Parallelogram Triangle Parallelogram π΄πππ = πβ = ππ π πππ Trapezoid You should remember this… a π = 3.1415 π = 2.7182 2 = 1.414 3 = 1.732 h π 1 πππ = 1808 /π = 57. 38 b (π₯ ± π¦)? = π₯ ? ± 2π₯π¦ + π¦ ? π₯ ? − π¦ ? = (π₯ + π¦)(π₯ − π¦) Triangle 1 π΄πππ = πβ 2 h π! = 1 β 2 β 3 β βββ β π 0! = 1 b Circle Trapezoid 1 π΄πππ = π + π β 2 a Quadratic Equation ππ₯ ? + ππ₯ + π = 0 Sector h π₯E,? Sphere Circle π΄πππ = ππ ? πππππππ‘ππ = 2ππ b Sector Cylinder E π΄πππ = π ? π ? π = ππ π - in rad. r Sphere 4 ππππ’ππ = ππ O 3 Cone π΄πππ = 4ππ ? r Cone 1 ππππ’ππ = ππ ? β 3 π = ππ π ? + β? (without base) −π ± π ? − 4ππ = 2π S r π r Cylinder Pyramid ππππ’ππ = ππ ? β π = 2ππβ (without circles) r h Pyramid 1 ππππ’ππ = π΄β 3 h r h A 1 TRIGONOMETRIC FUNCTIONS tan πΌ = sin πΌ cos πΌ cot πΌ = cos πΌ sin πΌ 1 cos πΌ sec πΌ = π¦ = sin π₯ csc πΌ = 1 sin πΌ sin(−π₯) = − sin π₯ cos(−π₯ ) = cos π₯ x sin π₯ cos π₯ o 30 1/2 √3/2 o 45 √2/2 √2/2 60 1/2 √3/2 π¦ = cos π₯ tan(−π₯ ) = − tan π₯ cot(−π₯ ) = − cot π₯ π₯ tan π₯ cot π₯ o 30 √3/3 √3 o 45 1 1 60 √3 √3/3 π¦ = tan π₯ O sin π₯ = π₯ − π¦ = cot π₯ Y Z π₯ π₯ π₯ + − +β― 3! 5! 7! cos π₯ = 1 − π₯? π₯\ π₯] + − 2! 4! 6! sin α ± β = sin α cos β ± cos α sin β sin 2πΌ = 2 sin πΌ cos πΌ cos α ± β = cos α cos β β sin α sin β cos 2πΌ = cos ? πΌ − sin? πΌ Sines sin? α + cos ? α = 1 Law of Sines π π π = = sin πΌ sin π½ sin πΎ Law of Cosines π ? = π? + π ? − 2ππ cos πΎ b πΌ c π½ πΎ a complex Complex Numbers π = −1 π₯ + ππ¦ = π(cos π + π sin π) = ππ de π df = −1 (Euler’s Formula) r y π x De Moivre’s Theorem (m-real number) π cos π + π sin π g = π g cos ππ + π sin ππ 2 f (x) f (x) = f (x) = 5 x f (x) = 2 x 3. The relationship between exponential functions and logarithm functions EXPONENTIAL & LOGARITHMIC We FUNCTIONS can see the relationship between the exponential function f(x) = e x and the logarithm function f(x) = ln x by looking at their graphs. π₯ j π₯ k = π₯ jlk (π₯ j )k = π₯ jk (π₯π¦)j = π₯ j π¦ j q π₯ = π₯ E/r q q π₯/π¦ = π₯ / q π¦ mn = π₯ jpk π₯ = 1/π₯ j q π₯ g = π₯ g/r π₯ s = 1 (π₯ ≠ 0) mo pj ln π₯π¦ = ln π₯ + ln π¦ π₯ ln = ln π₯ − ln π¦ π¦ ln π₯ r = π ln π₯ π x f (x) = x f (x) ln x axis. In general, f(x) = (1/a)x = a−x is a reflection of f(x) = ax inf (x) the=f(x) x e A particularly important example off (x) an =exponential function arises when a = e. You might recall that the number e is approximately equal to 2.718. The function f(x) = ex is often called ‘the’ x exponential function. Since e > 1 and 1/e < 1, we can sketch the graphs of the exponential x −x x functions f(x) = e and f(x) = e = (1/e) . ±de = cos π ± π sin π π de − π pde sin π = 2π de π + π pde cos π = 2 f (x) You can see straight away that the logarithm function is a reflection of the exponential function in the line represented by f(x) = x. In other words, the axes have been swapped: x becomes f(x), and f(x) becomes x. x −x f (x) = e f (x) = e HYPERBOLIC Key Point FUNCTIONS x f(x) = ln x. The exponential function f(x) = ex is the inverse of the logarithm function π₯? π₯O π =1+π₯+ + +β― HYPERBOLIC FUNCTIONS 2! 3! HYPERBOLIC FUNCTIONS GRAPHS OF HYPERBOkfC m 29 FUNCltONS 29 Exercises 1. Sketch the graph of the function f(x) =4a www.mathcentre.ac.uk HYPERBOLIC FUNCTIONS x GRAPHS pm πm − π 2 sinh x sinh π₯ = 8.49 8.49 y = y = sinh x π +π 8.50 2 axes. tanh π₯ =(a) cosh π₯ = 8.50 y = coshx y = coshx 8.52 π¦ = sinh π₯ 1 77 10 8.53 -1 10 π¦ -1 = tanh π₯ X /i 8.51 X 7 1 10 0 sinh 8.54 π₯± π¦ = sinh π₯ cosh π¦ ± cosh π₯ sinh π¦ y = csch x cosh π₯ ± π¦ = cosh π₯ cosh π¦ ± sinh π₯ sinh π¦ 0 sinh y π₯ = Fig. 8-5 tanh x c mathcentre 2009 β Fig. 8-2 8.53 Fig. 8-3 y = sech x siny ππ₯ = π sinh π₯ 8.54 cos ππ₯ = cosh π₯ sech x y = csch 8.54 csch x tan ππ₯ = π tanh π₯y = cot ππ₯— π coth π₯ Y sinh ππ₯ = π sin π₯ cosh ππ₯ = cos π₯ Y X tanh ππ₯ = π tan π₯ coth ππ₯ = X −π cot π₯ \ 0 \ 0 X X 0 y = csch x 8.54 Fig. 8-3 Fig. 8-3 -1 Fig. 8-4 Fig. 8-4 Fig. 8-4Y 1 6 1 3 8.51 y = tanh x π¦ = coth π₯ Fig. 8-3 \ coth(d)π₯a == (c) a = Fig. 8-2 8.53 y = sech x y = y y 1 π¦y== cosh π₯ coth x Fig. 8-2 /i 1 sinh(−π₯) = − sinh π₯ cosh −π₯ = cosh π₯ www.mathcentre.ac.uk 9 tanh −π₯ = − tanh π₯ coth −π₯ = − coth π₯ 8.52 y x= coth x y = coth /i X a = 3 (b) a = 6 cosh π₯ Fig. S-l y = tanh x Fig. S-l Fig. S-l 8.52 1 (e) a = 6 3 GRAPHS HYPERBOkfC 8.51 8.49 y OF = sinh x 8.50FUNCltONS y = coshx y = tanh x OF HYPERBOkfC FUNCltONS 2. Sketch the graph of the function f(x) = log x for the following values of a, on the same a m pm sinh π₯ cosh π₯ FUNCltONS 8.51 for the following of a, on 2009 the same axes. c values β mathcentre (a) a = 3 (b) a = 6 (c) a = 1 (d) a = 29 x x () ( 21 ) x f (x) = NCTIONS C 1 x 5 L X0 x \ L 0 L X Fig. 8-6 Fig. 8-5 sinh 2π₯ = 2 sinh π₯ cosh π₯ cosh 2π₯ = cosh? π₯ + sinh? π₯ Fig. Fig. 8-5 Fig. 8-6 cosh? π₯ − sinh? π₯ = 1 FUNCTIONS iNVERSE HYPERROLIC Y 8-6 L If x = sinhHYPERROLIC g, then y = sinh-1 xFUNCTIONS is called the inverse hyperbolic sine of x. Similarly we iNVERSE iNVERSE FUNCTIONS X otherHYPERROLIC The inverse hyperbolic functions are multiple-valued and. inverse hyperbolic functions. 3 case of inverse trigonometric functions [sec page 171 we restrict ourselves to principal values they ean be considered as single-valued. If x = sinh g, then y = sinh-1 x the is called inverse sine hyperbolic sine of we x. define Similarly If x = sinh g, then y = sinh-1 x is called inverse the hyperbolic of x. Similarly the we define the TheThe following list shows the principal values [unless otherwise indicated] of the h inverse hyperbolic functions are multiple-valued as inverse in the other inverse hyperbolic functions. The inverse hyperbolic functions are multiple-valued and. as in the and. other inverse hyperbolic functions. Line ANALYTIC GEOMETRY line y π΄π₯ + π΅π¦ + πΆ = 0, π¦ = ππ₯ + π (m-slope, b- y intercept) π₯ π¦ + =1 π π Planeb x a Plane z π΄π₯ + π΅π¦ + πΆπ§ + π· = 0 Circle c π₯ π¦ π§ + + =1 π π π b a y x Circle y (π₯ − π₯8 )? + (π¦ − π¦8 )? = π ? R (π₯π , π¦π) Ellipse x Ellipse (π₯ − π₯8 )? π¦ − π¦8 + ? π π? y ? =1 2a ("# , %# ) 2b x Parabola y π¦ − π¦8 ? − 4π π₯ − π₯8 = 0 a Focus (π₯π , π¦π ) x Hyperbola (π₯ − π₯8 )? π¦ − π¦8 − ? π π? ? =1 (π₯8 , π¦8 ) – center 2a, 2b –major and minor axes 4 DERIVATIVES π¦=π π₯ ππ¦ π π₯ + βπ₯ − π π₯ = lim ππ₯ βm→s βπ₯ ; π¦ = π π₯, π¦ ππ π π₯ + βπ₯, π¦ − π(π₯, π¦) = lim ππ₯ βm→s βπ₯ ππ’ ππ£ π£ − π’( ) π π’ ππ₯ ππ₯ = ππ₯ π£ π£? π ππ£ ππ’ π’π£ = π’ +π£ ππ₯ ππ₯ ππ₯ ππ(π’) ππ ππ’ = (πβπππ ππ’ππ) ππ₯ ππ’ ππ₯ π sin π₯ = cos π₯ ππ₯ π cos π₯ = − sin π₯ ππ₯ π 1 tan π₯ = ππ₯ cos ? π₯ π 1 cot π₯ = − ? ππ₯ sin π₯ π sinh π₯ = cosh π₯ ππ₯ π cosh π₯ = sinh π₯ ππ₯ π 1 tanh π₯ = ππ₯ cosh? π₯ π 1 coth π₯ = − ππ₯ sinh? π₯ Taylor Series π π₯ = π π + π… π π₯ − π + π ππ₯ r = πππ₯ rpE ππ₯ π 1 ππ’ ln π’ = ππ₯ π’ ππ₯ π „ ππ’ π = π„ ππ₯ ππ₯ π „ ππ’ „ π = π ln π ππ₯ ππ₯ π′′(π)(π₯ − π)? π′′′(π)(π₯ − π)O + +β― 2! 3! INTEGRALS π’ π₯ π£′(π₯) ππ₯ = π’π£ − π£ π₯ π’… π₯ ππ₯ Integration by parts: π₯ rlE π+1 sin π₯ = − cos π₯ sinh π₯ = cosh π₯ π₯ r ππ₯ = cos π₯ = sin π₯ cosh π₯ = sinh π₯ ππ₯ = ln π₯ π₯ tan π₯ = −ln cos π’ tanh π₯ = ln cosh π₯ π m ππ₯ = π m cot π₯ = ln sin π₯ coth π₯ = ln sinh π₯ ln π₯ ππ₯ = π₯ ln π₯ − π₯ Mean Value Theorem k π π₯ ππ₯ = π − π π π πππ π < π < π j Leibnitz Rule π ππ₯ k(m) j(m) ππ ππ π( π₯, π‘)ππ‘ = π π₯, π − π π₯, π + ππ₯ ππ₯ k(m) j(m) ππ(π₯, π‘) ππ‘ ππ₯ 5 ORDINARY DIFFERENTIAL EQUATIONS Linear 1st Order ππ¦ +π π₯ π¦ =π π₯ ππ₯ πΌππ‘πππππ‘πππ πΉπππ‘ππ = π π π¦(π₯)π •Œm = π π₯ π ππ₯ π¦ π₯ = πp •Œm π π •Œm ππ₯ + π ‹Œm ‹Œm Bernoulli’s Equation ππ¦ + π π₯ π¦ = π (π₯)π¦ r ππ₯ π¦(π₯)Epr = (1 − π)π p(Epr) For n=1, ln π¦ = Linear, Homogeneous, 2nd Order π? π¦ ππ¦ +π + ππ¦ = 0 ? ππ₯ ππ₯ a, b – real constants •Œm π π (Epr) •Œm ππ₯ + π π − π ππ₯ + π πE πππ π? πππ ππππ‘π ππ π? + ππ + π = 0 when πE πππ π? πππ ππππ π¦ π₯ = πΆE π gΕ½ m + πΆ? π g• m when πE = π? = π πππ ππ ππππ π¦ π₯ = πΆE π gm + πΆ? π₯π gm when πE = π + ππ, π? = π − ππ π¦ π₯ = π ‹m πΆE sin ππ₯ + πΆ? cos ππ₯ Linear, nonHomogeneous, 2nd Order π? π¦ ππ¦ + π + ππ¦ = π (π₯) ππ₯ ? ππ₯ a,b – real constants Same cases for πE and π? as above π gΕ½ m π¦ π₯ = πΆE π gΕ½ m + πΆ? π g• m + πE − π? π¦ π₯ = πΆE π gm + πΆ? π₯π gm + π₯π gm π pgΕ½ m π ππ₯ + π pgm π ππ₯ − π gm π g• m π? − πE π₯π pgm π ππ₯ π¦ π₯ = π ‹m πΆE sin ππ₯ + πΆ? cos ππ₯ π ‹m sin ππ₯ π ‹m cos ππ₯ + π p‹m π cos ππ₯ ππ₯ − π π Euler Equation π? π¦ ππ¦ π₯ ? ? + ππ₯ + ππ¦ = π (π₯) ππ₯ ππ₯ π pg• m π ππ₯ π p‹m π sin ππ₯ ππ₯ Substitute π₯ = π ’ then the equation becomes π? π¦ ππ¦ + (π − 1) + ππ¦ = π (π ’ ) ? ππ‘ ππ‘ See solutions above for 2nd order equations. If π π₯ = 0 use π¦ π₯ = π₯ g and follow Homogeneous 2nd Order solution 6 VECTOR ANALYSIS VECTOR ANALYSIS VECTOR ANALYSIS DotProduct Product Dot Dot Product |&||'|cos cosπ + 0 0≤≤π +≤≤π / π!β πβ #==|π΄||π΅| ! β # = |&||'| cos + 10 ≤ + ≤ / ? 1++ ?1 ? π΄&== π΄m& 0 π΄& — 1++π΄& ˜3 1 1 & = &0 + &1 + &13 @ 7 @ 7 CrossProduct Product Cross Cross Product !×#==π΄&|π΅| |'|sin sinπ +π§7 0 0≤≤π +≤≤π / π×π !×# = & |'| sin + 7 0 ≤ + ≤ / π’ π£ π€ 8 8β ! = & ; + & ? + π×π = π΄0m π΄—1 π΄˜&3 = 8β = &0 ; + &1 ? + &3 = ! π΅ π΅ π΅ m — + + z z ! 8β = &: ; + &> ? + &< = 88β ! &< = 8!β &< = k j k y j & ? > i &: ; y &> ? i &: ; 88β # 8β # 88β ! 8!β x x π×π = −π×π π β π = π΄m π΅m + π΄— π΅— + π΄˜ π΅˜ ˜ Derivative ππΉ— ππ ππΉm ππΉ˜ = π’+ π£+ π€ ππ₯ ππ₯ ππ₯ ππ₯ Del Operator π π π ∇= π’ +π£ +π€ ππ₯ ππ¦ ππ§ Gradient of a scalar ππ ππ ππ ∇π = π’+ π£+ π€ ππ₯ ππ¦ ππ§ Divergent of a Vector π π π ππΉm ππΉ— ππΉ˜ ∇βπ = π’ +π£ +π€ β πΉm π’ + πΉ— π£ + πΉ˜ π€ = + + ππ₯ ππ¦ ππ§ ππ₯ ππ¦ ππ§ Curl of a Vector π’ π π π π ∇×π = π’ +π£ +π€ × πΉm π’ + πΉ— π£ + πΉ˜ π€ = ππ₯ ππ₯ ππ¦ ππ§ πΉm Laplacian ∇ β ∇π = ∇? π = π?π π?π π?π + + ππ₯ ? ππ¦ ? ππ§ ? ∇× ∇π = 0 π£ π ππ¦ πΉ— π?π π?π π?π ∇ π = ?+ ?+ ? ππ₯ ππ¦ ππ§ ? ∇ β ∇×π = 0 Μ π Divergence Theorem ∇ β π ππ = V π€ π ππ§ πΉ˜ π β ππ S C S V S (π×π ) β ππ S S dS V Stokes Theorem π β ππ = Μ π dS S C dS Μ π dS Μ π C 7 COORDINATE SYSTEMS Cartesian z % $ π π π ∇= π’ +π£ +π€ ππ₯ ππ¦ ππ§ P(x,y,z) #Μ !Μ ∇? Φ = π?Φ π?Φ π?Φ + + ππ₯ ? ππ¦ ? ππ§ ? z y Po;ar x y x Cylindrical Μπ π z π₯ = π cos π π¦ = π sin π π§=π§ Μπ½ π π(π, π, π§) Μπ π Spherical ∇= ππ ∇? Φ = π 1 π π + ππ½ + ππ ππ π ππ ππ§ z π ? Φ 1 πΦ 1 π ? Φ π ? Φ + + + ππ ? π ππ π ? ππ ? ππ§ ? x π½ y r z !#" Spherical π₯ = π sin π cos π π¦ = π sin π sin π π§ = π cos π !#$ ' & ∇= ππ ∇? Φ = π 1 π 1 π + ππ½ + ππ ππ π ππ π sin π ππ !#% ((*, ', &) r y x 1 π ? πΦ 1 π πΦ 1 π?Φ π + sin π + π ? ππ ππ π ? sin π ππ ππ π ? π ππ? π ππ ? 8