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Chapter 3 70

# Chapter 3 70 - 220 C CHAPTEH3 DIFFERENTIATION RULES(b If a...

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Unformatted text preview: 220 C CHAPTEH3 DIFFERENTIATION RULES (b) If a is the angle between the tangent line and the :1: axis then tana —— slope of the line: sinh7 55. so a : tan‘1(sinh 56) z 0.343 rad~ ~ 19.66°. Thus. the angle between the line and the pole 1s 0 : 90o — a m 7034". 50. We differentiate the function twice. then substitute into the differential equation: y = z cosh ﬂ => pg T dy T paw pg . pgm 6122/ 19996 pg a: ——=—-sinh(—— )—= 11— :> ——= (—)—— ”—9 p9 dm pg T T s1n T dw2 cosh T T T cosh— T. 2 We evaluate the two sides separately: LHS— — %— — & cosh L555 RHS— — T 1 + (:1ch y) —pT—g1 I1 + sinh2—— pgm %——cosh pgzc ,by the identity proved 1n Example 1(a). 51. (a) y : Asinhmx + Bcoshmcc => y : mAcoshma: + mB sinhmm => y” : m2A sinh mm + m2B cosh ma: 2 m2 (A sinh mm + B cosh mm) = (b) From part (a), a solution of y” : 93/ is y(m) = Asinh 3x + B cosh 33:. So —4 : y(0) = AsinhO + BeoshO = B, so B : '4. Now y'(ac) = 3Acosh 3x — 12 sinh3a: :> 6 y'(0) 3A A 2. so y 2 sinh 3x — 4 cosh 3x. sinha: e“ # 67\$ 1 — e-2m 1 — 0 1 52.1' 21' —————:1' ———:———:_ 53. The tangent to y : cosha: has slope 1 when 3/ = sinhm = 1 => as : sinh—1 1 : ln(1 + ﬂ), by Equation 3. Since sinhx : 1 and y = coshm = V 1 —I— sinh2 m, we have coshzv : x/2— The point is (ln(1 + ﬂ) ,x/i). 54. coshm— — cosh[ln(sec 6 -I- tan 0)]: -[e Msec 9““ 9) +6 ”(sec “can 6)] -1 0+t 0+——i——— #1 sec6+tan0+———§E&-M—‘ _ 2 see an sect9+tan9 _ 2 (sec0+tan9)(sec0 —tan 9) — :[sec0+tan0+%] : é(sec0+tan9+sec9—tan0) :sec0 55. If (161 + be‘m = a cosh(:c + B) [or a sinh(m + [3)], then (16" —I— be” : m%(ez+ﬂ :E 647:) : %(ee m eﬂ :I: (fie—ﬂ) = (%eﬁ)el :I: (%e’ﬂ)e_1. Comparing coefﬁcients of e” and e we have a — —e’3 (1) and b— — 2E5 “e B (2). We need to ﬁnd oz and ﬂ. Dividing equation (1) by equation (2) gives us 2 = :lze2B => (*) 2ﬂ— 1n(;l:% ) => B : % ln(:t%). SOIVing equations (1) and (2) b 2 fore'B givesuseﬁzzsandeﬁzig—b so£=i% :—> a2=:l:4ab => oz:2\/::ab. (*) Ifa b > 0, we use the + sign and obtain a cosh function, whereas if % < 0, we use the — sign and obtain a sinh function. In summary. if a and b have the same sign, we have aem + be’aE : 2v ab cosh (m + % 1n 95), whereas, if a and b have the opposite sign, then (16“ + be” : 2\/ —ab sinh (x + % ln(# %)). 10 Related Rates dV dV d1: 2 d3: _ 3 d _ —— _ 1’V’a” :> dt dm dt 3“” dt dA dA (17' dr dA dr 2 7 2 —=-——: — —=2 —:2 30 1 =60ms 2. (a) A — 7r?" :> dt dr dt 27W dt (b) dt 71'1“ dt 7r( m)( m/s) 7r / ...
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