Pg The perimeter is P = 3x The area of a triangle is. where b is the base, h is the height. In our case b = x, then the area is
|
|
- Οὐρανός Ζώσιμος Σπανός
- 5 χρόνια πριν
- Προβολές:
Transcript
1 Pg. 9. The perimeter is P = The area of a triangle is A = bh where b is the base, h is the height 0 h= btan 60 = b = b In our case b =, then the area is A = = 0. By Pythagorean theorem a + a = d a a = d d = a = d a = d a = d The area of the square is d d A= a = = = d
2 . From the triangle ABC by Pythagorean theorem AC = AB + BC = a + a = a From the triangle ACD by Pythagorean theorem AD = AC + CD = a + a = a Thus, d = a d a = d a = d a = d a = The surface area is d d A 6a 6 = = = 6 = d The volume is d d d V = a = = = 7 9. (a) y = This equation can be rewritten as y, y< 0 = y, y 0 The first line passes through the points (, ) and (5, 5). The second line passes through the points (, ) and (5, 5). The graph is not graph of a function because we can draw a vertical line that intersects it more than once.
3 (b) y = This equation can be rewritten as y = 0 (y )(y + ) = 0 y = 0 or y + = 0 y = or y = The graph is not graph of a function because we can draw a vertical line that intersects it more than once.. (a) + y = In the Quadrant I > 0 and y > 0, so we have + y = y = + This is a line with the slope m = and the y-intercept (0, ). In the Quadrant II < 0 and y > 0, so we have + y = y = + This is a line with the slope m = and the y-intercept (0, ). In the Quadrant III < 0 and y < 0, so we have y = y = This is a line with the slope m = and the y-intercept (0, ). In the Quadrant IV > 0 and y < 0, so we have y = y = This is a line with the slope m = and the y-intercept (0, ). The graph is not graph of a function because we can draw a vertical line that intersects it more than once.
4 (b) + y = We have two equations: + y = and + y = or y = + and y = The first graph is a line with the slope m = and the y-intercept (0, ). The second graph is a line with the slope m = and the y-intercept (0, ). The graph is not graph of a function because we can draw a vertical line that intersects it more than once. 7., F( ) = +, > When then F() =, so the part of the graph of F that lies to the left of the vertical line = must coincide with the parabola y =. The -coordinate of the verte is b 0 = = = 0 a ( ) Find y for = 0: y = 0 =. The verte is (0, ). Find few points: y = 0 5 When > then F() = +, so the part of the graph of F that lies to the right of the vertical line = must coincide with the parabola y = +. The -coordinate of the verte is b = = = a Find y for = : y = + () =
5 The verte is (, ). Find few points: y = , < 0 G ( ) =, > 0 When < 0 then G() = /, so the part of the graph of F that lies to the left of the vertical line = 0 must coincide with the graph of y = /. Find few points: y = / When > 0 then G() =, so the part of the graph of F that lies to the right of the vertical line = 0 must coincide with the graph of y =. This line have slope m = and y-intercept (0, 0).
6 9. (a) On the interval [0, ] the line passes through the point (0, 0) and (, ), so the equation of the graph on this interval is y =. On the interval [, ] the line passes through the point (, ) and (, 0). The slope is y y 0 m = = = The equation of the line is y y = m( ) y = ( ) y = + y = + The equation of the function is, 0 < f( ) = +, (b) The equation of the function is, 0 < 0, < f( ) =, < 0,
7 0. (a) On the interval (0, ] the line passes through the point (0, ) and (, 0). The slope is y y 0 m = = = 0 The equation of the line is y y = m( ) y = ( 0) y = + On the interval (, 5] the line passes through the point (, ) and (5, 0). The slope is y y 0 m = = = 5 The equation of the line is y y = m( ) y 0 = ( 5) 5 y = + The equation of the function is +, 0 < f( ) = 5 +, < 5 (b) On the interval (, 0] the line passes through the point (, 0) and (0, ). The slope is y y 0 m = = = 0 ( ) The equation of the line is y y = m( ) y 0 = ( ( )) y = On the interval (0, ] the line passes through the point (0, ) and (, ). The slope is y y m = = = 0 The equation of the line is y y = m( ) y = ( 0) y = + The equation of the function is, 0 f( ) = < +, 0 <
8 Pg.9. f() =, g( ) =, h() =, j() = (a) y = y = ( f o g)( ) (b) y = y = ( jo g)( ) (c) y = y = ( go g)( ) (d) y = y = ( jo j)( ) (e) y = ( ) ( ho f)( ) = ( ) y = ( goho f)( ) (f) y = ( 6) ( jo f)( ) = ( 6) y = ( ho jo f)( ). (a) y = y = ( f o j)( )
9 (b) y = y = ( ho g)( ) (c) (d) y = 9 y = ( ho h)( ) y = 6 h ( ) = f( ) = 6 y = ( f o h)( ) (e) y = ( go f)( ) = y = ( hogo f)( ) (f) y = ( f o h)( ) = y = ( go f oh)( ) 7. y =, left, down If c > 0 then y = f( + c) shifts the graph of y = f() a distance c units to the left y = f() c shifts the graph of y = f() a distance c units downward Therefore, y = ( + ) y = ( + ) shifts the graph of y = one unit to the left shifts the graph of y = ( + ) one unit downward Find few points: 0 y =
10 8. y =, right, down If c > 0 then y = f( c) shifts the graph of y = f() a distance c units to the left y = f() c shifts the graph of y = f() a distance c units downward Therefore, y = ( ) shifts the graph of y = ( ) shifts the graph of Find few points: y = one unit to the right y= y = ( ) one unit downward
11 5. y = The graph of y = is the graph of y = moved by unit downward. shifted by unit to the right and then Find few points: y = 0
12 6. y = ( + ) / + The graph of y = ( + ) / + is the graph of y = / shifted by unit to the left and then moved by unit upward. Find few points: / y = y = +, compressed vertically by a factor of If c > then y = f ( ) compress the graph of y = f( ) vertically by a factor of c. c
13 Then y = + y = y = +, stretched horizontally by a factor of If c > then y = f stretch the graph of ( ) c y = f horizontally by a factor of c. Then y = + y = y = + Pg.8 9. π cos =,,0 + = sin = cos sin cos sin =± cos Since is in Quadrant IV then sin = cos 8 8 sin = = = = = 9 9 sin tan = = = cos
14 5 π 0. cos =,, π + = sin cos sin = cos sin =± cos Since is in Quadrant II then sin = cos 5 5 sin = = = = sin tan = = = cos 5 5. tan =, π π, sin + cos = sin cos tan + = cos cos + = = cos + tan cos =± tan + cos =± tan + Since is in Quadrant III then cos = tan +
15 5 cos = = = = = = sin = tan cos = = 5 5 π 9. y = cos π The graph of y = cos is the graph of cos Find the y-intercept: π π y = cos 0 = cos = 0 The y-intercept is (0, 0). Find the -intercepts: π cos = 0 y = shifted by π units to the right. π π = + π k k Z = π + π k The -intercepts are ( π + π k, 0). Find few points: 0 π 6 y = cos π π π π π π 0-0 y = cos is an even function so its graph is symmetric about the y-ais.
16 The period is π. π 0. y = sin + 6 π The graph of y = sin + is the graph of 6 Find the y-intercept: π π y = sin 0 + = sin = 6 6 The y-intercept is 0,. Find the -intercepts: π sin + = 0 6 π + = π k k Z 6 π = + π k 6 π The -intercepts are + π k,0 6. Find few points: π y = sin shifted by units to the left. 6 0 π 6 π π π π π π
17 y = sin y = sin is an odd function so its graph is symmetric about the origin. The period is π. π. cos = sin ( ) cos α β = cosαcos β + sinαsin β Transform the left side: π π π cos = cos cos + sin sin = cos 0 + sin = sin π. cos + = sin
18 7. ( ) cos α + β = cosαcos β sinαsin β Transform the left side: π π π cos + = cos cos sin sin = cos 0 sin = sin π cos 8 + cosα cos α = π + cos + π + cos = = = 8 8. π cos 8 + cosα cos α = 5π π cos cos π π + + π cosπ cos sinπsin 5π cos = = = ( ) + 0 = = = 5. sin θ cosθ = 0
19 sin θ cosθ = 0 sinθcosθ cosθ = 0 cos θ(sinθ ) = 0 cosθ = 0 or sinθ = 0 cosθ = 0 or sinθ = π m θ = + πk k Z or θ = ( ) sin + mπ m Z π m π θ = + πk or θ = ( ) + πm 6 5. cos θ + cosθ = 0 cos θ sin θ + cosθ = 0 θ cos θ ( cos θ) + cos = 0 cos θ + cos θ + cosθ = 0 cos θ + cosθ = 0 cos θ + cosθ = 0 cos θ + cosθ cosθ = 0 cos θ(cosθ + ) (cosθ + ) = 0 (cosθ + )( cosθ ) = 0 cosθ + = 0 or cosθ = 0 cosθ = or cosθ = cosθ = or cosθ = θ = π + πk k Z or θ =± cos + πm m Z π θ = π + πk or θ =± + πm Pg.6. ht () = cott
20 The average rate of change of a function y = f() over the interval [a, b] is π π (a), π π cot cot = = = π π π π π π π (b), 6 π π cot cot 6 0 = = = π π π π π 6 6 f ( b) f( a) b a.. gt () = + cost The average rate of change of a function y = f() over the interval [a, b] is (a) [ ] 0,π + cos π (+ cos0) + ( ) (+ ) = = = π 0 π π π (a) [ π, π ] + cos π (+ cos( π )) + ( ) ( + ( )) = = = 0 π ( π) π π f ( b) f( a) b a. 8. Eample is needed 9. Eample is needed. Eample is needed. Eample is needed Pg.7
21 + lim( + ) lim = = = + 6 lim( + 6) lim s(s ) = lim lims lim(s ) s / s / s / s / = = = =.. 7. lim = = h 0 lim h 0 h+ + lim( h+ + ) lim h+ + lim h 0 h 0 h 0 = = = lim( h) + lim lim h 0 h 0 5h+ ( 5h+ )( 5h+ + ) = lim h h( 5h+ + ) h 0 h 0 5h+ 5h = lim = lim h 0 h( 5h+ + ) h 0 h( 5h+ + ) 5 5 = = lim h 0 5 h = = + t + t t + t t t( t+ ) ( t+ ) lim = lim = lim t t t t t ( t )( t+) ( t+ )( t ) t+ = lim = lim t ( t )( t+ ) t t+ + = = + 8.
22 t + t+ t + t+ t+ t( t+ ) + ( t+ ) lim = lim = lim t t t t+ t t( t ) + ( t ) t t t ( t+ )( t+ ) t+ + = lim = lim = t ( t )( t+ ) t t =. ( )( ) ( )( )( + ) u ( u )( u + u+ ) ( u )( u + u+ ) u u u + u u+ u lim = lim = lim u u u ( u+ )( u + ) (+ )( + ) lim u = u + u+ = + + =. 5. v 8 ( v )( v + v+ ) ( v )( v + v+ ) lim = lim = lim v v 6 v ( v )( v ) v + ( v )( v+ )( v + ) v + v+ + + = = lim v ( )( ) ( )( v+ v ) = = 8 lim sec = lim = = = 0 0 cos cos 0. 5 ( 5)(+ 5) lim = lim + ( + )(+ 5) ( 5) + 5 = lim = lim ( + )(+ 5) ( + )(+ 5)
23 9 ( ) ( +) = lim = lim ( + )(+ 5) ( + )(+ 5) ( ) = = lim 5 ( ) = = = + 9. lim + cos( + π ) = lim + lim cos( + π ) π π π = π + cos( π + π) = π cos0 = π = π 56. lim p ( ) =, lim r ( ) = 0, lim s ( ) = (a) lim( p ( ) + r ( ) + s ( )) = lim p ( ) + lim r ( ) + lim s ( ) = ( ) = (b) lim( p ( ) r ( ) s ( )) = lim p ( ) lim r ( ) lim s ( ) = 0 ( ) = 0 (c) ( p ( ) + 5 r ( )) lim( p( ) + 5 r( )) lim( p( )) + lim(5 r( )) lim = = s ( ) lim s ( ) lim p ( ) + 5 lim r ( ) = = = 57. f() =, = f ( + h) f( ) ( + h) + h+ h m ( ) = lim = lim = lim h 0 h h 0 h h 0 h h + h = lim = lim( + h) = h 0 h h 0 At = we have m () = () =
24 58. f() =, = f ( + h) f( ) ( + h) + h+ h m ( ) = lim = lim = lim h 0 h h 0 h h 0 h h + h = lim = lim( + h) = h 0 h h 0 At = we have m( ) = ( ) = 6. f ( ) =, = 7 f ( + h) f( ) + h ( + h )( + h+ ) m ( ) = lim = lim = lim h 0 h h 0 h h 0 h( + h+ ) + h h = lim = lim = lim h 0 h( + h+ ) h 0 h( + h+ ) h 0 + h+ = = + At = 7 we have 7 m (7) = = 7 6. f ( ) = +, = 0 f( + h) f( ) ( + h) + + m ( ) = lim = lim h 0 h h 0 h ( ( + h) + + )( ( + h) ) = lim h 0 h( ( + h) ) ( + h) + (+ ) + h+ = lim = lim h 0 h( ( + h) ) h 0 h( ( + h) ) h = lim = lim h 0 h( ( + h) ) h 0 ( + h) = = At = 0 we have m (0) = = 0+
25 Pg.8 9. It follows from the graph that 9 if < < 5 5 then < < The interval, is not symmetric about = : = = 6 6 We can choose δ to be the smaller of these numbers, that is 7 δ = It follows from the graph that if.6 < <. then.8 < + <. The interval (.6,.) is not symmetric about = :. = 0..6 = 0.9 We can choose δ to be the smaller of these numbers, that is δ = 0.9
26 0. f( ) = 7, L =, 0 =, ε = Step : Solve the inequality f( ) L < ε to find the interval about 0 = on which the inequality holds for all 0 : 7 < < 7 < < 7 < 5 9< 7< 5 6 < < Step : Find a value of δ > 0 that places the centered interval 0 δ < < 0 + δ inside the interval 6 < < : = 9 6 = 7 The distance from 0 = to the nearer endpoint of (6, ) is 7. If we take δ = 7 or any smaller positive number, then 0 < < 7 7 <. f( ) =, L =, 0 =, ε = 0.05 Step : Solve the inequality f( ) L < ε to find the interval about 0 = on which the inequality holds for all 0 : < < < < < 0. 0 < < 5 Step : Find a value of δ > 0 that places the centered interval 0 δ < < 0 + δ inside the interval 0 < < 5 : 5 = 0 =
27 The distance from 0 = to the nearer endpoint of 0,5 is. If we take δ = or any smaller positive number, then 0 < < < f( ) = 5, L =, 0 =, ε = Step : Solve the inequality f( ) L < ε to find the interval about 0 = on which the inequality holds for all 0 : 5 < < 6< 5 < < 7 5 < < 7 Step : Find a value of δ > 0 that places the centered interval 0 δ < < 0 + δ inside the interval 5 < < 7 : The distance from 0 = to the nearer endpoint of ( ) take δ = 7 or any smaller positive number, then 0 < < 7 5 = 6 < 5, 7 is 7. If we. f( ) =, 0 =, ε =0.05 ( )( + ) L= lim f( ) = lim = lim = lim( + ) = + = 0 Step : Solve the inequality f( ) L < ε to find the interval about 0 = on which the inequality holds for all 0 : < < < 0.05
28 .95 < <.05 ( )( + ).95 < < < + < < <.05 Step : Find a value of δ > 0 that places the centered interval 0 δ < < 0 + δ inside the interval.95 < <.05:.05 = = 0.05 The distance from 0 = to the nearer endpoint of (.95,.05) is If we take δ = 0.05 or any smaller positive number, then 0 < < 0.05 < f( ) = , 0 = 5, ε = ( + ) + 5( + ) L= lim f( ) = lim = lim = lim ( + )( + 5) = lim = lim( + ) = 5 + = Step : Solve the inequality f( ) L < ε to find the interval about 0 = on which the inequality holds for all 0 : ( ) < < + < ( + 5)( + ).05 < < < + < < <.95 Step : Find a value of δ > 0 that places the centered interval 0 δ < < 0 + δ inside the interval 5.05 < <.95: 5 ( 5.05) = ( 5) = 0.05
29 . lim = The distance from 0 = 5 to the nearer endpoint of ( 5.05,.95) is If we take δ = 0.05 or any smaller positive number, then < + 5 < < Given ε > 0, we want to find δ > 0 such that if 0 < < δ then < ε We have = = < ε We find a positive constant C such that < C < C and we can make ε C < ε by taking < = δ C We restrict to lie in the interval <, then < < < < < ε So C = is suitable and we should choose δ = min,. Show that δ works. ε Given ε > 0, we let δ = min,. If 0 < < δ then < < < < Also ε < ε < ε < so ε = < = ε Therefore, lim =
30 . lim = Given ε > 0, we want to find δ > 0 such that if 0 < < δ then < ε We have ( ) ( + ) ) + = = = < ε Let s assume that δ. This is a valid assumption to make since, in general, once we find a δ that works, all smaller values of δ also work. Then < δ implies that < < + < < + + < + < + Then ) + < ) (+ ) < ε ε < ( + ) ε Now choose δ = min, ( +. This guaranties that both assumptions made about δ in ) the course of this proof are taken into account simultaneously. Thus, if 0 < < δ, it follows that < ε. This completes the proof.
31 Pg (a), f( ) = 0, = Find the -intercept: = 0 = 0 The -intercept is (0, 0). Find few points: y = (b) lim f( ) = lim f( ) = +
32 (c) Since lim f ( ) = lim f( ) then lim f( ) = + lim f ( ) eists: 8., f( ) =, = (a) Find the -intercepts: = 0 ( )( + ) = 0 = 0 or + = 0 = or = The -intercept is (, 0) (the point (, 0) is not on the graph of f). Find the y-intercept: f(0) = 0 = The y-intercept is (0, ). The -coordinate of the verte is b 0 = = = 0 a ( ) The verte is (0, ). Find few points: y =
33 (b) lim f( ) = 0 lim f( ) = 0 + (c) Since lim f( ) = 0 lim f ( ) = lim f( ) then + lim f ( ) eists: lim = = = lim As, we have ( ) 0 and ( + ),so lim = (+ 5) (+ 5) lim lim lim + = + = ( + )( + ) ( + )( + ) + 5 ( ) + 5 = lim = = = + ( + ) ( + ) ( ) sin tan sin sin lim lim cos = = lim = lim cos 0 cos sin = lim lim = = 0 0 cos cos 0 = 6. t t tcost cost lim = lim = lim = lim t 0 tan t t 0 sin t t 0 sin t t 0 sin t cost t limcost t 0 cos0 = = = sin t lim t 0 t
34 9. sin sin tan cos sin lim lim lim lim = 0 sin 8 0 sin 8 = 0 8 sin8 = cos 0 8 sin8 8 cos sin sin sin lim lim = = 0 8 sin8 lim cos 0 8 sin8 = cos 8 0 sin8 cos sin sin lim 0 lim lim = 8 0 sin8 = 0 cos 8 sin8 lim cos = = cos5y sin ycot 5y sin y cot 5y sin y sin 5y lim = lim = lim y 0 ycot y y 0 y cot y y 0 y cos y sin y sin y cos5y sin y sin y sin y cos5y = lim = lim y 0 y sin5y cosy y 0 y sin5y cosy sin y y sin y sin y cos5y y cos 0 = lim lim lim = lim y 0 y y 0sin 5y y 0cos y y 0 sin 5y 5y cos0 5y sin y sin y sin y y y y y = lim = lim = lim y 0 5y sin5y y 0 5 sin5y 5 y 0 sin5y 5y 5y 5y sin y lim y 0 y = = = 5 sin 5y lim 5 5 y 0 5 y
35 . sinθ tanθ sin sin sin sin lim lim cosθ θ θ θ θ = = lim = lim θ cot θ cos θ θ θ θ θ cosθ cos θ θ 0 θ θ θ cosθ cos θ sin θ sinθ sin θ sin θ = lim lim lim = lim θ 0 θ θ 0 θ θ 0cosθcosθ θ 0 θ cos 0cos 0 = = Pg. 0. y = The first term is a rational function so it is continuous everywhere ecept = (it is not defined at = ), the second term is a polynomial function so it is continuous everywhere. Therefore, y is continuous on the intervals (, ) (, ).. y = ( + ) + The first term is a rational function so it is continuous everywhere ecept = (it is not defined at = ), the second term is a constant function so it is continuous everywhere. Therefore, y is continuous on the intervals (, ) (, ) y = + Find the domain: + = 0 + = 0 ( ) ( ) = 0 ( )( ) = 0 = or = The domain is the intervals (, ) (, ) (, ). y is a rational function so it is continuous on its domain or on the intervals (, ) (, ) (, ).
36 5. π π π lim cos cos cos t 0 = = 9 sec t 9 sec 0 9 π π = cos = cos = 6 π π π π cos = cos = cos = cos = 9 sec0 9 6 Since lim f ( t) = f(0) then the function is continuous at t = 0. t 0 6. lim csc + 5 tan = + 5 = + 5 = π 6 csc + 5 tan = + 5 = + 5 = π Since lim f( ) = f π 6 then the function is continuous at π = lim sin π e sin π e sin π + = = 0 = π 0 π sin e = sin = Since lim f( ) = f 0 then the function is continuous from the right at = ( ) 9. 9 g ( ) = Find one-sided limits at = : 9 ( )( + ) lim = lim = lim( + ) = + = 6 9 ( )( + ) lim = lim = lim( + ) = + = Thus, 9 lim = 6 To be continuous at = we must define g() = 6.
37 t 0. ht () = + t 0 t Find one-sided limits at t = : t + t 0 t + 5t t 0 t( t+ 5) ( t+ 5) lim = lim = lim t t t t t t ( t+ 5)( t ) = lim = lim( t + 5) = + 5 = 7 t t t t + t 0 t + 5t t 0 t( t+ 5) ( t+ 5) lim = lim = lim t t t t t t ( t+ 5)( t ) = lim = lim( t + 5) = + 5 = t t t Thus, t + t 0 lim = 7 t t To be continuous at t = we must define h() = 7.. s f() s = s Find one-sided limits at s = : s ( s )( s + s+ ) s + s+ + + lim = lim = lim = = s s s ( s )( s+ ) s s+ + s ( s )( s + s+ ) s + s+ + + lim = lim = lim = = s s s ( s )( s+ ) s s+ + Thus, s lim s = s To be continuous at s = we must define f () = = 0 Consider the function f() = 5 +. f( ) = ( ) 5( ) + = f( ) = ( ) 5( ) + = 9 f( ) = ( ) 5( ) + = f( ) = ( ) 5( ) + = 5 f(0) = 0 5(0) + = f() = 5() + = f() = 5() + =
38 f() = 5() + = 7 f() = 5() + = 5 f is a polynomial function, so it is continuous everywhere. f changes its sign on the intervals (, ), (0, ) and (, ). By the Intermediate Value Theorem f must have three zeros on these intervals. 56. F( ) = ( a) ( b) + Fa ( ) = ( a a) ( a b) + a= 0+ a= a Fb ( ) = ( b a) ( b b) + b= 0+ b= b a+ b The number lies between a and b, so by the Intermediate Value Theorem there eists a a+ b number c between a and b such that f() c =. Pg f( ) = (a) (b) lim lim lim = = = = lim lim lim + = = = = h ( ) = (a) 7 7 lim lim lim = = = =
39 (b) = = = lim lim lim = 7 9. g ( ) = (a) (b) lim = lim = lim = = lim = lim = lim = = h ( ) = (a) (b) lim = lim = lim = = lim = lim = lim = =
40 . h ( ) = (a) (b) lim lim = = lim = = lim lim = = lim = = lim lim lim = = = = = lim lim lim lim lim = = = = = = lim = lim = lim = lim =. lim = 7 ( 7) because ( 7) 0 as 7.. lim = 0 ( + ) because 0 and + as 0.
41 55. (a) lim = lim lim = 0 lim = (b) lim = lim lim = 0 lim = ( ) 8 (c) lim 0 = = = = = (d) ( ) lim = = + = 56. (a) (b) (c) (d) lim = lim = ( + ) + because and as. + lim = lim = + ( + ) because and as. + ( )( + ) lim = lim = because 0, + and + 6 as lim = = ( )( ) lim ( ) = lim ( + ) + 9 = lim = lim = lim = because as.
42 8. ( ) ( + 5 )( + 5+ ) lim + 5 = lim ( ) = lim = lim = lim = because as. 0. y = The domain is the intervals (, ) (, ). ( ) y( ) = = + The function is neither odd nor even. Find the -intercepts: = 0 = 0 ( )( + ) = 0 = or = The -intercepts are (, 0) and (, 0). Find the y-intercept: 0 y(0) = = 0 The y-intercept is (0, ). The denominator is zero at =, so the line = is a vertical asymptote. To find the equation of a slant asymptote we use the long division: +
43 Thus, = + The slant asymptote is y = + Find few points: y =
44 0. y = + The domain is the intervals (, ) (, ). ( ) y( ) = = ( ) + + The function is neither odd nor even. Find the -intercepts: = 0 + = 0 ( )( + ) = 0 = or = The -intercepts are (, 0) and (, 0). Find the y-intercept: 0 y(0) = = (0) + The y-intercept is 0,. The denominator is zero at =, so the line = is a vertical asymptote. To find the equation of a slant asymptote we use the long division: + + Thus, = The slant asymptote is y =. Find few points: y = +
45
Practice Exam 2. Conceptual Questions. 1. State a Basic identity and then verify it. (a) Identity: Solution: One identity is csc(θ) = 1
Conceptual Questions. State a Basic identity and then verify it. a) Identity: Solution: One identity is cscθ) = sinθ) Practice Exam b) Verification: Solution: Given the point of intersection x, y) of the
Διαβάστε περισσότεραSection 8.3 Trigonometric Equations
99 Section 8. Trigonometric Equations Objective 1: Solve Equations Involving One Trigonometric Function. In this section and the next, we will exple how to solving equations involving trigonometric functions.
Διαβάστε περισσότεραSection 9.2 Polar Equations and Graphs
180 Section 9. Polar Equations and Graphs In this section, we will be graphing polar equations on a polar grid. In the first few examples, we will write the polar equation in rectangular form to help identify
Διαβάστε περισσότερα3.4 SUM AND DIFFERENCE FORMULAS. NOTE: cos(α+β) cos α + cos β cos(α-β) cos α -cos β
3.4 SUM AND DIFFERENCE FORMULAS Page Theorem cos(αβ cos α cos β -sin α cos(α-β cos α cos β sin α NOTE: cos(αβ cos α cos β cos(α-β cos α -cos β Proof of cos(α-β cos α cos β sin α Let s use a unit circle
Διαβάστε περισσότεραCRASH COURSE IN PRECALCULUS
CRASH COURSE IN PRECALCULUS Shiah-Sen Wang The graphs are prepared by Chien-Lun Lai Based on : Precalculus: Mathematics for Calculus by J. Stuwart, L. Redin & S. Watson, 6th edition, 01, Brooks/Cole Chapter
Διαβάστε περισσότεραPARTIAL NOTES for 6.1 Trigonometric Identities
PARTIAL NOTES for 6.1 Trigonometric Identities tanθ = sinθ cosθ cotθ = cosθ sinθ BASIC IDENTITIES cscθ = 1 sinθ secθ = 1 cosθ cotθ = 1 tanθ PYTHAGOREAN IDENTITIES sin θ + cos θ =1 tan θ +1= sec θ 1 + cot
Διαβάστε περισσότεραSection 7.6 Double and Half Angle Formulas
09 Section 7. Double and Half Angle Fmulas To derive the double-angles fmulas, we will use the sum of two angles fmulas that we developed in the last section. We will let α θ and β θ: cos(θ) cos(θ + θ)
Διαβάστε περισσότεραHomework 8 Model Solution Section
MATH 004 Homework Solution Homework 8 Model Solution Section 14.5 14.6. 14.5. Use the Chain Rule to find dz where z cosx + 4y), x 5t 4, y 1 t. dz dx + dy y sinx + 4y)0t + 4) sinx + 4y) 1t ) 0t + 4t ) sinx
Διαβάστε περισσότεραANSWERSHEET (TOPIC = DIFFERENTIAL CALCULUS) COLLECTION #2. h 0 h h 0 h h 0 ( ) g k = g 0 + g 1 + g g 2009 =?
Teko Classes IITJEE/AIEEE Maths by SUHAAG SIR, Bhopal, Ph (0755) 3 00 000 www.tekoclasses.com ANSWERSHEET (TOPIC DIFFERENTIAL CALCULUS) COLLECTION # Question Type A.Single Correct Type Q. (A) Sol least
Διαβάστε περισσότεραAreas and Lengths in Polar Coordinates
Kiryl Tsishchanka Areas and Lengths in Polar Coordinates In this section we develop the formula for the area of a region whose boundary is given by a polar equation. We need to use the formula for the
Διαβάστε περισσότεραInverse trigonometric functions & General Solution of Trigonometric Equations. ------------------ ----------------------------- -----------------
Inverse trigonometric functions & General Solution of Trigonometric Equations. 1. Sin ( ) = a) b) c) d) Ans b. Solution : Method 1. Ans a: 17 > 1 a) is rejected. w.k.t Sin ( sin ) = d is rejected. If sin
Διαβάστε περισσότεραTrigonometric Formula Sheet
Trigonometric Formula Sheet Definition of the Trig Functions Right Triangle Definition Assume that: 0 < θ < or 0 < θ < 90 Unit Circle Definition Assume θ can be any angle. y x, y hypotenuse opposite θ
Διαβάστε περισσότεραReview Test 3. MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Review Test MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Find the exact value of the expression. 1) sin - 11π 1 1) + - + - - ) sin 11π 1 ) ( -
Διαβάστε περισσότεραCHAPTER 25 SOLVING EQUATIONS BY ITERATIVE METHODS
CHAPTER 5 SOLVING EQUATIONS BY ITERATIVE METHODS EXERCISE 104 Page 8 1. Find the positive root of the equation x + 3x 5 = 0, correct to 3 significant figures, using the method of bisection. Let f(x) =
Διαβάστε περισσότεραMock Exam 7. 1 Hong Kong Educational Publishing Company. Section A 1. Reference: HKDSE Math M Q2 (a) (1 + kx) n 1M + 1A = (1) =
Mock Eam 7 Mock Eam 7 Section A. Reference: HKDSE Math M 0 Q (a) ( + k) n nn ( )( k) + nk ( ) + + nn ( ) k + nk + + + A nk... () nn ( ) k... () From (), k...() n Substituting () into (), nn ( ) n 76n 76n
Διαβάστε περισσότερα2 Composition. Invertible Mappings
Arkansas Tech University MATH 4033: Elementary Modern Algebra Dr. Marcel B. Finan Composition. Invertible Mappings In this section we discuss two procedures for creating new mappings from old ones, namely,
Διαβάστε περισσότεραAreas and Lengths in Polar Coordinates
Kiryl Tsishchanka Areas and Lengths in Polar Coordinates In this section we develop the formula for the area of a region whose boundary is given by a polar equation. We need to use the formula for the
Διαβάστε περισσότεραSection 8.2 Graphs of Polar Equations
Section 8. Graphs of Polar Equations Graphing Polar Equations The graph of a polar equation r = f(θ), or more generally F(r,θ) = 0, consists of all points P that have at least one polar representation
Διαβάστε περισσότεραTrigonometry 1.TRIGONOMETRIC RATIOS
Trigonometry.TRIGONOMETRIC RATIOS. If a ray OP makes an angle with the positive direction of X-axis then y x i) Sin ii) cos r r iii) tan x y (x 0) iv) cot y x (y 0) y P v) sec x r (x 0) vi) cosec y r (y
Διαβάστε περισσότεραCHAPTER 101 FOURIER SERIES FOR PERIODIC FUNCTIONS OF PERIOD
CHAPTER FOURIER SERIES FOR PERIODIC FUNCTIONS OF PERIOD EXERCISE 36 Page 66. Determine the Fourier series for the periodic function: f(x), when x +, when x which is periodic outside this rge of period.
Διαβάστε περισσότεραExercises 10. Find a fundamental matrix of the given system of equations. Also find the fundamental matrix Φ(t) satisfying Φ(0) = I. 1.
Exercises 0 More exercises are available in Elementary Differential Equations. If you have a problem to solve any of them, feel free to come to office hour. Problem Find a fundamental matrix of the given
Διαβάστε περισσότεραIntegrals in cylindrical, spherical coordinates (Sect. 15.7)
Integrals in clindrical, spherical coordinates (Sect. 5.7 Integration in spherical coordinates. Review: Clindrical coordinates. Spherical coordinates in space. Triple integral in spherical coordinates.
Διαβάστε περισσότεραChapter 6 BLM Answers
Chapter 6 BLM Answers BLM 6 Chapter 6 Prerequisite Skills. a) i) II ii) IV iii) III i) 5 ii) 7 iii) 7. a) 0, c) 88.,.6, 59.6 d). a) 5 + 60 n; 7 + n, c). rad + n rad; 7 9,. a) 5 6 c) 69. d) 0.88 5. a) negative
Διαβάστε περισσότεραSolution to Review Problems for Midterm III
Solution to Review Problems for Mierm III Mierm III: Friday, November 19 in class Topics:.8-.11, 4.1,4. 1. Find the derivative of the following functions and simplify your answers. (a) x(ln(4x)) +ln(5
Διαβάστε περισσότεραHomework 3 Solutions
Homework 3 Solutions Igor Yanovsky (Math 151A TA) Problem 1: Compute the absolute error and relative error in approximations of p by p. (Use calculator!) a) p π, p 22/7; b) p π, p 3.141. Solution: For
Διαβάστε περισσότεραIf we restrict the domain of y = sin x to [ π, π ], the restrict function. y = sin x, π 2 x π 2
Chapter 3. Analytic Trigonometry 3.1 The inverse sine, cosine, and tangent functions 1. Review: Inverse function (1) f 1 (f(x)) = x for every x in the domain of f and f(f 1 (x)) = x for every x in the
Διαβάστε περισσότεραMatrices and Determinants
Matrices and Determinants SUBJECTIVE PROBLEMS: Q 1. For what value of k do the following system of equations possess a non-trivial (i.e., not all zero) solution over the set of rationals Q? x + ky + 3z
Διαβάστε περισσότεραIf we restrict the domain of y = sin x to [ π 2, π 2
Chapter 3. Analytic Trigonometry 3.1 The inverse sine, cosine, and tangent functions 1. Review: Inverse function (1) f 1 (f(x)) = x for every x in the domain of f and f(f 1 (x)) = x for every x in the
Διαβάστε περισσότεραSolutions to Exercise Sheet 5
Solutions to Eercise Sheet 5 jacques@ucsd.edu. Let X and Y be random variables with joint pdf f(, y) = 3y( + y) where and y. Determine each of the following probabilities. Solutions. a. P (X ). b. P (X
Διαβάστε περισσότεραC.S. 430 Assignment 6, Sample Solutions
C.S. 430 Assignment 6, Sample Solutions Paul Liu November 15, 2007 Note that these are sample solutions only; in many cases there were many acceptable answers. 1 Reynolds Problem 10.1 1.1 Normal-order
Διαβάστε περισσότεραMath221: HW# 1 solutions
Math: HW# solutions Andy Royston October, 5 7.5.7, 3 rd Ed. We have a n = b n = a = fxdx = xdx =, x cos nxdx = x sin nx n sin nxdx n = cos nx n = n n, x sin nxdx = x cos nx n + cos nxdx n cos n = + sin
Διαβάστε περισσότεραExample Sheet 3 Solutions
Example Sheet 3 Solutions. i Regular Sturm-Liouville. ii Singular Sturm-Liouville mixed boundary conditions. iii Not Sturm-Liouville ODE is not in Sturm-Liouville form. iv Regular Sturm-Liouville note
Διαβάστε περισσότεραb. Use the parametrization from (a) to compute the area of S a as S a ds. Be sure to substitute for ds!
MTH U341 urface Integrals, tokes theorem, the divergence theorem To be turned in Wed., Dec. 1. 1. Let be the sphere of radius a, x 2 + y 2 + z 2 a 2. a. Use spherical coordinates (with ρ a) to parametrize.
Διαβάστε περισσότερα9.09. # 1. Area inside the oval limaçon r = cos θ. To graph, start with θ = 0 so r = 6. Compute dr
9.9 #. Area inside the oval limaçon r = + cos. To graph, start with = so r =. Compute d = sin. Interesting points are where d vanishes, or at =,,, etc. For these values of we compute r:,,, and the values
Διαβάστε περισσότερα3.4. Click here for solutions. Click here for answers. CURVE SKETCHING. y cos x sin x. x 1 x 2. x 2 x 3 4 y 1 x 2. x 5 2
SECTION. CURVE SKETCHING. CURVE SKETCHING A Click here for answers. S Click here for solutions. 9. Use the guidelines of this section to sketch the curve. cos sin. 5. 6 8 7 0. cot, 0.. 9. cos sin. sin
Διαβάστε περισσότεραDifferential equations
Differential equations Differential equations: An equation inoling one dependent ariable and its deriaties w. r. t one or more independent ariables is called a differential equation. Order of differential
Διαβάστε περισσότεραderivation of the Laplacian from rectangular to spherical coordinates
derivation of the Laplacian from rectangular to spherical coordinates swapnizzle 03-03- :5:43 We begin by recognizing the familiar conversion from rectangular to spherical coordinates (note that φ is used
Διαβάστε περισσότεραMathCity.org Merging man and maths
MathCity.org Merging man and maths Exercise 10. (s) Page Textbook of Algebra and Trigonometry for Class XI Available online @, Version:.0 Question # 1 Find the values of sin, and tan when: 1 π (i) (ii)
Διαβάστε περισσότεραHOMEWORK 4 = G. In order to plot the stress versus the stretch we define a normalized stretch:
HOMEWORK 4 Problem a For the fast loading case, we want to derive the relationship between P zz and λ z. We know that the nominal stress is expressed as: P zz = ψ λ z where λ z = λ λ z. Therefore, applying
Διαβάστε περισσότεραk A = [k, k]( )[a 1, a 2 ] = [ka 1,ka 2 ] 4For the division of two intervals of confidence in R +
Chapter 3. Fuzzy Arithmetic 3- Fuzzy arithmetic: ~Addition(+) and subtraction (-): Let A = [a and B = [b, b in R If x [a and y [b, b than x+y [a +b +b Symbolically,we write A(+)B = [a (+)[b, b = [a +b
Διαβάστε περισσότεραFourier Series. MATH 211, Calculus II. J. Robert Buchanan. Spring Department of Mathematics
Fourier Series MATH 211, Calculus II J. Robert Buchanan Department of Mathematics Spring 2018 Introduction Not all functions can be represented by Taylor series. f (k) (c) A Taylor series f (x) = (x c)
Διαβάστε περισσότεραCBC MATHEMATICS DIVISION MATH 2412-PreCalculus Exam Formula Sheets
System of Equations and Matrices 3 Matrix Row Operations: MATH 41-PreCalculus Switch any two rows. Multiply any row by a nonzero constant. Add any constant-multiple row to another Even and Odd functions
Διαβάστε περισσότεραST5224: Advanced Statistical Theory II
ST5224: Advanced Statistical Theory II 2014/2015: Semester II Tutorial 7 1. Let X be a sample from a population P and consider testing hypotheses H 0 : P = P 0 versus H 1 : P = P 1, where P j is a known
Διαβάστε περισσότεραEE512: Error Control Coding
EE512: Error Control Coding Solution for Assignment on Finite Fields February 16, 2007 1. (a) Addition and Multiplication tables for GF (5) and GF (7) are shown in Tables 1 and 2. + 0 1 2 3 4 0 0 1 2 3
Διαβάστε περισσότεραParametrized Surfaces
Parametrized Surfaces Recall from our unit on vector-valued functions at the beginning of the semester that an R 3 -valued function c(t) in one parameter is a mapping of the form c : I R 3 where I is some
Διαβάστε περισσότεραDESIGN OF MACHINERY SOLUTION MANUAL h in h 4 0.
DESIGN OF MACHINERY SOLUTION MANUAL -7-1! PROBLEM -7 Statement: Design a double-dwell cam to move a follower from to 25 6, dwell for 12, fall 25 and dwell for the remader The total cycle must take 4 sec
Διαβάστε περισσότερα2. THEORY OF EQUATIONS. PREVIOUS EAMCET Bits.
EAMCET-. THEORY OF EQUATIONS PREVIOUS EAMCET Bits. Each of the roots of the equation x 6x + 6x 5= are increased by k so that the new transformed equation does not contain term. Then k =... - 4. - Sol.
Διαβάστε περισσότεραSpherical Coordinates
Spherical Coordinates MATH 311, Calculus III J. Robert Buchanan Department of Mathematics Fall 2011 Spherical Coordinates Another means of locating points in three-dimensional space is known as the spherical
Διαβάστε περισσότεραRectangular Polar Parametric
Harold s Precalculus Rectangular Polar Parametric Cheat Sheet 15 October 2017 Point Line Rectangular Polar Parametric f(x) = y (x, y) (a, b) Slope-Intercept Form: y = mx + b Point-Slope Form: y y 0 = m
Διαβάστε περισσότεραLecture 34 Bootstrap confidence intervals
Lecture 34 Bootstrap confidence intervals Confidence Intervals θ: an unknown parameter of interest We want to find limits θ and θ such that Gt = P nˆθ θ t If G 1 1 α is known, then P θ θ = P θ θ = 1 α
Διαβάστε περισσότεραPhys460.nb Solution for the t-dependent Schrodinger s equation How did we find the solution? (not required)
Phys460.nb 81 ψ n (t) is still the (same) eigenstate of H But for tdependent H. The answer is NO. 5.5.5. Solution for the tdependent Schrodinger s equation If we assume that at time t 0, the electron starts
Διαβάστε περισσότερα26 28 Find an equation of the tangent line to the curve at the given point Discuss the curve under the guidelines of Section
SECTION 5. THE NATURAL LOGARITHMIC FUNCTION 5. THE NATURAL LOGARITHMIC FUNCTION A Click here for answers. S Click here for solutions. 4 Use the Laws of Logarithms to epand the quantit.. ln ab. ln c. ln
Διαβάστε περισσότεραMath 6 SL Probability Distributions Practice Test Mark Scheme
Math 6 SL Probability Distributions Practice Test Mark Scheme. (a) Note: Award A for vertical line to right of mean, A for shading to right of their vertical line. AA N (b) evidence of recognizing symmetry
Διαβάστε περισσότεραProblem 3.1 Vector A starts at point (1, 1, 3) and ends at point (2, 1,0). Find a unit vector in the direction of A. Solution: A = 1+9 = 3.
Problem 3.1 Vector A starts at point (1, 1, 3) and ends at point (, 1,0). Find a unit vector in the direction of A. Solution: A = ˆx( 1)+ŷ( 1 ( 1))+ẑ(0 ( 3)) = ˆx+ẑ3, A = 1+9 = 3.16, â = A A = ˆx+ẑ3 3.16
Διαβάστε περισσότεραChapter 6: Systems of Linear Differential. be continuous functions on the interval
Chapter 6: Systems of Linear Differential Equations Let a (t), a 2 (t),..., a nn (t), b (t), b 2 (t),..., b n (t) be continuous functions on the interval I. The system of n first-order differential equations
Διαβάστε περισσότεραΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 19/5/2007
Οδηγίες: Να απαντηθούν όλες οι ερωτήσεις. Αν κάπου κάνετε κάποιες υποθέσεις να αναφερθούν στη σχετική ερώτηση. Όλα τα αρχεία που αναφέρονται στα προβλήματα βρίσκονται στον ίδιο φάκελο με το εκτελέσιμο
Διαβάστε περισσότεραNowhere-zero flows Let be a digraph, Abelian group. A Γ-circulation in is a mapping : such that, where, and : tail in X, head in
Nowhere-zero flows Let be a digraph, Abelian group. A Γ-circulation in is a mapping : such that, where, and : tail in X, head in : tail in X, head in A nowhere-zero Γ-flow is a Γ-circulation such that
Διαβάστε περισσότεραCYLINDRICAL & SPHERICAL COORDINATES
CYLINDRICAL & SPHERICAL COORDINATES Here we eamine two of the more popular alternative -dimensional coordinate sstems to the rectangular coordinate sstem. First recall the basis of the Rectangular Coordinate
Διαβάστε περισσότεραQuadratic Expressions
Quadratic Expressions. The standard form of a quadratic equation is ax + bx + c = 0 where a, b, c R and a 0. The roots of ax + bx + c = 0 are b ± b a 4ac. 3. For the equation ax +bx+c = 0, sum of the roots
Διαβάστε περισσότεραSection 7.7 Product-to-Sum and Sum-to-Product Formulas
Section 7.7 Product-to-Sum and Sum-to-Product Fmulas Objective 1: Express Products as Sums To derive the Product-to-Sum Fmulas will begin by writing down the difference and sum fmulas of the cosine function:
Διαβάστε περισσότεραA Two-Sided Laplace Inversion Algorithm with Computable Error Bounds and Its Applications in Financial Engineering
Electronic Companion A Two-Sie Laplace Inversion Algorithm with Computable Error Bouns an Its Applications in Financial Engineering Ning Cai, S. G. Kou, Zongjian Liu HKUST an Columbia University Appenix
Διαβάστε περισσότεραis like multiplying by the conversion factor of. Dividing by 2π gives you the
Chapter Graphs of Trigonometric Functions Answer Ke. Radian Measure Answers. π. π. π. π. 7π. π 7. 70 8. 9. 0 0. 0. 00. 80. Multipling b π π is like multipling b the conversion factor of. Dividing b 0 gives
Διαβάστε περισσότεραAREAS AND LENGTHS IN POLAR COORDINATES. 25. Find the area inside the larger loop and outside the smaller loop
SECTIN 9. AREAS AND LENGTHS IN PLAR CRDINATES 9. AREAS AND LENGTHS IN PLAR CRDINATES A Click here for answers. S Click here for solutions. 8 Find the area of the region that is bounded by the given curve
Διαβάστε περισσότεραFinite Field Problems: Solutions
Finite Field Problems: Solutions 1. Let f = x 2 +1 Z 11 [x] and let F = Z 11 [x]/(f), a field. Let Solution: F =11 2 = 121, so F = 121 1 = 120. The possible orders are the divisors of 120. Solution: The
Διαβάστε περισσότεραComplete Solutions Manual for Calculus of a Single Variable, Volume 1. Calculus ELEVENTH EDITION
Complete Solutions Manual for Calculus of a Single Variable, Volume Calculus ELEVENTH EDITION Cengage Learning. All rights reserved. No distribution allowed without epress authorization. Ron Larson The
Διαβάστε περισσότεραOrdinal Arithmetic: Addition, Multiplication, Exponentiation and Limit
Ordinal Arithmetic: Addition, Multiplication, Exponentiation and Limit Ting Zhang Stanford May 11, 2001 Stanford, 5/11/2001 1 Outline Ordinal Classification Ordinal Addition Ordinal Multiplication Ordinal
Διαβάστε περισσότεραMath 446 Homework 3 Solutions. (1). (i): Reverse triangle inequality for metrics: Let (X, d) be a metric space and let x, y, z X.
Math 446 Homework 3 Solutions. (1). (i): Reverse triangle inequalit for metrics: Let (X, d) be a metric space and let x,, z X. Prove that d(x, z) d(, z) d(x, ). (ii): Reverse triangle inequalit for norms:
Διαβάστε περισσότεραStatistical Inference I Locally most powerful tests
Statistical Inference I Locally most powerful tests Shirsendu Mukherjee Department of Statistics, Asutosh College, Kolkata, India. shirsendu st@yahoo.co.in So far we have treated the testing of one-sided
Διαβάστε περισσότεραSCHOOL OF MATHEMATICAL SCIENCES G11LMA Linear Mathematics Examination Solutions
SCHOOL OF MATHEMATICAL SCIENCES GLMA Linear Mathematics 00- Examination Solutions. (a) i. ( + 5i)( i) = (6 + 5) + (5 )i = + i. Real part is, imaginary part is. (b) ii. + 5i i ( + 5i)( + i) = ( i)( + i)
Διαβάστε περισσότεραw o = R 1 p. (1) R = p =. = 1
Πανεπιστήµιο Κρήτης - Τµήµα Επιστήµης Υπολογιστών ΗΥ-570: Στατιστική Επεξεργασία Σήµατος 205 ιδάσκων : Α. Μουχτάρης Τριτη Σειρά Ασκήσεων Λύσεις Ασκηση 3. 5.2 (a) From the Wiener-Hopf equation we have:
Διαβάστε περισσότεραUniform Convergence of Fourier Series Michael Taylor
Uniform Convergence of Fourier Series Michael Taylor Given f L 1 T 1 ), we consider the partial sums of the Fourier series of f: N 1) S N fθ) = ˆfk)e ikθ. k= N A calculation gives the Dirichlet formula
Διαβάστε περισσότεραReview of Essential Skills- Part 1. Practice 1.4, page 38. Practise, Apply, Solve 1.7, page 57. Practise, Apply, Solve 1.
Review of Essential Skills- Part Operations with Rational Numbers, page. (e) 8 Exponent Laws, page 6. (a) 0 + 5 0, (d) (), (e) +, 8 + (h) 5, 9. (h) x 5. (d) v 5 Expanding, Simplifying, and Factoring Algebraic
Διαβάστε περισσότεραLecture 15 - Root System Axiomatics
Lecture 15 - Root System Axiomatics Nov 1, 01 In this lecture we examine root systems from an axiomatic point of view. 1 Reflections If v R n, then it determines a hyperplane, denoted P v, through the
Διαβάστε περισσότεραD Alembert s Solution to the Wave Equation
D Alembert s Solution to the Wave Equation MATH 467 Partial Differential Equations J. Robert Buchanan Department of Mathematics Fall 2018 Objectives In this lesson we will learn: a change of variable technique
Διαβάστε περισσότεραAnswer sheet: Third Midterm for Math 2339
Answer sheet: Third Midterm for Math 339 November 3, Problem. Calculate the iterated integrals (Simplify as much as possible) (a) e sin(x) dydx y e sin(x) dydx y sin(x) ln y ( cos(x)) ye y dx sin(x)(lne
Διαβάστε περισσότεραPartial Differential Equations in Biology The boundary element method. March 26, 2013
The boundary element method March 26, 203 Introduction and notation The problem: u = f in D R d u = ϕ in Γ D u n = g on Γ N, where D = Γ D Γ N, Γ D Γ N = (possibly, Γ D = [Neumann problem] or Γ N = [Dirichlet
Διαβάστε περισσότεραSecond Order Partial Differential Equations
Chapter 7 Second Order Partial Differential Equations 7.1 Introduction A second order linear PDE in two independent variables (x, y Ω can be written as A(x, y u x + B(x, y u xy + C(x, y u u u + D(x, y
Διαβάστε περισσότεραMATH423 String Theory Solutions 4. = 0 τ = f(s). (1) dτ ds = dxµ dτ f (s) (2) dτ 2 [f (s)] 2 + dxµ. dτ f (s) (3)
1. MATH43 String Theory Solutions 4 x = 0 τ = fs). 1) = = f s) ) x = x [f s)] + f s) 3) equation of motion is x = 0 if an only if f s) = 0 i.e. fs) = As + B with A, B constants. i.e. allowe reparametrisations
Διαβάστε περισσότεραLecture 2: Dirac notation and a review of linear algebra Read Sakurai chapter 1, Baym chatper 3
Lecture 2: Dirac notation and a review of linear algebra Read Sakurai chapter 1, Baym chatper 3 1 State vector space and the dual space Space of wavefunctions The space of wavefunctions is the set of all
Διαβάστε περισσότεραSCITECH Volume 13, Issue 2 RESEARCH ORGANISATION Published online: March 29, 2018
Journal of rogressive Research in Mathematics(JRM) ISSN: 2395-028 SCITECH Volume 3, Issue 2 RESEARCH ORGANISATION ublished online: March 29, 208 Journal of rogressive Research in Mathematics www.scitecresearch.com/journals
Διαβάστε περισσότεραProblem Set 9 Solutions. θ + 1. θ 2 + cotθ ( ) sinθ e iφ is an eigenfunction of the ˆ L 2 operator. / θ 2. φ 2. sin 2 θ φ 2. ( ) = e iφ. = e iφ cosθ.
Chemistry 362 Dr Jean M Standard Problem Set 9 Solutions The ˆ L 2 operator is defined as Verify that the angular wavefunction Y θ,φ) Also verify that the eigenvalue is given by 2! 2 & L ˆ 2! 2 2 θ 2 +
Διαβάστε περισσότεραDerivations of Useful Trigonometric Identities
Derivations of Useful Trigonometric Identities Pythagorean Identity This is a basic and very useful relationship which comes directly from the definition of the trigonometric ratios of sine and cosine
Διαβάστε περισσότεραΑπόκριση σε Μοναδιαία Ωστική Δύναμη (Unit Impulse) Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο. Απόστολος Σ.
Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο The time integral of a force is referred to as impulse, is determined by and is obtained from: Newton s 2 nd Law of motion states that the action
Διαβάστε περισσότεραOther Test Constructions: Likelihood Ratio & Bayes Tests
Other Test Constructions: Likelihood Ratio & Bayes Tests Side-Note: So far we have seen a few approaches for creating tests such as Neyman-Pearson Lemma ( most powerful tests of H 0 : θ = θ 0 vs H 1 :
Διαβάστε περισσότεραSrednicki Chapter 55
Srednicki Chapter 55 QFT Problems & Solutions A. George August 3, 03 Srednicki 55.. Use equations 55.3-55.0 and A i, A j ] = Π i, Π j ] = 0 (at equal times) to verify equations 55.-55.3. This is our third
Διαβάστε περισσότεραNotes on the Open Economy
Notes on the Open Econom Ben J. Heijdra Universit of Groningen April 24 Introduction In this note we stud the two-countr model of Table.4 in more detail. restated here for convenience. The model is Table.4.
Διαβάστε περισσότεραReminders: linear functions
Reminders: linear functions Let U and V be vector spaces over the same field F. Definition A function f : U V is linear if for every u 1, u 2 U, f (u 1 + u 2 ) = f (u 1 ) + f (u 2 ), and for every u U
Διαβάστε περισσότερα1. If log x 2 y 2 = a, then dy / dx = x 2 + y 2 1] xy 2] y / x. 3] x / y 4] none of these
1. If log x 2 y 2 = a, then dy / dx = x 2 + y 2 1] xy 2] y / x 3] x / y 4] none of these 1. If log x 2 y 2 = a, then x 2 + y 2 Solution : Take y /x = k y = k x dy/dx = k dy/dx = y / x Answer : 2] y / x
Διαβάστε περισσότεραEcon 2110: Fall 2008 Suggested Solutions to Problem Set 8 questions or comments to Dan Fetter 1
Eon : Fall 8 Suggested Solutions to Problem Set 8 Email questions or omments to Dan Fetter Problem. Let X be a salar with density f(x, θ) (θx + θ) [ x ] with θ. (a) Find the most powerful level α test
Διαβάστε περισσότεραSecond Order RLC Filters
ECEN 60 Circuits/Electronics Spring 007-0-07 P. Mathys Second Order RLC Filters RLC Lowpass Filter A passive RLC lowpass filter (LPF) circuit is shown in the following schematic. R L C v O (t) Using phasor
Διαβάστε περισσότεραPrinciples of Mathematics 12 Answer Key, Contents 185
Principles of Mathematics Answer Ke, Contents 85 Module : Section Trigonometr Trigonometric Functions Lesson The Trigonometric Values for θ, 0 θ 60 86 Lesson Solving Trigonometric Equations for 0 θ 60
Διαβάστε περισσότεραTRIGONOMETRIC FUNCTIONS
Chapter TRIGONOMETRIC FUNCTIONS. Overview.. The word trigonometry is derived from the Greek words trigon and metron which means measuring the sides of a triangle. An angle is the amount of rotation of
Διαβάστε περισσότεραF19MC2 Solutions 9 Complex Analysis
F9MC Solutions 9 Complex Analysis. (i) Let f(z) = eaz +z. Then f is ifferentiable except at z = ±i an so by Cauchy s Resiue Theorem e az z = πi[res(f,i)+res(f, i)]. +z C(,) Since + has zeros of orer at
Διαβάστε περισσότερα11.4 Graphing in Polar Coordinates Polar Symmetries
.4 Graphing in Polar Coordinates Polar Symmetries x axis symmetry y axis symmetry origin symmetry r, θ = r, θ r, θ = r, θ r, θ = r, + θ .4 Graphing in Polar Coordinates Polar Symmetries x axis symmetry
Διαβάστε περισσότερα10/3/ revolution = 360 = 2 π radians = = x. 2π = x = 360 = : Measures of Angles and Rotations
//.: Measures of Angles and Rotations I. Vocabulary A A. Angle the union of two rays with a common endpoint B. BA and BC C. B is the vertex. B C D. You can think of BA as the rotation of (clockwise) with
Διαβάστε περισσότεραFractional Colorings and Zykov Products of graphs
Fractional Colorings and Zykov Products of graphs Who? Nichole Schimanski When? July 27, 2011 Graphs A graph, G, consists of a vertex set, V (G), and an edge set, E(G). V (G) is any finite set E(G) is
Διαβάστε περισσότεραECE Spring Prof. David R. Jackson ECE Dept. Notes 2
ECE 634 Spring 6 Prof. David R. Jackson ECE Dept. Notes Fields in a Source-Free Region Example: Radiation from an aperture y PEC E t x Aperture Assume the following choice of vector potentials: A F = =
Διαβάστε περισσότεραLecture 2. Soundness and completeness of propositional logic
Lecture 2 Soundness and completeness of propositional logic February 9, 2004 1 Overview Review of natural deduction. Soundness and completeness. Semantics of propositional formulas. Soundness proof. Completeness
Διαβάστε περισσότεραCHAPTER 48 APPLICATIONS OF MATRICES AND DETERMINANTS
CHAPTER 48 APPLICATIONS OF MATRICES AND DETERMINANTS EXERCISE 01 Page 545 1. Use matrices to solve: 3x + 4y x + 5y + 7 3x + 4y x + 5y 7 Hence, 3 4 x 0 5 y 7 The inverse of 3 4 5 is: 1 5 4 1 5 4 15 8 3
Διαβάστε περισσότεραJesse Maassen and Mark Lundstrom Purdue University November 25, 2013
Notes on Average Scattering imes and Hall Factors Jesse Maassen and Mar Lundstrom Purdue University November 5, 13 I. Introduction 1 II. Solution of the BE 1 III. Exercises: Woring out average scattering
Διαβάστε περισσότερα