Chapter 3 54

# Chapter 3 54 - 204 52 53 54 55 56 57 58 59.f 60 61 CHAPTER...

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Unformatted text preview: 204 52. 53. 54. 55. 56. 57. 58. 59.f 60. 61. CHAPTER 3 DIFFERENTIATION RULES d—[email protected]§—@(t)——(t)d—”Thd t dd h h dt (15 dt — dsv —’Ut d5 e erivaive v/ tist erateofc angeofthe velocity with respect to time (in other words. the acceleration) whereas the derivative d1) / ds is the rate of change of By the Chain Rule a(t): the velocity with respect to the displacement. Let P03) 2 (13:2 + bx + G. Then P’(1:)= 2m: + b and P”(\$) = 2a. P”(2) = 2 : 2a 2 2 => (1 = 1. P'(2) =3 2 2(1)(2)+b=3 :> 4+b=3 2 b: —1. P(2) :5 => 1(2)2+(—1)(2)+c=5 => 2+c=5 => c=3.SoP(m):a:2—:c+3. Let Q(;c) = (1:123 + bx2 + car + d. Then Q/(m) = 3azt2 + 2ba: + c. Q"(:E) 2 Guns + 2b and Q/”(cc) : 6a. Thus, Q(1) :a+b+c+d: 1,Q' (1) =3a+2b+c:3. Q"(1) =6a+2b:6andQ'"(1) = 6a 2 12. Solving these four equations in four unknowns a, b. c and d we get a = 2. b : —3, c : 3 and d = —1, so Q(:::) = 2x3 — 3x2 + 3x 4 1. y = Asinx + B cosm :> y' : Acosm — B sin\$ :> y” : AAsinzc — Bcos 1;. Substituting into y” + y' — 2y : sinac gives us (73A — B) Sinm + (A — 3B) cosm = 1 sin :13. so we must have —3A — B : 1 and A * 3B : 0. Solving for A and B , we add the ﬁrst equation to three times the second to get B = — 1—10 and A = #%. y = A9102 + Bcc + C => 3/ : 2A9: + B => 3;" : 2A. We substitute these expressions into the equation y" + yI # 2y : \$2 to get (2A) + (2Acc + B) — 2(Aa:2 -I- Bic + C) : m2 2A+2Am+Ba2Arc2 —2Bm—2C’=a:2 (—2A)ac2 + (2A # 2B):c + (2A + B — 2C) : (1)32 + (on + (0) Thecoefﬁcients ofac2 oneach side mustbeequal. so 72A=1 => A=*%.Similarly.2A;2B=0 :> A:B:—§and2A+B—2C_0 4 —1 20—0 —> C——§. NIH y i e” —> y' — re” —,> y" # 7'26”, so y" + 51/ ~ 614- — r26” + 5%” — 68” = e” (r2 + 57‘ — 6) : e”(r + 6)(r — 1): 0 => (r+6)(r—1):O => rzlor—S. y # 6A2: _> y! _ A6)“ _> yll —6)\2 Ax. Thus,y+y' : yII <2} 6A3: + Aekm : AQGAz <:> (MEW- —1)=0 c» A: —:_——51‘/-sincee’\m7é0. ﬂat): 069(962) => f(\$)—\$ g’w( 2)296+9(m 2) 1:9(96 2)+29I:2g'(2102) => f”(w) : g'\$( 2)-2\$+2m2 -2-g”(:1:) 2m+g'2~(m) 4w26mg (m2 )+4x3g "(9:2) (93) I ; wg’(\$)-g(\$) mix— => f(:c)— \$2 u \$2I9’(x) + a39”(06) - 9'(\$)I ' 29: [\$9’(w) - 9(IB)I_ \$2 9"(\$) - 2909 (w) + 29(96 ) f (at) ' \$4 \$3 f(\$)=g(\/E) => f’(x):g’(\/E).%\$71/2_9'2(£) :> u 2x/E-g”(x/E)~émrlﬂ—g’k/ﬂ-Z-gw‘m m’1/2[\/Eg"(\/§)7g'(\/E)] _Ml‘_9/_([email protected] _ 4m\/:E :> ...
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