms of sinnD2 x dx. Check with Section 9-4 to make sure your formula is correct. 3. The graph shows y H sin3 x. Find numerically the volume of the solid generated by rotating about the x-axis the region under this graph from x H 0 to x H π. y 1 x 5. What did you learn as a result of doing this Exploration that you did not know before? 3 130 / Exploration Masters Calculus: Concepts and Applications Instructor’s Resource Book ©2005 Key Curriculum Press Calc2_IRB_ExplM_09-12.qxd 11/17/06 Name: 3:53 PM Page 131 Group Members: Exploration 9-5a: Integrals of Special Powers of Trig Functions Date: Objective: Integrate odd powers of sine and cosine, and even powers of secant and cosecant. 1. Integrate cos7x sin x dx as a power. 5. Integrate tan5 x sec2 x dx as a power. 2. Explain why cos7x dx cannot be integrated as a power. 6. Explain why sec8 x dx cannot be integrated as a power. 3. Associate all but one of the cosines in the integral in Problem 2, change them to sines using the Pythagorean properties, then evaluate the integral as a sum of powers. 7. Associate all but two of the secants in the integral in Problem 6, change them to tangents using the Pythagorean properties, then evaluate the integral as a sum of powers. 8. Explain why sec7 x dx cannot be integrated by the method of Problem 7. 4. Explain why cos6 x dx cannot be integrated by the method of Problem 3. 9. What did you learn as a result of doing this Exploration that you did not know before? Calculus: Concepts and Applications Instructor’s Resource Book ©2005 Key Curriculum Press Exploration Masters / 131 Calc2_IRB_ExplM_09-12.qxd Name: 11/17/06 3:53 PM Page 132 Group Members: Exploration 9-5b: Other Special Trigonometric Integrals Date: Objective: Find the antiderivatives sin2 x dx and cos2 x dx and cos ax sin bx dx. 1. You recall that cos (A C B) H cos A cos B D sin A sin B Use this property to write the double-argument property for cos 2x. 6. Products of sine and cosine with two different arguments can be integrated by parts. Integrate cos 7x sin 5x dx. 2. Transform the double-argument property so that cos 2x is expressed in terms of cos x alone. 3. Transform the double-argument property so that cos 2x is expressed in terms of sin x alone. 4. Transform the properties in Problems 2 and 3 so that cos2 x is expressed in terms of cos 2x and so that sin2 x is expressed in terms of cos 2x. 7. Use the result of Problem 6 to write a formula for cos ax sin bx dx. 5. Use the results of Problem 4 to find the integrals cos2 x dx and sin2 x dx. 8. What did you learn as a result of doing this Exploration that you did not know before? 132 / Exploration Masters Calculus: Concepts and Applications Instructor’s Resource Book ©2005 Key Curriculum Press Calc2_IRB_ExplM_09-12.qxd 11/17/06 3:53 PM Name: Page 133 Group Members: Exploration 9-6a: Introduction to Integration by Trig Substitution Date: Objective: Integrate a square root of a quadratic by making a rationalizing substitution. Write the integral in Problem 2 in terms of trig functions of θ. 8 y x –2 7 4. Using the techniques you have learned for integrating powers of trig functions, evaluate the integral in Problem 3 algebraically. 1. The diagram above shows the circle for which y H J 64 D x 2 Use the most time-efficient method to find the area of the zone of this circle from x H D2 to x H 7. 2. In Problem 1, you evaluated the definite integral 7 D2 5. Do the reverse substitution to get the answer in Problem 4 into terms of x. 64 D x 2 dx numerically. Although the radical can be written as (64 D x 2)1/2, explain why the indefinite integral cannot be found using the power rule. 3. The radical in Problems 1 and 2 looks like the third side of a right triangle (diagram below) with one leg x, hypotenuse 8, and θ in standard position. 6. Use the answer to Problem 5 to find exactly the area of the zone in Problem 1. Show that a decimal approximation of this exact answer agrees with the answer found numerically. v 8 x u Calculus: Concepts and Applications Instructor’s Resource Book ©2005 Key Curriculum Press 7. What did you learn as a result of doing this Exploration that you did not know before? Exploration Masters / 133 Calc2_IRB_ExplM_09-12.qxd 11/17/06 Name: 3:53 PM Page 134 Group Members: Exploration 9-7a: Integrals of Rational Functions by Partial Fractions Date: Objective: Integrate algebraically a function of the form (polynomial)/(polynomial). 1. The integral 10x D 32 dx x D 4x D 5 2 has an integrand that is a rational algebraic function. It can be transformed to 10x D 32 (x D 5)(x C 1) dx The integrand can then be broken up into two partial fractions, A B C xD5 xC1 Find the constants A and B. 3. Ask your instructor to show you the way to find the partial fractions in one step, in your head. 4. Study Heaviside’s method in Section 9-7 to see why the shortcut works. Check here when you feel you understand it well enough to explain it to other members of your study group. 5. Heaviside’s method and its shortcut can be extended to integrals with any number of distinct linear factors in the denominator, as long as the numerator is of degree at least one lower than the denominator. Do this integration: 11x D 22x D 13 dx x D 2x D 5x C 6 2 3 2 2. The integral in Problem 1 can be evaluated as a sum of two terms, each of which has the form of the reciprocal function. Find the indefinite integral. 6. What did you learn as a result of doing this Exploration that you did not know before? 134 / Exploration Masters Calculus: Concepts and Applications Instructor’s Resource Book ©2005 Key Curriculum Press Calc2_IRB_ExplM_09-12.qxd 11/17/06 3:53 PM Name: Page 135 Group Members: Exploration 9-8a: Integrals of Inverse Trigonometric Functions Date: Objective: Use integration by parts to learn how to integrate an inverse trigonometric function. For Problems 1 and 2, let I H sinD1 x dx. 1. The integral I can be written as a product in two different ways: (1)(sin D1 x dx) or (sin D1 3. Find I H secD1 x dx. Assume that x > 0. To do this, you will have to recall sec x dx and integration by trigonometric substitution, as well as integration by parts. x)(1 dx) Why is the first form not useful for integration by parts, while the second form is? 2. Evaluate the integral to find I. Check your answer with the answer in the text. If it does not agree, go back and find the source of your differences. 4. What did you learn as a result of doing this Exploration that you did not know before? Calculus: Concepts and Applications Instructor’s Resource Book ©2005 Key Curriculum Press Exploration Masters / 135 Calc2_IRB_ExplM_09-12.qxd Name: 11/17/06 3:53 PM Page 136 Group Members: Exploration 9-9a: Introduction to Hyperbolic Functions Date: Objective: Starting with two combinations of exponential functions, show that they have properties similar to those of trigonometric functions. For this exploration, let 1 u H 2(e x C eDx ) 1 and v H 2(ex D eDx) 1. On the same screen, plot the graphs of y1 H u, y2 H v, and y3 H (1/2)e x using a window with [D2, 2] for x and [D4, 4] for y. Sketch the results. Tell how the graphs of u and v are related to the graph of (1/2)e x. 4. The equation u2 D v 2 H 1 that you found in Problem 3 plots as a unit equilateral hyperbola in a uv-coordinate system, as shown here. Use the two parametric equations to find u(1.3) and v(1.3). Show that the point (u, v) is on the hyperbola. u 1 v 1 2. Show that uQ H v, uP H u, uQP H v, and u(4) H v, where uQP and u (4) are the third derivative and the fourth derivative of u, respectively. How do these derivatives compare to the first four derivatives of y H cos x? 5. Tell why the branch of the hyperbola for negative values of u is extraneous. 6. The figure shows the unit circle in a uv-coordinate system. Show that the point (u, v) H (cos 1.3, sin 1.3) is on this unit circle. 1 v u 3. Functions u and v are parametric functions of x. Eliminate the parameter x by finding u 2 D v 2. −1 1 −1 7. Why do you suppose that cosine and sine are called circular functions and u and v at the beginning of this Exploration are called hyperbolic functions? 8. What did you learn as a result of doing this Exploration that you did not know before? 136 / Exploration Masters Calculus: Concepts and Applications Instructor’s Resource Book ©2005 Key Curriculum Press Calc2_IRB_ExplM_09-12.qxd 11/17/06 3:53 PM Name: Page 137 Group Members: Exploration 9-9b: Chain Experiment Date: Objective: Find the particular equation for a hanging chain, and verify it by actual measurement. 4. Calculate y for each 20 cm from the vertex, from the left end to the right end of the chain. Use a time-efficient method. Round to one decimal place. y x y x y x 1. Hang a chain from the frame around the chalkboard, as shown in the diagram. Let x be the horizontal distance (cm) from the vertex to a point on the chain, and let y be the vertical distance (cm) from the chalk tray to the point. Measure x and y for the vertex and for the two endpoints. Vertex: Ends: Left: Right: 2. From Problem Set 9-9, you found that the general equation for a hanging chain is h w y H wcosh hx C C where h is the horizontal tension in the chain and w is the weight of the chain per unit length. Let k H h/w. Calculate the constants k and C. Show your work. Store the answers in your grapher. 5. Remove the chain from the board. Plot the data on the board as accurately as you can. Then hang the chain again. How closely do the calculated data fit the shape of the actual chain? 6. Use the equation to calculate the length of the chain between the two endpoints. Then measure the chain to see how close your calculated value is to the actual value. 7. On the back of this sheet, derive the equation h w y H wcosh hx C C 3. Weigh and measure the chain. Use the results to find values of h and w. using the fact that the tension in the chain is equal to the vector sum of the horizontal tension (of magnitude h) and the vertical tension, and is always directed along the chain. 8. What did you learn as a result of doing this Exploration that you did not know before? Calculus: Concepts and Applications Instructor’s Resource Book ©2005 Key Curriculum Press Exploration Masters / 137 Calc2_IRB_ExplM_09-12.qxd 11/17/06 Name: 3:53 PM Page 138 Group Members: Exploration 9-10a: Introduction to Improper Integrals Date: Objective: Find the limit of a definite integral as the upper limit of integration becomes infinite. Calvin is driving at 80 ft/s (about 55 mi/h). At time t H 0 s, he lifts his foot from the accelerator and coasts. His velocity decreases as shown in the graph. 3. Phoebe figures that her velocity is given by v(t) H 80eD0.1t How far has she gone after 10 s, 20 s, and 50 s? v (t) 80 Area = distance traveled. t 0 10 b 20 1. Calvin figures that his velocity is given by 4. Find algebraically the distance Phoebe has gone after t H b s. Use the result to find out how long it will take her to go a distance of 1000 ft. v(t) H 320(t C 4)D1 How far has he gone after 10 s, 20 s, and 50 s? 2. Find algebraically the distance Calvin has gone after t H b s. Use the result to find out how long it will take him to go a distance of 1000 ft. Phoebe does the same thing Calvin did. Her velocity-time graph is shown below. v (t) 80 Area = distance traveled. 5. Show that Phoebe’s distance approaches a limit as b approaches infinity, but Calvin’s distance does not. 6. How do you reconcile the fact that Phoebe’s distance approaches a limit, and the fact that her velocity never reaches zero? t 0 10 b 20 7. What did you learn as a result of doing this Exploration that you did not know before? 138 / Exploration Masters Calculus: Concepts and Applications Instructor’s Resource Book ©2005 Key Curriculum Press Calc2_IRB_ExplM_09-12.qxd 11/17/06 Name: 3:53 PM Page 139 Group Members: Exploration 9-11a: Miscellaneous Integration Practice! Date: Objective: Evaluate an antiderivative when the particular technique is not specified. 1. 2. tan 5 4x dx x 3eDx dx 1 1 C t dt 2 5. 3. ∞ 4. x (x D 2)(x D 3)(x D 4) dx tanh x dx (Over) Calculus: Concepts and Applications Instructor’s Resource Book ©2005 Key Curriculum Press Exploration Masters / 139 Calc2_IRB_ExplM_09-12.qxd Name: 11/17/06 3:53 PM Page 140 Group Members: Exploration 9-11a: Miscellaneous Integration Practice! continued 6. sin 7. (x 5 4 x dx 9. x e ax cos bx dx C 2)3 dx 10. 8. e Date: 2 x3 sin D1 ax dx dx 11. What did you learn as a result of doing this Exploration that you did not know before? 140 / Exploration Masters Calculus: Concepts and Applications Instructor’s Resource Book ©2005 Key Curriculum Press Calc2_IR_656_SE_09-12.qxd 10/26/06 5:51 PM Page 200 Exploration 8-7a 1. The graph agrees with the given figure. 10 2. r(0.3) H 3 D 2 c os 0.3 H 9.1799... x(0.3) H r(0.3) • cos 0.3 H 8.7699... y(0.3) H r(0.3) • sin 0.3 H 2.71286... The sample point shown is at about (8.8, 2.7). The angle measures about 17−, which is 0.296... radian. So all three values agree with the graph. 7. The graph shows that at an apparent intersection point in the second quadrant, r1, is positive, but r2 is negative, meaning that the rose is being traced in the fourth quadrant when the circle is being traced in the second quadrant. By the time the rose gets to the second quadrant points, the circle is being traced in the fourth quadrant. Apparent intersection r1 is positive. 3. At θ H 0.3, the radius (of the sector of circle that the wedge approximates) is r(0.3), and the central angle is dθ H 0.1, so 0.1 2 the approximate area is 2π • π • 9.1799... H 0.4589... unit. r2 is negative. dq of a circle of radius r and area 4. The sector is approximately 2π dq 1 • πr 2 H r 2 dθ. πr 2, so dA H 2π 2 5. A H θH2π θH0 2 10 1 2 3 D 2 cos θ dθ H 84.2977... (numerically) 6. The y-radius is approximately 4.5, so the area formula predicts that the area will be π • 6 • 4.5 H 84.8230.... Close! (y-radius is exactly 6 sin (cosD1 23) H 4.4721..., making the area π • 6 • 4.4721... H 84.2977..., which is precisely the numerical answer.) 8. Answers will vary. Exploration 8-7c 1. The graph agrees with the given figure. 2. r(0.8) H 35.9554... M 36.0 thousand miles 1 3. dA H 2r 2 dθ H 3528(10 D 11 cos θ)D2 dθ 7. Answers will vary. Exploration 8-7b 1. The graph agrees with the given figure. 5.5 AH 0.8 3528(10 D 11 cos θ)D2 dθ H 308.9953... M 309 million square miles 2 4. dL H (dr/dθ C r 2 dθ ) 2 H [D84(1 0 D 11 cos θ )211 sin θ] C [84 (10 D 1 1 cos θ )1]2 dθ LH 5.5 θH0.8 dL H 82.4852... M 82.5 thousand miles 5. Check: r(0.8) M 36 as the comet approaches Earth, and r(5.5) H 38.1015... M 38 as the comet recedes from Earth. So the distance traveled on the curved path should be a bit greater than 36 C 38 H 72 thousand miles. Thus the 82.5 thousand miles is reasonable. 6. Answers will vary. 2. 6 sin 2θ H 4 θ H 0.5 sinD1 (4/6) H 0.5(0.7297... C 2πn) or 0.5(π D 0.7297... C 2πn) H 0.3648... or 1.2059... in Quadrant I (Store as a and b.) 3. dA H 0.5r 22 D r 21 dθ A H 0.5 b a (36 sin2 20 D 16) dθ H 5.3132... square units 4. Counting squares in the region gives about 5.3, thus confirming the answer. 5. A H 0.5 π/2 36 sin2 θ dθ 0 H 14.1371... H 4.5π, an interesting multiple of π. Chapter 9 Exploration 9-3a 1. x 2 sin 3x dx u dv x2 + sin 3x – – 3 cos 3x 2x 2 0 + – 1 1 – 9 sin 3x 1 27 cos 3x 1 2 2 H D3x 2 cos 3x C 9x sin 3x C 27 cos 3x C C 6. dL H (dr)2 C (r dθ)2 H 144 co s2 (2θ ) C 36 sin2 (2θ) dθ LH π/2 θH0 dL H 14.5326… units 14.5326.../π H 4.6258..., so the answer is not an interesting multiple of π. 200 / Solutions for the Explorations Calculus: Concepts and Applications Instructor’s Resource Book ©2005 Key Curriculum Press Calc2_IR_656_SE_09-12.qxd 10/26/06 5:51 PM Page 201 2. x 5 Exploration 9-4a ln 4x dx u ln 4x dv x5 + 1. sin 10 x dx 1 u x6 x –1 6 ------- ---------1 1 6 – –9 sin8 x cos x x5 1 + 0 sin 9 x 36 1 6 1 x 6 ln 4x D H6 36 x C C 3. e 5x –cos x sin x cos x C 9 sin x cos x C 9 sin 8 x cos2 x dx H Dsin9 8 x (1 D sin2 x) dx sin x dx 10 sin x dx H Dsin x cos x C 9 sin x dx sin x dx H D110 sin x cos x C 190 sin x dx nD1 1 2. sin x dx H Dn sin x cos x C n sin x dx H Dsin9 cos 6x dx 8 10 u e5x 5e 5x 25e 5x dv cos 6x + 1 sin 6x – 6 + – 36 cos 6x e cos 6x dx 5 1 e5x sin 6x C e5x cos 6x C C e cos 6x dx H 6 36 6 5x 5 e sin 6x C e5x cos 6x C C1 e5x cos 6x dx H 61 61 3 3. V H dx 0 π(sin3 x)2 dx π 4. V H π 0 π (sin3 x)2 dx π Hπ 6 x cos2 x dx H Dsin 8 x (1 D sin x) dx 10 10 9 9 Exploration 9-5a 2 sin x dx 10 sin x dx H Dsin x cos x C 9 sin x dx sin x dx H D110 sin x cos x C 190 sin x dx 8 x dx 5. Answers will vary. –cos x 8 H Dsin9 2 sin x H Dsin9 x cos x C 9 9 3 dv sin x cos x C 9 sin x cos x C 9 sin 4 1 5 3 H D6 sin5 x cos x D 24 sin x cos x 15 1 1 x C 24 D2 sin x cos x C 2 C C 1 5 3 H D6 sin5 x cos x D 24 sin x cos x 5 5 x D 16 sin x cos x C 16 C C 1 5 3 V H π D6 sin5 x cos x D 24 sin x cos x π 5 5 x D 16 sin x cos x C 16 0 5 π Hπ 0D0D0C 16 C 0 C 0 C 0 C 0 5 2 π H 3.0842..., which agrees with Problem 3. H 16 x dx –9 sin8 x cos x sin x dx 5 1 3 x cos x C 6 D4 sin x cos x C 4 sin 1 5 x dx H D6 sin5 x cos x C 6 1 H D6 sin5 16 + – sin6 x dx 0 sin 1 3 3 3 H 2x2 (ln x)3 D 4x 2 (ln x)2 C 4x2 ln x D 8x 2 C C 1 11 5. sin10 x (cos x dx) H 11 sin x C C u sin6 x dx 0 1 sin9 x nD2 H 3.0842... (numerically) 2 6(ln x) x –1 4x ------------------------------------1 x 6 ln x 4 + 1 2 –1 x 6x 8 ------------------------------------1 x 8 6 – 1 2 + x 0 10 8 nD1 π Hπ u dv (ln x)3 + x 1 2 x 3(ln x)2 x –1 2 ------------------------------------1 x 3(ln x)2 2 – 6. sin 8 (n ≠ 0) 5x 5x 4. x (ln x) 10 9 n 1 x dx D 9 9 10 5 1 25 e5x sin 6x C e5x cos 6x D H6 36 36 61 36 sin x H Dsin x cos x C 9 9 x6 dv + – x dx D9 1. cos 7 1 x(sin x dx) H 8 cos8 x C C 10 8 8 7. Answers will vary. 2. The differential sin x dx of cos x does not appear in the integrand. 3. cos 7 (cos x) cos x dx H (1 D sin x) cos x dx H (1 D 3 sin x C 3 sin x D sin x) cos x dx H cos x dx D 3 sin x cos x dx C 3 sin x cos x dx D sin x cos x dx x dx H 2 3 2 3 2 4 6 2 4 6 3 1 H sin x D sin3 x C 5 sin5 x D 7 sin7 x C C Calculus: Concepts and Applications Instructor’s Resource Book ©2005 Key Curriculum Press Solutions for the Explorations / 201 Calc2_IR_656_SE_09-12.qxd 10/26/06 5:52 PM Page 202 4. The power of cos x is even, so converting to powers of (1 D sin2 x) leaves no cos x dx to be the differential of sin x. 5. 1 tan6 x C C tan x (sec x dx) H 6 5 2 6. The differential sec x tan x dx of sec x does not appear in the integrand. 7. sec 8 (sec x) sec x dx H (tan x C 1) sec x dx H (tan x C 3 tan x C 3 tan x C 1) sec x dx H tan x sec x dx C 3 tan x sec x dx C 3 tan x sec x dx C sec x dx x dx H 2 3 2 2 3 2 6 4 6 2 2 2 4 2 2 2 2 1 3 H 7 tan7 x C 5 tan5 x C tan3 x C tan x C C Exploration 9-6a 1. A H 2 7 D2 64 D x 2 dx H 126.9622... (numerically) 2. The differential 2x dx of the “inside function” 64 D x 2 does not appear in the integrand, so the integral cannot be done as the antiderivative of a power. x 3. 8 H sin θ ⇒ x H 8 sin θ dx H 8 cos θ dθ By the Pythagorean theorem, the horizontal leg of the right triangle is 64 D x 2. Thus, 8 cos θ H 64 D x 2. 64 D x dx H (8 cos θ) • 8 cos θ dθ H 64 cos θ dθ 4. 64 cos θ dθ H 32 (1 C cos 2θ) dθ N 2 2 2 8. The power of sec x is odd, so converting to powers of (tan2 x C 1) does not leave sec2 x dx to be the differential of tan x. H 32θ C 16 sin 2θ C C 5. 64 9. Answers will vary. cos 2 Exploration 9-5b 1. cos 2x H cos (x C x) H cos2 x D sin2 x 6. A H 2 2. cos 2x H cos x D sin x H cos2 x D (1 D cos2 x) H 2 cos2 x D 1 2 2 7 D2 θ dθ H 32θ C 16(2 sin θ cos θ) C C 64 D x x x H 32 sinD1 8 C 32 • 8 • CC 8 1 D1 x 2 H 32 sin 8 C 2x 64 D x C C 2 64 D x 2 dx H 64 sinD1 8 C x 64 D x 2 D2 x 7 7 D1 H 64 sinD1 8 C 715 D 64 sinD1 4 C 260 H 126.9622..., which agrees with the answer found numerically in Problem 1. 3. cos 2x H cos2 x D sin2 x H (1 D sin2 x) D sin2 x H 1 D 2 sin2 x 7. Answers will vary. 4. cos 2x H 2 cos x D 1 1 cos2 x H 2(1 C cos 2x) cos 2x H 1 D 2 sin2 x 1 sin2 x H 2(1 D cos 2x) 2 5. 6. Exploration 9-7a 10x D 32 A B 1. x D 5 C x C 1 H (x D 5)(x C 1) ⇒ A(x C 1) C B(x D 5) H 10x D 32 ⇒ A C B H 10, A D 5B H D32 ⇒ A H 3, B H 7 1 1 C cos 2x dx cos x dx H 2 2 1 1 sin 2x C C H 2x C 4 1 1 D cos 2x dx sin2 x dx H 2 2 1 1 sin 2x C C H 2x D 4 2 2. H 3 ln x D 5 C 7 lnx C 1 C C 3. (Instructor input on Heaviside method.) cos 7x sin 5x dx u cos 7x –7 sin 7x –49 cos 7x + – + 4. See Section 9-7 for reasons behind the Heaviside method. 5. x 3 D 2x 2 D 5x C 6 H (x D 1)(x C 2)(x D 3) 11x 2 D 22x D 13 dx (x D 1)(x C 2)(x D 3) 4 5 2 H x D 1 dx C x C 2 dx C x D 3 dx dv sin 5x 1 – 5 cos 5x – 1 25 sin 5x 1 49 7 H D5 cos 7x cos 5x D 25 sin 7x sin 5x C 25 cos 7x sin 5x dx 24 D 25 cos 7x sin 5x dx 1 7 H D5 cos 7x cos 5x D 25 sin 7x sin 5x C C cos 7x sin 5x dx 5 7 H 24 cos 7x cos 5x C 24 sin 7x sin 5x C C1 7. 10x D 32 3 7 (x D 5)(x C 1) dx H x D 5 dx C x C 1 dx cos ax sin bx dx H 4 lnx D 1 C 5 lnx C 2 C 2 lnx D 3 C C 6. Answers will vary. Exploration 9-8a 1. In the first form, dv would equal sinD1 x dx, and you don’t know how to do this integration. In the second form, dv H 1 dx, which you can integrate. a a H sin ax sin bx C cos ax cos bx C C a2 D b2 a2 D b2 8. Answers will vary. 202 / Solutions for the Explorations Calculus: Concepts and Applications Instructor’s Resource Book ©2005 Key Curriculum Press Calc2_IR_656_SE_09-12.qxd 10/26/06 5:52 PM Page 203 1 (e2x C 2 C eD2x) 3. u 2 H 4 2. u dv sinD1 x + dx (1 D x 2)D1/2 – x (1 D x ) 1 x C 2 (1 D x ) I H x sinD1 x D H x sinD1 2 D1/2 1 (e2x D 2 C eD2x) u2 H 4 1 (4) u2 D v 2 H 4 2 2 u Dv H1 4. u(1.3) H 1.9709..., v(1.3) H 1.6903... x dx 2 D1/2 v (D2x dx) 1 2 H x sinD1 x C 2 • 1(1 D x2)1/2 C C H x sinD1 x C 1 D x2 H C which agrees with the text. Here 1 u 1 3. u sin D1 dv x + 1 2 1/2 x(x D 1) – D1 I = x sec xD For x I 0, I H x secD1 x D Let I1 H dx x 5. u is always positive. The extraneous negative branch arises from squaring in Problem 3. x 2 1/2 dx x(x D 1) 6. cos 1.3 H 0.2674..., sin 1.3 H 0.9635... 1 dx (x D 1) 2 1 1/2 v 1 dx. (x D 1) 2 Here 1/2 Let x H sec θ. N dx H sec θ tan θ dθ and (x 2 D 1)1/2 H tan θ. sec q tan q dq I1 H H sec θ dθ tan q u –1 1 –1 H lnsec θ C tan θ C C = ln x C (x 2 D 1)1/2 C C N I H x secD1 x D ln x C (x 2 D 1)1/2 C C, which agrees, for x I 0, with the text’s x secD1 x D sgn x lnx C x 2 D 1 C C. 7. The coordinates u and v defined here have the same relationship to a unit hyperbola as u = cos x and v H sin x have to a unit circle, hence the names circular functions and hyperbolic functions. 4. Answers will vary. 8. Answers will vary. Exploration 9-9a Exploration 9-9b 1. u and v are asymptotic to y3. 1. (The following is simulated data, agreeing with the figure shown in the Exploration. Actual data will depend on the chain used.) Vertex: (0, 20) Left end: (D90, 120), Right end: (90, 120) 4 y u v –2 x 2 –4 1 2. uQ = 2(e x D eDx ) H v cosQ x = Dsin x 1 x Dx uP = 2(e C e ) H u cosP x = Dcos x 1 x uPQ = 2(e D eDx ) H v cosPQ x = sin x 1 x (4) Dx u = 2(e C e ) H u cos(4) x = cos x The u and v derivatives follow the same pattern, but without any minus signs. Calculus: Concepts and Applications Instructor’s Resource Book ©2005 Key Curriculum Press 1 2. y H k cosh kx C C 20 H k cosh 0 C C ⇒ 20 H k C C 90 120 H k cosh k C C Substitute 20 D k for C. 90 120 H k cosh k C 20 D k 90 0 H k cosh k D 100 D k Solving numerically gives k H 51.7801.... C H 20 D 51.7801... H D31.7801... Equation is 1 x y H 51.7801... cosh 51.7801... D 31.7801... 3. A typical medium-weight tow chain weighs about 0.015 lb/cm, which is the value of w. h H (51.7801...)(0.015) M 0.8 lb Solutions for the Explorations / 203 Calc2_IR_656_SE_09-12.qxd 10/26/06 5:52 PM Page 204 4. x y 0 20 20 23.9 40 36.2 6. It is a seeming paradox that the velocity remains positive but the distance approaches a limit! The Greeks, including Zeno of Elea (ca. 490–430 B.C.E.), wrestled with such paradoxes. It was the invention of calculus that allowed people to find out what happens in such situations. 60 58.8 7. Answers will vary. 80 95.1 Exploration 9-11a 5. The chain fits closely the data points. y 1. tan 5 4x dx H (tan 3 4x)(sec2 4x D 1) dx (tan 4x) dx 4x D (tan 4x)(sec 4x D 1) dx 1 4x D 8 tan 4x C tan 4x dx 1 4 H 16 tan 4x D 1 4 H 16 tan 1 4 H 16 tan 3 2 2 1 1 1 4 2 H 16 tan 4x D 8 tan 4x C 4 lnsec 4x C C 2. 1 C t dt 2 Let t/1 H tan θ. Thus dt H sec θ dθ and 1 C t 2 H sec θ. x v dx 2 C dy 2 6. dL H 1 H 1 C sinh2 kx dx L H dL H 285.3490... M 285 cm √1 + t 2 90 θ D90 u 1 The measured length should be close to this. 1 C t dt H sec θ sec θ dθ H sec 7. See the text derivation. 2 8. Answers will vary. 2 Exploration 9-10a 1. For Calvin, v(t) H 320(t C 4)D1, so distance d(t) is d(t) H 320(t C 4)D1 dt H 320 ln (t C 4) C C for t L 0. d(0) H 0 ⇒ C H D320 ln 4 d(t) H 320(ln (t C 4) D ln 4) d(10) H 320(ln 14 D ln 4) H 400.8841... M 401 ft d(20) H 320(ln 24 D ln 4) H 573.3630... M 573 ft d(50) H 320(ln 54 D ln 4) H 832.8606... M 833 ft 2. d(b) H 320(ln (b C 4) D ln 4) for b L 0 1000 H 320(ln (b C 4) D ln 4) b H 87.0395... M 87 s (Exactly 4(e3.125 D 1)) 3. For Phoebe, v(t) H 80eD0.1t, so distance d(t) is d(t) H 80eD0.1t dt H D800eD0.1t C C. d(0) H 0 ⇒ C H 800 d(t) H D800eD0.1t C 800 H 800(1 D eD0.1t ) d(10) H 800(1 D eD1) H 505.6964... M 506 ft d(20) H 800(1 D eD2) H 691.7317... M 692 ft d(50) H 800(1 D eD3) H 794.6096... M 795 ft 4. d(b) H 800(1 D eD0.1b ) 1000 H 800(1 D eD0.1b ) 1.25 H 1 D eD0.1b eD0.1b H D0.25, which is impossible. Phoebe never reaches 1000 ft! 5. lim 800(1 D eD0.1b) H 800 D 800 lim eD0.1b H 800 b→∞ t b→∞ lim 320(ln (b C 4) D ln 4) H ∞, since ln x is unbounded. b→∞ 204 / Solutions for the Explorations 3 θ dθ 1 1 H 2 sec θ tan θ C 2 ln sec θ C tan θ C C 1 1 H 2 t 1 C t 2 C 2 ln 1 C t 2 C t C C tanh x dx H ln (cosh x) C C 4. x e dx 3. 3 u x3 3x 2 6x 6 0 Dx + – + – + dv e –x –e –x e –x –e –x e –x H DeDx(x 3 C 3x 2 C 6x C 6) C C ∞ 1 x 3 eDx dx H lim b b→∞ 1 3 2 x 3 eDx dx H lim (DeDb(b C 3b C 6b C 6) C 6) b→∞ b3 ∞ lim eDb b3 H lim b → ∞ b→∞ b→∞ e 2 3b ∞ H lim b → ∞ b→∞ e 6b ∞ H lim b → ∞ b→∞ e 6 6 H lim b → ∞ b→∞ e H0 Similarly, each power of b multiplied by eDb approaches zero as b approaches infinity. N ∞ 1 x 3 eDx dx H 0 C 0 C 0 C 0 C 6 H 6 Calculus: Concepts and Applications Instructor’s Resource Book ©2005 Key Curriculum Press Calc2_IR_656_SE_09-12.qxd 10/26/06 5:52 PM Page 205 5. x (x D 2)(x D 3)(x D 4) dx 1 D3 2 H x D 2 dx C x D 3 dx C x D 4 dx 4. t H ln x D 2 D 3 lnx D 3 C 2 lnx D 4 C C 6. sin 5 (1 D cos x) sin x dx H (1 D 2 cos x C cos x) sin x dx x dx H 2 2 2 7. (x 4 2 1 H Dcos x C 3 cos3 x D 5 cos5 x C C C 2) dx H 4 3 (x 12 4 1 13 2 9 12 5 x C x C x C 8x C C H 13 3 5 x 9. e 8. 2 3 1 e x dx H 3 ax x3 ae ax 2 ax 1 – + b sin bx 1 – b2 e ax cos bx dx 11 203 1910 30 D4 238 4115 40 D13 153 6070 50 D20 D12 6775 60 0 D112 6155 ax dx H x sinD1 ax D ax dx H 0 C C 17 1 (6t D 0.3t 2) dt H (3t 2 D 0.1x 3) 17 1 1 4. No. 2[v(1) C v(17)] H 10.5, not 23.3. 5. v(t ) ax dx 1 D (a x)2 1 H x sinD1 ax C a1 D (ax )2 C C 37 2.8 ft 3. Average velocity H 17 D 1 s H 23.3 ft/s cos bx dx b a eax sin bx C eax cos bx C C1 H a2 C b2 a2 C b2 sin 20 H 372.8 ft ax D1 455 2. Displacement H e D1 88 1. v(1) H 6 D 0.3 H 5.7 v(17) H 6 • 17 D 0.3 • 289 H 15.3 Both (1, 5.7) and (17, 15.3) appear to lie on the graph. cos bx sin 12 Exploration 10-3a a ax a2 1 H beax sin bx C e cos bx D 2 2 b b a2 C b2 eax cos bx dx 2 b 1 ax a ax H e sin bx C e cos bx C C b b2 10. 0 10 7. Answers will vary. dv cos bx + d 3 6. The object did not go back beyond its starting point since the displacement at t H 60 is 6155 m, which is still positive. 1 3 (3x 2 dx) H 3e x C C cos bx dx u e ax a e e v 5 5. The fact that v changes sign somewhere between t H 0 and t H 60 shows that the object stops and begins going backward. C 6x C 12x C 8) dx 8 a 0 30 (a ≠ 0) (a H 0) 20 11. Answers will vary. Chapter 10 10 Exploration 10-2a 1. Assume the average acceleration for the first time interval is 1 a M 2(5 C 12) H 8.5. If the initial velocity is v H 3, then the velocity at the end of the first time interval is v H 3 C 8.5(10) H 88 m/s at t H 10. 2. t a v 0 5 3 10 12 88 20 11 203 30 D4 238 40 D13 153 50 D20 D12 60 0 D112 t 0 10 20 6. See the graph in Problem 5, showing that the area of the rectangle equals the area of the shaded region. 7. The shaded area above the rectangle equals the unshaded region within the rectangle. 8. There are about 26 squares under the curve for 0 K t K 10, so the displacement is about 26 units, and the average total disp. 26 H H 2.6. velocity is 10 time (The equation is v(t) H 6eD0.2t, so the area is 25.9399..., and the average is 2.5939....) 9. Answers will vary. 3. Assume the average velocity for the first time interval is 1 v H 2(3 C 88) H 45.5. If the initial displacement is d H 0, then the displacement at the end of the first time interval is d H 0 C 45.5(10) H 455 m at t H 10. Calculus: Concepts and Applications Instructor’s Resource Book ©2005 Key Curriculum Press Solutions for the Explorations / 205