SECTION CHAPTER 7 SECTION f 1 (x) = 1 (x 5) 1. Suppose f(x 1 )=f(x 2 ) x 1 x 2. Then 5x 1 +3=5x 2 +3 x 1 = x 2 ; f is one-to-one

Transcription

1 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 CHAPTER 7 SECTION SECTION 7.. Suppose f( )=f( ). Then 5 +3=5 +3 = f is one-to-one f(y) = 5y +3= 5y = 3 y = 5 ( 3) f () = 5 ( 3) dom f =(, ). f () = ( 5) 3 dom f =(, ) 3. f is not one-to-one e.g. f() = f( ) 4. f () = /5 dom f =(, ) 5. f () =5 4 on(, ) and f () = only at =f is increasing. Therefore, f is one-to-one. f(y) = y 5 += y 5 = y =( ) /5 f () =( ) /5 dom f =(, ) 6. not one-to-one e.g. f() = f(3) 7. f () =9 on (, ) and f () = only at =f is increasing. Therefore, f is one-to-one. f(y) = +3y 3 = y 3 = 3 ( ) y = [ 3 ( )] /3 f () = [ 3 ( )] /3 dom f =(, ) 8. f () =( +) /3 dom f =(, )

2 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 34 SECTION f () = 3( ) on(, ) and f () = only at =f is increasing. Therefore, f is one-to-one. f(y) = ( y) 3 = y = /3 y = /3 f () = /3 dom f =(, ). not one-to-one e.g. f() = f().. f () =3( +) on (, ) and f () = only at = f is increasing. Therefore, f is one-to-one. f(y) = (y +) 3 += (y +) 3 = y +=( ) /3 y =( ) /3 f () =( ) /3 dom f =(, ) 3. f () = 3 > for all 5/5 f is increasing on (, ) f(y) = y 3/5 = y = 5/3 f () = 5/3 dom f =(, ). f () = 4 (/3 +) dom f =(, ) 4. f () =( ) 3 + dom f =(, ) 5. f () = 3( 3) for all and f () = only at =/3 f is decreasing f(y) = ( 3y) 3 = 3y = /3 3y = /3 y = 3 ( /3 ) f () = 3 ( /3 ) dom f =(, ) 6. not one-to-one e.g. f() = f( )

3 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 SECTION f () = cos on [ π/,π/] and f () = only at = π/, π/. Therefore f is increasing on [ π/,π/] and f has an inverse. The inverse is denoted by arcsin this function will be studied in Section 7.7. The domain of arcsin is [, ] = range of sin on [ π/,π/]. 8. f is not one-to-one on ( π/,π/). For eample, f( π/4) = f(π/4) =. 9. f () = < for all f is decreasing on (, ) (, ). f () = dom f : f(y) = = y = = y = f () = dom f :. f is not one-to-one e.g. f ( ) = f(). not one-to-one e.g. f() = f() 3. f () = 3 ( 3 for all +) f is decreasing on (, ) (, ) 4. f () = + dom f : f(y) = y 3 + = y 3 += y 3 = dom f : f () = ( y = ( ) /3 ) /3 5. f () = < for all ( +) f is decreasing on (, ) (, ) 6. not one-to-one e.g. f() = f( 3) f(y) = y + y + = y +=y + y( ) = y = f () = dom f :

4 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 344 SECTION They are equal (a) Suppose f and g are one-to-one, and that f (g( )) = f (g( )). Then since f is one-to-one, g( )= g( ), and since g is one-to-one this implies =. (b) Since g (f (f (g()))) = g (g()) =, we have (f g) = g f. 33. (a) f () = + + k: f will be increasing on (, ) iff does not change sign. This will occur if the discriminant of f, namely 4 4k is non-positive. 4 4k = k (b) g () =3 +k +: g will be increasing on (, ) ifg does not change sign. This will occur if the discriminant of g, namely 4k is non-positive. 4k = k 3 = 3 k (a) ( f ) (5) = f () = 4 3 (b) g = (f ) (f ) g (3) = /3 = f () =3 oni =(, ) and f () = only at = f is increasing on I and so it has an inverse. f() = 9 and f () = (f ) (9) = f () = 36. f () = 3 (f ) (4) = f (f (4)) = f ( ) = 5

5 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 37. f () =+ > oni =(, ) f is increasing on I and so it has an inverse. 38. f () = + cos (f ) () = f(4) = 8 and f (4) = + = 3 ( f ) (8) = f (4) = 3/ = 3 f (f ( /)) = f ( π/6) = 39. f () = sin >oni =(, ) f is increasing on I and so it has an inverse. f(π/) = π and f (π/)= (f ) (π) = 4. f () = 4 ( ) (f ) (3) = f (f (3)) = f (3) = / = f (π/) = 4. f () = sec >oni =( π/, π/) f is increasing on I and so it has an inverse 4. f () = (f ) ( 5) = f(π/3) = 3 and f (π/3) = 4 (f ) ( 3) = f (f ( 5)) = f ( ) = 3 f (π/3) = 4 SECTION f () =3+ 3 > on I =(, ) f is increasing on I and so it has an inverse. 4 f() = and f () = 6 f () = f () = f () = sin oni =[,π], with f() = for only one value on I and so it has an inverse. f(π) = and f (π) = f ( ) = f (π) =. 45. Let dom (f ) and let f(z) =. Then 46. (f ) () = (f ) () = f (z) = f(z) = f (f ()) = +[f (f ())] = Let dom (f ) and let f(z) =. Then (f ) () = f (z) = [f(z)] = 48. (a) The figure indicates that f is one-to-one. (b) { ( +) /3 f, < () =,

6 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 346 SECTION (a) (b) f (c + d)a (a + b)c ad bc () = (c + d) = (c + d), Thus, f () iff ad bc. at + b ct + d = at + b = ct + d (a c)t = d b t = d b a c d/c f () = d b a c 5. f = f = a + b c + d = d b a c = a + ab ac = cd + d bd = a = d as long as either b or c. If b = c =, then a = ±d. 5. (a) f () = + >, so f is always increasing, hence one-to-one. (b) (f ) () = f (f ()) = f () = (a) f () = 6+() 4 () = > for all. (b) Since f(/) =, we have (f ) () = f (/) = 7 7 = (a) g () = f [g()] g () = (f [g()]) f [g()]g () = f [g()] (f [g()]) 3 (b) If f is increasing, then the graphs of f and g have opposite concavity if f is decreasing then the graphs of f and g have the same concavity. 54. (a) No. If p is a polynomial of even degree, then lim p() = or lim p() =. ± ± (b) Yes, for instance P () = 3 has an inverse. P () = 3 does not have an inverse. 55. Let f() = sin and let y = f (). Then sin y = cos y d = d = cos y = (y ± π/) sin y = ( ±)

7 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 56. Let y = f (). Then tan y =, so sec y d =. Thus d = sec y = cos y = f () = 8 +5, f () = SECTION f () = f () = + = f() 6. f () =3 +3> for all f is increasing on (, ) 6. f () = 3 5 /5 > ( ) f is increasing on (, ) 63. f () = 8 cos >, ( π/4,π/4) f is increasing on [ π/4, π/4] 64. f () = 3 sin 3 >, (,π/3) f is increasing on [, π/3] SECTION 7.. ln = ln + ln =.99. ln 6 = ln 4 = 4 ln = ln.6 =ln 6 = ln 4 ln = ln 3 4 = 4 ln 3 = ln. =ln =ln ln =.3 6. ln.5 =ln 4 =ln ln 4 =.9 7. ln 7. =ln 7 =ln8+ln9 ln = ln 63 = ln(9 7 ) = (ln 9 + ln 7 + ln ) = 3.

8 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 348 SECTION ln = ln =.35. ln.4 =ln 4 =ln4 ln =.9. For any positive integer k: k k k ( ) k d = [ln ]k k =lnk ln k =ln =ln. k. Fi a positive integer m. For any positive integer k: km ( ) km d =[ln]km k =lnkm ln k =ln =lnm. k 3. [L [ f (P )+U f (P )] = ] 69 = ln.5 = [L f (P )+U f (P )] =.9 5. (a) ln 5. = ln (.) = (a) ln.3 = ln + (.3) =.33 (b) ln 4.8 = ln 5 5 (.) =.57 (b) (b) ln 9.6 = ln + (.4) =.6 (c) ln 5.5 = ln (.5) =.7 (c) ln = ln + () =.4 7. = e 8. = e 9. ln = or ln =. Thus = e or =.. ln / ln( )== ln = =. ln[( + )( + )] = ln( +) ln[( + )( + )] = ln[( +) ] ( + )( +)=( +) + = ( + )( )= =, = = = We disregard the solution = since it does not satisfy the initial equation. Thus, the only solution is =.. ln( +) ln 4 =ln ( +) ==lne = ( +) = e = = ( ± e) 3. See Eercises 3., (3..6). lim ln = d d (ln ) = = = = 4. Let n> be a positive integer. Let P be the partition {,,,n}. Let f() =/. Then the lower sum < ln n. n Therefore, k = n.

9 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 5. Continuing Eercise 4, the upper sum > ln n. n Therefore, k = n. SECTION (a) ln g() = >, ln g() = ln <, so by the intermediate-value theorem ln r g(r) = for some r [, ]. (b) r =.795 y 7. (a) Let G() =ln sin. Then G() = ln sin =. < and G(3) = ln 3 sin 3 =.96 >. Thus, G has at least one zero on [, 3] which implies that there is at least one number r [, 3] such that sin r =lnr. (b) r =.9 8. (a) ln = <, ln =.69 >, so by the intermediate-value theorem 4 ln r = for some r [, ]. r (b) r = L = 3. L =

10 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 35 SECTION 7.3 SECTION 7.3. dom (f) =(, ), f () = 4 (4) =. dom (f) = ( ),, f () = + 3. dom (f) =(, ), f () = 3 + d ( 3 + ) = 3 d dom (f) =(, ), f () = dom (f) =(, ), f() = ln ( + ) so f () = [ ] + () 6. dom (f) =(, ), f () = 3 (ln ) 7. dom (f) ={ ±}, f () = d 4 d (4 ) = dom (f) =(, ), f () = ln 9. dom (f) = (, ), = + f () =( +) d [ln( + )] + ( + )() ln ( +) d =( +) + 4( +)ln( +) + = ( +)+4( +)ln( + ) = ( +)[+ln( + )]. dom(f) =(, ) (, ) (, ), f ()= (rewrite f() asln + ln 3 ). dom (f) =(, ) (, ), f() = (ln ) so f () = (ln ) d d (ln ) = (ln ). dom (f) =(, ), f () = ( +) 3. dom (f) =(, ), f () = cos (ln ) 4. dom (f) =(, ), f () = 5. d =ln + + C + sin(ln ) (rewrite f() as 4 ln( + ).) ( ) cos (ln ) =

11 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5: d 3 = { } u =3 du = d d = ln 3 + C 3 ( + d = + ) d =ln + C { } u =3 du =3d du 3 d = u = ln u + C = ln 3 + C tan 3d= 3 sec π d= π ln sec π + tan π + C { } u = du =d { } u = + cot du = csc d { } u =3 du = d u =ln( + a) du = + a d { } u = + cos du = sin d { } u =4 tan du = sec d { u =ln, du = d } { } u = 3 du =6 d { u =ln, du = d } sec d = tan udu= 3 ln sec u + C = ln sec 3 + C 3 sec udu= ln sec u + tan u + C = ln sec + tan + C SECTION csc du + cot d = = ln u + C = ln + cot + C. u (3 ) d = ln( + a) + a d = du u = u + C = (3 ) + C udu= u + C = [ln( + a)] + C sin + cos d = du = ln u + C = ln + cos + C u sec du 4 tan d = u = ln u + C = ln 4 tan + C d du ln = =lnu + C =ln ln + C u 3 d = du 6 u = 6 ln u + C = 6 ln 3 + C d (ln ) = du u = u + C = ln + C

12 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 35 SECTION { } u = + sec du = sec tan d sec tan + sec d = +u du = ln +u + C = ln + sec + C { } u = sin + cos 3. du = (cos sin ) d sin cos sin + cos d = du = ln u + C = ln sin + cos + C u u = du = d { u =+, du= 3 } / d d = ( + ) du +u =ln +u + C =ln + + C + d = du 3 u = ln u + C = 3 ln + + C {u =ln, du = } tan(ln ) d d = tan udu=ln sec u + C =ln sec(ln ) + C (+sec ( + sec ) d = + sec ) d = +ln sec + tan + tan + C 36. (3 csc ) d = (9 6 csc + csc ) d =9 6ln csc cot cot + C e e e e 5 4 /3 /4 d = [ln ]e =lne ln = = d d =[ln ]e =lne ln= =[ln ]e e =lne ln e = = ( + ) [ ] d = ln =ln 3 ln =ln4 3 [ ] 5 d = ln = 4 (ln 4 ln 5) = ln 8 5 tan π d = π [ln sec π ]/3 /4 = ( ) ln ln ln = π π. { } u = + sin = π/6 = u =3/ du = cos d = π/ = u = π/ π/6 cos + sin d = du 3/ u = [ln u] 3/ =ln4 3

13 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 SECTION π/ π/4 π/ π/4 ( + csc ) d = π/ π/4 ( + csc + csc ) d =[ +ln csc cot cot ] π/ π/4 = π 4 + ln( ) cot d=[ln sin ] π/ π/4 =ln ln =ln = ln { u =ln, du = d } e ln d = [ (ln ) ] e = 47. The integrand is not defined at =. 48. Let f() =ln. Then f () = and f () =. By the definition of the derivative at =,wehave f ln( + h) ln() ln( + h) ln( + ) () = lim = lim = lim = h h h h 49. ln g() = ln( +)+5ln +3ln g ( ) () g() = ( g () =( +) ( ) ln g() =ln +ln + a +ln + b +ln + c g () g() = + + a + + b + g () =( + a)( + b)( + c) ) g () = + c ( + + a + + b + + c 5. ln g() =4ln +ln ln + ln( +) g () g() = g () = 4 ( ) ( +)( +) + ( ) g () = + ) g ( b) = b(a b)(c b) 5. ln g() = (ln +ln ln 3 ln 4 ) g () g() = ( + 3 ) 4 ( ) g () = ( ( ) ( ) ( 3) ( 4) + 3 ) g () = 4

14 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 354 SECTION A = π/6 ( sec ) d =[ ln sec + tan ] π/6 = π 3 ln = π 3 ln A = = / (csc π ) d [ π ln csc π cot π ] / = π ln( ) 3 8 = π ln( + ) A = π/4 ( tan ) d =[ ln sec ] π/4 = π 4 ln = π 4 ln 56. A = π/4 (sec cos ) d = [ln sec + tan sin ] π/4 =ln(+ ) 57. A = 58. A = 59. V = 4 8 [ 5 ] [ 5 d = ln ( 3 ) d = [3 ln π 6. By shells: V = ] 4 ] = 5 8 ln 4 = 3 ln ( ) 8 d = π + + d = π [ln + ]8 = π ln 9 3 π 3 + d = [ 3π ln( + ) ] 3 =3πln

15 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 SECTION V = =π π/3 π/3 π/3 π ( sec ) d sec d =π [ln sec + tan ] π/3 =π ln ( + 3) v (t) = a (t) dt = V = = π π/4 π/4 π tan d (sec ) d = π [tan ] π/4 ( = π π ) 4 (t +) dt = t + + C. Since v () =, we get = + C so that C =. Then s = The particle traveled ln 5 ft. 4 v(t) dt = 4 dt t + = [ ln (t +)]4 =ln v(t) = a(t) dt = (t +) dt = + C, v() = = v(t) = t + t ( ) Then s = v(t) dt = +t + dt = [ln(t +)+t ] 4 =4+ln5 ft 65. d d (ln ) = d d (ln ) = d 3 d 3 (ln ) = 3 d 4 3 (ln ) = d4 4. d n (n )! (ln ) =( )n dn n 66. d (ln( )) = d d (ln( )) = d d 3 (ln( )) = d3 ( ) ( ) 3 d 4 3 (ln( )) = d4 ( ) 4. d n (n )! [ln( )] = dn ( ) n

16 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 356 SECTION csc (csc cot ) csc csc cot csc d= d = d csc cot csc cot { } u = csc cot du =( csc cot + csc ) d du csc d= u =ln u + C =ln csc cot + C [ ] g 68. (a) If g() =g ()g (), (7.3.7) gives g () =g() () g () + () g g () [ ] g = g () =g ()g () () = g () g ()+g () g (). g () + () g g () (b) If g() = g ( () g () = g () g () ), then ( ) = g g () =g () () g () g () + d d (/g ()) = ()g g () g ()g () /g () [g ()] 69. f() = ln (4 ), < 4 f () = 4 f () = ( 4) (i) domain (, 4) (ii) (iii) (iv) decreases throughout no etreme values concave down throughout: no pts of inflection 7. f () = f () = (i) domain (, ) (ii) decreases on (, ], increases on [, ) (iii) f() = local and absolute min (iv) concave up on (, ) no pts of inflection

17 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 SECTION f() = ln, > f () =ln + f () =ln +3 (i) domain (, ) (ii) decreases on (, / e], increases on [/ e, ) (iii) f(/ e)= /e local and absolute min (iv) concave down on (, /e 3/ ), concave up on (/e 3/, ) pt of inflection at (/e 3/, 3/e 3 ) 7. f () = 4 (i) domain (, ) f () = (4 + ) (4 ) (ii) increases on (, ], decreases on [, ) (iii) f() = ln 4 local and absolute ma (iv) concave down on (, ) no pts of inflection [ 73. f() =ln f () = f () = 4 4 ( + 3 ) (i) domain (, ) ],> (ii) increases on (, ], decreases on [, ) (iii) f() = local and absolute ma (iv) concave down on (,.58), concave up on (.58, ) pt of inflection at (.58,.9338) (appro.) - - y 5

18 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 358 SECTION 7.3 [ ] f() =ln f () = 3 f () = ( 6 +3 ) ( ) (i) domain (, ) (, ) (ii) increases on [ 3, ), decreases on (, ) ( ], 3 (iii) f(3/) =.9 local and absolute min (iv) concave down on (, ) (.366, )), concave up on (,.366) pt of inflection at (.366,.7) (appro.) y Average slope = b b a a d = (b a) ln b a 76. (a) f () = () = g () = 3 (3) = (b) F () = k (k) = (c) F () =lnk =lnk +ln, so F () =+ d d (ln ) = intercept: abs min at =/e -intercept at = abs min at =.765 abs ma at = abs ma at =

19 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 SECTION intercepts:, 3.47 abs min at = abs ma at =4.85 -intercept at = π/ abs ma at = π/ local min at =.769 abs min at = 8. (a) v(t) v() = v(t) = t t a(u)du, t 3 [ 4 (u +)+ 3 ] du + u + = [ u u +3ln u + ] t =+t t + 3 ln(t +) (b) (c) ma velocity at t =.58 min velocity at t = 8. (a) v(t) = a(t) dt = [ cos (t +)+ ] dt = sin (t +)+ln(t +)+C t + v() = = v(t) = sin (t +)+ln(t +)+ sin y 6 (b) (c) ma at t = min at t = (b) -coordinates of the points of intersection: =, 3.38 (c) A =.344

20 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 36 SECTION (b) -coordinates of the points of intersection: =.68, =.68, 3 =.68 (c) A = 3 [f() g()] d = (a) f() = ln, f () = ln 3, f () = 5+6ln 4 (b) f() =, f (e / )=, f (e 5/6 )= (c) f() > on (, ) f () > on (,e / ) f () > on (e 5/6, ) f() < on (, ) f () < on (e /, ) f () > on (,e 5/6) (d) f(e / ) local and absolute maimum 86. (a) f() = +ln ln, f () = ln 4(ln ), f 3 4(ln ) () = 3/ 8 (ln ) 5/ (b) f(), f (e / )=, f (e 3/ )= (c) f() > on (, ) f () > on (e /, ) f () > on (,e ) 3/ f () < on (,e / ) f () < on ( e 3/, ) (d) f(e / ) local and absolute maimum SECTION 7.4. d = e d d ( ) = e. d =3e+ =6e d = d e d ( ) = e d = d ( ) d e (ln )+ln d d (e )=e +ln 4. d =e 4 ( 4) = 8e 4 6. d =e + e d = d d (e )+e d d ( )= e e = ( + )e ( ) d = e + d = (e e ). d = (e e ( )) = (e + e ) d = e d ( ) ( ) d ln +ln d d (e )=e +ln e = ( e + ln ) d = 3(3 e ) ( e ( )) = 6e (3 e )

21 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 SECTION d =(e +) d d (e +)=(e +)e d =(e e ) (e +e )=4(e 4 e 4 ) d =( +) d d (e )+e ( ) = e d d ( )=4e (e +) d =e + e e e =( +)e ( +)e d = (e +)e (e ) e (e +) = e (e +) 8. d = e (e +) (e )e 4e (e +) = (e +) 9. y = e 4ln =(e ln ) 4 = 4 so d =43.. y =lne 3 =3 = d =3. f () = cos(e ) e =e cos(e ). f () =e sin cos 3. f () =e ( sin )+e ( ) cos = e ( cos + sin ) 4. f () = cos e sin e (e )= e tan(e ) 5. e d = e + C 6. e d = e + C 7. e k d = k ek + C 8. e a+b d = a ea+b + C 9. { u =, du =d } e d = e u du = eu + C = e + C 3. e d = e u du = eu + C = e + C 3. {u =, du = d } e / d = e u du = e u + C = e / + C 3. e d = e + C 33. ln e d = d= + C 34. e ln d = d= + C d = e 4e / d = 8e / + C

22 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 36 SECTION e du e + d = u =ln u + C = ln(e +)+C { } u = e + du = e d { } u = e a + du =ae a d { } u =e +3 du =4e d { } u = e du = e d 4. { u = sin, du = cos d} e e + d = du = u u / du =u / + C = e + +C e a e a + d = du a u = ln u + C = a a ln(ea +)+C e e +3 d = du 4 u = 4 ln u + C = 4 ln(e +3)+C sin(e ) e d = cos e sin d = sin udu= cos u + C = cos (e )+C e u du = e u + C = e sin + C 4. { u = e, du = e d } e [ + cos (e )] d = ( + cos u) du = u sin u + C = e sin (e )+C 43. e d =[e ] = e ln π ln e k d = k [e k ] = k ( e k ) [ e 6 d = 6 e 6] ln π = 6 e 6lnπ + 6 e = ( π 6 ) 6 e d = [e ] = e + e d = 4 e e d = ( ) e ( + e ) d =[ e ] = ( e ) ( ) = e (4e ) d =[ 4e ] =3 4e e e + d = [ ln (e +)] ln =ln(e ln +) ln (e + ) = ln 3 ln=ln 3 e ( ) 3 4 e d =[ ln e 4 ] =ln 4 e

23 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 SECTION ln π 4 (e +)d = (e +) d = [ ] + e e sec e d = [ln sec e + tan e ] ln π 4 =ln ( ) +. = ( e +) ( +) = (e +) 53. (a) f() =e a,f () =ae a,f () =a e a,...,f (n) () =a n e a (b) f() =e a,f () = ae a,f () =a e a,...,f (n) () =( ) n a n e a 54. (a) (t) =kae kt kbe kt (t) = = kae kt kbe kt = = e kt = B A = t = ln B ln A k The particle is closest to the origin at time t = ln B ln A. k (b) (t) =k Ae kt + k Be kt = k (t) k is the constant of proportionality. 55. A =e A =( e )+e =e ( )= = = ± and y =. ( e Put the vertices at ± ),. e 56. y = e = =lny. The area of the rectangle is given by: A = y ln y. da dt =(+lny) dt. At y =3, dt =. da Thus dt = ( + ln 3) square units per minute. 57. f() =e f () = e f () =(4 )e (a) symmetric with respect to the y-ais f( ) =f() (b) increases on (, ], decreases on [, ) (c) f() = local and absolute ma (d) concave up on (, / ) ( /, ), concave down on ( /, / ) 58. (a) V = (b) V = π (e ) d = π π e d =π [ ] e d = π e = π [ e ] e d.

24 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 364 SECTION (a) V = (b) V = πe d = π [ ] e = π[ ] e π [e ] d = π e d 6. A = ln 4 (4 e y ) + ln 4 (4 e y ) = [ 4y + e y] [ ln 4 + 4y ] ln 4 ey =ln 9 6. A = (e e ) d + 4 = [ e e ] + [ e 4 e ] 4 (e 4 e ) d = ( e4 e +) +(4e 4 e 4 e 4 + e ) = (3e4 +) 6. A = triangle upper left corner = e (e e ) d = e [e e ] = e 63. A = (e y ) =[e y y] = e e

25 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 SECTION f () = e f () = e e = e ( + ) (i) domain (, ) (ii) increases on (, ], decreases on [, ) (iii) f() = local and absolute ma (iv) concave up on (, ), concave down on (, ) pt of inflection (, /e) 65. f() =e (/) f () = 3 e(/) f () = e (/) (i) domain (, ) (, ) (ii) increases on (, ), decreases on (, ) (iii) no etreme values (iv) concave up on (, ) and on (, ) 66. f () =( )e f () =e ( 4 + ) (i) domain (, ) (ii) decreases on (, ], and on [, ), increases on [, ] (iii) f() = local and absolute min, f() = 4e local ma (iv) concave up on (, ) and ( +, ), concave down on (, + ) points of inflection at =±.

26 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 366 SECTION f() = ln f () = ln + f () =3+ln (i) domain (, ) (ii) decreases on (, e / ), increases on (e /, ) (iii) f(e / )= /e is a local and absolute min (iv) concave down on (, e 3/ ) and concave up on (e 3/, ) (e 3/, 3/e 3 ) is a point of inflection 68. f() =( )e f () =( 3 +)e f () = ( 5 +4)e (i) domain (, ) (ii) decreases on (r,r ), where r = 3 5,r = 3+ 5 increases on (,r ) (r, ) (iii) f (r )) is a local and absolute ma f (r ) is a local min (iv) concave down on (, ) (4, ) concave up on (, 4) pts of inflection at (, ) and (4, e 4 ) y n e d =[e ] n = e n e n =n = n =ln(n + ). 7. (a) f() = k ln, f () = k + k k ln = k ( + k ln ). f () =: +k ln = = ln = /k = = e /k. f () < on (,e /k ) and f () > on (e /k, ) f has a local and absolute minimum at = e /k. (b) f() = k e, f () = k e + k k e = k e (k ). f () =: k = = = k. f () > on (,k) and f () < on (k, ) f has a local and absolute maimum at = e /k.

27 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 SECTION (a) For y = e a we have /d = ae a. Therefore the line tangent to the curve y = e a at an arbitrary point (,e a ) has equation y e a = ae a ( ). The line passes through the origin iff e a =(ae a ) iff =/a. The point of tangency is (/a, e). This is point B. By symmetry, point A is ( /a, e). (b) The tangent line at B has equation y = ae. By symmetry /a [ A I = (e a ae) d = a ea ae (c) The normal at B has equation y e = ( ). ae a This can be written ] /a = (e ). a Therefore A II = /a y = ae + a e + a. e ( ae + a e ) + a e a e d = +a e a 3. e 7. By induction. True for n =:e > for >. Assume true for n. Then e =+ e t dt > + (+t + t! t t3 tn+ =+ [t ! (n + )! =+ +! + 3 n ! (n + )! So the result is true for n + ] ) tn + + dt n! 73. For >(n + )! e > [ ] n+ (n + )! > n+ (n + )! = n (n + )! > n. 74. (a) (b) Intersect at = ±.753 (c) Area =.98

28 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 368 SECTION (a) (b) =.9646, =.58 (c) A = [ 4 e ] d = [4 3 ].58 3 e.9646 = f (g()) = e ln = e (/) ln = e ln = f (g()) = e( ln ) = e ln = 78. f (g()) = e +ln = e ln = 79. (a) f() = sin (e ) f() = = e = nπ = =lnnπ, n =,,. (b) y - π π π - π -

29 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 SECTION (a) f() =e sin f() = = sin = = = nπ, n any integer. (b) y -π -π π π - 8. (a) y (b) = (c) f (.398) =.6987 g (.398) = (d) the tangent lines are not perpendicular 8. (a) y (a) (b) The -coordinates of the points of intersection are: = ln and =ln5. ln 5 [ (c) A = 7 e e ] d =.44 (b) ln e d = ln e + C (c) e ( e e e tan cos d = etan + C ) 4 d = 5 e 5 + e 4 e 3 +e e + C PROJECT 7.4 ( Step. ln + ) = n + n dt + t n dt = n (since throughout the interval of integration) =. t ( ln + ) + n dt + = n t n dt + = n + n = n +. n (since t + n throughout the interval of integration)

30 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 37 SECTION 7.5 Step. From Step, we get + n e/n = ( + ) n e and e /n+ + n n = e ( + ) n+ n Combining these two inequalities, we have ( + n) n ( e + ) n+ n SECTION 7.5. log 64 = log ( 6 )=6. log 64 = log 6 = 6 3. log 64 (/) = ln (/) ln 64 = ln 6ln = 6 4. log. = log = 5. log 5 = log 5 ( 5 ) = 6. log 5. = log 5 5 = 7. log 5 (5) = log 5 ( 5 3 ) =3 8. log 4 3 = log 6 =6 9. log p y = ln y ln p = ln +lny ln p = ln ln p + ln y ln p = log p + log p y. log p = ln ln p = ln ln p = log p.. log p y = ln y ln p = y ln ln p = y log p. log p y = ln y ln p = ln ln y ln p = log p log p y 3. = e = ( e ln ) = e = e ln = e = ln = = (ln ) = = Thus, =. 4. log 5 =.4 = =5.4 ln ln 5. log = log 4 = = ln ln 4 = ln = ln Thus, =. ln = ln = ln ln 6. log = log 3 = ln ln = ln ln 3 = ln = ± (ln )(ln 3) = = e ± (ln )(ln 3) 7. The logarithm function is increasing. Thus, e t <a<e t = t =lne t < ln a<ln e t = t.

31 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 SECTION Since the eponential function is increasing, e ln <e b <e ln, so <e b < 9. f () =3 (ln 3)() = (ln 3)3. g () =4 3 (ln 4) 6. f () = 5 (ln )(5)3 ln ln (ln 3) ( =5 3 ln 5ln+ ln 3 ). F () =5 + (ln 5)( 4 +) ( ) 3. g () = (log 3) / ln 3 = (ln 3) log 3 4. h () =7 sin (ln 7)(cos ) 5. f () = sec (log 5 ) (ln 5) = sec (log 5 ) ln 5 6. g () = ( ) ln () ln ln 4 = 4 = ln ln 3 ln. 7. F () = sin ( + )[ ln ln ] = ln ( ) sin ( + ) 8. h () =a ln a( ) cos b + a ( sin b)b = (ln a)a cos b ba sin b 9. 3 d = 3 ln 3 + C 3. d = ln + C ( 3 +3 ) d = ln 3 + C d = d ln 5 = ln 5 d u du = u ln + C = ln + C = ln ln 5 + C = log 5 + C 34. log5 d = ln ln 5 d = ln 5 (ln ) + C = (ln ) ln5 + C 35. log 3 d = ln 3 ln d = 3 ln ln d = 3 [ ] (ln ) + C = 3 ln ln 4 (ln ) + C 36. Write c = b log b c Then log a c = log a ( b log b c ) = (log b c)(log a b). 37. f () = ln 3 so f (e) = e ln 3

32 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 37 SECTION f() = log 3 f () = log 3 + ln 3 ln + = f (e) = ln 3 ln f () = ln so f (e) = e ln e = e 4. f() = log 3 (log )= ln ( ) ln ln = ln 3 ln(ln ) ln(ln ) f () = ln 3 ln 3 ln = f (e) = e ln 3 4. f() =p ln f() = ln p f () f() =lnp f () =f()lnp 4. f() =p g() ln f() =g()lnp f () f() = g ()lnp f () =f()g ()lnp = p g() g ()lnp f () =p ln p 43. y =( +) ln y = ln ( +) y d = + =( +) d +ln( +) [ ] +ln( +) y = (ln ) ln y = ln(ln ) = ln(ln )+ y d ln [ = (ln ) ln(ln )+ ] d ln 45. y = (ln ) ln ln y =ln [ln (ln )] [ ] y d =ln + [ln (ln )] ln = (ln )ln d [ ] +ln(ln) 46. y = ( ) ln y = ln = ln y d = ln = ln ( ) d = [ + ln ] 47. y = sin ln y = (sin )(ln ) ( ) = (cos )(ln ) + sin y d [ d = sin (cos )(ln )+ sin ] 48. y = (cos ) + ln y =( + ) ln(cos ) ( y d =ln(cos )+( +) sin ) cos d = (cos + [ ln(cos ) ( + ) tan ] )

33 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 49. y = (sin ) cos ln y = (cos )(ln[sin ]) =( sin )(ln[sin ]) + (cos ) y d = (sin )cos d [ cos sin ( ) (cos ) sin ] (sin )(ln[sin ]) 5. y = ln y = ln y d =ln + d = + ( ln +) SECTION y = ln y = ln y d = ln ln + ( ) [ ] d = ln ln + 5. y = (tan ) sec ln y = sec ln(tan ) y d = sec tan ln(tan ) + sec sec tan d = (tan )sec [sec tan ln(tan ) + sec 3 cot ] 53. From the definition of the derivative, the derivative of f() = ln at = is Since f () =, we have f ln ( + h) ln () = lim. h h ln ( + h) ln h = h ln ( + h) = ln ( + h)/h ash. Set =/h. Then h = and ( ln + ) as Therefore ( + ) = e ln (+/) e = e as

34 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 374 SECTION d = ] [ = ln 4ln 6. [ 4 4 d = ln 4 ] = 3 ln d ln = [log ] 4 = log 4 = 6. [ + d = ln + 5p [ + d = + p ] + ln p ( + ) [ d = ln ] = 45 ( ) = ln ln = ( p p ) ln p ] = 3 + ln [ p p / / d = ln p ] = (p ) ln p / ln 7 = / ln 7 = ( e ln 7) / ln 7 = e = appro ln 7/ ln 5 = ( e ln 5) ln 7/ ln 5 = e ln 7 =7 68. appro / ln = ( e ln 6) / ln = e ln 6/ ln = e 4ln/ ln = e 4 = (b) the -coordinates of the points of intersection are: =.98, = 3 and 3 = 3.48 (c) for the interval [.98, 3], A = for the interval [3, 3.48], A = (b) the -coordinates of the points of intersection are: =.7667, = and 3 =4 (c) The area of the region bounded by the two graphs is: A = 4 ( ) d =.5 SECTION 7.6. We begin with A(t) =A e rt and take A = $5 and t =. The interest earned is given by A() A = 5 ( e r ). Thus, (a) 5 ( e.6 ) = $4.6 (b) 5 ( e.8 ) = $6.77 (c) 5 (e ) = $ We want A(t) =A e rt =A, so e rt = = rt =ln = t = ln r (a) t = ln.6 =.55 years. (b) t = ln.8 = 8.66 years. (c) t = ln. = 6.93 years.

35 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 SECTION In general We set A(t) =A e rt. and solve for r: 3A = A e r 3=e r, ln 3 = r, r = ln 3 = 5 %. 4. We want A() = A e r =A, so r = ln = 6.9% 5. P (t) = 9 e ( t ln (4/3) = 9 eln (4/3)t/ = 9 4 ) t/ P (t) =P e kt. P(4) = P e k4 =3P = k = ln 3 4 (a) P () = P e ln 3 4 = P e 3ln3 =7P == P =.37 square inches (b) P (t) =P e ln 3 4 t =P = ln 3 4ln t =ln = t = 4 ln 3 =.5 hours. 7. (a) P (t) =, e t ln =, () t (b) P (6) =, () 6, P(5) =, () 5 8. qc = Ce kp = q = e kp = p = k ln q 9. (a) P () = P ()e.35()t = P ()e.35t. Thus it increases by e.35. (b) P () = P ()e 5k = k = ln 5.. Let P (t) be the world population at time t years, and let 99 correspond to t =. Then P (t) = 49 e kt P () = 49 e k = 8 = e k = 8 49 = k =.. Thus P (t) = 49 e. t. According to this model, the population in 98 was P ( ) = 49 e. = million.. Using the data from E., the growth constant k =.. Therefore P () = 49 e k = 49 e.4 = 37. million. P () = 49 e k = 49 e.33 = 84.4 million.

36 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 376 SECTION 7.6. Pe kt =P = kt =ln. Since k = ( ) 8 ln ln, we get t = ( ) = 57.3 years ln e.43t =3 =.43t =ln 3 = t 5.7 years. 4.5 Thus maimum population will be reached in ds d = s V = s() =Ce /V = s e /V. We want s() = s, so e /V =, hence = V ln = V ln V ln =, ln = 693 gallons V (t) =kv (t) V (t) ktv (t) = e kt V (t) ke kt V (t) = d [ e kt V (t) ] = dt e kt V (t) =C V (t) =Ce kt. Since V () = C =, V (t) = e kt. Since V (5) = 6, e 5k = 6, e 5k = 4 5, ek = ( 4 5 and therefore V (t) = ( 4 t/5 5) liters. ) /5 6. A(t) =A e kt. A(5) = A e 5k = 3 A = k = ln(/3) 5 A(t) = A = ln(/3) 5 t =ln 7. Take two years ago as time t =. In general ( ) A(t) =A e kt. We are given that Thus, = t = 5 ln(/) ln(/3) = 8.55 years. A = 5 and A() = 4. = A(t) =A e ln(/3) 5 t 4=5e k 4 so that 5 = ek or e k = ( 4 / 5). We can write A(t) =5 ( ) 4 t/ 5 and compute A(5) as follows: A(5) = 5 ( ) 4 5/ 5 5 =5e ln(4/5) = 5e.56 =.86. About.86 gm will remain 3 years from now.

37 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 SECTION Let t = correspond to a year ago. Then A(t) =4e kt, and A() = 3 = 4e k =3= k = ln(3/4) Therefore, A(t) =4e ln(3/4)t =4 ( 3 t ( 4). Ten years ago, t = 9 A( 9) = ) = 53.7 grams. 9. A fundamental property of radioactive decay is that the percentage of substance that decays during any year is constant: [ ] [ A (t) A (t +) A e kt A e k(t+) ] = A (t) A e kt = ( e k ) If the half-life is n years, then [ Thus, ( A = A e kn so that e k = ( /n ). ) /n ] % of the material decays during any one year.. A(t) =ne kt. A(5) = ne 5k = m = 5k = ln(m/n) = k = 5 ( m ) A() = ne ln(m/n) m = n = n n grams. ln(m/n) and A(t) =ne ln(m/n) 5 t.. (a) A(6) = A e 6k = ln A = k = Thus A(5) = A e 5k =.87A. Hence 8.7% will remain. (b).5a = A e kt = t = 34 years.. Ae 5.3k = A = k.38. (a) A(8) = A e 8k.35A. Thus 35.% will remain. (b) = A e 3k = A 48 grams. 3. (a) (t) = 6 t, (t) =e t (b) d dt [ (t) (t)] = d dt [6 t (e t )] = 6 e t This derivative is zero at t = 6 ln = 3.8. After that the derivative is negative. (c) (5) <e 5 =(e 3 ) 5 = 5 = 5 ( 5 )=3.( 6 ) < 5( 6 )= (5) (8) = e 8 =(e 3 ) 6 = 6 = 64( 6 ) > 8( 6 )= (8) (8) (8) = 64( 6 ) 8( 6 ) = 46( 6 ) (d) If by time t EXP has passed LIN, then t > 6 ln. For all t t the speed of EXP is greater than the speed of LIN: for t t > 6ln, v (t) =e t > 6 = v (t).

38 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 378 SECTION (a) (t) =t, 3 (t) = 6 ln(t +) (b) d dt [ 3(t) (t)] = d dt [ 6 ln(t +) t ] = 6 t + This derivative is at t = 6. After that the derivative is negative. (c) ( 7 ) = 7 < 7(ln ) 6 = 6 ln 7 = 3 ( 7 ) (d) 3 ( 8 )= 6 ln 8 = ( 6 )8 ln < ( 6 )4 < 8 = ( 8 ) If by time t LIN had passed LOG, then t > 6. For all t t the speed of LIN is greater than the speed of LOG: for t t > 6, v (t) =t> 6 t + = v 3(t). 5. Let p (h) denote the pressure at altitude h. The equation dp = kp gives dh ( ) p (h) =p e kh where p is the pressure at altitude zero (sea level). Since p = 5 and p () =, Thus, ( ) can be written (a) p (5) = 5 ( / 3) =.5 lb/in.. (b) p (5) = 5 ( 3/ 3) = 8.6 lb/in.. = 5e k, 3 = ek, ln 3 = k. p (h) =5 ( 3) h/. 6. P =, e (.6)(4) = $5, From Eercise 6, we have 6 =,e 8r. Thus e 8r = 6, = 3 5 8r = ln(3/5) and r =.64 or r =6.4% 8. (a) P =5, e (.4)() = $, (b) P =5, e (.6)() = $5, 59.7 (c) P =5, e (.8)() = $, The future value of $5, at an interest rate r, t years from now is given by Q(t) =5, e rt. Thus (a) For r =.5 : P (3) = 5, e (.5)3 = $9, (b) For r =.8 : P (3) = 5, e (.8)3 = $3, (c) For r =. : P (3) = 5, e (.)3 = $35,

39 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 SECTION dv dt = kv = v = ce kt, v() = ce = c, so c is velocity when power is shut off. 3. By Eercise 3 ( ) v (t) =Ce kt, t in seconds. We use the initial conditions v () = C = 4 mph = 9 to determine e k : mi/sec and v (6) = = 8 mi/sec Thus, ( ) can be written 8 = 9 e 6k, e 6k =, e k = /6. v (t) = 9 t/6. The distance traveled by the boat is s = 6 9 t/6 dt = 9 [ ] 6 6 ln t/6 = 3 ln mi = 76 ln ft (about 54 ft). 3. Since the amount A(t) of raw sugar present after t hours decreases at a rate proportional to A, wehave A(t) =A e kt. We are given A = and A() = 8. Thus, so that 8 = e k 4, 5 = ek, e k = ( 4 5 ) / A(t) = ( 4 t/ 5). Now, ( ) / 4 A() = = 64 5 after more hours of inversion there will remain 64 pounds. 33. Let A(t) denote the amount of 4 C remaining t years after the organism dies. Then A(t) =A()e kt for some constant k. Since the half-life of 4 C is 57 years, we have = e57k k = ln 57 =. and A(t) =A()e.t If 5% of the original amount of 4 C remains after t years, then.5a() = A()e.t t = ln.5. =, 4 (years) 34. A(t) =A e ln 57 t A() = A e ln 57 =.78 A 78% remains

40 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 38 SECTION f (t) =tf(t) f (t) tf(t) = e t / f (t) te t / f(t) = d dt [e t / f(t)] = e t / f(t) =C f(t) =Ce t / 36. f (t) = sin tf(t) f (t) sin tf(t) = e cos t f (t) sin te cos t f(t) = d [ e cos t f(t) ] = dt e cos t f(t) =C f(t) =Ce cos t 37. f (t) = cos tf(t) f (t) cos tf(t) = e sin t f (t) cos te sin t f(t) = d [ e sin t f(t) ] = dt e sin t f(t) =C f(t) =Ce sin t 38. Write the equation as f (t) g(t)f(t) = and set h(t) = g(t) dt. Then f (t) g(t)f(t) = e h(t) f (t) g(t)e h(t) f(t) = [e h(t) f(t)] = e h(t) f(t) =C f(t) =Ce h(t) = Ce g(t) dt SECTION 7.7. (a) (b) π/3. (a) π/3 (b) π/3 3. (a) π/3 (b) 3π/4 4. (a) / 3 (b) 5. (a) / (b) π/4 6. (a) π/6 (b) π/4

41 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 SECTION (a) does not eist 8. (a) 4/5 (b) 5/3 9. (a) 3/ (b) 7/5 (b) does not eist. (a) arc cosine domain: [, ], range: [, π] (b) arc cotangent: domain: (, ), range: [, π]. d = +( +) = + +. d = +( ) = ( + ) 3. f () = d ( ) = ( ) d 4 4 ] 4. f () =e arcsin + e = [arcsin e + 5. f d () = arcsin + () = arcsin + () d 4 6. f () =e arctan + 7. du d arcsin = (arcsin ) (arcsin ) = d d d = +(e ) e = e +e ( ) d = + () arctan = ( ) + arctan ( + ) d = ( + + ) + = ( +) +. f () = (arctan ) / d d (arctan ) = (arctan ) / +() = (+4 ) arctan. f () = arctan + = ( + ) arctan 3. d = + (ln ) d d (ln ) = [ + (ln ) ] 4. g sin () = cos + (cos +)

42 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 38 SECTION dθ dr = d ( r ) dr ( r )= r = r r r r r dθ dr = [r/(r + )] (r +) = (r +) r + 7. g () = arcsec ( ) + ( ) ( ) = sec 8. dθ dr = +[/( + r )] r ( + r ) = r r 4 +r + 9. d = cos [arcsec (ln )] ln (ln ) = cos [arcsec (ln )] ln (ln ) 3. f () =e arcsec 3. f () = c + = c (/c) earcsec ( ) = c c c c = c + 3. d = = ( ) c () c ( c ) (/c) c (c ) 3/ (c ) / = (c ) 3/ ( ) c 33. (a) sin (arcsin ) = (b) cos (arcsin ) = (c) tan (arcsin ) = (d) cot (arcsin ) = (e) sec (arcsin ) = (f) csc (arcsin ) = 34. (a) tan (arctan ) = (b) cot (arctan ) = (c) sin (arctan ) = + (d) cos (arctan ) = + (e) sec (arctan ) = + (f) csc (arctan ) = +

43 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 SECTION { } au = + b adu = d { } au = + b adu = d d = a ( + b) d a +( + b) = a adu a a u = du u ( ) + b = arcsin u + C = sin + C a adu +u = a arctan u + C = a arctan ( + b a ) + C { } au = + b 37. adu = d d = ( + b) ( + b) a adu au a u a = a du u u = a arcsec u + C = ( ) + b a arcsec + C a 38. (a) sec (arcsec ) = sec (arccos /) = = csc (arccsc ) = csc (arcsin /) = / / = (b) arc secant: range: [, π/) (π/, π) (c) arc cosecant: range: [ π/, ) (, π/] 39. d + = [arctan ] = π 4 4. d + = [arctan ] = π ( 4 π ) = π 4 4. / d = [arcsin ]/ = π 4 4. d [ = arcsin ] = π [ d 5 + = 5 arctan ] 5 = π 5 8 d 5 6 = [ arcsec ] 8 = ( arcsec arcsec 5 ) = π arcsec 5 4 { } 3u = = = u = 3/ d 3 du =d =3/ = u = 9+4 = du 6 +u = 6 [arctan u] = π 4 5 d 9+( ) = [ ( )] 5 arctan = ( π ) = π { } u =4 =3/ = u =6 du =4d =3 = u = 3 3/ 6 4 d 6 9 = 6 du/4 (u/4) u 9 = [ arcsec 3 ( )] u = arcsec 4 π 9 d 6 ( 3) 6 +8 = d 4 ( 3) ( 3) = [arcsec ( 3)]6 4 = arcsec 3

44 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 384 SECTION [ ( d +3 = arcsin 4 ( +3) )] 3 = π ln 3 ln e d = [ arcsin e ] ( ) ( ) ln 3 e ln = arcsin arcsin = π ( ) 3 6 arcsin 3 { } u = e = = u = du = e d =ln= u = ln / { u = e d = +e d 3 4 = 3 / du =d } sec 9 tan d = du +u = [arctan u] = arctan π 4 =.3 d = 4 3 d = 4 [ arcsin 3 ] / ( ) = 3 arcsin 3 du u = arcsin u + C = arcsin + C du ( u ) = arcsin + C = arcsin 9 u 3 ( tan 3 ) + C 55. { u = du =d } + 4 d = du +u = arctan u + C = arctan + C 56. d = 4 ( ) d = arcsin + C 4 ( ) { } u = tan du = sec d cos 3 + sin d = sec 9 + tan d = du 9+u = ( u ) 3 arctan + C = ( ) tan 3 3 arctan + C 3 du 3+u = ( ) u arctan + C = ( ) sin arctan + C u = arcsin du = d arcsin d = udu= u + C = (arcsin ) + C arctan + d = u =ln du = d udu= u + C = (arctan ) + C d (ln ) = du = arcsin u + C = arcsin(ln )+C u

45 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5: (ln ) d = du = arctan u + C = arctan(ln )+C +u SECTION y 63. A = = d = d 4 4 [ ( )] arcsin = π A = 3 3 [ 3 ( ) ] 3 d = arctan = π π 4 = π = 4 = ± ( 8 A = +4 ) 4 d = [ = V = π arctan ( ( 8 +4 ) 4 d ) ] 3 =π d = π [ ] arctan(/) = π 8 y 67. V = 68. V = 6 3 π d = π 4+ π 9 6 d =π 3 [ ] d =π 4+ =4π( ) 4+ [ d ( ) ] 6 9 =π 3 arcsec 3 = π Let be the distance between the motorist and the point on the road where the line determined by the sign intersects the road. Then, from the given figure, ( ) s + k θ = arctan arctan s, << ( and dθ d = s + k ) (s + k) + ( s ) + s = (s + k) +(s + k) + s + s = s k + sk k [ +(s + k) ][ + s ] Setting dθ/d = we get = s + sk. Since θ is essentially when is close to and when is large, we can conclude that θ is a maimum when = s + sk. 7. y = arctan 3 dt = +(/3) 3 d dt If dt = 6 and = 5 then dt = (5) 6= 9 7 rad/sec

46 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 386 SECTION 7.7 a [ ( 7. (b) a d = a a + a ) ] a arcsin = πa a a The graph of f() = a on the interval [ a, a] is the upper half of the circle of radius a centered at the origin. Thus, the integral gives the area of the semi-circle: A = πa. 7. (a) f () = + ( a+ a ) +a ( a) = +a ( + a )( + ) = + (b) lim f() = π/ f() =π/ (/a) lim (/a) + (c) Let g() = arctan. For =<, a g() =, f() = arctan(a), so f() =g() + arctan a for < a, i.e., C = arctan a For >, note that a lim arctan = π, ( ) a + lim arctan = arctan( /a) so a f() =g() + arctan( /a) π for > a, i.e., C = arctan( /a) π 73. Set y = arccot. Then cot y = and, by the hint, tan ( π y) =. Therefore π y = arctan, arctan + y = π, arctan + arccot = π. 74. Set y = arccsc. Then csc y = and sec ( π y) [ ( = sec π θ)] = csc θ ]. Therefore π y = arcsec, arcsec + y = π, arcsec + arccsc = π. 75. The integrand is undefined for. 76. Numerical work suggests limit =. One way to see this is to note that the limit is the derivative of f() = arcsin at = and this derivative is : f () =, f () = 77. I =.5 d = [f(.5) + f(.5) + f(.5) + f(.35) + f(.45)] =.53 and sin(.53) =.499. Eplanation: I = arcsin(.5) and sin [arcsin(.5)] = (a).5698,.574,.576,.577 (b).578 = π (c) = π

47 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 PROJECT 7.7 SECTION (a) n sin θ = n sin θ = n sin θ. (a) Think of n and θ as functions of altitude y. Then Differentiation with respect to y gives n sin θ = C n cos θ dθ + dn sin θ =, dθ cot θ + dn n = and so when α = π θ, Now, α = arctan ( ) d and dα d = dn dθ = cot θ n = ( dα ) = dα d d d y d ( ). + d (b) + (c) n(y) = ( ) = + tan α = + cot θ = csc θ = n d C = (a constant) [n(y)]. k, with b, k constants, k>. y + b SECTION d ( = cosh ) =cosh. d d d = (cosh a) / (a sinh a) = a sinh a cosh a d = a cosh a + a sinh a = a(cosh a + sinh a) = sinh( + a) d (cosh ) (cosh ) sinh (sinh ) = d (cosh ) = cosh cosh sinh = d

48 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 388 SECTION = ab cosh b ab sinh a = ab (cosh b sinh a) d d = e (cosh + sinh )+e (sinh + cosh ) =e (cosh + sinh ) d = sinh a (a cosh a) =a coth a. d = a sinh a ( sinh a)a = cosh a cosh a. 3. d = cosh (e )e () = e cosh (e ). d = e cosh +e sinh 4. d = sinh (ln 3 ) 3 3 = 3 sinh (ln 3 ) d = + sinh cosh = cosh 5. d = cosh (sinh ) = tanh 6. d = cosh = coth sinh 7. ln y = ln sinh y cosh = ln sinh + d sinh and d = (sinh ) [ ln sinh + coth ] 8. y = cosh = ln y = cosh ln = y d = sinh ln + cosh = ( d = cosh sinh ln + cosh ) ( e 9. cosh t sinh t + e t ) ( e t e t t = ) = 4 {( e t ++e t) ( e t +e t)} = 4 4 =. sinh t cosh s + cosh t sinh s = (et e t ) (es + e s )+ (et + e t ) (es e s ) = ( e s+t e (s+t)) = sinh(s + t) ( e t + e t )( e s + e s ) ( e t e t )( e s e s ). cosh t cosh s + sinh t sinh s = + = { e t+s +e (t+s)} = et+s + e (t+s) = cosh (t + s) 4. Follows from Eercise, with s = t. 3. Set s = t in cosh (t + s) = cosh t cosh s + sinh t sinh s to get cosh(t) = cosh t + sinh t. Then use Eercise 9 to obtain the other two identities. 4. cosh( t) = ( e t + e ( t)) = ( e t + e t) = cosh t 5. sinh( t) = e( t) e ( t) = et e t = sinh t

49 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 SECTION y = 5 cosh + 4 sinh = 5 (e + e )+ 4 (e e )= 9 e + e d = 9 e e = e (9e ) d = = e = 9 = = ln 3. d y d = 9 e + e > for all, so abs min occurs at = ln 3. At = ln 3, y = 9 ( 3 )+ (3) = y = 5 cosh + 4 sinh = 5 (e + e )+ 4 (e e )= e 9 e d = e + 9 e = e (9 e ) d y d = = e =3or =ln3 d = e 9 e < for all so abs ma occurs at =ln3. The abs ma is y = eln 3 9 e ln 3 = (3) ( 9 ) 3 = y = 4 cosh + 5 sinh = 4 (e + e )+ 5 (e e )= 9 e e d = 9 e + e > always increasing, so no etreme values. [ 9. [cosh + sinh ] n e + e = + e e ] n =[e ] n = e n = en + e n + en e n = cosh n + sinh n 3. y = A cosh c + B sinh c, y = Ac sinh c + Bccosh c, y = Ac cosh c + Bc sinh c = y = c y 3. y = A cosh c + B sinh c y () = = =A. y = Ac sinh c + Bccosh c y () = = =Bc. y = Ac cosh c + Bc sinh c = c y y 9y = = ( c 9 ) y =. Thus, c =3, B = 3, and A =. 3. From Eercise 3, y = c y, so c =. =y() = A cosh + B sinh = A = A = =y () = Ac sinh + Bccosh = Bc = B = c =4 33. a sinh a + C 34. a cosh a + C 35. 3a sinh3 a + C 36. 3a cosh3 a + C 37. a ln (cosh a)+c 38. ln sinh a + C a

50 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 39 SECTION a cosh a + C 4. sinh d = 4 (e +e ) d = ( 4 e ) e + C = 4 sinh + C = sinh cosh + C 4. From the identity cosh t = cosh t (Eercise 3), we get cosh t = ( + cosh t). Thus, cosh d = ( + cosh ) d = ( + ) sinh + C = ( + sinh cosh )+C 4. sinh e cosh d = e u du = eu + C = e cosh + C 43. { u = du = d/ } sinh d = sinh udu= cosh u + C = cosh + C 44. sinh + cosh d = du =ln +u + C = ln( + cosh )+C +u 45. A.V. = 46. A.V. = A = 48. A = 49. V = ( ) ln ln 5 ln 5 4 cosh d= [sinh ] = e e sinh d= 8 [cosh ]4 = 8 [ e 8 + e 8 =.75 ] = sinh d= [ cosh ] ln = eln + e ln = 8 cosh d= 5 [sinh ]ln ln 5 = 3 5 π ( cosh sinh ) d = πd= π

51 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 5. V = ln 5 π[sinh ] d = π ln 5 sinh d ln 5 = 4 π ( e +e ) d = 4 π [ e e ] ln 5 = 4π [sinh( ln 5) ln5] SECTION V = ln 5 ln 5 π[cosh ] d =π ln 5 cosh d ln 5 = π ( e 4 ++e 4) d = π [ 4 e e 4] ln 5 = π [ 4 sinh(4 ln 5) + ln 5] 5. (a) sinh e e ( ) lim e = lim e = lim e = (b) cosh e + e ( lim e a = lim e a = lim e a + e a) For <a<, limit =. For a>, limit =. 53. (a) (.6935,.5) (b) A = (a) (±.68,.68) (b) A =.388 SECTION d = tanh sech. d = tanh sech = sech csch 4. d = tanh 3 sech 3 3 = 6 tanh 3 sech 3 d = sech (ln ) d = cosh ( arctan e ) d ( arctan e ) = e cosh ( arctan e ) d +e 4 d = sech (3 + ) tanh(3 + )(6) = 6 sech (3 + ) tanh(3 +) d = csch( + ) d ( + ) = d + csch( + ) d = ( sech )(tanh ) = tanh sech

52 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 39 SECTION ( + cosh )( sech tanh ) sech (sinh ) = d ( + cosh ) = sech (tanh + cosh tanh + sinh ) sech (tanh + sinh ) ( + cosh ) = ( + cosh ) sinh ( + sech ) cosh ( sech ) tanh sinh + tanh = d ( + sech ) = ( + sech ). d d (coth ) = d d. [ ] cosh sinh (sinh ) cosh (cosh ) = sinh sinh = cosh sinh sinh = sinh = csch ( ) d d ( sech ) = = d d cosh (cosh ) sinh = cosh sinh = sech tanh cosh 3. [ ] d d ( csch ) = d d sinh = cosh sinh = csch coth 4. tanh(t + s) = sinh(t + s) sinh t cosh s + cosh t sinh s tanh t + tanh s = = cosh(t + s) cosh t cosh s + sinh t sinh s + tanh t tanh s 5. (a) By the hint sech = 9 5. Take sech = 3 since sech = > for all. 5 cosh (b) cosh = = 5 ( )( ) 5 4 (c) sinh = cosh tanh = = 4 sech (d) coth = cosh sinh = 5/3 4/3 = 5 4 (e) csch = sinh = sech t = tanh t = 5 44 = = sech t = cosh t = sinh t = cosh t tanh t = 9 5 = 5 9 coth t = csch t = 9 =. sinh t 5 = tanh t 5 = sech t 9 7. If, the result is obvious. Suppose then that >. Since, we have. Consequently = += and therefore. 8. We will show that [ ( )] + tanh ln = for all [, ].

53 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 First we observe that It follows that tanh s = es e s e s + e s and therefore tanh(ln t) = t /t t +/t = t t + [ tanh ln ( )] ( + = tanh ln ) + = SECTION = =. 9. By Theorem 7.9., d ( sinh ) = d [ ( ln + )] d d + = + + ( + ) = d d (cosh )= d [ ln( + ] d ) = + ( + ) =.. By Theorem 7.9. [ ( )] d d + (arctan ) = d d ln = ( ) ( ) () ( + )( ) ( ) + ( ) = ( + ) = ( ). y = sech = sech y = = cosh y = ( ) = y = cosh, ( ) so d = (/) =.. 3. Let y = csch. Then csch y = and sinh y =. sinh y = cosh y d = d = cosh y = +(/) = + 4. y = coth = coth y = = tanh y = ( ) = y = tanh, so d = ( (/) ) = =

54 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 394 SECTION (a) d =at = d d > if< = sech tanh = sinh cosh d < if> f is increasing on (, ] and decreasing on [, ) f() = is the absolute maimum of f. d y (b) d = cosh sinh cosh 3 = sinh cosh 3 d y = sinh = ± d sinh = e e = e e = =ln(+ ) =.88 sinh = e e = e +e = = ln ( + ) =.88 (c) The graph of f is concave up on (,.88) (.88, ) and concave down on (.88,.88) points of inflection at = ± (a) y (b) y y = sinh d = cosh d y = sinh. d d y = sinh = =. d y = sinh =ln ( + + ) d y d = = =. ( 3/ +) d = d y d = ( +). 3/

55 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 It is easy to verify that (, ) is a point of inflection for both graphs. SECTION (a) (b) 9. (a) tan φ = sinh (b) sinh = tan φ 3. V = φ = arctan(sinh ) = sinh (tan φ) dφ d = cosh + sinh =ln ( tan φ + tan φ + ) = cosh cosh = = sech = ln(tan φ + sec φ) cosh = ln(sec φ + tan φ) (c) = ln (sec φ + tan φ) d dφ = sec φ tan φ + sec φ = sec φ tan φ + sec φ ( ) π sech d=[πtanh ] e e = π e + e e e e =π + e ( e ) e + 3. sinh tanh d= cosh d { } u = cosh du = sinh d sinh cosh d = du =ln u + C = ln cosh + C u

56 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 396 SECTION coth d= cosh du sinh d = =ln u + C =ln sinh + C u 33. sech d= cosh d = { } u = e du = e d e d = + e e e d = + e e + d u + du = arctan u + C = arctan (e )+C csch d= e d = e { } u = sech du = sech s tanh d e e d = { tanh e + C, e < = coth e + C, e >. du u (u = e ) sech 3 tanh d= u du = 3 u3 + C = 3 sech 3 + C 36. sech d = sech udu= tanh u + C = tanh + C 37. { } u = ln (cosh ) du = tanh d tanh ln (cosh ) d = udu= u + C = [ln (cosh )] + C 38. ( + tanh cosh d = sech + sinh ) cosh 3 d = tanh cosh + C = tanh sech + C 39. { } u = + tanh du = sech d sech + tanh d = du =ln u + C =ln + tanh + C u 4. tanh 5 sech d= 6 tanh6 + C { } = a sinh u d = a cosh udu a d = a d a + d = (/a) d = a cosh u a + a sinh u du = ( du = u + C = sinh ) + C a du ( u = cosh (u)+c = cosh ) + C a

57 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 REVIEW EXERCISES Suppose <a. { } = a tanh u d d = a sech udu a d = a sech u a a tanh u du = du = u a a + C = ( ) a tanh + C a The other case is done in the same way. mg 44. (a) v() = k tanh = ( )( ) ( ) mg gk gk gkm v (t) = k sech m t = g sech m t ( ) ( )] mg kv = mg k mg gk gkm k tanh m t = mg [ tanh t ( ) gk = mg sech m t = m dv dt ( ) mg gk mg (b) lim v(t) = lim t t k tanh m t = k. REVIEW EXERCISES. f () = 3 /3 > ecept at = f is increasing, it is one-to-one f () =( ) 3.. f is not one-to-one f(3) = f( ) = 3. Suppose f( )=f( ). Then + = + + = + = Thus f is one-to-one. f () = +. = 4. f () = 6( +) > ecept at = f is increasing, it is one-to-one f () = /3. 5. f () = e < ecept at = f is decreasing, it is one-to-one f () = ln. 6. f() is not one-to-one f(π/) = f(3π/) =,. 7. f is not one-to-one. Reason: f () =+ln = f decreases on (, /e] and increases on [/e, ). There eist horizontal lines that intersect the graph in two points.

58 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD7-7 JWDD7-Salas-v November 5, 6 5:4 398 REVIEW EXERCISES 8. Suppose f( )=f( ). Then + = = =8 = Thus f is one-to-one. f () = f () = e ( + e < for all f is one to one and has an inverse function. ) Since f() =, (f ) ( ) = f () = 4. f () =3+ 3 > for all > f is one-to-one and has an inverse function. 4 f() = therefore (f ) () = f () = 6. f () = 4+ > = f has an inverse. ( Since f() =, ) f () = f () =. f = sin for all f has an inverse function. Since f(π) = (f ) ( ) = f (π) = 3. f () =3 ( ( ln ) ) ln 4 (ln ) () =6 = 4. y = (cos e 3 )(e 3 )(3) = 6e 3 cos e 3 5. g () = ( ) +e e e ( e ) ( + e ) = e ( e ) ( + e ) 6. ln f() = sinh ln( f () + ) f() = sinh + +ln( + ) cosh [ ] f () =( +) sinh cosh ln( sinh +) d = ( ln 3 ) = 3 +3 ln g () = sinh + cosh