Click here for answers. f x CD 1 2 ( BC AC AB ) 1 2 C. (b) Express da dt in terms of the quantities in part (a). can be greater than.

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1 CHALLENGE PROBLEM CHAPTER 3 A Click here for answers. Click here for solutions.. (a) Find the domain of the function f x s s s3 x. (b) Find f x. ; (c) Check your work in parts (a) and (b) by graphing f and f on the same screen. CHAPTER 34 A Click here for answers. Click here for solutions. C D A FIGURE FOR PROBLEM B. Find the absolute maximum value of the function. (a) Let ABC be a triangle with right angle A and hypotenuse. (ee the figure.) If the inscribed circle touches the hypotenuse at D, show that C (b) If, express the radius r of the inscribed circle in terms of a and. (c) If a is fixed and varies, find the maximum value of r. 3. A triangle with sides a, b, and c varies with time t, but its area never changes. Let be the angle opposite the side of length a and suppose always remains acute. (a) Express ddt in terms of b, c,, dbdt, and dcdt. (b) Express dadt in terms of the quantities in part (a). 4. Let a and b be positive numbers. how that not both of the numbers a b and b a can be greater than. 4 f x CD ( BC AC AB ) 5. Let ABC be a triangle with BAC and AB AC. (a) Express the length of the angle bisector AD in terms of x AB. (b) Find the largest possible value of. x x AD a BC tewart: Calculus, ixth Edition. IBN: Brooks/Cole. All rights reserved. CHAPTER 54 A Click here for answers. Click here for solutions.. how that. 7 y x dx uppose the curve y f x passes through the origin and the point,. Find the value of the integral f x dx. x 3. In ections and we used the formulas for the sums of the kth powers of the first n integers when k,, and 3. (These formulas are proved in Appendix E.) In this problem we derive formulas for any k. These formulas were first published in 73 by the wiss mathematician James Bernoulli in his book Ars Conjectandi. (a) The Bernoulli polynomials B n are defined by B x, B nx B nx, and x Bnx dx for n,, 3,.... Find B nx for n,, 3, and 4. (b) Use the Fundamental Theorem of Calculus to show that B n B n for n. tewart: Calculus, eventh Edition. IBN: (c). Brooks/Cole, Cengage Learning. All rights reserved.

2 CHALLENGE PROBLEM (c) If we introduce the Bernoulli numbers b n n! B n, then we can write B x b B x x! b! x! b! B x x! b! B 3x x 3 3! b! x! b! x! b3 3! and, in general, where [The numbers are the binomial coefficients.] Use part (b) to show that, for n, and therefore B nx n! k n kx n kb nk ( n k ) b n n n b b n k n n k kb n b n k n! k!n k! n b n n b n This gives an efficient way of computing the Bernoulli numbers and therefore the Bernoulli polynomials. (d) how that B n x n B nx and deduce that b n for n. (e) Use parts (c) and (d) to calculate b 6 and b 8. Then calculate the polynomials B 5, B 6, B 7, B 8, and B 9. ; (f) Graph the Bernoulli polynomials B, B,..., B 9 for x. What pattern do you notice in the graphs? (g) Use mathematical induction to prove that B kx B kx x k k!. (h) By putting x,,,..., n in part (g), prove that k k 3 k n k k! B kn B k k! y n B kx dx (i) Use part (h) with k 3 and the formula for B 4 in part (a) to confirm the formula for the sum of the first n cubes in ection (j) how that the formula in part (h) can be written symbolically as k k 3 k n k k n bk b k where the expression n b k is to be expanded formally using the Binomial Theorem and each power b i is to be replaced by the Bernoulli number b i. (k) Use part (j) to find a formula for n 5.equator that have exactly the same temperature. CHAPTER 56 A Click here for answers. Click here for solutions.. A solid is generated by rotating about the x-axis the region under the curve y f x, where f is a positive function and x. The volume generated by the part of the curve from x to x b is b for all b. Find the function f. tewart: Calculus, eventh Edition. IBN: (c). Brooks/Cole, Cengage Learning. All rights reserved. tewart: Calculus, ixth Edition. IBN: Brooks/Cole. All rights reserved.

3 CHALLENGE PROBLEM 3 CHAPTER 86 A Click here for answers. Click here for solutions. ;. The Chebyshev polynomials T n are defined by T nx cosn arccos x, n,,, 3,... (a) What are the domain and range of these functions? (b) We know that T x and T x x. Express T explicitly as a quadratic polynomial and T 3 as a cubic polynomial. (c) how that, for n, T nx xt nx T nx. (d) Use part (c) to show that T n is a polynomial of degree n. (e) Use parts (b) and (c) to express T 4, T 5, T 6, and T 7 explicitly as polynomials. (f) What are the zeros of T n? At what numbers does T n have local maximum and minimum values? (g) Graph T, T 3, T 4, and T 5 on a common screen. (h) Graph T 5, T 6, and T 7 on a common screen. (i) Based on your observations from parts (g) and (h), how are the zeros of T n related to the zeros of T n? What about the x-coordinates of the maximum and minimum values? (j) Based on your graphs in parts (g) and (h), what can you say about x Tnx dx when n is odd and when n is even? (k) Use the substitution u arccos x to evaluate the integral in part ( j). (l) The family of functions f x cosc arccos x are defined even when c is not an integer (but then f is not a polynomial). Describe how the graph of f changes as c increases. CHAPTER 9 A Click here for answers. Click here for solutions. tewart: Calculus, ixth Edition. IBN: Brooks/Cole. All rights reserved. r y (i) FIGURE FOR PROBLEM r P=P b. A circle C of radius r has its center at the origin. A circle of radius r rolls without slipping in the counterclockwise direction around C. A point P is located on a fixed radius of the rolling circle at a distance b from its center, b r. [ee parts (i) and (ii) of the figure.] Let L be the line from the center of C to the center of the rolling circle and let be the angle that L makes with the positive x-axis. (a) Using as a parameter, show that parametric equations of the path traced out by P are x b cos 3 3r cos, y b sin 3 3r sin. Note: If b, the path is a circle of radius 3r; if b r, the path is an epicycloid. The path traced out by P for b r is called an epitrochoid. ; (b) Graph the curve for various values of b between and r. (c) how that an equilateral triangle can be inscribed in the epitrochoid and that its centroid is on the circle of radius b centered at the origin. Note: This is the principle of the Wankel rotary engine. When the equilateral triangle rotates with its vertices on the epitrochoid, its centroid sweeps out a circle whose center is at the center of the curve. (d) In most rotary engines the sides of the equilateral triangles are replaced by arcs of circles centered at the opposite vertices as in part (iii) of the figure. (Then the diameter of the rotor is constant.) how that the rotor will fit in the epitrochoid if b 3( s3)r. x tewart: Calculus, eventh Edition. IBN: (c). Brooks/Cole, Cengage Learning. All rights reserved. y (ii) P P x (iii)

4 4 CHALLENGE PROBLEM CHAPTER Click here for solutions.. (a) how that, for n,, 3,..., (b) Deduce that The meaning of this infinite product is that we take the product of the first n factors and then we take the limit of these partial products as n l. (c) how that This infinite product is due to the French mathematician Francois Viète (54 63). Notice that it expresses in terms of just the number and repeated square roots.. uppose that a cos,, b, and Use Problem to show that s sin cos cos 4 cos 8 sin n sin cos n cos 4 cos 8 cos s s a n a n b n s s s b n sb na n lim an lim n l n l bn sin n tewart: Calculus, eventh Edition. IBN: (c). Brooks/Cole, Cengage Learning. All rights reserved. tewart: Calculus, ixth Edition. IBN: Brooks/Cole. All rights reserved.

5 CHALLENGE PROBLEM 5 ANWER CHAPTER Chapter 3 olutions. (a) [ ] (b) CHAPTER Chapter 34 olutions (a) tan + 5. (a) = +, (b) + + sec (b) + cos CHAPTER Chapter 54 olutions 3. (a) () =, () = +, 3() = , 4() = (e) 6 = 4, 8 = 3 ; 5 () = 7 () = 54 9 () = 36, , 6 () = , , 6 8 () = 4, , (f ) There are four basic shapes for the graphs of (excluding ), and as increases, they repeat in a cycle of four. For =4, the shape resembles that of the graph of cos ;for =4 +,thatof sin ; for =4 +,thatofcos ;andfor =4 +3,thatofsin. (k) ( +) ( + ) tewart: Calculus, ixth Edition. IBN: Brooks/Cole. All rights reserved. Chapter CHAPTER 65 Chapter CHAPTER 76 olutions. () = olutions. (a) [ ]; [ ] for (b) () =, 3 () =4 3 3 (e) 4() = , 5() = , 6 () = , 7 () = (f ) =cos +, an integer with ; =cos(), an integer with tewart: Calculus, eventh Edition. IBN: (c). Brooks/Cole, Cengage Learning. All rights reserved.

6 6 CHALLENGE PROBLEM (g) (h) (i) The zeros of and + alternate; the extrema also alternate (j) When is odd, and so () =;when is even, the integral is negative, but decreases in absolute value as gets larger. if is even (k) cos() sin= if is odd (l ) As increases through an integer, the graph of gains a local extremum, which starts at = and moves rightward, compressing the graph of as continues to increase. Chapter CHAPTER 9 olutions. (b) = 5 = 5 = 3 5 = 4 5 tewart: Calculus, eventh Edition. IBN: (c). Brooks/Cole, Cengage Learning. All rights reserved. tewart: Calculus, ixth Edition. IBN: Brooks/Cole. All rights reserved.

7 CHALLENGE PROBLEM 7 OLUTION E Exercises CHAPTER Chapter 3. (a) () = 3 = 3, 3, 3 = 3, 3, 3 = 3, 4 3, 3 = 3,, 3 = { 3,, 3 } = { 3,, } = { } =[ ] (b) () = 3 ()= 3 3 = = (c) Note that is always decreasing and is always negative. tewart: Calculus, ixth Edition. IBN: Brooks/Cole. All rights reserved. E Exercises Chapter CHAPTER 43. () = ( ) = + + ( ) + + +( ) if if if ( ) + (3 ) if () = ( + ) + (3 ) if ( + ) ( ) if We see that () for and () for. For, wehave () = (3 ) ( +) = ( ) (3 ) ( +) = (3 ) ( +),so () for, () = and () for. Wehaveshownthat () for ; () for ; () for ; and () for. Therefore, by the First Derivative Test, the local maxima of are at =and =,where takes the value 4. Therefore, 4 is the absolute maximum value of. 3 3 tewart: Calculus, eventh Edition. IBN: (c). Brooks/Cole, Cengage Learning. All rights reserved.

8 8 CHALLENGE PROBLEM 3. (a) = with sin =,so = sin. But is a constant, so differentiating this equation with respect to,weget cos == cos = sin + sin + sin + = tan +. (b) We use the Law of Cosines to get the length of side in terms of those of and, and then we differentiate implicitly with respect to : = + cos = + ( sin ) + cos + cos = + + sin cos cos. Now we substitute our value of from the Law of Cosines and the value of from part (a), and simplify (primes signify differentiation by ): = + + sin [ tan ( + )] ( + )(cos ) + cos = + [sin ( + )+cos ( + )] cos + cos = + ( + )sec + cos 5. (a) Let =, =,and =,sothat =. We compute the area A of 4 in two ways. First, A = sin 3 = 3 A =(area of 4)+(area of 4) = 3 4. econd, = sin + sin = 3 + () 3 = ( +) Equating the two expressions for the area, we get 3 + = 3 = = +,. Another method: Use the Law of ines on the triangles and. In4, wehave + + = =8 =. Thus, = sin( ) = sin cos cos sin = sin sin 3 cos + sin sin = 3 cot +,and by a similar argument with 4, 3 cot = +. Eliminating cot gives = + + = +,. (b) We differentiate our expression for with respect to to nd the maximum: = + () = ( +) ( =when =. This indicates a maximum by the First Derivative Test, +) since () for and () for, so the maximum value of is () =. tewart: Calculus, eventh Edition. IBN: (c). Brooks/Cole, Cengage Learning. All rights reserved. tewart: Calculus, ixth Edition. IBN: Brooks/Cole. All rights reserved.

9 CHALLENGE PROBLEM 9 E Exercises Chapter CHAPTER 54. For, wehave 4 4 =6,so+ 4 7 and Thus, 7 = 7.Also+4 4 for, so + 4 and = = = 7. Thus, we have the estimate 4 3. (a) To nd (), we use the fact that () = () () = () = = +.Nowwe impose the condition that () = = ( + ) = + = + =.o () =. imilarly () = () = () = = + = + = 6 4 () = +. 3() = () = + = +. But,so = But 3() = = = + + =.o () = () = 3 () = = But 4() = = = 48 7 =. 7 o 4() = (b) By FTC, () () = () = () =for, by denition. Thus, () = () for. (c) We know that () =! = () = () =! for, weget =.Ifweset =in this expression, and use the fact that =. Now if we expand the right-hand side, we get tewart: Calculus, ixth Edition. IBN: Brooks/Cole. All rights reserved. = the LH and divide by = : = +. We cancel the terms, move the term to for, as required. (d) We use mathematical induction. For =: ( ) =and () () =,sothe equation holds for =since =. Now if ( ) =() (), then since +( ) = +( ) ( ) = ( ), wehave +( ) =()() () =() + (). Integrating, we get + ( ) =() + + ()+. But the constant of integration must be, since if we substitute =in the equation, we get + () = () + + () +, and if we substitute =we get + () = () + + () +, and these two equations together imply that + () = () + () + + () + + = + () + =. o the equation holds for all, by induction. Now if the power of is odd, then we have tewart: Calculus, eventh Edition. IBN: (c). Brooks/Cole, Cengage Learning. All rights reserved.

10 CHALLENGE PROBLEM + ( ) = +(). In particular, +() = +(). But from part (b), we know that () = () for. The only possibility is that + () = +() = for all, and this implies that + =( +)! + () = for. (e) From part (a), we know that =! () =, and similarly =, = 6, 3 =and 4 = 3. We use the formula to nd 6 = 7 = The 3 and 5 terms are,sothisisequalto = = 6 4 imilarly, 8 = = = = 3 Now we can calculate 5 () = ! = = = 6 () = = 7 7 () = = 54 8 () = 4,3 9 () = = 4, , = 36, tewart: Calculus, eventh Edition. IBN: (c). Brooks/Cole, Cengage Learning. All rights reserved. tewart: Calculus, ixth Edition. IBN: Brooks/Cole. All rights reserved.

11 CHALLENGE PROBLEM (f ) = = =3 =4 =5 =6 =7 =8 =9 There are four basic shapes for the graphs of (excluding ), and as increases, they repeat in a cycle of four. For =4, the shape resembles that of the graph of cos ; For =4 +,thatof sin ; for =4 +,thatofcos ;andfor =4 +3,thatofsin. (g) For =: ( +) () = + =,and! =, so the equation holds for =. We tewart: Calculus, ixth Edition. IBN: Brooks/Cole. All rights reserved. now assume that ( +) () = [ ( +) ()] =. We integrate this equation with respect to : ( )!. But we can evaluate the LH using the denition ( )! + () = (), and the RH is a simple integral. The equation becomes + ( +) + () = ( )! =!, since by part (b) + () + () =,andsothe constant of integration must vanish. o the equation holds for all, by induction. (h) The result from part (g) implies that =![ + ( +) + ()]. If we sum both sides of this equation from =to = (note that is xed in this process), we get =! [ + ( +) + ()]. Butthe tewart: Calculus, eventh Edition. IBN: (c). Brooks/Cole, Cengage Learning. All rights reserved. = =

12 CHALLENGE PROBLEM RH is just a telescoping sum, so the equation becomes =![ + ( +) + ()]. But from the denition of Bernoulli polynomials (and using the Fundamental Theorem of Calculus), the RH is equal to! + (). (i) If we let =3and then substitute from part (a), the formula in part (h) becomes =3![ 4 ( +) 4 ()] =6 ( 4 +)4 ( +)3 + ( 4 +) (j) =! = ( +) [ + ( +) ( +)] 4 =! + + ()! [by part (h)] = = = ( +) [ ( +)] 4 + Now view as ( + ), as explained in the problem. Then = = + ( + ) = ( + ) + + = + ( +) = = ( ++)+ + + (k) We expand the RH of the formula in (j), turning the into, and remembering that + =for : = 6 ( +) 6 6 = 6 ( +) 6 +6( +) ( +) ( +) 4 = 6 ( +) 6 3( +) ( +)4 ( +) = ( +) ( +) 4 6( +) 3 +5( +) = ( +) [( +) ] ( +) ( +) = ( +) ( + ) E Exercises CHAPTER Chapter 65. The volume generated from =to = is [()]. Hence, we are given that = [()] for all. Differentiating both sides of this equation using the Fundamental Theorem of Calculus gives = [()] () =,since is positive. Therefore, () =. E Exercises CHAPTER Chapter 86. (a) () =cos( arccos ). The domain of arccos is [ ], and the domain of cos is R, so the domain of () is [ ]. As for the range, () =cos=,sotherangeof () is {}. But since the range of arccos is at least [] for, and since cos takes on all values in [ ] for [], the range of () is [ ] for. tewart: Calculus, eventh Edition. IBN: (c). Brooks/Cole, Cengage Learning. All rights reserved. tewart: Calculus, ixth Edition. IBN: Brooks/Cole. All rights reserved.

13 CHALLENGE PROBLEM 3 (b) Using the usual trigonometric identities, () = cos( arccos ) = [cos(arccos )] =,and 3 () = cos(3 arccos ) = cos(arccos +arccos) = cos(arccos ) cos( arccos ) sin(arccos ) sin( arccos ) = sin(arccos ) [ sin(arccos ) cos(arccos)] = 3 sin (arccos ) = 3 cos (arccos ) = 3 =4 3 3 (c) Let =arccos.then + ()=cos[( +)] =cos( + ) =cos cos sin sin =cos cos (cos cos +sin sin ) = () cos( ) = () () (d) Here we use induction. () =, a polynomial of degree. Now assume that () is a polynomial of degree. Then + () = () (). By assumption, the leading term of is, say, so the leading term of + is = +,andso + has degree +. (e) 4 () = 3 () () = = , 5 () = 4 () 3 () = = , 6 () = 5 () 4 () = = , 7 () = 6 () 5 () = = (f ) The zeros of () =cos( arccos ) occur where arccos = + for some integer, since then cos( arccos ) =cos + =. Note that there will be restrictions on,since arccos. We tewart: Calculus, ixth Edition. IBN: Brooks/Cole. All rights reserved. continue: arccos = + arccos = +. This only has solutions for + +.[Thismakessense,becausethen () has zeros, and it is a polynomial of degree.] o, taking cosines of both sides of the last equation, we nd that the zeros of () occur at =cos + = () = sin( arccos ), an integer with. To nd the values of at which () has local extrema, we set sin( arccos ) = sin( arccos ) = arccos =, some integer arccos =. This has solutions for, but we disallow the cases =and =, since these give =and = respectively. o the local extrema of () occur at =cos(), an integer with. [Again, this seems reasonable, since a polynomial of degree has at tewart: Calculus, eventh Edition. IBN: (c). Brooks/Cole, Cengage Learning. All rights reserved.

14 4 CHALLENGE PROBLEM most ( ) extrema.] By the First Derivative Test, the cases where is even give maxima of (),sincethen arccos [cos()] = is an even multiple of,sosin ( arccos ) goes from negative to positive at =cos(). imilarly, the cases where is odd represent minima of (). (g) (h) (i) From the graphs, it seems that the zeros of and + alternate; that is, between two adjacent zeros of,there is a zero of +, and vice versa. The same is true of the -coordinates of the extrema of and + : between the -coordinates of any two adjacent extrema of one, there is the -coordinate of an extremum of the other. (j) When is odd, the function () is odd, since all of its terms have odd degree, and so () =.When is even, () is even, and it appears that the integral is negative, but decreases in absolute value as gets larger. (k) () = cos( arccos ). We substitute =arccos =cos = sin, = =,and = =. o the integral becomes cos() sin= [sin( )+sin( + )] = cos[( )] + = + if is even = if is odd (l ) From the graph, we see that as increases through an integer, the graph of gains a local extremum, which starts at = and moves rightward, compressing the graph of as continues to increase. cos[( + )] + if is even + if is odd + tewart: Calculus, eventh Edition. IBN: (c). Brooks/Cole, Cengage Learning. All rights reserved. tewart: Calculus, ixth Edition. IBN: Brooks/Cole. All rights reserved.

15 CHALLENGE PROBLEM 5 E Exercises CHAPTER Chapter 9. (a) ince the smaller circle rolls without slipping around, the amount of arc traversed on ( in the gure) must equal the amount of arc of the smaller circle that has been in contact with. ince the smaller circle has radius, it must have turned through an angle of =. In addition to turning through an angle, the little circle has rolled through an angle against. Thus, has turned through an angle of 3 as shown in the gure. (If the little circle had turned through an angle of with its center pinned to the -axis, then would have turned only instead of 3. The movement of the little circle around adds to the angle.) From the gure, we see that the center of the small circle has coordinates (3 cos 3 sin ). Thus, has coordinates ( ),where =3 cos + cos 3 and =3 sin + sin 3. (b) = 5 = 5 = 3 5 = 4 5 (c) The diagram gives an alternate description of point on the epitrochoid. moves around a circle of radius,and rotates one-third as fast with respect to atadistanceof3. Place an equilateral triangle with sides of length 3 3 so that its centroid is at and one vertex is at.(the distance from the centroid to a vertex is 3 times the length of a side of the equilateral triangle.) As increases by, the point travelsoncearoundthecircleofradius 3, returning to its original position. At the same time, (and the rest of the triangle) rotate through an angle of 3 about,so s position is tewart: Calculus, ixth Edition. IBN: Brooks/Cole. All rights reserved. occupied by another vertex. In this way, we see that the epitrochoid traced out by is simultaneously traced out by the other two vertices as well. The whole equilateral triangle sits inside the epitrochoid (touching it only with its vertices) and each vertex traces out the curve once while the centroid moves around the circle three times. (d) We view the epitrochoid as being traced out in the same way as in part (c), by a rotor for which the distance from its center to each vertex is 3, so it has radius 6. To show that the rotor ts inside the epitrochoid, it sufces to show that for any position of the tracing point, there are no points on the opposite side of the rotor which are outside the epitrochoid. But the most likely case of intersection is when is on the -axis, so as long as the diameter of the rotor (which is 3 3) is less than the distance between the -intercepts, the rotor will t. The -intercepts occur tewart: Calculus, eventh Edition. IBN: (c). Brooks/Cole, Cengage Learning. All rights reserved.

16 6 CHALLENGE PROBLEM when = or = 3 = ±(3 ), so the distance between the intercepts is 6, and the rotor will t if ( 3 ). E Exercises CHAPTER Chapter. (a) sin = sin cos = sin 4 cos cos 4 = sin 8 cos cos cos 8 4 = = sin cos cos = sin cos cos 4 cos 8 cos cos cos cos 8 4 (b) sin = sin cos cos 4 cos cos sin 8 sin ( ) =cos cos 4 cos cos 8. sin Now we let,usinglim =with = sin lim = lim sin ( ) : cos cos 4 cos 8 cos sin (c) If we take = in the result from part (b) and use the half-angle formula cos = (see Formula 7a in Appendix D), we get sin =cos 4 = + = cos cos = cos + 4 =cos cos 4 cos 8. ( + cos ) tewart: Calculus, eventh Edition. IBN: (c). Brooks/Cole, Cengage Learning. All rights reserved. tewart: Calculus, ixth Edition. IBN: Brooks/Cole. All rights reserved.