# 9 - SECTION3.9 HYPERBOLICFUNCTIONS 3 215 3.9 Hyperbolic...

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Unformatted text preview: SECTION3.9 HYPERBOLICFUNCTIONS 3 215 3.9 Hyperbolic Functions 1. (a) sinhO : 5(e0 — .20) = 0 (b) coshO = §(e0 +e0) = §(1+1)= 1 O_ —O)/2 e1‘e~1 e2_1 (6 6 _ : * = ~0.76159 2. (a) tanhO = m — 0 tanh 1 el + e_1 62 +1 1,12 71n2 1n2 V ln2 —1 _ 71 2_ l 3 . e —e e (6 ) 2 2 _ 2 _ _ 3. (a) Sinh(ln2) : 2 : 2 _ 2 u 2 7 4 (b) sinh2 = %(62 — 6—2) % 3.62686 ln3 — ln3 1 e + e 3 + - 5 4. (a) cosh3 = §(e3 + e”) z 10.06766 (b) cosh(ln3) = ‘2 z 2 3 = g 1 1 _1 I = — 2 1 b cosh 1: 0 because coshO = 1. 5. (a) sechO COShO 1 ( ) 6. (a) sinhl = §(e]L — erl) z 1.17520 (b) Using Equation 3, we have sinh‘1 1 =In(1+ «12 +1) : 1n(1+ x/i) a 0.88137. 7. sinh(—m) = %[e’”E — e_(_1)] = %(e_1 - em) 2 —%(er 7 e”) : —sinhm 8. cosh(—:c) : He” + 6_(72)] = ﬂea—z + e“) 2 ﬂex + e“) = coshm 9. coshm +sinhsc : %(ez +e‘z) + 10. coshm — sinha: = %(ez +e‘w) — 11. sinhxcoshy + coshmsinhy 2 (ez — 6—1)] (ey + 8—9)] + (ex + 671)] [%(ey ~ Big” 2 iKEHy + ex‘y r €71” — fry) + (erﬂ’ — eky + err+y _ 6—24)] 2 ﬁ(2e2+y _ 2e—r—y) = Hera _ e—(z+y)] : Sinh<m +y) 12. coshz'coshy + sinhxsinhy : (el + 6—38)] (6y + 6—11)] + (ex _ e—z)] (621 ﬂ 8711)] : iﬂemﬂ’ + 61W + 6"” + e‘ry) + (eE‘Ly — em-y _ e~w+y + e-z—y)] I %(2ez+y + 2e—z—y) : %[ez+y + e—(z+y)] : coshov + y) 13. Divide both sides of the identity cosh2 x — sinh2 x = 1 by sinh2 z: cosh2 x sinh2 ac 1 _ 2 , 2 . 2 <¢ coth2zc—lzcsch2x. smh cc smh 5c smh ac sinh x cosh y cosh :1: sinh y 14. tanhm + y) I _ sinhxcoshy + coshmsinh y R W cosh(x + y) cosh .7: cosh y + sinh ac sinh y COSh J? cosh y smh m Slnh y cosh .73 cosh y cosh m cosh y ~ tanh m + tanh y _ 1+ tanha: tanh y 15. Putting y : ac in the result from Exercise 1 1. we have sinh 2x = sinh(:2: + :c) : sinh ac cosh ac + cosh av sinh :1: = 2 sinh x cosh x. 16. Putting y = m in the result from Exercise 12, we have cosh 2x : cosh(:c + m) 2 cosh z cosh at + sinhz sinh :L' : cosh2 a: + sinh2 ac. 216 CHAPTER 3 DIFFERENTIATION RULES 17 tanh(1nx)_ _ w _ 5” 1(elm)_1 _ w i 3311 i _ cosh(ln:c) _ (elnx +e‘1“\$)/2 _ \$+(elnx)’1 _ \$4.134 _m—1/m: (\$2—1)/x:w2—1 \$+1/ac (w2+1)/;L' \$2+1 18 1 + tanha: 1 + (sinhx)/coshx 7 coshx +sinhx _ ﬂex +64) + ﬂex — e—I . 1— tanhx 1 — (sinh x) /c0sh\$ _ coshsc — sinh a: _ ﬂex + 6—3”) — %(€I # 6’1) (2“8 + 61x + e“: — 6—1 26m 2x 1 ~ — e e“6 + e—E — 6m + cfm 2e” coshzv + sinhm 6”: 2w 0r: Using the results of Exercises 9 and 10, ———— : = e cosh x — sinh ac e‘“E 19. By Exercise 9, (coshx + sinh as)" : (can : em = cosh ms + sinh me. 20. sinhsc 2% => cschac = l/sinhm = g. cosh2a: : sinh2m+ 1 : % + 1 : %% => cosha: = 2 (since cosh an > 0). sechx = l/coshsc : §,tanh :c = sinhz/coshz = gﬁ = %,and cotha: = l/tanhm = 21. tanhm = g > O,so:c > 0. cothm : 1/tanhm 2 asechzm : 1 #tanh2zc = 1 — (\$2 = 595 :> sechm : % (since sechsc > 0), coshw :1/sechw : sinhx = tanhxcoshm = g . g : §,and cschx = 1/sinhac : %. 22. y: “Cm: sinlhx ex — e‘“0 @—m 1 — 6’29” 1 — 0 ' = ' ——. = 1' _—’ z __ z 1 23. (a) Eli-n; tanhx 215130 er + (rm 6% 1 +e_2\$ 1+ 0 . h 1. eI—e’z mﬂ l. ezw—1_O—1_ 1 mkrgwtan (ax—+671: . ex e2m+1 _ _ L!) _‘ '(E (c) lim sinhm : lim g—e— : 00 (E—’OO IE—>OO 2 (d) lim sinhsc = lim 1:6— : —oo za~oo are—o0 2 2 ' : i '— : 0 (e) x1320 sechw \$1320 em + 8% (3m + 6'30 €71 . 1 + 6-22 1 + 0 ‘ :' . = ———:———:1 0r.-Use arta] (f) mleréocothx \$1330 em # eﬂu e_m \$1330 1 _ 64\$ 1 _ 0 [ p ( ) (g) lim cothsc : lim 00811 x : 00, since sinh cc —> 0 through positive values and coshm —> 1. :EaO‘l' 3—»0+ sinha; (h) lim cothw : lim Céshm z —00, since sinhx —> 0 through negative values and coshx ——> 1. z—»0’ 55—»0- smha: 2 (i) lim cschzc : lim ——j = 0 m—v—oo maroo 6z —8 SECTION3.9 HYPEHBOLICFUNCTIONS 3 217 d d 24. (a) % coshzc 2 53; [%(em +e_’:)] 2 %(ex — (fr) 2 sinhm b d t h _ d sinhm ‘ coshac coshac — sinharsinh :5 _ cosh2 m — sinh2 ac _ 1 : sech2 a: ( ) E an m _ E coshw cosh2 9: cosh2 ac cosh2 cc (1 d 1 cosh a: 1 cosh x _ 2— :_ :— - . 2~cschxc0thzc (C) dac CSCh :6 dz [sinh a: sinh2 m sinh a: smh a: d d 1 sinh a: 1 sinh m _ 2— 2— :— ~ 2‘sechxtanhac (d) dm seCh m dz [cosh 1‘] cosh2 ac cosh m cosh x ( ) d th _ d coshzz: _ sinhmsinh 3: — coshx coshx a sinh2 x — cosh? x _ _ 1 e da: co m T dx sinh cc _ sinh2 av sinh2 a: sinh2 :c 2 — csch2 z 25. Let y 2 sinh‘1 .73. Then sinh y 2 an and, by Example 1(a). cosh2 y — sinh2 y 2 1 2> [with coshy > 0] coshy2\/1+sinhr’2 2\/1+m2.SobyExercise9,ey2sinhy+coshy2x+ 1+9:2 2 y2ln(x+\/1+x2). 26. Let y 2 cosh’1 ac. Then coshy 2 ac and y 2 0, so sinhy 2 x/cosh2 y — 2 \/\$2 — 1. So. by Exercise 9, ey 2 coshy+sinhy 2 w+ \/\$2 —1 => y 2 ln(w+ x/w2 — 1). Another method: Write m 2 cosh y 2 % (6y + 6‘9) and solve a quadratic. as in Example 3. sinhy _ (ey — e‘y)/2 ey 62y —1 coshy (ey + e‘y)/2 e?! _ 629 +1 mezy+\$2e2y—1 2> 1+I=82y—.’L‘€2y 2 1+\$2629(1~m) 2> 1+m 1+\$ 1+:c 2y: :1 :1 e 1_m 2 2y n<1_\$> 2 y 21n(1_x). (b) Let y 2 tanh_1 3:. Then as 2 tanh 3/, so from Exercise 18 we have 1+tanhy 1+\$ 1+1‘ 1+2: 2yz‘: : :1 e 1—tanhy 1—3: : 2y ln(1—x> :> y 2ln(1—:1c ‘ 27. (a) Let y 2 tanh‘1 m. Then a: — tanhy — :> 281 (a) (i) y 2 csch‘1 :2 <2 cschy 2 x (2: 2 0) (ii) We sketch the graph of csch‘1 by reﬂecting the graph of csch (see Exercise 22) about the line y 2 m. (iii) Let y 2 csch_1 :51 Then :5 2 cschy 2 L 2 \$631 2 me‘” 2 2 e?! —_ e—y /2 2 x(ey)2—2ey—:c20 => 642%. m ‘/ 2 _ 2 Butey>0,sofor:c>0,ey2Mandforag<uey=1 x +1. 37 x 2 Thus, csch‘1 x 2111(1 + 75 [\$1 218 C CHAPTER3 DIFFERENTIATION RULES (b) (i) y = sech‘1 ac <=> sechy = x and y > 0. (ii) We sketch the graph of sech‘1 by reﬂecting the graph of sech (see Exercise 22) about the line y = 1:. 2 (iii) Let y = sech‘1 as. so x = sechy : —— => xey + mew : 2 ey+e—y / _ 2 => m(ey)2-Zey+m=0 42> eyzl—ji—l—L. a: _./ __ 2 Buty>0 :> ey>1.Thisrulesouttheminussignbecauseg—i>1 ® lﬂx/l—x2>x :c <=> 1—\$>\/1—x2 <:> 1’2\$+w2>1—\$2 <=> x2>m <-7 w>1,butm:sechy§1.Thus, 1+\/1——_a? _1 1+\/1——_x_2 :——\$——— => sech len —————. cc ell (c) (i) y = coth’1 ac ¢> cothy = :c (ii) We sketch the graph of coth‘1 by reﬂecting the graph of coth (see Exercise 22) about the line y : so. y —y (iii) Let y 2 coth’1 cc. Then as : cothy : e——+—e—— :> e?! _ e—y grey —ace’y 2 e” +e_y => (:6 — 1)ey : (w+ Me” :> 1 e2y2\$+1 => 2y=lnm+ => cothflsc=§1nw+ m—l m—l w—l ,1 . dy 29. (a) Lety : cosh cc. Then coshy : m and y 2 0 => smhy E = 1 => 1 gy— _ . 1 _ 1 — ————— (since sinhy Z 0 for y 2 0). 0r: Use Formula 4. d1} Slnhy ‘ (cosh2 y _ 1 «E2 — 1 _ dy dy 1 1 1 1 2 (b) Let y tanh m hen tan 3; :1: => sec 3; dx div sech2y 1 # tanhz y 1 _ 332 0r: Use Formula 5. dy _ 1 dy _ __ z 71 _ z _ h h —— : 1 2 — — — - (c) Lety csch so Then cschy m i 650 yCOt y dm dw CSChyCOthy By Exercise 13, cothy = :I: csch2 y + 1 : ivy? + 1.1f3: > 0. then cothy > 0, so cothy : V352 + 1. If a: < 0, then cothy < 0. so cothy : ~x/x2 + 1. In either case we have dy 1 1 dscr cschycothy Tlmh/mz—t—l' d (d) Lety : sech‘1 :13. Then sechy = a: => —sechytanhy (1—: = 1 : dy 1 1 1 ._ :___.__.:_ :—————.(Notethat >Oandsotanhy>0.) dm seCh?! tanh y sech y ‘/ 1 — sech2y a: v 1 — x2 y d (e) Let y : coth‘1 3:. Then cothy = :5 => — csch2 y 8% : 1 => dy _ 1 1 _ 1 by Exercise 13. dm _ cschzy : 1~coth2y _ 1—132 30. f(:c) : tanh4a: => f'(ac) = Alsech2 4x 31. f(zc) : xcosha: => f'(:c) : m (cosh m)’ + (cosh = wsinhx + coshsc SECTION3.9 HYPERBOLICFUNCTIONS 3 219 32. 9(36) : sinh2 :0 => 9’ : 2 sinh a: cosh x 33. h(:c) = sinh(w2) => h'(:c) : cosh(ac2) ‘ 2m : 2w cosh(\$2) 34. : sinh x tanh a: => F'(a:) = sinhsc sech2 as + tanh x cosh a: _ 1— coshsc _ 1+ coshw 0,66) 2 (1+ cosh 2:) (— sinh :13) — (1 ; cosh w) (sinh x) (1 +coshzz) 35. G(m) _ ~sinhac ~sinhmcosh\$ — sinhm—i—sinhmcoshx —2 sinhm (1 + cosh 202 (1+ cosh an)2 36. f(t) : etsecht => f’(t) 2 et(— sechttanh t) + (secht) at = etsecht(1 — tanh t) _ tcsch2 «1 + t2 31m = tin/H152 => h’t :— h2x/1+t2-11+t2 1/2 2t :—E 0 co (> 2( ) < ) m 38. f(t) = ln(sinh t) => f’(t) : cosht : cotht sinh t 39. H(t) = tanh(et) :> H'(t) : sech2(e‘) -e’ = etsech2(et) 40. y = sinh(cosh w) :> y' = cosh(c0sh 2v) - sinhm 41. y 2 ewsh3m => 3/ : ewSW ~sinh 3x - 3 = 3eC°Sh 31' sinh 3:1: 1 W ' 2 + sinh—1(2cc) ~ 211: Z 215 + sinh—1(2x)] 42. y = x2 sinh—1(2m) :> y' = 3:2- , 1 1 43.y=tanh 1 ac 2 ’zﬁim-lﬂ: f y 1—<¢E>2 2 win—m) 44. y 2 xtanhilx+ln\/1—x2 = mtanh_1x+ %1n(1—\$2) :> _ 1 1 y’:tanh 130+ x +—( >(—2\$):tanh‘1w 45. y = xsinh_1(\$/3) — 9+ac2 :> , . _1 x 1/3 2:1: . _ m as a: x y =smh — +\$\—%:smh 1 _ + _ I. _1(_) (3) ./1+(x/3)2 2x/9+2:2 (3) t/9+' 232 x/9+x2 smh 3 e 1 —2x :3 46.y=sech lx/l—xz’ => 3/27 x:‘ x/l—x2\/1—(1—\$2)2m (l—x2)fxl — 1 2:1: 1 47. y=coth lx/ccz—f—l —> ’— ¥ﬁ y 1‘(372+1)2v.z2+1 x\/:1:2+1 48. 20 a=12 3 4 For y = a cosh(m/a) with a > 0. we have the y—intercept equal to a. As a increases. the graph ﬂattens. _10 ‘ur 0 10 49. (a) y = 20 cosh(a:/20) — 15 => 3/ = 20 sinh(ac/20) - 2—10 2 sinh(m/20). Since the right pole is positioned at :v = 7, we have y’(7) = sinh 2—70 z 0.3572. 220 C CHAPTER3 DIFFERENTIATION RULES (b) If a is the angle between the tangent line and the x—axis. then tan a = slope of the line = sinh 2—70 , so a = tan—1(sinh 2—70) m 0.343 rad x 19.660. Thus. the angle between the line and the pole is 6 = 900 — a x 70.349 50. We differentiate the function twice, then substitute into the differential equation: y = E cosh ELLE => 09 T dy T . p996 pg p927 dzy m w _:_ h(__)_:»h__ __ (PQ)PQZP9 P9 dm pg sm T 8111 T —> (1932 cosh T T T cosh T . 2 We evaluate the two sides separately: LHS : (1—?! = @ cosh ﬂ, dun? T T 9 2 pg dy pg . 2 pgw p pgw . . . RHS : — 1 — : _ I1 __ 2 _ __ T + ( dm) T —I— smh T T cosh T , by the identity proved in Example 1(a). 51. (a) y : Asinh mm + Bcosh mm => y' = mA coshmsc + m3 sinh mac => y" : mzAsinh mm + szcoshmw = m2(Asinh ms: + Bcosh mm) = m2y (b) From part (a), a solution of y” : 9y is y(\$) = Asinh 3m + B cosh 336. So —4 : y(0) = AsinhO + BcoshO = B, so B : '4. Now y'(m) = 3Acosh 3x — 12 sinh 356 => 6=y'(0)=3A : A:2.soy:2sinh3x—4cosh3w. sinhx emﬂe’x 1—6—2“c 1—0 1 52.1. :1. ————-—:l' ———:———:— 53. The tangent to y : cosha: has slope 1 when y’ = sinhm = 1 :> as = sinh—1 1 = ln(1 + by Equation 3. Since sinhx : 1 and y = coshx = V 1 + sinh2 m, we have coshsc 2: The point is (ln(1 + , 54. coshac : cosh[1n(sec 9 + 1221116)] 2 % [emsec 9H“ 0) + 6— Ms“ 9H“ 9)] = %[sec6+tan0+ : \$[sec6+tan0+ sec0 — tan6 seCG + tanG (sec 6 + tan 6) (sec 9 — tan 6) sec (9 — tan 0 sec2 0 - tan2 6 1 =§[se06+tan0+ ]:%(sec9+tan6+sec0—tan6):sec0 55. If aez + be” 2 a cosh(\$ + ﬂ) [or a sinh(x + m], then aeI + be"c : g (624-6 :I: 65373) : % (emeﬂ :I: (fie—3) = (geek:I :I: (ge’ﬁk—m. Comparing coefﬁcients of e” and 6”, we have a : 92—65 (1) and b = :Ege"3 (2). We need to ﬁnd a and ﬂ. Dividing equation (1) by equation (2) gives us 5; = :l:e2ﬂ => 2B 2 ln(i%) : B 2 % ln(:l:%). SOIVing equations (1) and (2) foreﬁ gives useﬂ : 29 and (3‘3 = \$2.50 2—9 = i9- :> a2 = :I:4ab => (1 : 2\/:I:ab. (1 2b oz 2b (*) If g > 0, we use the + sign and obtain a cosh function, whereas if g < 0, we use the — sign and obtain a sinh function. In summary. if a and b have the same sign, we have ae“c + be’3E : 2v ab cosh (cc + % ln 96) , whereas, if a and b have the opposite sign, then aem + be” : 2V —ab sinh (x + % ln(#%)). 3.10 Related Rates dV dV dav dm _ 3 —— : —— z 2 __ LVTSC :> dt dzc dt 356 dt dA dAdr dr dA d7" 2 —’ ‘——‘ - —— = —— 2 1 z 60 dt _ dr dt 27”" dt (‘3) dt 27”“ dt 27430 m)( m/S) wm/s 2. (a) A — 71'7'2 —> ...
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## This note was uploaded on 12/08/2009 for the course MATH 101 taught by Professor Dr.tahir during the Fall '08 term at King Fahd University of Petroleum & Minerals.

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9 - SECTION3.9 HYPERBOLICFUNCTIONS 3 215 3.9 Hyperbolic...

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