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1 CHLLENGE PROBLEM Challenge Problems Chapter 3 Click here for answers.. (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 4 Click here for answers. C D. Find the absolute maximum value of the function. (a) Let BC be a triangle with right angle 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. f x x x CD ( BC C B ) a BC FIGURE FOR PROBLEM B 3. 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). Chapter 5 Click here for answers.. In ections 5. and 5. 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 ppendix 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 rs Conjectandi. (a) The Bernoulli polynomials are defined by,, and x Bn Bx Bnx Bnx 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. (c) If we introduce the Bernoulli numbers b n n! B n, then we can write B x b B x x! and, in general, b! x b!! B x x! B 3x x 3 3! b! b! x b!! x b3! 3! B nx n! k n kx n kb nk ( n k ) where n k n! k!n k! [The numbers are the binomial coefficients.] Use part (b) to show that, for n, b n k n n k kb

2 CHLLENGE PROBLEM and therefore b n n n b n b 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 5.. (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. Chapter 6 Click here for answers.. 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. Chapter Click here for answers. y. circle C of radius r has its center at the origin. circle of radius r rolls without slipping in the counterclockwise direction around C. 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. y P P=P r r b x P x (i) (ii) (iii) (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.

3 CHLLENGE PROBLEM 3 ; (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. Chapter. (a) how that, for n,, 3,..., (b) Deduce that sin n sin cos n cos 4 cos 8 cos n 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. sin cos cos 4 cos 8 s s s s s s. uppose that a cos,, b, and a n a n b n b n sb na n Use Problem to show that lim an lim n l n l bn sin

4 4 CHLLENGE PROBLEM nswers Chapter 3 olutions. (a), (b) (8s3 x s s3 x s s s3 x ) Chapter 4 olutions (a) tan dc (b) c dt db b dt b db dt c dc dt b dc db c sec dt dt sb c bc cos Chapter 5 olutions Chapter 6. (a) (e) b 6 4, b 8 3; (f) There are four basic shapes for the graphs of (excluding B ), and as n increases, they repreat in a cycle of four. For n 4m, the shape resembles that of the graph of cos x ; for n 4m, that of sin x; for n 4m, that of cos x; and for n 4m 3, that of sin x. (k) B x x, B x x x, B 3x 6 x 3 4 x x, B 4x 4 x 4 x 3 4 x 7 B 5x (x 5 5 x x 3 6 x), B 6x 7 (x 6 3x 5 5 x 4 x 4), B 7x 54 (x 7 7 x 6 7 x x 3 6 x, B 8x 4,3 (x 8 4x x x 4 3 x 3), B 9x 36,88 (x 9 9 x 8 6x 7 5 x 5 x 3 3 x) n n n n B n olutions. f x sx Chapter olutions. (b) b = 5 r b = 5 r b = 3 5 r b = 4 5 r

5 CHLLENGE PROBLEM 5 olutions E Exercises Chapter 3. (a) f(x) = 3 x D = x 3 x, 3 x, 3 x = = x 3 x, 4 3 x, 3 x = x x 3,x, 3 x x 3 x, 3 x, 3 x = {x x 3,x, 3 x} = {x x 3,x,x } = {x x } =[, ] (b) f(x) = 3 x f (x) = 3 x d dx 3 x = 3 x d 3 x 3 x dx = 8 3 x 3 x 3 x (c) Note that f is always decreasing and f is always negative. E Exercises Chapter 4. f(x) = + x + + x x + (x ) = +x + (x ) +x + +(x ) if x< if x< if x ( x) + if x< (3 x) f (x) = ( + x) + if <x< (3 x) ( + x) if x> (x ) We see that f (x) > for x< and f (x) < for x>. For <x<,wehave f (x) = (3 x) (x +) = (x +x +) (x 6x +9) 8(x ) = (3 x) (x +) (3 x) (x +),sof (x) < for <x<, f () = and f (x) > for <x<. Wehaveshownthatf (x) > for x<; f (x) < for <x<; f (x) > for <x<; andf (x) < for x>. Therefore, by the First Derivative Test, the local maxima of f are at x =and x =,wheref takes the value 4. Therefore, 4 is the absolute maximum value of f. 3 3

6 6 CHLLENGE PROBLEM 3. (a) = bh with sin θ = h/c,so = bc sin θ. But is a constant, so differentiating this equation with respect to t,weget d dt == bc cos θ dθ dt + b dc db sin θ + dt dt c sin θ bc cos θ dθ dt = sin θ b dc dt + c db dθ dt dt = tan θ dc c dt + b (b)weusethelawofcosinestogetthelengthofsidea in terms of those of b and c, and then we differentiate implicitly with respect to t: a = b + c bc cos θ a da db dc =b +c dt dt dt bc( sin θ) dθ dt + b dc db cos θ + dt dt c cos θ da dt = b db a dt + c dc dθ + bc sin θ dt dt b dc db cos θ c dt dt cos θ. Now we substitute our value of a from the Law db. dt of Cosines and the value of dθ/dt from part (a), and simplify (primes signify differentiation by t): da dt = bb + cc + bc sin θ [ tan θ (c /c + b /b)] (bc + cb )(cos θ) b + c bc cos θ = bb + cc [sin θ (bc + cb )+cos θ (bc + cb )]/ cos θ b + c bc cos θ = bb + cc (bc + cb )secθ b + c bc cos θ E Exercises Chapter 5. (a) To find B (x),weusethefactthatb (x) =B (x) B (x) = B (x) dx = dx = x + C. Now we impose the condition that B (x) dx = = (x + C) dx = x +[Cx] = + C C =. o dx = x x + D. But B (x) =x. imilarly B (x) = B (x) dx = x B (x) dx = = x x + D dx = + D 6 4 D =,so B (x) = x x +.B3 (x) = B (x) dx = x x + dx = 6 x3 4 x + x + E. But B 3 (x) dx = = 6 x3 4 x + x + E dx = + + E E =. o 4 4 B 3 (x) = 6 x3 4 x + x. B 4 (x) = B 3 (x) dx = 6 x3 4 x + x dx = 4 x4 x3 + 4 x + F. But B 4 (x) dx = = 4 x4 x3 + 4 x + F dx = + + F F =. o B 4 (x) = 4 x4 x3 + 4 x 7. (b) By FTC, B n () B n () = B n (x) dx = B n (x) dx =for n, bydefinition. Thus, B n () = B n () for n. (c) We know that B n (x) = n n b k x n k.ifwesetx =in this expression, and use the fact that n! k= k B n () = B n () = bn n! for n, wegetbn = n n b k. Now if we expand the right-hand side, we get k k=

7 CHLLENGE PROBLEM 7 b n = n b + n b + + n n and divide by n n = n: bn = n bn + n bn + n n bn. We cancel the b n terms, move the b n term to the LH b + n b + + n bn for n, as required. n n (d) We use mathematical induction. For n =: B ( x) =and ( ) B (x) =,sothe equation holds for n =since b =.NowifB k ( x) =( ) k B k (x), thensince d B dx k+ ( x) =Bk+ ( x) d ( x) = B dx k ( x), we have d dx B k+ ( x) =( ) ( ) k B k (x) =( ) k+ B k (x). Integrating, we get B k+ ( x) =( ) k+ B k+ (x)+c. But the constant of integration must be, since if we substitute x =in the equation, we get B k+ () = ( ) k+ B k+ () + C, and if we substitute x =we get B k+ () = ( ) k+ B k+ () + C, and these two equations together imply that B k+ () = ( ) k+ ( ) k+ B k+ () + C + C = B k+ () + C C =. o the equation holds for all n, by induction. Now if the power of is odd, then we have B n+ ( x) = B n+ (x). In particular, B n+ () = B n+ (). But from part (b), we know that B k () = B k () for k>. The only possibility is that B n+ () = B n+ () = for all n>, and this implies that b n+ =(n +)!B n+ () = for n>. (e) From part (a), we know that b =!B () =,andsimilarlyb =, b = 6, b 3 =and b 4 = 3. We use the formula to find b 6 = b 7 = b + b + b + b 3 + b 4 + b The b 3 and b 5 terms are, so this is equal to n = = 6 4 imilarly, b 8 = b + b + b + b 4 + b = = = 3 Now we can calculate B 5 (x) = 5 5 b k x 5 k 5! k k= = x 5 +5 x x 5 5 x x3 x 6 = x 3 +5 x 6 3

8 8 CHLLENGE PROBLEM (f ) B 6 (x) = 7 = 7 x 6 +6 x x 6 3x x4 x B 7 (x) = x 7 +7 x x 7 7 x6 + 7 x5 7 6 x3 + x 6 = 54 B 8 (x) = 4,3 B 9 (x) = = 4,3 x 8 +8 x x 8 4x x6 7 3 x4 + 3 x 3 x 9 +9 x ,88 x x x x 3 +7 x x = 36,88 x 9 9 x8 +6x 7 5 x5 +x 3 3 x x x x x x 3 +9 x n = n = n =3 n =4 n =5 n =6 n =7 n =8 n =9 There are four basic shapes for the graphs of B n (excluding B ), and as n increases, they repeat in a cycle of four. For n =4m, the shape resembles that of the graph of cos πx;forn =4m +,thatof sin πx;forn =4m +,thatof cos πx;andforn =4m +3,thatofsin πx.

9 CHLLENGE PROBLEM 9 (g) For k =: B (x +) B (x) =x + x =,and x! =, so the equation holds for k =.Wenow assume that B n (x +) B n (x) = [B n (x +) B n (x)] dx = xn. We integrate this equation with respect to x: (n )! x n dx. But we can evaluate the LH using the (n )! definition B n+ (x) = B n (x) dx, and the RH is a simple integral. The equation becomes B n+ (x +) B n+ (x) = (n )! n xn = n! xn, since by part (b) B n+ () B n+ () =, andsothe constant of integration must vanish. o the equation holds for all k, by induction. (h) The result from part (g) implies that p k = k![b k+ (p +) B k+ (p)]. If we sum both sides of this equation from n n p =to p = n (note that k is fixed in this process), we get p k = k! [B k+ (p +) B k+ (p)]. But the RH is p= just a telescoping sum, so the equation becomes k + k +3 k + + n k = k![b k+ (n +) B k+ ()]. But from the definition of Bernoulli polynomials (and using the Fundamental Theorem of Calculus), the RH is equal to k! n+ B k (x) dx. (i) If we let k =3and then substitute from part (a), the formula in part (h) becomes n 3 =3![B 4 (n +) B 4 ()] =6 (n 4 +)4 (n +)3 + (n 4 +) p= = (n +) [ + (n +) (n +)] 4 = (n +) [ (n +)] 4 n(n +) = (j) k + k +3 k + + n k = k! n+ B k (x) dx [by part (h)] n+ k k n+ k = k! b j x k j k dx = b j x k j dx k! j= j j= j Now view k k j b j x k j as (x + b) k, as explained in the problem. Then j= k + k +3 k + + n k = n+ (x + b) k dx = n+ (x + b) k+ = (n ++b)k+ b k+ k + k + (k) We expand the RH of the formula in (j), turning the b i into b i, and remembering that b i+ =for i>: n 5 = 6 (n ++b) 6 b 6 = 6 (n +) 6 +6(n +) 5 b (n +)4 b (n +) b 4 = 6 (n +) 6 3(n +) (n +)4 (n +) = (n +) (n +) 4 6(n +) 3 +5(n +) = (n +) [(n +) ] (n +) (n +) = n(n +) (n +n )

10 CHLLENGE PROBLEM E Exercises Chapter 6. The volume generated from x =to x = b is b π [f (x)] dx. Hence, we are given that b = b π [f (x)] dx for all b>. Differentiating both sides of this equation using the Fundamental Theorem of Calculus gives b = π [f (b)] f (b) = b/π,sincef is positive. Therefore, f (x) = x/π. E Exercises Chapter. (a) ince the smaller circle rolls without slipping around C, the amount of arc traversed on C (rθ in the figure) must equal the amount of arc of the smaller circle that has been in contact with C. ince the smaller circle has radius r, it must have turned through an angle of rθ/r =θ. In addition to turning through an angle θ, the little circle has rolled through an angle θ against C. Thus, P has turned through an angle of 3θ asshowninthe figure. (If the little circle had turned through an angle of θ with its center pinned to the x-axis, then P would have turned only θ instead of 3θ. The movement of the little circle around C adds θ to the angle.) From the figure, we see that the center of the small circle has coordinates (3r cos θ, 3r sin θ). Thus, P has coordinates (x, y), wherex = 3r cos θ + b cos 3θ and y =3rsin θ + b sin 3θ. (b) b = 5 r b = 5 r b = 3 5 r b = 4 5 r (c) The diagram gives an alternate description of point P on the epitrochoid. Q moves around a circle of radius b,andp rotates one-third as fast with respect to Q at a distance of 3r. Place an equilateral triangle with sides of length 3 3 r so that its centroid is at Q and one vertex is at P. (The distance from the centroid to a vertex is 3 times the length of a side of the equilateral triangle.) s θ increases by π,thepointq travels once around the circle of radius b, 3 returning to its original position. t the same time, P (and the rest of the triangle) rotate through an angle of π 3 about Q,soP s position is occupied by another vertex. In this way, we see that the epitrochoid traced out by P 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.

11 CHLLENGE PROBLEM (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 3r, so it has radius 6r. To show that the rotor fits inside the epitrochoid, it suffices to show that for any position of the tracing point P, 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 P is on the y-axis,soaslongasthediameteroftherotor(whichis3 3 r)is less than the distance between the y-intercepts, the rotor will fit. The y-intercepts occur when θ = π or θ = 3π y = ±(3r b), so the distance between the intercepts is 6r b, and the rotor will fit if3 3r 6r b b 3( 3) r. E Exercises Chapter. (a) sin θ =sin θ cos θ = sin θ 4 cos θ 4 = = sin θ cos θ cos n n = n sin θ n cos θ cos θ 4 cos θ 8 cos θ n cos θ = sin θ 8 cos θ cos θ cos θ 8 4 θ cos θ cos θ cos θ n 8 4 (b) sin θ = n sin θ cos θ n cos θ 4 cos θ θ cos sin θ θ/ n 8 n θ sin (θ/ n ) =cosθ cos θ 4 cos θ θ cos 8. n sin x Now we let n,usinglim x x =with x = θ : n sin θ θ/ n lim = lim cos θ n θ sin (θ/ n ) n cos θ 4 cos θ 8 cos θ n sin θ θ =cos θ cos θ 4 cos θ 8 (c) If we take θ = π in the result from part (b) and use the half-angle formula cos x = ( + cos x) (see Formula 7a in ppendix D), we get sin π/ π/ cos π =cos π π = = cos π cos π =

AP Calculus BC Chapter 4 AP Exam Problems. Answers

AP Calculus BC Chapter 4 AP Exam Problems Answers. A 988 AB # 48%. D 998 AB #4 5%. E 998 BC # % 5. C 99 AB # % 6. B 998 AB #80 48% 7. C 99 AB #7 65% 8. C 998 AB # 69% 9. B 99 BC # 75% 0. C 998 BC # 80%.