Other Coordinates System Polar coordinates. When you have to work curves defined by a Point that is moving around the origin using Cartesian Coordinates not efficient since the equations that describe the curve is not a function. To solve this reason Bernoulli revive an old system used by the Greek astronomer Hipparchus (190-120 BC). The polar coordinates use an origin that call Pole and an axis and a horizontal semi-line oriented positive to the right. A point has coordinates are Theta and r that is the distance from the origin to the point P an angle Theta is the angle of sides OP and the polar axis (r, π) Pole O (origin) Polar Axis Ex 1: Graph r=sin( ) π/2+ π/3 Let do a table for (r, ) π/2+π/4 x x π − π/3 π/2 x π/3 π/4 x π/6 x x π π x x 0 Type equation here. r = sin π 0 0.0 π/6 π/4 1/2 = 0.5 π/3 3/2= 0.9 π/2 + 0 1.0 2/2= 0.7 π/2 +π /6 3/2= 0.9 + π/4 2/2= 0.7 +π/3 3/2= 0.9 π/2 +π/2 + 0.0 π 0 Ex 1: Graph r=sin(π ) Let do a table for (r, π ) π/2 + π/3 x x π/2+π/4 π − π/3 x Type equation here. Type equation here. π/6 x x π/4 2/2= 0.7 π/2 + 0 π/4 x x π π/3 x π/6 0 1/2 = 0.5 π/3 π/2 0 r = sin π 3/2= 0.9 1.0 π/2 +π /6 + π/4 3/2= 0.9 +π/3 3/2= 0.9 π/2 +π/2 + 0.0 2/2= 0.7 Trick #3: PARTIAL FRACTIONS. Let P(x)/Q(x) a polynomial fraction where Q(x) is completely factored and deg P(x) < deg Q(x) then the fraction can be written as sum of simples fraction where the denominators are the factors involved in the factorization of Q(x) using the following two rules. π΄1 π΄2 π΄π 1) For each linear factor of type (ax+b)π we add n simple fractions (ππ₯+π)1 + (ππ₯+π)2 + … . + (ππ₯+π)π π 2) For each linear factor of type (aπ₯ 2 + ππ₯ + π) we add n simple fractions π΄1π₯ +π΅1 ( aπ₯ 2 +ππ₯+π Ex 1. Ex ) 1 + π΄2 π₯+π΅2 ( aπ₯ 2 +ππ₯+π 3π₯ 2 −π₯+5 π₯−1 π₯+4 ( 2π₯−1)2 πΈπ₯+πΉ π΄π₯+π΅ 2. 2 π₯ +4 π₯ 2 +1 + ) = 2 + β―+ π΄ π₯−1 πΆπ₯+π· (π₯ 2 +1)2 + + π΄ππ₯+π΅π (aπ₯ 2+ππ₯+π) π΅ π₯+4 πΈπ₯+πΉ π₯ 2 +4 + πΆ 2π₯−1 π + π· (2π₯−1)2 Ex 3. Integrate 2π₯ 2 +7π₯+15 (x+2)2 (x+3) dx , first decompose 2π₯ 2 +7π₯+15 (x+2)2 (x+3) = π΄ (π₯+2) + π΅ π₯+2 2 + πΆ (π₯+3) Multiplying both sides by (x + 2)2 (x + 3) we get 2π₯ 2 − 7π₯ + 5 = π΄ π₯ + 2 + π΅( π₯ + 3 + πΆ π₯ + 2 2 Take x= - 2 8-14+15= 9 B(1) = 9 ο B = 9 Take x= -3 A( -1) + C(-1) = - 3 ο - A - C= -3 ο A + C= 3 Take x=0 A(2) + B(3) + C( 4) =5 ο 2A +3B =5 and since B=9 5 So Ex. 4 Ex. 5 πππ +ππ+ππ (π±+π)π (π±+π) 4π₯ + 5 ππ¦ = π₯2 + π₯ − 2 ππ± = −ππ π+π dx + 4π₯ + 5 ππ₯ = π₯+2 π₯−1 π π π π+π π − ππ dx (π+π) 1 3 + ππ₯ = π₯+2 π₯−1 4π₯ 2 + 17π₯ − 4 π΄ π΅π₯ + πΆ ππ₯ = + π₯ + 5 π₯2 + 1 π₯+5 π₯−1 Type equation here. A +C = 3 ο C = 14 2A + 27 = 5 ο A = -11 = - 11ln|x+2|- π (π₯+2)2 + 14 ln|x+3| + C ππ π₯ + 2 + 3ππ π₯ − 1 + πΆ PARAMETRIC EQUATIONS When “the equation in Cartesian Coordinates is to complex” mathematician use an extra variable t called “parameter” such that both x and y are functions of t like x= x(t) and y= y(t) . We say then that the original function is now defined by x=x(t) , y= y(t), y(t) Ex 1. Find the Cartesian equation corresponding to the curve defined by x = 3t y = π‘2 + 1 π₯ From the first parametric equations we get t = and replacing on the second we get y =( π₯ 3 )2 + 1= π₯2 9 3 +1 and we recognize that is a parabola. Ex 2. Find the length of the graph defined by x=4cos(t) and y =4sin(t) It is clear that π₯ 2 + π¦ 2 = 4 1 that is a circle with radio 2. So the length is 2 pi Type equation here. Type equation here.Type equation here. LENGH OF AN ARC IN PARAMETRIC Let have a curve defined in parametric form =x(t) ; y=y(t) , we get that the arc length π‘2 π‘1 of an arc is given by s = ds = ππ₯ ππ 2 ππ¦ + ππ ππ₯ 2 ππ 1+( π₯′ π‘ 2 ) (π¦ ′ (π‘) ππ‘ 2 ππ₯ = ππ₯ 2 ππ + ππ¦ 2 ππ ππ₯ ππ ππ₯ = ππ₯ 2 ππ + Ex 1: Find the length of the curve defined by x = 2t-1 and y=π‘ 2 +1 from t = 0 to t= pi/6 Let find dx/dt = 2 and dy/dt =2t … ππ/6 ds= 0 2 2 + 2π‘ 2 ππ‘ = 4 1 + π‘ 2 ππ‘ =4 sec3 π dπ = Use t =tanπ dt=sec2 π dπ 1 (sec 2 ππ¦ 2 ππ ππ Type equation here. . π π‘πππ + ln |secπ+tanπ|= πΈπ₯. 3 π₯2 π¦2 πΉπππ π‘βπ ππππβπ‘ π ππ π‘βπ πππππ π 2 + 2 = 1 π π 2 ππ¦ ππ = 1 + π¦′ 2 = 1+ ππ ππ₯ ππ₯ = ππ ππ₯ ππ 2 (π’π π πππππππ‘πππ πππ’ππ‘ππππ ππ¦ + ππ 2 ππ = = π = 4π π 2 0 1+ π 2 π‘ππ2 π π₯ = π sin π π¦ = π cos π ππ , where π2 π = π π cos π 1+ 2 <1 2 + −π sin π −π sin π 2 π cos π 2 ππ 2 ππ Type equation here. This integral cannot be solved since the there are not exist a function whose derivative is REMARK: Niels Abel (1813-1829) study 3 types of family functions that he call Elliptic Integrals and prove that there have not anti-derivative and integral is one them ,this means that the ellipsis length can not be calculate by integration. There are tables for Elliptic integrals. where you find he value. Ex 1. A function Y =f(x) is defined by the parametric equations x= 2t+1 and y = t+3. Find y = f( x ) In other words we need to eliminate t using the given equations. From the first 2t = x-1 From the second t=y-3 1 3 1 3 ο¨ X - 1 = 2(y-3) ο¨ x – 1 = 2y – 6 ο¨ y = 2 x − 2 ο¨y= f(x)= = 2 x − 2 . It is a line Ex 2. Get y=f(x) for x(t)= a cos(t) and y(t) = b sin(t)., where a,b are positive numbers. Since cos2 π‘ + π ππ2 (t) =1 ο¨ π₯(π‘) 2 π¦(π‘) 2 ( ) +( ) = 1 It is an ellipse. π π ARC LENGTH IN PARAMETRICS EQUATIONS. Since ds = 1 + π¦′ 2 dx and π¦ ′ = ππ¦ ππ‘ ππ₯ ππ‘ ο¨ ds = 1+ ππ¦ ππ‘ ππ₯ ππ‘ 2 ππ₯ = ππ₯ 2 ππ‘ + ππ¦ 2 ππ‘ dt Ex 1. A function Y =f(x) is defined by the parametric equations x= 2t+1 and y = t+3. Find y = f( x ) In other words we need to eliminate t using the given equations. From the first 2t = x-1 From the second t=y-3 1 3 1 3 ο¨ X - 1 = 2(y-3) ο¨ x – 1 = 2y – 6 ο¨ y = 2 x − 2 ο¨y= f(x)= = 2 x − 2 . It is a line Ex 2. Get y=f(x) for x(t)= a cos(t) and y(t) = b sin(t)., where a,b are positive numbers. π₯(π‘) 2 π¦(π‘) 2 ο¨ ( ) +( ) = 1 It is an ellipse. π π Since cos2 π‘ + π ππ2 (t) = 1 ARC LENGTH IN PARAMETRICS EQUATIONS. Since ds = 1 + π¦ ′ 2 ππ¦/ππ‘ ′ and π¦ = ο¨ ds = 1 + ππ₯/ππ‘ 2 dx = ππ₯2 ππ¦2 + ππ‘ ππ‘ 2 ππ₯ ππ‘ dx = ds = ππ₯ ππ‘ 2 ππ¦ + ππ‘ ππ₯ 2 ππ‘ + 2 ππ₯ ππ₯ ππ‘ ππ¦ 2 dt ππ‘ πΈπ₯. 3 π₯2 π¦2 πΉπππ π‘βπ ππππβπ‘ π ππ π‘βπ πππππ π 2 + 2 = 1 π π ππ¦ ππ = 2 1 + π¦′ 1+ = ππ₯ 2 ππ ππ = ππ ππ₯ ππ 2 ππ¦ + ππ ππ₯ 2 ππ (π’π π πππππππ‘πππ πππ’ππ‘ππππ π₯ = π sin π π¦ = π cos π 2 Remark. Mathematician Niels Abel (1810- 1828) study for do not exist anti-derivatives ππ = π ππππππππ ππ₯, ππ¦, ππ π€π βππ£π: ππ = π =4 π cos π π 2 0 2 + −π sin π ππππ π 2 ππ = 4 π 2 0 2 2 2 2 ππ = π πππ π + π ππ = π2 πππ 2 π π 1+ π π 1+ π 2 π‘ππ2 π ππ = ππ π πππ’π‘πππ ! 2 π‘ππ2 π ππ There do not exist a function whose derivative is integral function if k= π 2 π < 1. REMARK: Niels Abel (1813 - ?) study 3 types of family functions that he call Elliptic Integral and prove that there have not anti-derivative. He also publish a table giving the approximated value the for different limits. integral is one them, this means that the ellipsis length can not be calculate by integration. Sequences and Series A sequence { ππ } is a function whose domain is the non negative integer π + U {0} and the range a subset of the real numbers, meaning a(n) : Z+ U {0} βΆ −−→ R. The functional values : π1 , π2 , π3 , ……. are the n components and a(n) is called the general term of sequence. π 1 2 3 4 π Ex 1: { } = { , , , ,.. . , , . ...} π+1 2 3 4 5 π+1 Let’s graph the sequence …. 1.0 The graph says lim of seq = 1, e.g. Converge! Ex 2. Test for convergence π3 +2π2 −2 { 2 } π +4 dividing num. and denom. by π3 we get 0.5 1 2 1+π+−π^2− 1 +4/π π Taking lim we get infinity, so the sequence Diverge! π→∞ 1 2 3 4 5 6 7 8 Definition: Let { ππ } be a sequence we will call Series the sum adding of the terms of the sequence. Definition: Let { ππ } be a sequence we will call Series the sum adding of the terms of the sequence. Ex 1 aπ ={ 1, x ,π₯ 2 , π₯ 3 , π₯ 4 , ……… } then Series. Definition : A polynomial Series is ∞ π=1 ππ ∞ π=0 ππ = {1 + x + π₯ 2 + π₯ 3 + π₯ 4 + ……… } is a is convergent if lim π→∞ ∞ π=1 ππ exist. Ex 2. Test for convergence the series = 1-1+1-1+1-1 …… This series is Divergen , since the sum is alternating from -1 to +1 Remark: This sequence converge to different values in concordance with the way we regroups the terms of the series A) (1-1)+(1-1)+(1-1) …….. = 0 B) 1- ( (1-1)+(1-1) + ……... ) = 1 A) (1-1)+(1-1)+(1-1) …….. = 0 B) 1- ( (1-1)+(1-1) + ……... ) = 1 C) (-1+(1-1)+(1-1)+(1-1)….. =-1 This series is Divergent , since the sum is alternating from -1 to +1 This example shows us that we need a more strong definition for convergence of series that assures us that the limit be if exist be a unique real number. To get this we are going add that the series of the absolute values of the terms should be convergence, to assures that that the new adding be associative and commutative. Type equation here. GEOMETRIC SERIES Definition: The series series. ∞ π=0 ππ π where r is a fix real number are called a geometric Theorem: geometrics series converge if |r|<1 divergent if |r|≥1 2 Proof: Call S(n) = a + ar + aπ + ππ 3 + +ππ π multiplying by r r S(n )= ar + aπ 2 + ππ 3 + +ππ π + ππ π+1 + …….. ππ π+1 Substracting … S(n) (1-r) = So if |r | < 1 the series Converges if |r |≥ 1 Diverges. ο¨ π−ππ π+1 S(n)= 1−π , so lim S(n )= π π₯ = π −0 π = 1−π 1−π 1 Given β ABC area = 2 bh where h = CH = √ π΄πΆ 2 − ()AH 2 = = 3 2 a C So area Type equation here. So area= A = 3 4 π2 For second triangle a is replacing by So the series is A+ 1 4 π΄+ 1 42 π 2 1 4 and Area by = (first area) A+.... So the sum of all areas… is S = π΄ 1 1−( ) 4 a Geom. Series with r=1/4 = 4π΄ 3 = 4 3 3 4 π2 = 3 3 π2 A 0.5 a H B Definition: ∞ π=1 ππ π₯ is absolute convergent if ∞ π=1 |ππ π₯ | is convergent. Remark: The absolute convergent assures us that adding of a series is commutative and associative. Ratio Test : Let ∞ π=1 ππ π₯ a series and such that lim | π→∞ ππ+1 π₯ ππ π₯ | = r. 1) The series converges if r < 1 2) The series diverges if r > 1 3) There are no information if r =1 Proof: 1) The limit assures that exist an integer N such that | ππ+1 π₯ ππ π₯ | < 1 for every n>N So |ππ π₯ |>r |ππ+1 π₯ |>π 2 |ππ+2 π₯ |> …. Is a GEOMETRIC SERIES OF 0< r<1 is convergent π This implies that ∞ π=1 |ππ π₯ | = π=1 |ππ π₯ | + π‘βπ ππππππ‘πππ π ππππ is convergent. 2) and 3) is the similar argument. πΈπ₯. 1 π₯π π=1 π! ∞ π = lim π→∞ π₯ π+1 π₯ππ₯ π! π₯ π+1 π! π+1 ! ∗ ∗ = lim = lim π₯π π→∞ π + 1 π! π₯ π π→∞ π + 1 ! π₯ π π! 0 < 1 , π‘βπππππππ π‘βπ π ππππ ππππ£ππππ πππ πππ ππππ π₯. lim π→∞ π₯ π+1 =0 πΌππ‘πππ£ππ ππ πΆπππ£πππππππ ππ ππππππ πππ π‘ πππ ππππ£πππππππ … ∞ ππ (π₯ − π)π ππ ππππ£ππππππ‘ π=0 π·ππ: ππ π π‘βπ πππ‘ππ ππππ‘ππππ ππ+1 (π₯ − π)π+1 lim = π→∞ ππ (π₯ − π)π ππ π (π₯ − π) < 1 ο¨ ππ+1 (π₯ − π) ππ+1 π₯ − π π (π₯ − π) lim = π|π₯ − π| < 1 lim = π→∞ π π π→∞ π ππ (π₯ − π) −1 π 1 < π₯−π <r 1 1 ο¨ ( a -- π < x < a + π ) 1 Remark : The lector should check if for the series converge or diverge for x= a ± π and see if it is need to improve to (a , b) to [a , b) or (a , b ] or [a , b] by checking for convergence at x=a and x=b. πΈ2. πΉπππ π‘βπ πππ‘πππ£ππ ππππππ£πππππππ ππππ‘βπ π ππππ π₯ π₯3 π₯5 π₯7 π π₯ = − + − +β― 1! 3! 5! 7! Using πππ‘ππ π‘ππ π‘ π€π βππ£π π₯ π+1 π₯ π₯ππ₯ π! π+1 ! lim = lim ∗ π = π = lim = π→∞ π→∞ π + 1 π π + 1 π! π₯ π₯ π→∞ π! π‘βπππππππ π‘βπ π ππππ ππ ππππ£ππππππ‘ πππ ππ£πππ¦ ππππ π£πππ’π ππ π₯ π₯ lim π→∞ 1 π+1 =0 πΈ3. πΉπππ π‘βπ πππ‘ππ ππ ππππ£πππππππ¦. ∞ (π₯ − 2)π π=0 π₯ − 2 π (π₯ − 2) π₯−2 π (π₯ − 2)π+1 = π₯ − 2 lim = π₯−2 <1 π = lim = lim π→∞ π→∞ (π₯ − 2)π π→∞ (π₯ − 2)π (π₯ − 2)π −1 < π₯ − 2 < 1 , ππππ£πππππ ππ 1 < π₯ < 3 πΏππ‘ π’π π ππ ππ π‘βπ π ππππ ππππ£πππππ ππ π‘βπ ππ₯π‘πππππ π₯ = 1 πππ π₯ = 3 πβππ π₯ = 1 π‘βπ π ππππ ∞ (π₯ − 2)π ∞ = π=0 (1 − 2)π ∞ = π=0 (−1)π πππ£πππππ π=0 πβππ π₯ = 3 π‘βπ π ππππ ∞ (π₯ − 2)π ∞ = (3 − π=0 π=0 ∞ πβπππππππ π‘βπ π ππππ 2)π ∞ = (1)π πππ£πππππ π=0 (π₯ − 2)π ππππ£πππππ πππ π₯ ∈ (1,3) π=0 ππ πππ πππ‘ππππππ π‘π ππππ π€βπππ π π ππππ ππππ£πππππ ∞ ∞ πππππ πππ π(π₯π )βx = = π=0 π π₯π ππ₯ 0 πβπ π ππππ ππππ£πππππ ππ πππ ππππ¦ ππ π‘βπ πππ‘πππππ ππ₯ππ π‘π . ππ‘βπππ€ππ π π‘βπ π ππππ πππ£πππππ . ∞ πΈ1. 1 π = π₯ π₯=1 ∞ 1 π₯ −π+1 ∞ 1 ππ₯ = = π₯π −π + 1 1 πππ π > 1, π‘βπ π ππππ ππππ£πππππ ππ πππ πππ π < 1, π‘βπ π ππππ πππ£πππππ πππ π = 1, π‘βπ π ππππ πππ£πππππ πππ π > 1, π‘βπ πππ‘πππππ ππ₯ππ π‘π πππ π < 1, π‘βπ πππ‘πππππ ππππ πππ‘ ππ₯ππ π‘ πππ π = 1, π‘βπ πππ‘πππππ ππππ πππ‘ ππ₯ππ π‘ ππππ¦ππππππ ππππππ₯ππππ‘πππ ππ π ππ’ππ‘πππ π€ππ‘β πππππππ‘π πππππ£ππ‘ππ£ππ ππππ’π‘ π₯ = π ∞ ππ (π₯ − π)π π π₯ = π=0 1π π‘ ππ ππππ π‘π πππππ’πππ‘π ππ π π₯ = π0 + π1 π₯ − π + π2 (π₯ − π)2 +π3 π₯ − π π π = π0 + π1 0 + π2 (0)2 + β― = π0 3 π′ π₯ = + 3 ∗ 4π4 π₯ − π 1 π1 + π2 π₯ − π + 2 ∗ 3π3 π₯ − π π ′′ π₯ = 0+ π (3) π₯ = 0+ π (4) π₯ = 0+ 0 0 π=0 + + 3! π3 0 π (π) π₯ (π₯ − π)π π! 4 + 4! π4 π₯ − π + 4 +β― 0! π0 = π π 2! π2 + 3! π3 π₯ − π + 4! π₯ − π ∞ f(x) = 2 + π4 π₯ − π 4! π4 3 +β― π4 + … +... +β― 1! π1 = π ′( π) 2! π2 = π ′′ π 3! π3 = π (3) (π) 4! π4 = π (4) (π) ππ = π (π) π₯ π! in the same year Taylor publish the series using . Later in the same year Brooke Taylor publish the extension series in the vicinity of x=a with a≠ 0. Ex 1. Find the polynomial series of McLauring for f(x)= π π₯ 1 π₯ 0 f(x)= π ο¨ f(0) = π = 1 => π0 = =1 f’’(x)= π π₯ ο¨ f(‘’0)= We have ππ₯= 1+ π0 1 1! + =1 => π1 = 1 2! + 1 3! + 0! 1 =1 1! 1 + ...… 4! and since => ππ = . 1 π! + 1 =1 π! for all i=0,1,2,3….. 1 + ..... (π+1)! Let find the interval of convergence (using the Ratio Test ) π = lim π→∞ lim π→∞ 1 π+1 ! 1 π! = 1 lim | | π→∞ π+1 π₯π+1 π+1 ! π₯π π! = =0 So, The series converges when r =|x-0|=|x|0 < 1 meaning always, so interval of convergence is (-∞, ∞). Ex 1. Find Mclauring series for sinx Getting the f’ , f” … 1) f(x) = sin x = 0 + 1 1! x 0 +2! π₯ 2 f’(x) = cosx f’’(x)= sinx f’’’(x) = cosx start reppeting ! f(0) = f’(x)= cosx f”(x)=- -sinx f’’’(x)= -cosx f’’’’(x)= sinx + −1 3! ο¨ .... = x π₯3 − 3! Sin(x) = x π₯3 − 3! + π₯5 5! + π₯5 5! …… = x π₯3 − 3! + π₯5 5! …… f’( 0) =. 1 f’’’(0 )=0 f’’’( 0)= -1 f””(0)=0 start reppeting !. . . Remark: Since the set of series convergent is closed for + , - , * , / , including * by a real number and more Plus like the set of polynomials, we can see the convergent series as an extension of set of the polynonmial, e.g. polynomials of infinite degree . Further more if f(x) = ak-6∑_(k=0)^∞ Ex 2. Find Mcauling series for cosx , since we known the series for sinx then π€π πππ πππ‘ πππ π₯as ′ π πππ₯ = (x π₯3 − 3! + π₯5 5! ……)’ = 1 π₯2 − 2! + π₯4 4! ……) ‘ = −1 π₯ π−1 π π+1 π=1 (−1) 2π+1 ! sin π₯ π₯ Ex 3. The function f(x) =sin(x)/x do not has antiderivative. Solve 3 π ππ π₯ π₯ ππ₯ = π = π1 (−1) π=1 5 x − π₯3! + π₯5! …… π₯ π₯ π−1 2π + 1 ! dx = (1 − π₯2 3! + π₯4 5! ……) dx = 1- π₯3 (3!)3 + π₯5 5!.(3) ππ₯ π’π πππ π πππππ π₯7 − 7!.(3) … HYPERBOLIC FUNCTIONS. πΌπ πβπ¦π πππ ππ‘ ππ ππππ£ππ π‘βππ‘ π‘βπ πππ’ππ‘πππ ππ π‘βπ ππ’ππ£π πππππππ ππ¦ ππ πππππ‘πππ πππππ βππππππ ππππ π‘π€π π πππ − βππβ ππππππππ πππ π‘π π‘βπ π βπππ ππ π‘βπ ππ’ππ£π ππ ππππππ πππ‘πππππ¦ πππ πππ ππ πππππππ π‘βπππ’πβ π‘βπ πππ’ππ‘πππ π= π π₯ (π 2 π +π −π₯ π) π X= 2 (π πΌπ π€π ππππ π = 1 π‘βππ Y= π₯ π −π −π₯ π (π π₯ +π −π₯ ) 2 ) X= (π π₯ −π −π₯ ) 2 π€βππβ π ππ‘ππ ππ¦ π‘βπ πππ’ππ‘πππ π¦ 2 − π₯ 2 = 1 π·ππππ π‘πππ‘πππ: (π π₯ + π −π₯ ) 2 2 (π π₯ − π −π₯ ) + 2 2 π 2π₯ + 2 + π −2π₯ π 2π₯ + 2 + π −2π₯ 4 = + = =1 4 4 4 ππ π ππ (π π₯ +π −π₯ ) π‘βππ‘ 2 πΌπ π€π π’π π π‘βπ (π π₯ −π −π₯ ) πππ πππ π ππππππ π‘π cos 2 βπ¦πππππππ π¦ 2 − π₯ 2 = 1 πππ π‘πππ ππ π‘βπ (π π₯ + π −π₯ ) cosh π₯ = 2 π€βπππ π₯ πππ sin π₯ ππ π‘πππππππππ‘ππππ . ππππππ π₯ 2 + π¦ 2 = 1 π‘βππ π€π βππ£π (π π₯ − π −π₯ ) sinh π₯ = 2 cosh π₯ − sinh π₯ = 1 (π π₯ + π −π₯ ) π= 2 sinh(π₯) tanh π₯ = cosh(π₯) (π π₯ −π −π₯ ) = 2 ππ ππππππ 2 2 X Remark 1) sinh and tanh are odd functions and cosh is even. 2) For every relation or formula in Trigonometry there is an analog one in Hyperbolic and sometimes the one sign could be changed at most. Example: Cos2x=2coπ 2 x -1 ----ο Cosh2x=2coπ β2 -1 sinπ₯ 2 (x)+ cosπ₯ 2 (x) = 1 - --ο - sinπ₯β2 (x) + cosπ₯β2 (x) = 1 INVERSE OF HYPERBOLIC FUNCTION Type equation here. Type equation here. 1) sinh−1 (x) = ln|x|+ π₯ 2 +1 for all x 2) cosh−1 (x) = ln|x+ π₯ 2 −1 for all |x |≥1 1 1+π₯ 3) tanh−1 (x) = ln|x|=2 ln | 1−π₯ | for all x < 1 4) sech−1 (x) = 1 | π₯| π₯ 2 −1 for all |x| >1 1) Proof: Let y = sinh−1 (x) ο¨ sinh (sinh−1 (x)) =sinh(y) ο¨ x = So working you get 2x =π π¦ + π −π¦ ο¨ y = ln|x|+ π₯ 2 +1 Same procedure for 2) ,3) and 4) Type equation here. (π π¦ +π −π¦ ) 2 for all x. Type equation here. Type equation here. Type equation here. Type equation here. Ex 1. Let x = sinh (u) dx = cosh(u)du cosh π’ππ’ ( sinh π’ )2 +1 ππ₯ π₯ 2 +1 Ex 2. π ππ₯ dx = ln|x|+ π₯ 2 + 1 + C ο¨ −1 Find ( πππ β (sec(x)) π 1 ( πππ β −1 (secx))= 2 ππ₯ −1 du =ππ’ = u + C = π ππβ (x) + C π ππ (π₯)−1 5 2 ππ₯ π₯ 2 +1 π sec( x) tan(x) = sec(x) Remark: This example reinforce the correspondence Trigonometry ο¨ Hyperbolic functions dx = ln ln |5+ 26| ln | 2+ 5 | Ex 1. Solve 5 2 ππ₯ π₯ 2 +1 Let x = sinh (u) dx = cosh(u)du dx cosh π’ππ’ ( sinh π’ )2 +1 ππ₯ π₯ 2 +1 Ex 2. −1 du =ππ’ = u + C = π ππβ (x) + C dx = ln|x|+ π₯ 2 + 1 + C ο¨ 5 2 ππ₯ π₯ 2 +1 π Find ( πππ β−1 (sec(x)) ππ₯ π 1 ( πππ β −1 (secx))= 2 ππ₯ π ππ (π₯)−1 = sec( x) tan(x) = sec(x) Remark: This example reinforce the correspondence Trigonometry ο¨ Hyperbolic functions dx = ln ln |5+ 26| ln | 2+ 5 | VECTORS ON THE REAL PLANE Vectors are important and necessary to understand Science because work with oriented magnitudes, like force, velocity, acceleration , etc. The vague idea of a vector is an oriented arrow ( having a starting point an with a starting point and a length that represent the power or the magnitude of the vector. To added the vectors select a point O on the plane and we graph the two vectors starting from the point O, finally we complete a parallelogram adding the two missing sides . The diagonal from the point O is the adding the two given vectors. B B C C R R A C Type equation here. D A =O = C D A =O = C Definition: Let A=(π1 , π1 ), B=(π1 , π2 ) two point on the real plane we will call a vector v = AB = AB the oriented segment AB going from A to B (we will use the AB to avoid the arrow on top). And we will call magnitude of the vector to the length of the segment length of the segment AB. Ex 1: Let A=(2,3) and B=(5,4) then the length ||AB ||= ((2 − 5)2 +(3 − 4)2 = 10 and the orientation is going down and to the right. 10 going down and to the right. Ex 2: Let AB Y B A We see that the vector start in A and end in B. Furthermore that = AB = (π2 − π1 ) , π2 − π1 ) and in general for any two points we can define the vector for any point P and any point Q we can define The vector PQ = (ππ − ππ , ππ − ππ ) . We see too that || AB|| is given by Pythagoras Theorem (π΄πΆ)2+ (BC)2 ||= ||AB||. π2 O π2 − π1 C π2 − π1 π2 π1 π1 Ex 3: Find a unite vector EN LA DIRECTION π2 X Type equation here. Type equation here. Type equation here. Type equation here. equation here.Type Type equation here. Type equation here. Type equation here. Type equation here. STANDARD REPRESENTATION OF VECTOR IN THE PLANE. The unit vectors I =(1,0 )and j = (0,1) Find a direction vector pointing in the direction positive X axis and Yaxis respectively are called standard basis. Any vector v= (π£1 , π£2 ) can be expressed as a linear combination of the i and j because any (π£1 , π£2 ) = π£1 (1,0) + π£2 (0,1) = π£1 i + π£2 j Ex 1: Find a direction v in the direction of vector V= ( 3, - 4 ). Since ||v||= 32 + 42 = 5 v (π,π) 3 4 Let u = = = ( , ) |π½| ππ 5 5 Ex 2: Tree forces F1 = i – 2 j , F2 = 3 I – 7 j , F3 = i – 2 j act on an object. What additional force F4 should be applied keep the body at rest ? Total Force = ( 1+3-1) i + ( -2-7-3) j = 3i -11 j , so F4 = -3 i – 11 j Ex 3: Find standard representation of the vector PQ for points P(3,-4) and Q (-2,6). The component form of PQ= < (-2) -3 , 6 – (- 4] = (<5,10> , which means that PQ has standard representation PQ = -5 i + 10 j . VECTORS IN Rπ A vector v in space is a trio of real numbers) ( a, b , c) . We’ll define the dot product of a and b as ||v|| ||w|| cos( ) where is the angle (in space) defined by v and w. a h ß b a.b = ||a||.|| b|||cos( ) (Since h = a cos( ) a . b = a b cos(ß) ) = Area of the parallelogram ß If beta = Type equation here. a.b = 0 . This allows us to check if two vectors are perpendicular the angle defined by two vector are perpendicular. Ex. Are the vectors defined by a=(3,5,1) , b =(2,-3, 9) are perpendicular? Type equation here. a.b = 6 – 15 +9 = 0 , so the angle is aright one. CROSS PRODUCT Definition: The cross product of two vector a and b in R3 is defined as the determinant axb= I j k π1 π2 π3 π1 π2 π3 It is clear = that a x b = o if the third row is proportional to the second, e.g. vectors a is parallel to vector b. . Type equation here. Ex 1. Check vectors a = (2 ,-3 ,6) and b = (6,9,18) for parallelism . I j k I (3)(18)- j ( 2)(18) + k( 2)(9)-(6)(3) = 0 or quikly 3 times = axb= 2 3 6 second row = third row. 6 9 18 So second and third vectors are parallel. Remark : It is easy to prove that both dot and cross multiplication are distributives e.g. ( a + b ) . c = a . c + b . c and (a + b) X c = a X c + b X c . If we are using standard notation for vector we are going to need multiplication for I, j and k unitary vectors . Ex 3. Find the values of I , j ,k dot product I.I = j . j = k .k = 1 (Angle 0 ) I.J = J . K = I .K = 0 ( Angle 90 ) (π1 π + π₯2 π + x 3 k) .( π1 π + π¦2 π + y 3 k) = π₯1 π¦1 + π₯2 π¦2 + π₯3 π¦3 This a second way to calculate the dot product. Ex 4 Calculate (2,1,5) . (2,3,1) =2.2+1.3+5.1 = 12 Now we can calculate the angle as a.b = ||a|| ||b|| cos ( ) ο¨ cos ( ) = <(a,b) = πππ −1 [ 6 105 ]= 12 (30) (14) Type equation here. Ex 5. Find the proyection of v= 2i-3j+5k onto w=2i-2j+k. ( v).(w) = 4+6+5=15 , so the proyection over v onto w is (π£.π€) W= π€.π€ 15 9 (2 I -2 J+K). Ex 6. Prove that a x b =||a||.||b|| sin ß Type equation here. π.π π π sin ß= 1 − πππ 2 ß ο¨ 1 − a x b = ||a|| . || b||sin . 2 π| π ο¨ 2 − (π.π)2 (||a|| . || b||) 2 = ππ₯π ||a|| . || b|| π Remark: This probe that a x b = area of the parallelogram defined by a and b. Besides the triple product a.(b x c) measured the volumen of the parallepiped defined by a , b , c. the determinant: π1 π2 π3 π1 π2 π3 π1 π2 π3 π = a.(b x c) SURFACES IN Rπ I ) PLANES: A plane is the most simple in space , similar to the line in two dimension. The equation of a plane is a x + b y + c z =d where a , b , c , where a , b , c , d are real numbers Since a x + b y + c z = (a , b , c) . (x , y , z) we have that the vector (a , b , c ) is perpendicular to with every vector (x , y , z) in the plane. We will call this vector the normal vector to the plane. Ex1: Find the plane containing vectors v = (1,3, -2 ) , w = (2,5,1) and containing the point P= (1,2,4) We need to find a vector perpendicular to u and v and v x w do that. Type equation here. I j k 1 3 -2 = 13 I - 5 j – 8 k , this vector is normal to the plane , and the equation of the parallel plane 2 5 1 plane passing thought (0,0,0) is 13x-5y-8z = 0 . The plane parallel containing P = (1,2,4) is 13x-5y-8z = d, meaning 13(1) - 5(2) – 8(4)= - 29 , so d=-29 Finally the plane equation is is 13x-5y-8z = = -29. π1 Ex 2. Find the acute angle define by the planes 3x + 2y –z = 3 , 5x + 4y + z = 5 π2 We have to see that the angle define by the planes is equal to the corresponding angle defines by the normal to the planes meaning 7 7 Cos (π) =(3,2,-1)). (5,4,1)/( 14 42 = =ο¨ πππ −1 ( ). 2 3 2 3. Ex 3. Find the equation of the common line to the planes 2x+5y+3z = 4 and x+4y+z = 5 By eliminating z we get -3y+z =- 6. Now shifting y try to left 0 in the second member and the second member 3(y+2) – 6 +z = - 6 ο¨ 3(y+2)+ z=0 ο¨ calling y+2 = π we get z=-3π and since y = π − 2 , we can get x as function of 2 +5(π − 2) +3z = 0 ο¨ 2+ 5π – 4 = -3z so x= 5π -2. So we have. π₯+2 π¦+2 π§ Isolating π from each equations we get π = = = the equation of the common 5 1 3 line that is v = (5,1,3) is vector on the direction of the line that pass through P=(-2,-2,0) To get the parametric expression could be done in a different way . For example from -3y + z =-6 ο¨ z = 3y + 6 if we make y= π and then z=3 π+6 , and we calculate x from x+4y+z = 5 ο¨ x+4 π +3 π = 5 ο¨ x=5 -7 π . Now we need to write the equation of the line. π₯−5 π¦−0 π§−6 In summary x=5-7π , y = π , z = 3π finally the equation is = = −7 1 3 Type equation here. EX Find the equation of the line that passes through the point Q=(2-1,3) and is orthogonal to the plane 3 x- 7y +5z+ 55 = 0. Where does the line intersect the plane? We know that the orthogonal to the plane is n = (2,-7,5), the line has to be parallel to n and contains the the point Q(2,-1,3) , so we replace on the equation of the line we get: x = 2+ 3 t π₯−2 3 == π¦+1 −7 = π§−3 5 ο¨ y= - 1-7t and substituting on the equation of the plane z= 3 +5t n Q* 3(3t +2) - 7(-7t-1 ) +5 (5t +3) = -55 ο¨ 9t +6 +49t +7 +25t +15 = -55 ο¨ 83t =-83 ο¨ t=-1 P(-1,6,-2) Now using t= - 1 , we get x= -1, y-6 , z=-2 ο¨ so (-1,6,-2) Q=( 2,-1,3) MORE APPLICATION OF DOT AND CROSS MULTIPLICATION OF VECTORS . 1) TRIPLE SCALAR PRODUCT. 2) u x v =||u|| ||v||sin∝ ||h|||cosß= π΄ So | u x v .w| = Volume of parallelepide. 3) Exercise. Prove that axb.c = uxv π1 π2 π3 π1 π2 π3 π1 π2 π3 2 Type equation here. 2 w ∝ h * v u Y 2 5) Find the plane tangent to the sphere π₯ + π¦ +π§ = 10 at P=(1,0,3). p * The plane is perpendicular to the radius OP , and the equation of the radius is QP = (1, 0, 3 ) -- (0, 0, 1 ) = (1, 0 ,2 ) and OQ = (0 ,0 ,1 ) So, equation of the plane x + 2z = t , since P (1 , 0, 3 ) we get t=7 Finally the equation of the plane is 2z=7 O Z Type equation here. Q X 2 i 3 2 -30 2 j 3 k 2 0 0 0 2 2 Z T = PQ x F y P F X In Physics when a force F is applied to a point Q of an “arm” vector PQ , then the Torque of F around P is defined as the cross product of the arm PQ with the force F. The magnitude of the torque ||T|| provide a measure of the tendency of the arm PQ to rotate counter clock about the an axis perpendicular to the plane determinate by PQ and F. EXAMPLE. The half-open door open that is 3 ft. The horizontal force of 30 lb. is applied at the edge of the door. Find the cos torque of the force about the hinge on the door. 3 3 = 3(cos(45) i+sin45j) = 2 π + 2j+k Type equation here. Q 3 FT = 45 2 k y The magnitude of the torque is 45 2 ft-lb. F Force =-30i X