Calculus 2 Cheat Sheet Summary for Exams

NOTES  Rejectin Rejecting g a value for quadra quadratics: tics: as it is not in the range of __  Integratio Integration n of fractions: partial fractions or completing square then integration by substitution or just integration by substitution 1  When lim 2 , it becomes unbounded as x n, hence 𝑥→𝑛 (𝑥−𝑛) cannot be arbitrarily close to any real number as xn, therefore limit does not exist.  Derive of cosx = -sinx, coshx = sinhx LIMIT LAWS 1. lim[𝑓 (𝑥) + 𝑔(𝑥 )] = lim 𝑓(𝑥) + lim 𝑔(𝑥) 2. 3. 𝑥→𝑎 𝑥→𝑎 𝑥→𝑎 lim [ 𝑓(𝑥) 4. 𝑥→𝑎 𝑔(𝑥) 5. 𝑥→𝑎 ]= lim 𝑐 = 𝑐 𝑥→𝑎 lim 𝑓(𝑥) 𝑥→𝑎 lim 𝑔(𝑥) 𝑥→𝑎 𝑥→𝑎    provided that lim 𝑔(𝑥) ≠ 0 𝑥→𝑎 6. lim 𝑥 = 𝑎 𝑥→𝑎 SANDWICH TTHEOREM HEOREM Now lim −𝑥 = lim 𝑥= n by standard limits 𝑥→∞ 𝑥→∞ Hence by the sandwich rule, lim 𝑓(𝑥)= n 𝑥→∞ ∞ 𝑎𝑟 𝑛 ∑𝑛=0 GEOMETRIC SE SERIES: RIES: 1 1−𝑟𝑛 SN=a( 1−𝑟 ) S ∞=    𝑎 1−𝑟 HARMONIC P-SERIES P-SERIES:: ∑ ∞ 𝑛=1 𝑛𝑝 , p>1 converges, p<1 diverges TESTS  Divergence Divergence: lim 𝑎𝑛 ≠0, diverges  𝑛→∞ 4𝑛3 −2 ≈ = 1 4𝑛2 4𝑛 −2 4𝑛 −2𝑛 𝑤     5. 𝑤 6. 6𝑛+1 𝑛! 6𝑛 = √𝑥 2  𝑥→𝑛 For lim 𝑓(𝑥) to exist, we require 𝑥→𝑛 lim 𝑓(𝑥) = lim 𝑓(𝑥) 𝑥→𝑛− 𝑥→𝑛+  6 𝑛+1 CONTINUITY LLAWS AWS  Theorem 2 Combination of continuous functions  Theorem 3 nth root functions, trig (state the domain, and if it is being derived for a definite limit, state that continuity at x = n). poly (anything that is not just x), ex (x ∈ 𝑅 ), log(x) (x>0)  Definition o off continuity lim 𝑓(𝑥)= f(n) When c= a, f(n)= value = lim 𝑓(𝑥), so f 𝑥→𝑛 continuous at x = n Hence f continuous on (-∞, n]U[n, ∞) 1 , √𝑎2 −𝑥2  FIRST ODE C CONTINUED ONTINUED  Mixing pro problems blems 1. Find volume V(t) = initial amount of liquid + volume added per hour x t – volume removed per hour x t 2. Find conc. 𝑥 C= 3. Find 𝑑𝑥 𝑉 𝒅𝒙 𝒅𝒕 = rate thing added in – rate thing removed 𝑑𝑡 = flow rate in x conc (from Q) – flow rate out x conc (from prev step) − 3 3 1 , (𝑎 2 + 𝑥 2 )− 2 , etc. √𝑎2 +𝑥2 5 1 , (𝑥 2 − 𝑎2 ) 2 , etc. √𝑥 2 −𝑎2 1 1 , 2 2 2, etc. 2 2 𝑎 +𝑥 (𝑎 +𝑥 ) 𝑥 2 +1 = x – arctan(x) + C Integratin Integrating g ex y ∫ 𝑒 3𝑥 sin(2𝑥 ) 𝑑𝑥 I = 𝑒 3𝑥 sin(2𝑥 ) 𝑑𝑢 Let u = e3x 𝑑𝑥 𝑑𝑣 1 1+𝑥 2 −𝑒 3𝑥 2 x = acoshθ If ½ < y < 3, then 1 1+𝑥 2 dx  = 3e3x 𝑑𝑣 𝑑𝑥 3 Substitute into the first equation, I= I= cos(2x) + 2 −2𝑒 3𝑥 13 3 3x (e ½ 2 3 3x 2 3 2 𝑑𝑦 > 𝑑𝑡 0, so solutions which start with ½ < y(0) 𝑎 𝑎 𝑘 𝑝  𝑑𝑝 𝑑𝑡 = rate a = carrying capacity p = population Logistic mo model del with harves harvesting ting 𝑑𝑝 = kp - p2 = kp(1- )-h 𝑎 (2𝑥)𝑒 3𝑥 dx ∫ sin⁡ sin(2x) - 𝐼) cos(2x) + e sin(2x) + C 13 0, so solutions which start with ½ < y(0) 𝑑𝑡 𝑑𝑡 𝑑𝑝 = 3e3x 𝑑𝑥 (2𝑥)3𝑒 3𝑥 dx = e3x½ sin(2x) ∫ cos⁡ −𝑒 3𝑥 o 𝑑𝑢 v = ½ sin(2x) =0 𝑑𝑦 > 𝑑𝑡 Logistic mo model del 𝑑𝑝 𝑘 𝑝 = kp - p2 = kp(1- ) 𝑑𝑡 3 = cos(2x) 𝑑𝑥 < 3 increase and hence move further from y = 1 2 as t increases. 𝑑𝑦 If y < ½ , then < 0, so solutions which start with y(0) < ½ 𝑑𝑡 decrease and hence move further from y = ½ as t increase. So y = ½ is an unstable equilibrium. Population m models odels o Malthus’s Doomsday mod model el p(t) = p0ekt p0 (initial) k= constant of proportionality (net births/time) o Doomsda Doomsdayy model with harv harvesting esting 𝑑𝑝 = kp-h h>0 o cos(2x) + 2 ∫ cos⁡ (2𝑥)3𝑒 3𝑥 dx Let u = e3x 𝑑𝑝 Family of solutio solutions ns (p,t): lines curve in, p(0)=x Stability In terms of whether or not it moves closer/further from the point as t inc. Unrealistic: k>0 (unbounded exponential), k<0 (dies), k=0 (constant) Example of explanations: y = 3: < 3 increase and hence move closer to y = 3 as t increases. So y = 3 is a stable equilibrium. y=½ = sin(2x) v = -½ cos(2x) 𝑑𝑥 By int. by parts, I = e3x(-½cos(2x))+½ ∫ cos⁡ (2𝑥)3𝑒 3𝑥 dx I= Graphs 𝒅𝒑 Phase plot ( 𝒅𝒕 ,p): looks like a curve, gen solns o If ½ < y < 3, then x = atanθ dx = ∫ 1 − 𝑥 +P(x)y = Q(x)yn, substituting u=y1-n to make linear 𝑑𝑥 Substitution x = asinθ x = acosθ x = asinhθ Long divis division ion Numerator/denominator = result of fraction + remainder/denominator Integratin Integrating g quadratics 𝑥2 𝑥 Bernoulli’s eeqn qn 𝑑𝑦 𝑑𝑦 𝑎2 , ∫ 1+𝑥 2 dx = ∫ 1+𝑥 2 − 𝑑𝑥 2 If y > 3, then < 0, so solutions which start with y(0) > 3 𝑑𝑡 decrease and hence move closer to y = 3 as t inc. (𝑎 2 − 𝑥 2 ) 2 , etc. √𝑎2 + 𝑥2 , 3 2𝑛2 of form ∑∞𝑛=1𝑎𝑛 , where an>0 = (𝑛+1)! X 𝑥−𝑏 Integrand √𝑎2 − 𝑥2 , 𝑛→∞ 𝑎𝑛 𝑎𝑛 + P(x)y= Q(x) Homogeneous type ODE 𝑦 𝑦 =f( ), substituting u= to make a separable 𝑑𝑦 o o 𝐵 𝐶 = + + (𝑥−𝑎)(𝑥−𝑏)(𝑥−𝑐) 𝑥−𝑎 𝑥−𝑏 𝑥−𝑐 𝑝𝑥 2 +𝑞𝑥+𝑟 𝐴 𝐵 𝐶 = + + (𝑥−𝑎)2 (𝑥−𝑏) 𝑥−𝑎 (𝑥−𝑎)2 𝑥−𝑏 𝑝𝑥 2 +𝑞𝑥+𝑟 𝐴 𝐵𝑥+𝐶 = + (𝑥−𝑎)(𝑥 2 +𝑏𝑥+𝑐) 𝑥−𝑎 𝑥 2 +𝑏𝑥+𝑐 𝐴+𝐵𝑥 𝐶+𝐷𝑥 𝑞𝑥+𝑟 = 2 + 2 2 2 (𝑥 −𝑎)(𝑥 −𝑏𝑥) 𝑥 −𝑎 𝑥 −𝑏 4. arg(zw) = arg(z) + arg(w) 𝑧 arg( ) = arg(z) – arg(w) 𝐵 𝑝𝑥+𝑞 𝐴 𝐵 = + (𝑥−𝑎)2 𝑥−𝑎 (𝑥−𝑎)2 𝑝𝑥 2 +𝑞𝑥+𝑟 𝐴 3. Example: 6𝑛 is 𝑛! 𝑎𝑛+1 Partial frac fractions tions 𝐴 𝑝𝑥+𝑞 = + 1. (𝑥−𝑎)(𝑥−𝑏) 𝑥−𝑎 2. |𝑤| 4. Check (3) conv/div by harmonic p-series 5. Therefore f(n) conv/div by comparison test Ratio (has to be f(n)>0) 𝑎 L = lim 𝑛+1 , L<1 conv, L>1 div, L=1 inconclusive Linear 𝑑𝑦 𝐼(𝑥)= 𝑒 ∫ 𝑃𝑑𝑥 𝑢 E.g: differential of cosh(u) is double angle of cosh( ) √2 √2 De Moivre’s Theorem: zn=(reiθ)n= rneinθ Modulus Modulus:: |z| = √𝑥 2 + 𝑦2 Argument: Arg(z) Properties of Modulus and Argument 1. |zw| = |z||w| (for Im numbers, don’t square i) 𝑧 | | = |𝑧| 2. 𝑑𝑥 1 𝑑𝑦 =M(x) 𝑁(𝑦) 𝑑𝑥 1 ∫ 𝑁(𝑦)dy = ∫ 𝑀(𝑥)dx 𝑑𝑥 (conv.) For all n≥1 𝑛+2 𝑛+2𝑛 ≤ 3 3= 3 3.  𝑛 4𝑛3  multiply with conjugate to find Cartesian form Finding Re: only use the real values Finding Im: only use the imaginary values Complex num number: ber: z=r(cosθ + isinθ) Principle aargument: rgument: -π < θ < π Polar form: z = reiθ Polar to Cartesian: 1 r( - 1 i) 3. 4. Compariso Comparison n 1. an ≤ bn, converges 2. an ≥ bn, diverges Process: 1. Check if f(n)>0 (positive) 2. Find for large n + expectation (div/conv) 𝑛+2 FIRST ODE If the arbitrary constant is c ∈ R, the solution will give a more general solution than when A = ±ed ∈ R\{0}  Separable 𝑑𝑦 = M(x)N(y) (M(x)≠0, N(y)≠0) 𝑐+𝑖𝑑 lim[𝑓 (𝑥 )𝑔(𝑥)] = lim 𝑓(𝑥) ∙ lim 𝑔(𝑥 ) 𝑥→𝑎 INTEGRALS  Integratio Integration n of square roots Example: ∫ √𝑥 2 − 4 dx 𝑎+𝑖𝑏 𝑥→𝑎 lim[𝑐𝑓(𝑥)] = 𝑐 lim 𝑓(𝑥) 𝑥→𝑎 COMPLEX NUMBERS  Cartesian form: z= x +iy  Complex co conjugate: njugate: z= x – iy (Im changes sign) Properties: 1. z + z real 2. z – z imaginary 3. zz real 4. z+w=z+w 5. zw = zw  Division o off complex num numbers bers 𝑎 = rate a = carrying capacity p = population 𝑑𝑡 h = harvesting rate Example: Finding the value of h for a sustainable ecosystem Transient: decaying to 0 as t∞ St Steady eady state: constants (state that as lim 𝑥 = 0, thus at t∞ it will be at a fixed value) 𝑡→∞ Even: f(-x) = f(x) O dd: f(-x) = -f(x) N either: both ≠, prove by comparing points Domain ℝ ℝ 𝜋 ℝ \ { ±nπ} 2 ℝ \ {nπ} 𝜋 ℝ \ {2 ±nπ} cot(x) sec(x) ℝ \ {nπ} csc(x) arcsin(x) [-1,1] Range [-1,1] [-1,1] ℝ Odd/even Odd Even Odd (-∞, -1]U[1,∞) (-∞, -1]U[1,∞) Odd Even (-∞, -1]U[1,∞) 𝜋 𝜋 [- 2, ,] Odd Odd 2 arcos(x) arctan(x) [-1,1] ℝ [0,π] 𝜋 𝜋 (- 2, ,) arccot(x) arcsec(x) (-∞, -1]U[1,∞) [0, arccsc(x) (-∞, -1]U[1,∞) [- ℝ (0,π) ℝ ℝ ℝ sinh(x) cosh(x) tanh(x) coth(x) sech(x) csch(x) arcsinh(x) arccosh(x) arctanh(x) arccoth(x) arcsech(x) arccsch(x) Even Odd 2 Odd Even 𝜋 𝜋 )U( ,π] 2 2 𝜋 𝜋 ,0)U(0, ] 2 2 Odd ℝ Odd Even Odd Odd Even Odd Odd Neither Odd Odd Neither Odd [1,∞) (-1,1) (-∞,-1)U(1, ∞) (0,1] (-∞,0)U(0, ∞) ℝ (-∞,0)U(0, ∞) ℝ (-∞,0)U(0, ∞) ℝ [1,∞) (-1,1) (-∞, -1)U(1,∞) (0,1] (-∞,0)U(0, ∞) [0, ∞) ℝ (-∞,0)U(0, ∞) [0, ∞) (-∞,0)U(0, ∞) Radians sin cos tan csc sec cot 0 0 1 0 - 1 - 1 2 √2 √3 √3 2 2 √3 2 √2 √3 𝜋 6 𝜋 4 𝜋 3 𝜋 2 𝜋 2 √3 2 1 0 2 √2 2 1 √2 2 1 2 0 √3 - 2 √3 2 1 -1 0 - √3 3 0 -1 - NOTES  ALWAYS ST STATE ATE DOMAIN A AND ND RANGE WH WHEN EN USING A TRIG FU FUNCTION NCTION  Domains Example: g(x) = sin√𝑥 2 − 9 We require x2-9≥0 for this function to be defined. That is we require x≥3 or x≤-`3. Hence domain= (-∞, -3]U[3, ∞) Two exceptio exceptions: ns: {x|x ≠y,z} one exce exception: ption: [x,∞)  Fub ubini ini ini’’s theorem F: R2R be a continuous function over domain R=[a,b]x[c,d], where order of double integration is not important FUNCTIONS O OFF TWO VARIA VARIABLES BLES SECOND O ODE DE  Shapes ((h,k) h,k) center  Cases (𝑦−𝑘)2 (𝑥−ℎ)2 1 o b2 – 4ac > 0 (2 disti distinct) nct) Ellipses: 𝑎2 + 2 =1, where rx= √ o , same for y 𝑏 𝑥⁡ 𝑣𝑎𝑙𝑢𝑒 y(x) = Aeλ1x+Beλ2x 2 o Circle: (x-h) +(y-k)2=r2 Parabola: (y-k)2=4p(x-h) o b2 – 4ac = 0 (1 distinc distinct) t) OR (x-h)2=4p(y-k) Straight lines lines:: mx+c y(x) = Aeλx+Bxeλx  Drawing o b2 – 4ac < 0 (none (none)) o Level curv curves es αx αx αx y(x) = Ae cos(βx)+Bxe sin(βx) λ= α±iβ  e cos/sin(βx) Solve z = c, where (domain of c)  Particular ssolutions olutions e.g. GS(H) = yH(x) + yp(x) z = √1 − 𝑥 2 − 𝑦 2 | Let z = c, c∈R g(t) yp(t) guess So c = √1 − 𝑥 2 − 𝑦 2, c≥ 0 | c 2 = 1 – x2 – y 2 aeβt Aeβt x2+y2 = 1 – c2 | Note that 1-c2 ≥0 a A | So possible values of c∈ [0,1] c2≤1  |𝑐| ≤1 acos(βt) Acos(βt) + Bsin(βt) o Cross sections: find x-z by y=0, find y-z by x=0 bsin(βt) Acos(βt) + Bsin(βt) (or add t) o Intercepts ((x,y,z) x,y,z) acos(βt) + bsin(βt) Acos(βt) + Bsin(βt)  First order partial derivati derivatives ves ves: measures rate of change of one nth degree Antn + An-1tn-1 + …. + A1t + A0 𝜕𝑓 𝜕𝑓 variable while keeping the other constant (𝜕𝑥=fx, =fy) 𝜕𝑦 polynomial e.g. 2x  yp(x)= ax + b  Cartesian eeqn qn of tangent plan plane e 𝜕𝑓 𝜕𝑓 z-z0= |(x0,y0)(x-x0)+⁡ |(x0,y0)(y-y0), find fx and fy at (x,y) then sub 𝜕𝑥 𝜕𝑦 final eqn should be in the form of x + y + z = 0 o  1 - 2 Just find where the peak is (so in sech(x), the x has to be 0) Function sin(x) cos(x) tan(x)  At equilibrium T = ks, and acc is 0 so by Newton’s law: Net force = mass x acceleration mg – Ts = 0 mg – ks = 0  k = mg/s mg + F - R - T = my R=-βy (β=damping constant≥0) mg + F – βy – k(s+y) = my substitute mg=ks Upward = ((-), -), downwar downward d = (+) Damping o β = 0: λ = ±𝑖𝑏  simple harmonic o 0< β <2√𝑚𝑘 : λ = 𝑎 ± 𝑖𝑏  underdamped o β = 2√𝑚𝑘 0: λ = a,a  critically damped o β > 2√𝑚𝑘 0: λ = a,b  overdamped      𝜕𝑓 |(x0,y0)(x-x0)=0 𝜕𝑥/𝑦 Linear ap approx. prox. f(x,y)≈ f(x0,y0)+tangent plane  just input the new (x,y) values into existing tangent plane, unless you must estimate based on a specific linear approx. (find linear approx, follow equation) 𝜕𝑓 𝜕𝑓 Δf≈ 𝜕𝑥|(x0,y0)Δx+⁡ |(x0,y0)Δy  input (x,y) values ∙ change 𝜕𝑦 Second or order der partial deri derivatives vatives (fxy = fyx) 𝜕2 𝑓 𝜕 𝜕𝑓 𝜕 𝜕𝑓 fxy = ( ) = fxx = ( ) = 2 fyx =  Horizontal: if z constant, which is only if 𝜕𝑥 𝜕 𝜕𝑦 𝜕𝑥 𝜕𝑓 ( 𝜕𝑥) = 𝑑𝑧 𝜕𝑥 𝜕2 𝑓 𝜕𝑦𝜕𝑥 𝑑𝑥 𝜕𝑥 𝑑𝑡 Chain rul rule: e: =𝜕𝑧 𝑑𝑥 + 𝜕𝑧 𝑑𝑦 𝜕𝑦 𝑑𝑡 fyy = 𝜕𝑥 𝜕𝑦 𝜕 𝜕𝑓 𝜕𝑦 ( 𝜕𝑦) = 𝜕2 𝑓 𝜕𝑥𝜕𝑦 𝜕2 𝑓 𝜕𝑦 2 𝜕𝑓 Directional directive Dûf|P0= ( |P0, 𝜕𝑓 |P0)∙(u1,u2), where u=(x,y) 𝜕𝑦 𝜕𝑥 OR Dûf|P0=∇f|P0⁡ ∙⁡ û OR |∇f|cosθ (don’t use this one often) 𝜕𝑓 𝜕𝑓 Gradient vvectors ectors ∇f =( ,⁡ ) 𝜕𝑥 𝜕𝑦 Properties o off 𝛁𝒇 and Dûf  for fixed ∇f, Dûf max when cosθ=1 so inc rapidly, and min when cosθ=0 and Dûf=0 so dec rapidly as ∇f ⊥ û (level curves of f), just multiply the ∇f with -1 to find dec. Stationary points: ∇f = 0, tangent plane parallel/horizontal o Local max: f(x,y)≤f(x0y0) o Local min min:: f(x,y)≥f(x0,y0) o Saddle poin point: t: f(x,y) >/< f(x0,y0) 𝜕𝑓 𝜕𝑓 Critical poin point: t: ∇f(x0,y0)=0 (differentiable) or and/or o 𝜕𝑥    RLC series electric circuit circuit: 𝑑2 𝑞 𝑑𝑞 𝑞 L 2 +R + =V, 𝑑𝑡 𝑑𝑡 𝐶 where V=0 in closed loop  Physical in interpretation terpretation of cu current rrent as tt ∞ Current oscillates between ±𝑡 Amps (taken from i(t) eqn). That is,  there is unbounded oscillation leading to resonance Resonance: when external F has the same form as one of the terms in 1 the GS(H) e.g. 50tsin(50t) (from i(t) eqn) = sin(50t) 50 𝜕𝑦 does not exist  z=√𝑥 2 + 𝑦 2 so min at (0,0) but ∇f does not exist at (0,0) o Second de derivative rivative test (Hessian function) 𝑓𝑥𝑥 𝑓𝑥𝑦 ] H(x,y) = fxxfyy – (fxy)2 = det[ 𝑓 𝑦𝑥 𝑓𝑦𝑦 1. Local min H(x0,y0)>0 and fxx(x0,y0)>0 2. Local max H(x0,y0)>0 and fxx(x0,y0)<0 3. Saddle poin pointt H(x0,y0)<0 Partial inte integration: gration: same principle as derivatives, but if holding y constant, + C(y), and vice versa Double int integrals egrals (for f: RR2 co continuous ntinuous over D iin n R2 ) Domain R = [a,b]x[c,d] rectangular domain for a ≤ 𝑥 ≤ b, 𝑐 ≤ 𝑦 ≤d, to find volume, if ∫ ∫ 1 𝑑𝑥 𝑑𝑦= area of domain ∬ 𝑓(𝑥, 𝑦)𝑑𝑥 𝑑𝑦= volume