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HSC Extension 2 Mathematics
HSC Extension 2 Mathematics Topics • Complex Numbers • Curve Sketching • Polynomials • Integration • Conics • Volumes • Mechanics • Harder Extension 1
Maths Extension 2 - Complex Numbers Complex Number Rules ! z = x + iy ! A complex number is represented in this form ! The conjugate, z , is x – iy ! z = x – iy ! z = rcis _____OR ! z = r(cosθ + isinθ) ! r= x2 + y 2  y ! θ = tan-1   x ! |z1z2| = |z1||z2| ! ! The Modulus-Argument (Mod-Arg) form of representing a complex number ! The Modulus of a complex number Note: |z| = r ! The Argument of a complex number Note: arg z = θ ! (left) Multiplication and Division rules of the Modulus and the Argument z1 |z | = 1 z2 | z2 | ! arg(z1z2) = argz1 + argz2 ± 2π z  ! arg  1  = arg z1 − arg z2 ± 2π  z2  ! |z| = | z | = x2 + y 2 ! (below) More complex number rules ! z z = |z|2 = | z |2 = x2 + y2 ! z + z = 2x ! z1 z2 = z1 z2 ! z - z = 2yi z  z !  1 = 1 z  z  2 2 ! z1 ± z2 = z1 ± z2 ! arg z = –arg z ! z-1 = 1 z = 2 z z ! zn + 1 = (z + 1)(zn-1 – zn-2 + … - z + 1)_______________n is odd ! zn – 1 = (z - 1)(zn-1 + zn-2 + … + z + 1)_______________n is odd ! zn – 1 = (z - 1)(z + 1)(zn-2 + zn-4 + … + z + 1)_________-n is even ! arg (zn) = n arg z http://www.geocities.com/fatmuscle/HSC/ 1
Maths Extension 2 - Complex Numbers General Ideas of Complex Numbers Complex Numbers are written in the form of a real part and an imaginary part Complex Number = x + iy Real =x Imaginary =y i i2 i3 i4 = −1 = –1 = –i =1 P(x,y) Modulus |z| = r = r x2 + y 2 Argument  y arg z = θ = tan-1   x θ Modulus-Argument form of a Complex Number z = r(cosθ + isinθ) z = rcisθ x r x = rcosθ ∴ z = x + iy = rcisθ + isinθ = r(cosθ + isinθ) = rcisθ cosθ = Euler’s Formula eiθ = cosθ = isinθ DeMoivre’s Theorem y r y = rsinθ sinθ = zn = rncis nθ (cosθ + i sin θ ) n = cos nθ + i sin nθ http://www.geocities.com/fatmuscle/HSC/ 2
Maths Extension 2 - Complex Numbers Proof by Mathematical Induction (cosθ + i sin θ ) n = cos nθ + i sin nθ Let n = 1 LHS 1 _ (cosθ + i sin θ ) = cosθ + i sin θ RHS = cos(1)θ + i sin(1)θ = cosθ + i sin θ ∴true for n = 1 Assume true for n = k _ (cosθ + i sin θ ) k Let n = k + 1 _ (cosθ + i sin θ ) k +1 = (cosθ + i sin θ )1 (cosθ + i sin ϑ ) k = (cosθ + i sin θ )(cos kθ + i sin kϑ ) = cosθ cos kθ − sin θ sin kθ + i cosθ sin kθ + i sin θ cosθ = cos(θ + kθ ) + i sin(θ + kθ ) = cos(k + 1)θ + i sin(k + 1)θ = RHS = cos kθ + i sin kθ = cos(k + 1)θ + i sin( k + 1)θ ∴true for n = k + 1 True for n = 1 True for n = k True for n = k + 1 ∴True for all positive integer values of n http://www.geocities.com/fatmuscle/HSC/ 3
Maths Extension 2 - Complex Numbers Expressing Complex Numbers in Mod-Arg Form –3 + 3i Mod z = x2 + y 2 9+9 = 18 =3 2 = θ Arg  y θ = tan-1   x 3   −3 = tan-1  = tan-1 –1 = 135° 3π = 4 z = rcisθ 3π 3π   ∴ −3 + 3i = 3 2  cos + i sin  4 4   Expressing Complex Numbers in x + iy form 2cis 3π 2 3π 3π   = 2  cos + i sin  2 2   1   1 = 2 − +i  2 2  = –1 + i Express in x + iy form from a quadratic formula −1± 1− 4 x = 1 3 2 i =− + 2 2 − 1 ± 3 × −1 = 2 1 3 i =− − − 1 ± 3i 2 2 = 2 Conjugate pairs http://www.geocities.com/fatmuscle/HSC/ 4
Maths Extension 2 - Complex Numbers Theory (a + ib) + (c + id) (a + ib) – (c + id) Example Addition (3 + 4i) + (2 + 5i) = a + ib + c + id = (a + c) + i(b + d) = a + ib – c – id = (a – c) + i(b – d) Subtraction (3 + 4i) - (2 + 5i) (a + ib)(c + id) Multiplication = ac + iad + ibc – bd (3 + 4i)(2 + 5i) = (ac – bd) + i(ad + bc) a + ib c + id a + ib c − id = × c + id c − id Division 3 + 4i 2 + 5i = 3 + 4i + 2 + 5i = 5 + 9i = 3 + 4i – 2 – 5i =1–i = 6 + 15i + 8i – 20 = –14 + 23i = 3 + 4i 2 − 5i × 2 + 5i 2 − 5i = ac − iad + ibc + bd c2 + d 2 = 6 − 15i + 8i + 20 4 + 25 = (ac + bd ) (bc − ad ) +i 2 c2 + d 2 c − d2 = 26 7 −i 29 29 http://www.geocities.com/fatmuscle/HSC/ 5
Maths Extension 2 - Complex Numbers Finding Square Roots of Complex Numbers x + iy x + iy = a + ib = (a + ib)2 = a2 – b2 + 2aib ∴ x = a2 – b2 y = 2aib Eg. Find the Square Root of 5 + 12i 5 + 12i 5 + 12i = a + ib = a2 – b2 + 2aib 5 = a2 – b2 5= 12 = 2ab 6 = ab 6 a= b 36 – b2 2 b ! Simultaneous Equations ! Find b 5b2 = 36 – b4 0 = b4 + 5b2 – 36 0 = (b2 – 4)(b2 + 9) ∴ b = ±2 a= 6 2 =3 a= 6 −2 = –3 Because square roots of ! 3 + 2i complex numbers come in ____OR ! –3 –2i conjugate pairs, 5 + 12i = To test these, we just square them (3 + 2i)2 = (3 + 2i)(3 + 2i) = 9 + 12i – 4 = 5 + 12i (–3 –2i)2 = (–3 –2i)(–3 –2i) = 9 + 12i – 4 = 5 + 12i http://www.geocities.com/fatmuscle/HSC/ 6
Maths Extension 2 - Complex Numbers Complex Numbers on the Argand Diagram Z2 Z1 Subtraction of Vectors Flip the tail, and do a normal addition |z1 – zz| ≥ |z1| – |z2| Addition of Vectors Add tip to tail |z1 + z2| ≤ |z1| + |z2| z1 – zz z2 – z1 Z3 Z4 Z2 Z1 Multiplication of vectors arg z3 = arg z1 + arg z2 |z3| = |z1||z2| z4 = iz2 Multiplication of i, is a rotation through an π angle of in the anti-clockwise direction. 2 Division of i, is a rotation through an angle of π in the clockwise direction. 2 http://www.geocities.com/fatmuscle/HSC/ 7
Maths Extension 2 - Complex Numbers LOCUS |z – (a + ib)| = |z – (x + iy)| |z – (–4 + 3i)| = |z – (–2 – 5i)| by first principles: |x + iy + 4 – 3i| |(4 + x) + i(–3 + y)| -4+3i (4 + x )2 + (− 3 + y )2 = (2 + x )2 + (5 + y )2 x 2 + y 2 + 8 x − 6 y + 25 = x 2 + y 2 + 4 x + 10 y + 29 = 4 x + 10 y + 4 8x 6y + 25 = 4x − 4 16 y x 1 ∴y = − 4 4 -2-5i |z – (a + ib)| = k = |x + iy + 2 + 5i| = |(2 + x) + i(5 + y)| k is the distance |z – 2| =3 by first principles: |x + iy – 2| 3 (x − 2) 2 + y2 ( x − 2) + y 2 2 =3 2 =3 __= 9 arg( z − a) =θ arg( z − b) arg(z – 1) – arg(z + 1) = π ________OR 4 arg( z − 1) π = arg( z + 1) 4 A B Can either be 2 circles Check the angle ± ? Work out the locus by drawing a line and estimating the angle A = arg(z + 1) B = arg(z – 1) http://www.geocities.com/fatmuscle/HSC/ 8
Maths Extension 2 - Graphs Graphing y = f(x) y = f (x) y 10 2 (x-2) -3 ((x-2)^2)-3 5 x 10 5 5 10 5 x +C 10 5 10 y = f ( x) + C y ! C is positive, the function is shifted up ! C is negative, the function is shifted down 10 5 10 5 5 –C 10 y = f (x − C) y 10 ! C > 0, shift function right C units ! C < 0, shift function left C units The function becomes: y = f ( x + C ) 5 x 10 5 5 5 –C 10 C 10 http://www.geocities.com/fatmuscle/HSC/ 1
Maths Extension 2 - Graphs y = C. f (x) y ! C > 1, the function is ‘steeper’ ! 0 < C < 1, the function is ‘shallower’ 10 5 x 10 5 5 10 5 10 y = f (Cx ) ! Multiply the x co-ordinate by 1 C y 10 ! C > 1, thinner, steeper ! C < 1, fatter, shallower ! (x, 1) remains the same 5 x 10 5 5 10 5 10 http://www.geocities.com/fatmuscle/HSC/ 2
Maths Extension 2 - Graphs y = − f (x) y ! Flip about the x-axis ! All positive become negative ! All negative become positive 10 5 x 10 5 5 10 5 10 y = f (− x) y 10 ! Flip about the y-axis 5 x 10 5 5 10 5 10 http://www.geocities.com/fatmuscle/HSC/ 3
Maths Extension 2 - Graphs y = f (x) y 10 ! All negative y values become positive 5 x 10 5 5 10 5 10 y = f (x) y 10 ! Positive x values are ‘mirrored’ about the y-axis 5 x 10 5 5 10 5 10 y = f (x) y 10 ! Positive y values are ‘mirrored’ about the x-axis 5 x 10 5 5 10 5 10 http://www.geocities.com/fatmuscle/HSC/ 4
Maths Extension 2 - Graphs y= f (x) y ! If y value is > 1, y value is smaller but still > 1 ! If y value is = 1, y value is the same ! If y value is 0 < y < 1, y value is larger but still < 1 ! If y value is < 0, y value doesn’t exist 10 5 x 10 5 5 10 y 5 2 1. 5 1 10 0. 5 y 2 = f ( x) - 5 0. - 5 0. x 0. 5 1 y 10 sketch y= f (x) and y = − f (x) Both Brown and Green are y 2 = f ( x) 5 x 10 5 5 10 5 10 y = [ f (x)] 2 y = [ f ( x )] y C 20 If y value: ! = 0, y remains 0 ! > 1, y becomes steeper (above) ! 0 < y < 1, y becomes steeper (below) ! if C is even, y = [ f ( x)] ≥ 0 15 10 5 x C ! if C is odd, y = [ f ( x)] has the same sign as f (x) C 10 5 5 10 5 10 http://www.geocities.com/fatmuscle/HSC/ 5
Maths Extension 2 - Graphs y= 1 f ( x) y 5 or y = [ f ( x )] −1 x ! Asymptotes where y = 0 ! For f(x), y values > 1 becomes small _________ y values 0 < y < 1 becomes big ! Same case for y < 0, but negative ! Points along y = 1 and y = −1 stay the same 5 5 5 y = f −1 ( x ) y or 10 x = f ( y) ! INVERSE FUNCTION Flip about the line y = x 5 x 10 5 5 10 5 10 http://www.geocities.com/fatmuscle/HSC/ 6
Maths Extension 2 - Graphs Graphing y = f(x), y = g(x) y = g (x) y = sin x y 10 5 x 10 5 5 10 5 10 y = f ( x) + g ( x) y ! y co-ordinates are added 10 5 x 10 5 5 10 5 10 y = f ( x) − g ( x) y 10 ! g (x) co-ordinate is subtracted from the f (x) y co-ordinate 5 x 10 5 5 10 5 10 http://www.geocities.com/fatmuscle/HSC/ 7
Maths Extension 2 - Graphs y = f ( x).g ( x) y ! y co-ordinates are multiplied 10 y 5 5 x x 5 10 5 5 5 10 5 10 5 y= f ( x) g ( x) y ! Reciprocal, Multiplication rule 1 f ( x). g ( x) 5 x 5 5 5 y 10 5 x 10 5 5 10 5 10 http://www.geocities.com/fatmuscle/HSC/ 8
Maths Extension 2 - Graphs y = f [g (x)] ! Composite functions y = f (u ) , u = g (x) y 10 Take: f ( x) = e x g ( x) = sin x 5 x 10 5 5 10 5 10 y 10 10 5 y u 5 x u 10 5 5 10 x 10 5 5 5 5 10 10 10 http://www.geocities.com/fatmuscle/HSC/ 9
Maths Extension 2 - Polynomials Polynomials ! Definitions and properties of polynomials ! Factors & Roots ! Fields ~ Q Rational ~ R Real ~ C Complex ! Finding zeros over the complex field ! Factorization & Division of polynomials ! Remainder & Factor Theorem ! Rational Roots ! Multiplicity Theorem/ Repeated Roots ! Relationship between the roots and coefficients of a polynomial equation http://www.geocities.com/fatmuscle/HSC/ 1
Maths Extension 2 - Polynomials Definitions and properties of polynomials Polynomial Expression P(x) = p0xn + p1xn-1 + p2xn-2 + … + pn-1x + pn where p0 ≠ 0 Coefficients p0 , p1 , p2 , p3 , … Leading term pnxn Constant p0 If pn = 1 It is a monic If p0 = p2 = p3 = 0 Then P(x) is a zero polynomial Example 1 P(x) = 3x4 – x3 + 7x2 – 2x + 3 Coefficient of .x4 .x3 Leading term is 3x4 Constant is 3 Is ___3 Is ___– 1 More points: _ai Are the coefficients of the polynomial _a0 Is the constant term _anxn Is the leading term _an Is the leading coefficient _P(x) Polynomial of degree n _an = 1 The polynomial is a monic _P(x) ≡ 0 Null polynomial _P(x) Expression of polynomial _P(x) = 0 Polynomial equation http://www.geocities.com/fatmuscle/HSC/ 2
Maths Extension 2 - Polynomials Factors and Roots Factor A polynomial that divides into another and has remainder 0 (x – 2) is a factor of x2 – 4 Root P(2) is a root of x2 – 4 _x – 2 ∴x =0 =2 Fields Q Rational Integer numbers ± 1, 2, 3,… R Real Irrational Numbers Surdic roots occur in conjugate pairs If a + b is a root, so too will a − b C Complex Numbers over the complex field a ± ib Complex roots occur in conjugate pairs If a + ib is a root, so too will a − ib Example Factorize x4 – 2x2 – 15 over Q, R, C Y2 – 2Y - 15 = (Y – 5)(Y + 3) Q = (x2 – 5)(x2 + 3) 2 R = (x + 5 )(x – 5 )(x + 3) C = (x + 5 )(x – 5 )(x + 3i )(x – 3i ) http://www.geocities.com/fatmuscle/HSC/ 3
Maths Extension 2 - Polynomials Finding zeros over the complex field 1. If one root is complex, then one of the other roots is it’s conjugate 2. If (1 + i ) is a root, [x – (1 + i )] is the factor So [x – (1 + i )][x – (1 – i )] = x2 – 2x + 2 How do we get this??? (x – z)(x - z ) 2 = x – (z + z )x + z z Where: = x2 – 2x + 2 _z z _z + z zz =1+ i =1- i =2 =2 Example Find all the zeros of P(x) = x4 – x3 – 2x2 + 6x – 4 over C. (1 + i ) is a zero. If (1 + i ) is a root, then (1 – i ) is also a root. By multiplying out the factors: x2 – 2x + 2 2 _x – 2x + 2 4 _x _x4 3 –x – 2x3 _x3 _x3 _x2 – 2x2 +2x2 – 4x2 – 2x2 – 2x2 – 2x2 +x + 6x + 6x + 2x + 4x + 4x –2 –4 –4 –4 0 The factorized equation is [x – (1 + i )][x – (1 – i )](x + 2)(x – 1) The zeros are (1 + i ), (1 – i ), -2, 1 http://www.geocities.com/fatmuscle/HSC/ 4
Maths Extension 2 - Polynomials Factorization and Division of Polynomials Factorizing polynomials 1. Simple factorizing. Trinomials, Grouping, Difference of 2 squares, etc… 2. Quadratic Formula 3. Completing the Square Simple factorizing _x2 + x – 2 _x4 – 1 Quadratic formula _x2 + 2x + 3 = (x + 2)(x – 1) = (x2 – 1)(x2 + 1) = (x – 1)(x + 1)(x2 + 1) = (x – 1)(x + 1)(x + i )(x - i ) _x = − 2 ± 22 − 4(1)(3) 2(1) − 2± −8 2 −2±2 −2 = 2 = − 1 ± 2i ___________(x – a) = 2 _4x + 3x + 2 (x + 1 + _x = 2i )(x + 1 – 2i ) − 3 ± 32 − 4(4)(2) 2(4) − 3 ± − 23 8 − 3 ± 23i = 8 3 3 x + 8 + 23i x + 8 − 23i 8 8 We can only find the factors of the polynomial, not the constant outside = ( )( ) Completing the Square _x2 + 2x + 3 = x2 + 2x + 1 + 2 = (x + 1)2 + 2 = (x + 1)2 – 2i = [(x + 1) – 2i ][(x + 1) + 2i ] = (x + 1 – 2i )(x + 1 + 2i ) _4x2 + 3x + 2 = 4 x2 + 3 + 1 4 2 [ = 4[(x [ ] 2 + + 3 4 3 2 9 64 [ Complete the square, then add end term to satisfy the equation. ! ! i 2 = -1 Difference of two squares )+ ] = 4 (x + 8 ) − 23i 64 3 = 4 x2 + 8 + ! 2 23i 8 ] ][x 23 64 2 3 +8− 23i 8 ] http://www.geocities.com/fatmuscle/HSC/ 5
Maths Extension 2 - Polynomials Division of polynomials P(x) Dividend = A(x) = Divisor × Q(x) × Quotient + R(x) + Remainder 3x4 – x3 + 7x2 – 2x + 3 =x–2 × 3x3 + 5x2 +17x + 32 + 67 LONG DIVISION!!! x –2 4 3x 3x4 3x3 – x3 – 6x3 5x3 5x3 + 5x2 + 7x2 + 7x2 – 10x2 17x2 17x2 + 17x – 2x – 2x – 34x 32x 32x + 32 +3 +3 – 64 67 Example 2 Divide and find “a” such that R(x) = 0 x +2 For R(x) = 0, 3 x x3 x2 + ax2 + 2ax2 (a – 2)x2 (a – 2)x2 2a – 2 2a + (a – 2)x + ax + ax + (2z – 4)x (4 – a)x (4 – a)x + (4 – a) +6 +6 + 2(4 – a) 2a – 2 =0 =2 a =1 http://www.geocities.com/fatmuscle/HSC/ 6
Maths Extension 2 - Polynomials Factor and Remainder Theorems Remainder Theorem ! If a polynomial P(x) is divided by (x – a), then the remainder is P(a) Example 1 x – 2; a = 2 P(2) = 3(2)4 – (2)3 + 7(2)2 – 2(2) + 3 = 48 – 8 + 28 – 4 + 3 = 67 x –2 4 3x 3x4 3x3 – x3 – 6x3 5x3 5x3 + 5x2 + 7x2 + 7x2 – 10x2 17x2 17x2 + 17x – 2x – 2x – 34x 32x 32x + 32 +3 +3 – 64 67 Factor Theorem ! For any polynomial P(x), if P(a) = 0, then (x – a) is a factor of P(x) OR ! For any polynomial P(x), if (x – a) is a factor of P(x), then P(a) = 0 http://www.geocities.com/fatmuscle/HSC/ 7
Maths Extension 2 - Polynomials Rational Roots Let P(x) have degree n with integer coefficients. p Suppose P(x) has a rational root of q . _p, q, are prime integers. Then p|a0 and q|an | is ‘divides into’ Example Given that P(x) = 2x3 – 3x2 – 11x + 6 has a rational root. Find all the zeros of P(x). p Let the root be q . _p|6_;_q|2 The possible rational roots are: _p ± 6, ± 3, ± 2, ± 1 _q ± 3 , ± 1 2 2 P(-2) = 0 P(3) = 0 P( 1 ) = 0 2 The zeros of P(x) are –2, 3, 1 2 The factorized equation is (x + 2)(x – 3)(x - 1 2 ) OR (x + 2)(x – 3)(2x – 1) Given that P(x) = x3 + 4x2 + x – 6 has a rational root. Find all the zeros of P(x). p Let the root be q . _p|-6_;_q|1 The possible rational roots are: _p ± 6, ± 3, ± 2, ± 1 _q ± 1 ∴(x – 1) is a factor P(1) = 0 _x – 1 _x3 _x3 _x2 + 4x2 – x2 5x2 5x2 + 5x +x +6 –6 (x – 1)(x + 2)(x + 3) +x – 5x 6x 6x –6 –6 0 The zeros of P(x) are 1, -2, -3 The factorized equation is (x + 2)(x + 3)(x – 1) http://www.geocities.com/fatmuscle/HSC/ 8
Maths Extension 2 - Polynomials Multiplicity – Repeated Roots _x = a is a repeated root of P(x) or multiplicity r. so (x – a) / Q(x) | If P(x) = (x – a)r.Q(x) If x = a is of multiplicity r of P(x), then multiplicity (r – 1) or P`(x) Proof Let P(x) = (x – a)r.Q(x) P`(x) = (x – a)r.Q`(x) + Q(x).r(x – a)r-1.1 = (x – a)r-1[(x – a).Q(x) + r.Q(x)] = (x – a)r-1.R(x) ∴P`(x) has a root at x = a with multiplicity (r – 1) Product rule Example Find roots of P(x) = x3 + 3x2 – 4, given that the polynomial has a double root. P(x) = x3 + 3x2 – 4 P`(x) = 3x2 + 6x P(x) has a single root, so only need to differentiate once. 0 = 3x(x + 2) _x = 0 or –2______________Check both results by remainder theorem!!! P(0) ≠ 0 P(-2) = 0 _x = -2 is a double root. To find the other root, we can either use Long Division or Sum of the Roots Long Division (x + 2)2 = x2 + 4x + 4 x2 + 4x + 4 _x3 _x3 + 3x2 + 4x2 – x2 – x2 Sum of the Roots Sum of the roots one at a time = α + β +γ –2 + (–2) + γ γ _x +0 + 4x – 4x – 4x –1 –4 –4 –4 0 −b a = –3 = –3 =1 So the roots of P(x) are: -2, -2, 1 http://www.geocities.com/fatmuscle/HSC/ 9
Maths Extension 2 - Polynomials For polynomials which have: Double roots Differentiate one time Triple roots Differentiate two times Quadruple roots Differentiate three times If number of multiple roots are not given, keep differentiating until the P`(x), P``(x), etc.. can be factorized. Example Factorize and find the zeros of P(x) = x4 + x3 – 3x2 – 5x – 2 if it has multiple roots. P(x) = x4 + x3 – 3x2 – 5x – 2 P`(x) = 4x3 + 3x2 – 6x – 5 P``(x) = 12x2 + 6x – 6 = 6(2x – 1)(x + 1) P(-1) = 0 α + β + γ + δ = –1 –3 + δ = –1 δ =2 Example P(x) = x3 – 3x2 – 9x + C has a double root. Find C P`(x) = 3x2 – 6x – 9 = 3(x – 3)(x + 1) _x = 3 or –1 We can substitute both, but our C will be different. Both answers must be given P(3) = 27 – 27 – 27 + C C = 27 P(-1) = –1 – 3 + 9 + C C = –5 http://www.geocities.com/fatmuscle/HSC/ 10
Maths Extension 2 - Polynomials Relationship between the roots and coefficients of a polynomial equation Quadratic : ax2 + bx + c α+β αβ = −b a c = a Sum of roots 1 at a time Sum of roots 2 at a time (product of roots) Cubic : ax3 + bx2 + cx + d α + β +γ αβ + βγ + γα αβγ = −b a c = a = −d a Sum of roots 1 at a time Sum of roots 2 at a time Sum of roots 3 at a time (product of roots) Quartic : ax4 + bx3 + cx2 + dx + e α + β +γ +δ αβ + βγ + γδ + δα + αγ + βδ αβγ + βγδ + γδα + δαβ αβγδ = −b a c = a = −d a e = a Sum of roots 1 at a time Sum of roots 2 at a time Sum of roots 3 at a time Sum of roots 4 at a time (product of roots) If n is the number of roots 1 at a time, to find out how many combinations there are, we use: n Cr. For a quartic: α + β +γ +δ αβ + βγ + γδ + δα + αγ + βδ αβγ + βγδ + γδα + δαβ αβγδ = Σα = Σαβ = Σα i = Σα iα j 4 C1 4 C2 = Σαβγ = Σα iα jα k 4 C3 = Σαβγδ = Σα iα jα kα l 4 C4 http://www.geocities.com/fatmuscle/HSC/ 11
Maths Extension 2 - Polynomials Example Roots are α , β , γ for 2x3 – 4x2 – 3x – 1 1. (α − 1)( β − 1)(γ − 1) = αβγ − (αβ + βγ + γα ) + (α + β + γ ) − 1 = 1 − 1 + 2 −1 2 2 =3 2. ( β + γ − α )(γ + α − β )(α + β − γ ) If α + β + γ = 2 ∴ α + β − γ = 2 − 2γ ∴ β + γ − α = 2 − 2α ∴ γ + α − β = 2 − 2β = (2 − 2α )(2 − 2 β )(2 − 2γ ) = − 8(α − 1)( β − 1)(γ − 1) ____From part 1 = –8 × 3 = –24 3. Σα i 2 = (Σα i ) − 2(Σα iα j ) 2 = 22 - 2 (− 3 ) 2 =4+3 =7 4. Σα i 3 2α 3 2β 3 2γ 3 3 2(Σα i ) 2Σα i 3 2Σα i − 4α 2 − 4β 2 − 4γ 2 − 3α − 3β − 3γ –1 = 0 –1 = 0 –1 = 0 − 4(Σα i ) − 3(Σα i ) –3 = 0 –28 –6 –3 = 0 2 3 = 37 Σα i 3 = 37 2 http://www.geocities.com/fatmuscle/HSC/ 12
Maths Extension 2 - Polynomials 5. Σα i 4 P(x) is a cubic 2x3 – 4x2 – 3x – 1 2x4 – 4x3 – 3x2 – x x.P(x) is now a quadratic Sub it in 2α 4 2β 4 2γ 4 4 2(Σα i ) 2Σα i 4 2Σα i 3 − 4(Σα i ) − 4(37 ) 2 4 2Σα i − 4α 3 − 4β 3 − 4γ 3 4 –74 97 = 2 − 3α 2 − 3β 2 − 3γ 2 2 −α −α −α − 3(Σα i ) − Σα i − 3(7) −2 –21 –2 6. Σα 2 β 2 2 Σα i α j 2 = α 2 β 2 + β 2γ 2 + γ 2α 2 = (Σα iα j ) 2 − 2(Σα iα jα k )(Σα ) = (− 3 ) 2 − 2( 1 )(2) 2 2 1 = 4 7. Σ α12 = 1 1 1 + 2+ 2 2 α β γ 2 = ∑α i α j (∑α α α ) 2 i = 2 1 4 1 4 j k =1 8. ∑α 2 β 2 Σα i α j = α 2 β + α 2γ + β 2α + β 2γ + γ 2α + γ 2 β = (α + β + γ )(αβ + βγ + γα ) − 3αβγ = (Σα i )(Σα iα j ) − 3(Σα iα jα k ) = (2)(− 3 ) − 3( 1 ) 2 2 = −9 2 http://www.geocities.com/fatmuscle/HSC/ 13
Maths Extension 2 - Polynomials Relationship & Transformation methods Example If α , β , γ are the roots of 2x3 – 5x2 + 4x + 6 = 0, form the equation whose roots are: 2α , 2 β , 2γ Relationship Method: Sum of the roots one at a time 2α + 2 β + 2γ Sum of the roots two at a time (2α )(2 β ) + (2 β )(2γ ) + (2γ )(2α ) Sum of the roots three at a time (2α )(2 β )(2γ ) = 2 (α + β + γ ) = 2(5 ) 2 =5 = 4 (αβ + βγ + γα ) = 4(4 ) 2 =8 = 8(αβγ ) = 8(−3) = –24 ∴The new equation is x3 – 5x2 + 8x + 24 = 0 Transformation Method: x = α , β ,γ _y = 2x y y = 2α ,2 β ,2γ _x = 2 3 0= 0= 2  y  y  y 2  − 5  − 4  + 6 2 2 2 3 2  y   y   y 2  − 5  − 4  + 6  8   4      2 y3 5 y 2 − − 2y + 6 4 4 0 = y 3 − 5 y 2 − 8 y + 24 Change y for x 0 = x 3 − 5 x 2 − 8 x + 24 0= The Transformation Method is preferred http://www.geocities.com/fatmuscle/HSC/ 14
Maths Extension 2 - Polynomials Example If α , β , γ are the roots of 2x3 + 3x2 + x – 5 = 0, form the equation whose roots are: 1 1 C) 2 + α , 2 + β , 2 + γ B) 2α , 2 β , 2γ A) α , β , γ1 D) α − 2, β − 2, γ − 2 E) α 2 , β 2 , γ 2 F) − α , − β , − γ , A) 2x3 + 3x2 + x – 5 = 0 x=1 y =2 0 = ( ) + 3( ) + 1 3 y + 2 y3 1 2 y 3 y2 −5 1 y + 1 −5 y = 2 + 3 y + y 2 − 5 y3 0 = 5 x3 − x2 − 3x − 2 B) 2x3 + 3x2 + x – 5 = 0 y x= 2 =2 0 = ( ) + 3( ) + y 3 2 3 2y 8 3 + y 2 2 3y 4 2 y 2 −5 y + 2 −5 = y + 3 y 2 + 2 y − 20 0 = x 3 + 3 x 2 + 2 x − 20 C) 2x3 + 3x2 + x – 5 = 0 x = y−2 = 2(y – 2)3 + 3(y – 2)2 + (y – 2) – 5 = 2(y – 2)(y – 2)(y – 2) + 3(y – 2)(y – 2) + (y – 2) – 5 0 = 2(y3 – 6y2 + 4y – 8) + 3(y2 – 4y + 4) + (y – 2) – 5 = 2y3 – 9y2 + 13y – 11 0 = 2 x 3 − 9 x 2 + 13 x − 11 D) 2x3 + 3x2 + x – 5 = 0 x = y+2 = 2(y + 2)3 + 3(y + 2)2 + (y + 2) – 5 = 2(y + 2)(y + 2)(y + 2) + 3(y + 2)(y + 2) + (y + 2) – 5 0 = 2(y3 + 6y2 + 12y + 8) + 3(y2 + 4y + 4) + (y + 2) – 5 = 2y3 + 15y2 + 37y + 25 0 = 2 x 3 + 15 x 2 + 37 x + 25 http://www.geocities.com/fatmuscle/HSC/ 15
Maths Extension 2 - Polynomials E) 2x3 + 3x2 + x – 5 = 0 x= y 3 2 = 2 y 2  + 3 y 2  +  y 2  − 5             0 3 1 = 2 y 2 + 3y + y 2 − 5 1 1 1 3y – 5 = − y 2 (2 y + 1) Square both sides 9y2 – 30y + 25 = y(4y2 + 4y + 1) = 4y3 + 4y2 + y 0 = 4y3 – 5y2 + 31y – 25 0 = 4 x 3 − 5 x 2 + 31x − 25 1 F) 2x3 + 3x2 + x – 5 = 0 x = −y 0 = 2(-y)3 + 3(-y)2 + (-y) – 5 = –2y3 + 3y2 – y – 5 = 2y3 – 3y2 + y + 5 0 = 2 x3 − 3x 2 + x + 5 http://www.geocities.com/fatmuscle/HSC/ 16
Maths Extension 2 – Conics Conics ! General Equations of Conics ! Parabola, Circle, Ellipse, Hyperbola, Rectangular Hyperbola ! ! ! ! Equation Eccentricity Tangents Normals ! ! ! ! ! Foci Directrices Asymptotes Chord Chord of Contact ! General (Cartesian) ! (x1, y1) ! Parametric ! (a cosθ , a sin θ ) - Circle ! (a cosθ , b sin θ ) - Ellipse ! (a secθ , b tan θ ) - Hyperbola http://www.geocities.com/fatmuscle/HSC/ 1
Maths Extension 2 – Conics Shapes… From Phoenix Senior Maths Notepad Study Guides by George Fisher Eccentricity of Conics PS PM =e Eccentricities Circle e =0 Ellipse e 01 b2 = a2(e2 – 1) Rectangular Hyperbola e = 2 http://www.geocities.com/fatmuscle/HSC/ 2
Maths Extension 2 – Conics General Equation of Conics Circle Basic x2 + y 2 = r 2 (x − h )2 + ( y − k )2 = a 2 x 2 + y 2 + 2 gx + 2 fy + c = 0 ! Centre (-g, -f) ! General Radius g2 + f 2 − c Basic _r _r _y (h, k) _x Parametric General Form _x2 + y2 + 4x – 6y + 10 = 0 P(acosθ, asinθ) Complete the Square _a θ asinθ _acosθ (x + 2)2 + (y – 3)2 + 10 = 0 Centre = (-2, 3) Radius = ______= Parabola Basic 4 + 9 − 10 3 x 2 = 4ay ! ! ! Vertex (0, 0) Focus (0, a) Directrix y = –a (x − h )2 = 4a( y − k ) ! Vertex (h, k) SEE 3U Parabola NOTES http://www.geocities.com/fatmuscle/HSC/ 3
Maths Extension 2 – Conics Ellipse ! ! P M Foci (±ae, 0) Directrices x = ± a e ! ! x2 y2 + =1 a2 b2 Major Axis A(a, 0); A’(-a, 0) Minor Axis B(0, b); B’(0, -b) S and S’ are the two foci D and D’ are the two Directrices P is a point on the ellipse M is the perpendicular distance from D to P _b is the minor axis _a is the major axis _b _a S’ S D’ D y= a e B B (0, ae) A’ A’ A (-ae, 0) A (ae, 0) (0, -ae) x = −a e AA’ > BB’ _b2 = a2(1 – e2) B’ x= a e B’ y = −a e AA’ < BB’ _a2 = b2(1 – e2) http://www.geocities.com/fatmuscle/HSC/ 4
Maths Extension 2 – Conics Hyperbola ! ! D’ Foci (±ae, 0) Directrices x = ± a e ! x2 y2 − =1 a 2 b2 Asymptotes x2 − y2 = a2 D S’ D x= D x=− D D’ S’ S D D S S (ae,0) (− ae,0) xy = c 2 x+ y = a x + y = −a S a e a e (c, c ) (− c,−c ) http://www.geocities.com/fatmuscle/HSC/ 5
Maths Extension 2 – Conics Conics in Detail Circle Equation x2 + y 2 = r 2 (x − h )2 + ( y − k )2 = a 2 Eccentricity Tangents P(x1 , y1 ) P(a cosθ , a sin θ ) Normals P(x1 , y1 ) P(a cosθ , a sin θ ) x 2 + y 2 + 2 gx + 2 fy + c = 0 0 xx1 + yy1 = a 2 (x1 − h )(x − x1 ) + ( y1 − k )(y − y1 ) = a 2 (x − x1 )(x1 + g ) + ( y − y1 )( y1 + f ) = 0 x cosθ + y sin θ = a (x − h )cosθ + ( y − k )sin θ = a (x + g )cosθ + (y + f )sin θ = a xy1 − yx1 = 0 ( y1 − k )(x − x1 ) − (x1 − h )(y − y1 ) = 0 ??? x sin θ − y cosθ = 0 (x − h − a cosθ )sin θ − ( y − k − a sin θ )cosθ = 0 ??? http://www.geocities.com/fatmuscle/HSC/ 6
Maths Extension 2 – Conics Circle – Tangent x2 + y2 = a2 P(x1 , y1 ) OR P(a cosθ , a sin θ ) 2x + 2 y dy dx dy dx 2y dy dx x − 1 y1 − xx1 + x1 2 2 2 x1 + y1 =0 = − 2x x =− y y − y1 = x − x1 =− (x − h )2 + ( y − k )2 = a 2 dy 2(x − h ) + 2( y − k ) dx dy 2( y − k ) dx 2 Equation of tangent P(x1 , y1 ) OR P(h + a cosθ , k + a sin θ ) =0 = − 2(x − h ) x−h x −h cosθ =− 1 =− y−k y1 − k sin θ cosθ y − k − a sin θ − = sin θ x − h − a cosθ 2 ( y − k ) sin θ − a sin θ = − (x − h )cosθ + a cos 2 θ ( y − k ) sin θ + ( x − h) cosθ = a dy dx Gradient = yy1 − y1 = xx1 + yy1 x cosθ + y sin θ = a xx1 + yy1 = a 2 cosθ x1 =− sin θ y1 =− (x − h )cosθ + ( y − k )sin θ x 2 + y 2 + 2 gx + 2 fy + c = 0 (x − x1 )(x1 + g ) + ( y − y1 )( y1 + f ) = 0 (x + g )cosθ + (y + f )sin θ = a =a Gradient {(cos 2 θ + sin 2 θ = 1) Equation of tangent P(x1 , y1 ) OR P(− g + a cosθ ,− f + a sin θ ) Equation of tangent http://www.geocities.com/fatmuscle/HSC/ 7
Maths Extension 2 – Conics Circle – Normal x2 + y2 = a2 dy dx y1 x1 y1 x − x1 y1 xy1 xy1 − yx1 = 0 y x y − y1 = x − x1 = x1 y − x1 y1 = yx1 = P(x1 , y1 ) OR P(a cosθ , a sin θ ) Gradient y sin θ = 1 = x1 cosθ x sin θ − y cosθ = 0 P(x1 , y1 ) OR P(h + a cosθ , k + a sin θ ) (x − h )2 + ( y − k )2 = a 2 y−k x−h y1 − k y − y1 = x1 − h x − x1 (y1 − k )(x − x1 ) − (x1 − h )(y − y1 ) = 0 dy dx Equation of normal = (y1 − k )(x − x1 ) − (x1 − h )(y − y1 ) = 0 (x − h − a cosθ )sin θ − ( y − k − a sin θ )cosθ =0 = y1 − k sin θ = x1 − h cosθ Equation of normal http://www.geocities.com/fatmuscle/HSC/ 8
Maths Extension 2 – Conics Ellipse Equation x2 y2 + =1 a2 b2 b 2 = a 2 (1 − e ) x = (± ae,0 ) a x=± e xx1 yy1 + 2 =1 a2 b Eccentricity Foci Directrices Tangents P(x1, y1) P(a cosθ , b sin θ ) x cosθ y sin θ + =1 a b y = mx + c *c2 = a2m2 + b2 a2 x b2 y − = a2 − b2 x1 y1 Normals P(x1, y1) P(a cosθ , b sin θ ) Chord P (a cosθ , b sin θ ) & Q (a cosφ , b sin φ ) Chord of Contact T (x0 , y0 ) ax by − = a2 − b2 cosθ sin θ y θ + φ  x θ + φ  θ −φ  sin   + cos  = cos  b  2  a  2   2  xx0 yy0 + 2 =1 a2 b Other Stuff Angles in an ellipse The tangent at a point P on the ellipse is equally inclined to the focal chords through P The segment of the tangent to an ellipse between the point of contact and the directrix subtends at right angles at the corresponding focus. P S’ Area PS + PS’ = 2a S πab http://www.geocities.com/fatmuscle/HSC/ 9
Maths Extension 2 – Conics Area of the Ellipse by integration y2 b2 a A = ∫ y dx −a a b2 x 2 b − 2 a = 2∫ 2 −a a = 2b ∫ 1 − −a a = 2b ∫ −a x2 a2 y 2 y =1– x2 a2 b2 x2 =b − 2 a 2 = b2 − b2 x2 a2 a2 − x2 a2 2b a 2 2 ∫a a − x a − 2b 1 2 . πa = a 2 = a2 − x2 Is a Semicircle Area of Ellipse = πab Gradient of an Ellipse x2 y2 + a 2 b2 dy 2 x 2 y dx + 2 a2 b dy dx =1 =0 = − 2 x b2 . a2 2 y = Implicit Differentiation − b2 x a2 y http://www.geocities.com/fatmuscle/HSC/ 10
Maths Extension 2 – Conics Chords in an Ellipse A chord through the centre is a diameter Eg. PR C P A chord through a focus is a focal chord Eg. CD S’ S Q A focal chord that is perpendicular to the major axis is the Latus Rectum Eg. PQ R D Chord – P (a cosθ , b sin θ ) & Q (a cosφ , b sin φ ) b sin φ − b sin θ m = a cos φ − a cosθ ( =− 2a sin ( b cos( =− a sin ( φ +θ 2 φ +θ 2 φ +θ 2 φ +θ 2 2b cos )sin ( ) )sin ( ) φ −θ 2 φ −θ 2 ) OR = − b cot φ + θ    Gradient of Chord a  2  ) b cos( ) = y − b sin θ − x − a cosθ a sin ( ) − xb cos( ) + ab cosθ cos( ) = ay sin ( )− ab sin θ sin ( ) θ +φ 2 φ +θ 2 φ +θ 2 θ +φ 2 y θ + φ  x θ + φ  sin   + cos  b  2  a  2  θ +φ 2 θ +φ 2 θ + φ  θ + φ  = cosθ cos  + sin θ sin    2   2  θ + φ  = cos   2  y θ + φ  x θ + φ  θ −φ  sin   + cos  = cos  b  2  a  2   2  Equation of chord http://www.geocities.com/fatmuscle/HSC/ 11
Maths Extension 2 – Conics Ellipse – Tangent dy dx =− Gradient of Tangent xb 2 ya 2 At (x1, y1) x1b 2 y1a 2 b cosθ =− a sin θ 2 y − y1 b x − 2 1 = x − x1 a y1 =− − b 2 xx1 + b 2 x1 2 2 b 2 x1 + a 2 y1 2 At (a cosθ , b sin θ ) = a 2 yy1 − a 2 y1 = b 2 xx1 + a 2 yy1 2 x1 y + 12 2 a b xx1 yy1 + 2 a2 b xx1 yy1 1 = 2 + 2 a b = x cosθ y sin θ + =1 a b xx1 yy1 + 2 =1 a2 b Ellipse – Normal dy ya 2 = dx xb 2 y a2 = 1 2 x1b Gradient of Normal At (x1, y1) At (a cosθ , b sin θ ) a 2b sin θ b 2 a cosθ y − y1 y1a 2 = x − x1 x1b 2 = a 2 y1 x − a 2 y1 x1 a 2 y1 x + b 2 x1 y1 y1 a 2 x + b 2 x1 ( ) = b 2 x1 y − b 2 x1 y1 = b 2 x1 y + a 2 x1 y1 = x1 b 2 y + a 2 y1 ( 2 a x + b2 x1 a 2 x b2 y − x1 y1 a2 x b2 y − = a2 − b2 x1 y1 Equation of tangent ) 2 = b y + a2 y1 = a 2 − b2 ax by − = a2 − b2 cosθ sin θ Equation of normal http://www.geocities.com/fatmuscle/HSC/ 12
Maths Extension 2 – Conics Hyperbola Equation x2 y2 − =1 a 2 b2 b2 = a 2 e2 − 1 x = (± ae,0 ) a x=± e xx1 yy1 − =1 a 2 b2 ( Eccentricity Foci Directrices Tangents P(x1, y1) P(a secθ , b tan θ ) ) x secθ y tan θ − =1 a b Normals P(x1, y1) a 2 x b2 y + = a2 + b2 x1 y1 P(a secθ , b tan θ ) Chord P (a secθ , b tan θ ) & Q (a sec φ , b tan φ ) Chord of Contact T (x0 , y0 ) Gradient of a Hyperbola x2 y2 =1 − a 2 b2 dy 2 x 2 y dx =0 − 2 a2 b dy 2 x b2 = 2. dx a 2y = ax by + = a 2 + b2 secθ tan θ x θ − φ  y θ + φ  θ + φ  cos −   = cos  a  2   2  b 2  xx0 yy0 − 2 =1 a2 b Implicit Differentiation b2 x a2 y http://www.geocities.com/fatmuscle/HSC/ 13
Maths Extension 2 – Conics Hyperbola – Tangent dy b2 x = 2 dx a y Gradient of Hyperbola At (x1, y1) b 2 x1 a 2 y1 b secθ = a tan θ y − y1 = x − x1 = b 2 x1 a 2 y1 2 b 2 xx1 − b 2 x1 b 2 xx1 − a 2 yy1 xx1 yy1 − a 2 b2 xx1 yy1 − a 2 b2 At (a secθ , b tan θ ) 2 = a 2 yy1 − a 2 y1 2 2 = b 2 x1 − a 2 y1 2 2 x y = 12 − 12 a b =1 x secθ y tan θ − =1 a b xx1 yy1 − =1 a 2 b2 Hyperbola – Normal dy a2 y =− 2 dx b x a2 y =− 2 1 b x1 a tan θ =− b secθ 2 y − y1 a y − 2 1 = x − x1 b x1 − a 2 y1 x + a 2 y1 x1 a 2 x1 y1 + b 2 x1 y1 x1 y1 a 2 + b 2 ( ) a +b 2 a 2 + b2 = 2 a2 x b2 y + x1 y1 Equation of tangent Gradient of Normal At (x1, y1) At (a secθ , b tan θ ) = b 2 x1 y − b 2 x1 y1 = a 2 xy1 + b 2 x1 y = a 2 xy1 + b 2 x1 y = a2 x b2 y + x1 y1 a 2 + b2 = ax by + secθ tan θ Equation of normal http://www.geocities.com/fatmuscle/HSC/ 14
Maths Extension 2 – Conics Rectangular Hyperbola Equation Eccentricity Foci e= 2 a 2 * (± a,± a ) a x=± 2 * x + y = ±a y = ±x * x & y axis Directrices Asymptotes Tangents P(x1, y1) ( P cp, c p x2 − y2 = a2 * xy = c 2 xy1 + yx1 = 2c 2 ) x + p 2 y = 2cp Normals P(x1, y1) xx1 − yy1 = x1 − y1 P(a cosθ , b sin θ ) p 3 x − py = c p 4 − 1 Chord P(x1, y1) & Q(x2, y2) c 2 x + x1 x2 y = c 2 (x1 + x2 ) ( ) 2 ( ) 2 ( c P cp, c & Q cq, q p x + pqy = c( p + q ) Chord of Contact T (x0 , y0 ) ) xy0 + yx0 = 2c 2 Other Stuff The area of the triangle bounded by a tangent 2c2 and the asymptotes is a constant Intersection of 2 tangents P and Q  2 xpq 2c  =  p+q, p+q    http://www.geocities.com/fatmuscle/HSC/ 15
Maths Extension 2 – Conics Gradient of a Rectangular Hyperbola xy = c 2 y c2 = x dy c2 =− 2 dx x ( ) ( ) c Chord – P cp, c & Q cq, q p y− c p x − cp c (x − cp ) q − cp ( ) xc xc c 2 p − − + c2 q p q c q = − c p cq − cp = y − c (cq − cp ) p ( ) c2q = cqy − cpy − + c2 p 0 = pq 2 y − p 2 qy − cq 2 + cp 2 + qx − px = pqy (q − p ) − c q 2 − p 2 + x(q − p ) = pqy − c( p + q ) + x ( x + pqy = c( p + q ) ) Equation of Chord http://www.geocities.com/fatmuscle/HSC/ 16
Maths Extension 2 – Conics Rectangular Hyperbola – Tangent dy c2 =− 2 dx x 2 y1 =− 2 x1 1 =− 2 p − 1 p2 = y− x + p2 y xy1 + yx1 = 2c 2 ( At cp, c p x − cp ( c p ) = p 2 y − pc = 2cp Equation of tangent x + p 2 y = 2cp Rectangular Hyperbola – Normal dy x2 = 2 dx c 2 x = 12 y1 Gradient of Normal At (x1, y1) ( At cp, c p = p2 p2 p (x − cp ) 2 ) c p 2 − (x − cp ) = p y − − x + cp At (x1, y1) = y− ) c p x − cp = y− c p p 2 x − cp 3 = y− p 3 x − cp 4 = py − c 2 xx1 − yy1 = x1 − y1 2 c p ( ) p 3 x − py = c p 4 − 1 Equation of Normal http://www.geocities.com/fatmuscle/HSC/ 17
Maths Extension 2 - Integration ! ! ! ! Standard Integrals Function & Derivative Product – Integrate by Parts Partial Fractions ! A, B ! A, Bx + C ! Completing the Square ! Substitution ! Normal ! Trigonometry ! T - Formula ! Reduction Formula http://www.geocities.com/fatmuscle/HSC/ 1
Maths Extension 2 - Integration INTEGRALS *c is a constant x n +1 + c _________ n ≠ −1 = n +1 § ∫ x n dx §∫ 1 dx x = ln | x | +c ∫ (ax + b ) n ∫ dx f ' ( x) dx f ( x) 1 § Standard Integral = (ax + b )n +1 + c a(n + 1) = ln | f ( x) | + c = 1 ln | ax + b | + c a § ∫ e ax dx = 1 ax e +c a x ∫ a dx ax = +c ln a ∫ ax + b dx 1 §∫ x + a2 §∫ x −a 2 1 2 ∫x 2 ∫a 2 2 [ ] [ ] dx = ln x + x 2 + a 2 + c dx = ln x + x 2 − a 2 + c 1 dx − a2 = 1 x−a ln +c 2a  x + a  1 dx − x2 = 1 a+ x ln +c 2a  a − x  http://www.geocities.com/fatmuscle/HSC/ 2
Maths Extension 2 - Integration 1 sin ax + c a § ∫ cos ax dx = § ∫ sin ax dx 1 = − cos ax + c a § ∫ sec 2 ax dx = §∫ 1 a −x 2 2 1 1 tan ax + c a dx  x = sin −1   + c a dx  x  x = cos −1   + c __OR__ − sin −1   + c a a ∫− a 2 − x2 §∫ 1 dx 2 a + x2 = 1 = − cot ax + c a ∫ cos ec ax dx 2 ∫ sec ax. tan ax 1  x tan −1   + c a a dx ∫ cos ecax.cot ax dx = 1 sec ax + c a 1 = − cos ecax + c a Integration by special properties – Theorems *notes, all these are with respect to dx ∫ ∫ ∫ a 0 a −a a −a ∫ a f (a − x) dx f ( x) dx = f ( x) dx = 2 ∫ f ( x) dx ___if f (x) is an EVEN function f ( x) dx = 0 _________-__if f (x) is an ODD function 0 a 0 http://www.geocities.com/fatmuscle/HSC/ 3
Maths Extension 2 - Integration Trigonometric Identities sin2θ + cos2θ = 1 ! sin2θ = 1 – cos2θ ! cos2θ = 1 – sin2θ tan2θ = sec2θ – 1 ! sec2θ – tan2θ = 1 ! sec2θ = tan2θ + 1 1 + cot2θ = cosec2θ ! 1 = cosec2θ – cot2θ ! cot2θ = cosec2θ – 1 sin θ = tan θ cosθ cosθ = cot θ sin θ cos2θ = 1 (1 + cos 2θ ) 2 sin2θ = 1 (1 − cos 2θ ) 2 sinθ = 2t 1+ t2 cosθ = 1− t 1+ t2 tanθ = 1 1− t2 T-Formula Substitution 2 θ  t = tan   2 a 2 − x 2 = Let x = a sin θ x 2 − a 2 = Let x = a secθ Trigonometric Substitution a 2 + x 2 = Let x = a tan θ http://www.geocities.com/fatmuscle/HSC/ 4
Maths Extension 2 - Integration STANDARD INTEGRALS Basic Form n +1 ∫x 5 dx 1 ∫x dx or ∫ x −1 dx ∫ (ax + b ) dx ∫ (2x + 3) dx n 6 f ' ( x) dx f ( x) 2x ∫ x 2 + 1 dx ∫ 1 ∫x dx or ∫ x −1 dx 1 ∫ ax + b dx ∫ 3x + 2 dx 1 ∫e ax dx ∫e 3x dx ∫a ∫3 x dx x dx x + c _________ n ≠ −1 n +1 ! General Integral x6 +c = 6 = ln | x | +c ! 1st special function = ∫ x dx n = (ax + b )n +1 + c a(n + 1) ! Linear function (2 x + 3)7 + c = 14 = ln | f ( x) | + c ! Function and Derivative = ln x 2 + 1 + c Logarithm and Exponential = ln | x | +c ! 1st special function 1 = ln ax + b + c a 1 = ln 3 x + 2 + c 3 1 = e ax + c a 1 3x = e +c 3 ax = +c ln a 3x = +c ln 3 http://www.geocities.com/fatmuscle/HSC/ 5
Maths Extension 2 - Integration Trigonometric Functions ∫ sin ax dx ∫ sin 3x dx ∫ sin 3 x dx 1 = − cos ax + c a cos 3 x =− +c 3 = ∫ sin 2 x.sin x dx ∫ (1 − cos x ).sin x dx = ∫ sin x − cos x.sin x dx = 2 2 − ∫ cos 2 x.sin x dx Let u = cos x du = − sin x dx du ∴ dx = − sin x ∫ sin x dx = − cos x = − cos x + ∫ cos ax dx ∫ cos 3x dx ∫ cos x dx 3 = − ∫ u 2 . sin x du − sin x u3 = 3 cos 3 x +c 3 1 sin ax + c a sin 3 x = +c 3 = ∫ cos 2 x. cos x dx = ∫ (1 − sin x ). cos x dx = ∫ cos x − sin x. cos x dx = 2 2 − ∫ sin 2 x. cos x ∫ cos x dx = sin x Let u = sin x du = cos x dx du ∴ dx = cos x = − ∫ u 2 . cos x = du cos x u3 3 sin 3 x sin x − +c 3 http://www.geocities.com/fatmuscle/HSC/ 6
Maths Extension 2 - Integration ∫ sec 2 ax dx ∫ sec 2 3x dx ∫ tan x dx ∫ tan 2 ∫ tan 3 1 tan ax + c a 1 = tan 3 x + c 3 sin x = ∫ dx _________Function and Derivative cos x = - ln cos x + c = ∫ (sec 2 ) x −1 dx x dx = x dx = tan x − x + c = ∫ tan 2 x. tan x dx ∫ (sec x − 1). tan x dx = ∫ sec x. tan x − tan x dx_____Let tan x = u = 2 2 = tan 2 x + ln cos x + c 2 http://www.geocities.com/fatmuscle/HSC/ 7
Maths Extension 2 - Integration u = f (x) du = f ' ( x)dx Function and Derivative ∫ [ f ( x )] . f ' ( x ) n ∫ dx f ' ( x) dx f ( x) f ( x) n ∫ sin x.cos x dx f ' ( x) ∫ 1 + [ f ( x)] 2 dx 1 5 n +1 = e f ( x) + c = sin f ( x) + c sin n +1 x = +c n +1 = tan −1 f ( x) + c = ln ln x + c 4 x dx du = − sin x dx du dx = − sin x = ∫ sin 4 .sin x. cos 4 x dx ∫ (1 − cos x ) .sin x. cos x dx = ∫ ( − 2 cos x + cos x ).sin x. cos x dx 1 = ∫ cos x.sin x − 2∫ cos x.sin x + ∫ cos x.sin x dx x+3 ∫ x − 2 x − 4 dx 2 split the integral take out constants complete the square standard integrals 2 2 = 4 2 4 4 = ! ! ! ! [ f ( x)]n +1 + c = ln f ( x) + c ∫ f ' ( x).e dx ∫ f ' ( x).cos f ( x) dx ∫ x ln x dx ∫ sin x.cos = = = = = 4 6 8 cos5 x 2 cos7 x cos9 x − + − +c 5 7 9 1 (2 x − 2) + 4 ∫ 2x 2 − 2 x − 4 dx 1 2x − 2 1 ∫ x 2 − 2 x − 4 dx + 4∫ x 2 − 2 x − 4 dx 2 1 2x − 2 1 ∫ x 2 − 2 x − 4 dx + 4∫ (x − 1)2 − 5 dx 2 1 2  x −1− 5  +c ln x 2 − 2 x − 4 + ln 2 5  x −1+ 5    ( ) http://www.geocities.com/fatmuscle/HSC/ 8
Maths Extension 2 - Integration ∫ uv' dx = uv − ∫ vu ' dx Integrating by parts ∫ x cos x dx Let u = x du =1 dx ∫ ln x dx dv = cos x dx v = sin x Let u = ln x du 1 = dx x dv =1 dx v=x Let u = du = dx = x sin x − ∫ sin x dx Let v= dv = dx = x sin s + cos x + c = ∫ ln x.1 dx = x ln x − ∫1 dx = x ln x − x + c These integrals have only one step, other integrals have multi-step ∫e x Let du dx Let du dx cos x dx = I u = ex = ex u = ex = ex dv = cos x dx v = sin x dv = sin x dx v = − cos x = e x sin x − ∫ e x sin x dx [ = e x sin x − − e x cos x + ∫ e x cos xdx ] I = e x sin x + e x cos x − I 2I = e x sin x + e x cos x 1 ∴I = (e x sin x + e x cos x + c) 2 ∫e x Let du dx Let du dx sin x dx = I u = ex = ex u = ex =e x dv = sin x dx v = − cos x dv = cos x dx v = sin x = − e x cos x + ∫ e x cos x dx [ = − e x cos x + e x sin x − ∫ e x sin xdx ] I = − e x cos x + e x sin x − I 2I = e x sin x − e x cos x 1 ∴I = (e x sin x − e x cos x + c) 2 http://www.geocities.com/fatmuscle/HSC/ 9
Maths Extension 2 - Integration 1. 2. 3. 4. Split to Standard Integrals A, B A, Bx + C Completing the Square x +1 1 = ∫ + dx x +1 x +1 1 = ∫1 + dx x +1 = x + ln x + 1 + c Partial Fractions x+2 ∫ x + 1 dx 1 ∫ ( x − 2)( x + 1) dx **************** 1 ( x − 2)( x + 1) When: 1 1 x = –1 1 x =2 = **************** A B = + x − 2 x +1 = A( x + 1) + B( x − 2) = ___0___+__–3B = __-3A A B ∫ x − 2 + x + 1 dx 1 1 1 1 ∫ x − 2 − 3 ∫ x + 1 dx 3 1 = [ln x − 2 − ln x + 1 ] + c 3 1   x − 2  = ln  +c 3   x + 1   = P( x) ∫ Q( x) ______ P( x) ≥ Q( x) 3x − 2 ∫ 2 x + 1 dx = = = 2x2 − x + 1 ∫ x − 2 dx ******** LONG DIVISION ********** 2 x + 3 r7 2 x − 2 2x − x + 1 = = ∫ x=4 x=0 x =1 A=1 __17 __1 __5 (2 x + 1) − 3 1 2 dx 2x + 1 31 3 1 − ∫ 2 dx 2∫ 2x + 1 3 7 x − ln 2 x + 1 + c 2 2 7 ∫ (2x + 3)dx + ∫ x − 2 dx x 2 + 3x + 7 ln x − 2 + c ____ 2 x 2 − 4 x ___ _________ 3 x + 1 _________ 3x − 6 _____________7 1 + 4x ∫ (4 − x)( x 2 + 1) dx 1 + 4x 3 2 Bx + C dx x2 + 1 −1 x = −∫ +∫ 2 dx 4− x x +1 = − ln 4 − x + 1 ln x 2 + 1 + c 2 = = A( x 2 + 1) = 17A =A = 2A B=1 + ( Bx + c)(4 − x) +0 + C4 – 0 + Bx + 3 C=0 A ∫4−x +∫ http://www.geocities.com/fatmuscle/HSC/ 10
Maths Extension 2 - Integration Finding x in terms of u Let u = du ∴ = dx ∴ dx = Substitution 1 12 ∫ (4 + x ) x 4 dx Let u = x du 1 = dx 2 x dx = 2 x du x = u2 = = dx Let u = x + 1 du 2u = 2x dx udu dx = x 2 2 x du Changing the Limits Let u = 2 x2 = u 2 − 1  −1  x    tan   2     4   2 3 −1  2   tan −1   2  − tan  2      π π − 3 4 π 12 x 3 udu = ∫ u2 x IMPLICIT DIFFERENTIATION x x = 12 x=4 12 = x2 + 1 x 1 du 4 + u2 1  u  = 2  tan −1    2  2 u = tan −1   2 = ∫ 2 = 2∫ = x3 1 ∫ (4 + u ) u=2 3 u=2 2 3   u  =  tan −1    2  2  = π π − 3 4 = π 12 ∫ x du = ∫ (u −1)du 2 = 2 u3 −u +c = 3 1 = x2 + 1 x2 + 1 − x2 + 1 + c 3 ( ) http://www.geocities.com/fatmuscle/HSC/ 11
Maths Extension 2 - Integration Trigonometry Substitution a2 − x2 x −a 2 Let x = a sin θ Let x = a secθ Let x = a tan θ 2 a2 + x2 ∫ x 4 − x2 dx Let x = 2 sin θ dx = 2 cosθ dθ dx = 2 cosθ dθ Trig Identity ∫ 4 sin 2 θ 2c cosθ dθ 4 − 4 sin 2 θ sin 2 θ cosθ = 8∫ dθ 4 − 1 − sin 2 θ = ( = 4∫ ) sin θ cosθ x = 2 sin θ x = sin θ 2 2 dθ cos 2 θ = 4 ∫ 1 (1 − cos 2θ ) dθ 2  x θ = sin −1   2 sin 2θ   +c = 2θ − 2     x = 2 sin −1   − 2 sin θ cosθ + c 2 2  x  4 − x  −1  x   +c = 2 sin   − 2  2   2   2    x 2 θ 4 − x2 2  x x 4− x = 2 sin −1   − +c 2 2 http://www.geocities.com/fatmuscle/HSC/ 12
Maths Extension 2 - Integration t = tan (θ ) ________ − π < θ < π 2 T-Formula sin θ = 2t 1+ t2 1− t2 cosθ = 1+ t2 tan θ = ∫ π 2 2t 1− t2 4 dθ 3 + 5 cosθ dt 1 2  θ  = sec   dθ 2 2 2dt dθ = sec 2 (θ ) 2 2dt dθ = 1 + tan 2 (θ ) 2 4 2dt ∫ 3 + 5( ).1 + t = ∫ 3(1 + t ∫ 8 − 2t = 2dt dθ = 1+ t2 = = 0 2dt 1+ t2 dθ = ∫4−t 1− t 2 1+ t 2 2 8 dt . 2 ) + 5(1 − t ) 1 + t 2 1+ t2 2 8 4 2 2 dt dt *** = 4∫ 1 dt 4 − t2 π  1  2 + t  4 = 4. . ln   2(2)  2 − t  0 π   2 + t  4 = ln    2 − t  0 = ln 3 ***By Partial Fractions 4 = A(2 – t) 4= 0 B= 1 t = −2 = 4 ∫4−t 1 2 + B(2 + t) + 4B 4 = 4A A= 1 t=2 +0 dt 1 ∫2−t + ∫2+t = [− ln 2 − t + ln 2 + t ] = π 4 0 π   2 + t  4 = ln    2 − t  0 http://www.geocities.com/fatmuscle/HSC/ 13
Maths Extension 2 - Integration Reduction Formula Show: π 2 In = ∫0 cos n x dx_______n ≥ 2 In = (n – 1)In-2 – (n – 1)In In = ∫ cosn −1 x. cos1 x dx Let n −1 u = cos x du = (n − 1)cos n − 2 . − sin x dx dv = cos x dx v = sin x = Take out constant, and expand [sin x.cos n −1 π 2 ] x − − ∫ (n − 1)cos n − 2 x.sin 2 x dx 0 = 0 + ∫ (n − 1)cosn − 2 x.(1 − cos2 x )dx = (n − 1)∫ cosn − 2 x dx – ∫ (n − 1)cosn x dx = (n – 1)In-2 – (n – 1)In Find: I6 π 2 I6 = ∫0 cos6 x dx I6 = (6 – 1)I6-2 – (6 – 1)I6 I6 = 5I4 – 5I6 6I6 = 5I4 6I6 = 5(3I2 – 3I4) 6I6 = 15I2 – 15I4 6I6 = 15I2 - 15 × 3 I2 4 15 6I6 = 4 I2 I6 = 5 I2 8 I4 = 3I2 – 3I4 4I4 = 3I2 I4 = 3 I2 4 π 2 I2 = ∫0 cos 2 x dx π I2 = I2 = I6 =  sin 2 x  2 sin x −   2 0  1 2 5 16 http://www.geocities.com/fatmuscle/HSC/ 14
Maths Extension 2 Volumes Volumes Ø Slicing Ø Volume of Pyramid Ø Relationships & Ratios Ø Slicing Ø Revolution Ø Cross Sectional Area Ø Shells Ø Washer Method Ø Torus, Quadratic Equation Basic Volumes Formulae Basic b V= π ∫ y 2 dx a b V= π ∫ x 2 dy a General b V= ∫ A(z ).dz a http://www.fatmuscle.cjb.net 1
Maths Extension 2 Volumes Slicing Volume of a Pyramid h h h k k k β α β α b a b a ∆k α Find the volume of one slice α k β k = = a h b h α= ak h β = β Volume of one slice = αβ∆k ∴ V = αβ∆k bk h Volume of entire pyramid h V = ∫ αβ .dk 0 h ak bk . dk h h 0 = ∫ = ab h 2 k dk h2 ∫ 0 h ab  k 3  = 2  h  3 0 abh 3 3h 2 1 = abh 3 = http://www.fatmuscle.cjb.net 2
Maths Extension 2 Volumes Relationship 4 h 5 l 3 4 h 5 12 w l 3 h 10 5 w 12 10 Volume of One Slice y=4 h=0 y = 12 h=5 y = mx + b l = y=3 y = 10 m= rise run h=0 h=5 b = int ercept x=h 8h +4 5 w = 7h +3 5 V = w.l.∆h Volume of entire object 5 V  8h  7 h  + 3 dh = ∫  + 4   5  0 5 5 = ( ) 4 2 ∫ 14h + 65h + 75 dh 25 0 5  4 14h3 65h 2 = + + 75h   25  3 2 0 = 283 http://www.fatmuscle.cjb.net 3
Maths Extension 2 Slicing Shape The base of a certain solid is x 2 + y 2 = 9 . Plane sections are perpendicular to the y are equilateral triangles, on side being the solid base. Find the volume. Volume of 1 slice 2y Volumes axis and 2y 2y Volume of one slice V 1 = × 2 y × 2 y × sin 60 × ∆x 2 = 3 y 2 ∆x Total Volume x2 + y 2 3 V = ∫ 3 y 2 dx y −3 2 =9 = 9 − x2 ∫ 3 (9 − x )dx 3 = 2 −3 3 ( ) = 2 3 ∫ 9 − x 2 dx −3 3  x3  = 2 3 9 x −  3  −3  = 36 3 http://www.fatmuscle.cjb.net 4
Maths Extension 2 Shells y = f (x ) Volumes R r Area of one strip A = f ( x )dx ∆x f (x ) Volume of One Shell V = π R 2 − r 2 . f (x ) ( ) = π (( x + ∆x) − x ). f (x ) = π (x + 2 x∆x + (∆x) − x ). f (x ) 2 2 2 2 2 = π (2 x∆x ). f ( x ) = 2πx.∆x. f ( x ) Volume of Total b V = 2π ∫ x. f ( x ).dx a b = 2π ∫ xy.dx a http://www.fatmuscle.cjb.net 5
Maths Extension 2 Volumes Find the volume of the solid obtained by revolving the region x + y = 2 , x = 0 , y = 0 about the y-axis. x+ y y b V = 2π ∫ xy.dx =2 = 2−x a 2 = 2π ∫ x(2 − x).dx 0 2 ( ) = 2π ∫ 2 x − x 2 .dx 2 0 2  2 x 2 x3  = 2π  −  3 0  2  8 = 2π 4 −  3  8π cubic units = 3 -2 2 Find the volume bounded by the graphs y = x 3 , y = 0 within the range of 1 ≤ x ≤ 3 rotated about the y-axis. b V = 2π ∫ xy.dx a 3 = 2π ∫ xx3 .dx 1 3 = 2π ∫ x 4 .dx 1 3  x5  = 2π    5 1  243 1  = 2π  −  5  5 484 = cubic units 5 http://www.fatmuscle.cjb.net 6
Maths Extension 2 Find the volume of the solid obtained by revolving the ellipse V = 2π ∫ xy.dx a y2 b2 a = 2π ∫ x.2. b a 2 − x 2 .dx a 0 = V x2 y2 + = 1 about the y-axis. a2 b2 x2 y2 + a2 b2 b 4πb a 2 2 ∫ x a − x .dx a 0 u = a2 − x2 du = − 2x dx dx du = − 2x = 4πb 0 du ∫ x u. − 2x a a = Volumes y 2 x=a, u =0 x = 0 , u = a2 =1 = 1− x2 a2 = b2 − b2 x2 a2 b2a 2 − b2 x 2 a2 b2 = 2 (a 2 − x 2 ) a b 2 = ± a − x2 a = 2πb 0 u .du a a∫2 y 0 2πb  2u 2    = a  3  2  a 3 4πb  2 2  (a )  =   3a  4πba 2 cubic units = 3 3 http://www.fatmuscle.cjb.net 7
Maths Extension 2 Washer Method Area of one slice A(z ) = πR 2 − πr 2 = π R2 − r 2 = π (R + r )(R − r ) = π (x2 + x1 )(x2 − x1 ) ( Volumes ) R r Solve the quadratic formula to find x2 and x1 . Where x2 > x1 − b ± b 2 − 4ac 2a Washer Taurus Find the volume of the region bounded by the parabola y = x 2 and the lines y = 0, x = 2. Volume of One Slice V = π 4 − x 2 ∆y ( ) x = 2 is the larger radius 22 = 4 Total Volume 4 ( ) 2 V = π ∫ 4 − x dy 0 4 = π ∫ (4 − y )dy 0 4  y2  = π 4 y −  2 0  = 8π http://www.fatmuscle.cjb.net 8
4 Unit Maths – Mechanics Newton’s Second Law of Motion. 𝑭 𝒏𝒆𝒕 = 𝒎𝒙′′ Fnet is defined as the sum of forces. The direction of forces need to be taken into account. 𝒗(𝒙): Equations of motion. 𝒙′′ = 𝒗(𝒕): 𝒙′′ = 𝒗𝒅𝒗 𝒅𝒙 Horizontal resisted motion. 𝒅𝒗 𝒅𝒕 𝐹 𝑛𝑒𝑡 = 𝐷𝑟𝑖𝑣𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒 + 𝑅𝑒𝑠𝑖𝑠𝑡𝑖𝑣𝑒 𝑓𝑜𝑟𝑐𝑒 Vertical resisted motion. 𝐹 𝑛𝑒𝑡 = 𝐺𝑟𝑎𝑣𝑖𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑓𝑜𝑟𝑐𝑒(𝑚𝑔) + 𝑅𝑒𝑠𝑖𝑠𝑡𝑖𝑣𝑒 𝑓𝑜𝑟𝑐𝑒 Sometimes, It is necessary to describe two motions. Terminal velocity Terminal velocity is a constant velocity reached by a projectile moving vertically downwards. 𝒙′′ = 𝟎 𝒍𝒊𝒎 𝒗 𝒕→∞ 𝒍𝒊𝒎 𝒗 𝒙→∞ Circular motion. Angular 𝜃 𝑫𝒊𝒔𝒕𝒂𝒏𝒄𝒆 = 𝜽 Tangential 𝑨𝒏𝒈𝒖𝒍𝒂𝒓 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 = 𝒅𝜽 = 𝝎 𝒅𝒕 𝜃 Thompson Ly 1
4 Unit Maths – Mechanics Polar 𝑺 = 𝒓𝜽 𝒅𝑺 𝒅𝑺 𝒅𝜽 𝑻𝒂𝒏𝒈𝒆𝒏𝒕𝒊𝒂𝒍 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 = = × = 𝒓𝝎 𝒅𝒕 𝒅𝜽 𝒅𝒕 𝜃 𝒙 = 𝒓𝒄𝒐𝒔𝜽 𝒚 = 𝒓𝒔𝒊𝒏𝜽 𝒙′ = −𝒓𝒘𝒔𝒊𝒏𝜽 𝒚′ = 𝒓𝒘𝒄𝒐𝒔𝜽 Through calculations, it can be shown that: 𝑭 𝒏𝒆𝒕 = 𝒎𝒓𝝎 𝟐 𝒎𝒗 𝟐 𝒓 𝒐𝒓 𝑭 𝒏𝒆𝒕 = If there is more than one particle in the system, isolation is required. Pendulum motion. ℎ 𝛼 𝑇 𝑹𝒆𝒔𝒐𝒍𝒗𝒊𝒏𝒈 𝒉𝒐𝒓𝒊𝒛𝒐𝒏𝒕𝒂𝒍𝒍𝒚: 𝑹𝒆𝒔𝒐𝒍𝒗𝒊𝒏𝒈 𝒗𝒆𝒓𝒕𝒊𝒄𝒂𝒍𝒍𝒚: 𝑟 𝑚𝑔 𝑻𝒄𝒐𝒔𝜽 = 𝒎𝒓𝝎 𝟐 𝑻𝒔𝒊𝒏𝜽 = 𝒎𝒈 If the particle is touching any object, such as a sphere or a cone, the particle will be subjected to a normal force. Banked motion. Going quick • When an object moves quickly, it moves up the bank. 𝜃 𝑁 𝐹𝑟 𝜃 𝑹𝒆𝒔𝒐𝒍𝒗𝒊𝒏𝒈 𝒉𝒐𝒓𝒊𝒛𝒐𝒏𝒕𝒂𝒍𝒍𝒚: 𝑹𝒆𝒔𝒐𝒍𝒗𝒊𝒏𝒈 𝒗𝒆𝒓𝒕𝒊𝒄𝒂𝒍𝒍𝒚: Thompson Ly 𝑚𝑔 𝒎𝒗 𝟐 𝒓 𝑵𝒄𝒐𝒔𝜽 − 𝑭 𝒓 𝒔𝒊𝒏𝜽 = 𝒎𝒈 𝑵𝒔𝒊𝒏𝜽 + 𝑭 𝒓 𝒄𝒐𝒔𝜽 = 2
4 Unit Maths – Mechanics Going slow • When an object moves slowly, it moves down the bank. 𝑁 𝜃 𝑹𝒆𝒔𝒐𝒍𝒗𝒊𝒏𝒈 𝒉𝒐𝒓𝒊𝒛𝒐𝒏𝒕𝒂𝒍𝒍𝒚: 𝑹𝒆𝒔𝒐𝒍𝒗𝒊𝒏𝒈 𝒗𝒆𝒓𝒕𝒊𝒄𝒂𝒍𝒍𝒚: 𝐹𝑟 𝑚𝑔 𝒎𝒗 𝟐 𝒓 𝑵𝒄𝒐𝒔𝜽 + 𝑭 𝒓 𝒔𝒊𝒏𝜽 = 𝒎𝒈 𝑵𝒔𝒊𝒏𝜽 − 𝑭 𝒓 𝒄𝒐𝒔𝜽 = Ideal velocity Ideal velocity occurs when the object is subjected to no friction. Slipping 𝑭𝒐𝒓𝒄𝒆 𝒕𝒉𝒂𝒕 𝒘𝒆 𝒏𝒆𝒆𝒅 ≤ 𝑭𝒐𝒓𝒄𝒆 𝒕𝒉𝒂𝒕 𝒘𝒆 𝒉𝒂𝒗𝒆 𝑭𝒐𝒓𝒄𝒆 𝒕𝒉𝒂𝒕 𝒘𝒆 𝒏𝒆𝒆𝒅 ≤ 𝝁𝑵 Where, µ is the coefficient of friction. Thompson Ly 3
Maths Extension 2 – Harder Maths Extension 1 (3U) Harder Maths Extension 1 / Harder 3U ! Inequalities ! Induction http://fatmuscle.cjb.net 1
Maths Extension 2 – Harder Maths Extension 1 (3U) Inequalities Proof a − 2ab + b 2 ≥ 0 ≥0 a2 + b2 2 ( a 2 + b 2 ≥ 2ab ) True if True if Example 1 PROVE (a + b )2 (a + b )2 a=b a≠b ≥ 4ab = a 2 + 2ab + b 2 = (a − b ) + 4ab 2 ∴ (a + b )2 ≥ 4ab Example 2 If a, b, c, d are > 0, then a 4 + b 4 + c 4 + d 4 ≥ 4abcd PROVE a 2 + b 2 + c 2 ≥ ab + bc + ca (a 2 ) ( ) ( + b2 + b2 + c 2 + c2 + a 2 2 a 2 + b2 + c 2 ( ) ) ∴ a2 + b2 + c2 ≥ 2ab + 2bc + 2ca ≥ (ab + bc + ca ) ≥ ab + bc + ca Using a 2 + b 2 ≥ 2ab Example 3 The arithmetic mean = The geometric mean = PROVE a+b 2 (a + b )2 4 (a + b )2 a 2 + 2ab + b 2 a 2 − 2ab + b 2 (a − b )2 True if True if a+b 2 ab ≥ ab ≥ ab ≥ ≥ ≥ ≥ 4ab 4ab 0 0 a=b a≠b http://fatmuscle.cjb.net 2
Maths Extension 2 – Harder Maths Extension 1 (3U) Example 4 1 a+b 1 a+b Are they equal? ab a + 2ab + b − ab a 2 + ab + b 2 2 if 2 1 1 + a b 1 1 ≤ + a b b+a ≤ ab 2 ≤ (a + b ) ≥ 0 ≥ 0 = Prove by contradiction 1 1 1 + ≥ a b a+b b+a 1 ≥ ab a+b 2 (a + b ) ≥ ab Since (a + b )2 (a + b )2 ≥ 2ab ≥ ab http://fatmuscle.cjb.net 3
Maths Extension 2 – Harder Maths Extension 1 (3U) Induction Example 1 f (x ) = xe x f ' (x ) = xe x + e x f ' ' (x ) = xe x + e x + e x f ' ' ' (x ) = f n (x ) = Prove for n = 0 LHS = f 0 (x ) = xe x Assume n = k f k (x ) = e x (x + k ) = xe x + ke x f k +1 (x ) = xe x + e x + ke x = e x (x + 0) = e x (x + 1) = e x (x + 2) = e x (x + 3) = e x (x + n ) RHS = e x (x + 0) = xe x Assume n = k + 1 f k +1 (x ) = e x (x + k + 1) = xe x + e x + ke x http://fatmuscle.cjb.net 4
Maths Extension 2 – Harder Maths Extension 1 (3U) Example 2 Recurrence Induction A sequence of terms U n , n = 1, 2, 3, … Tn = U n U n = 4U n −1 − 5U n − 2 + 2U n − 3 Initial conditions U1 = 3 , U 2 = 1 , U 3 = 0 Show that by induction U n = 2n −1 − 3n + 5 n = 4, 5, … n = 1, 2, 3, … Consider S(n) = 2n −1 − 3n + 5 S(1) = 21−1 − 3(1) + 5 =1–3+5 =3 S(2) = 22 −1 − 3(2) + 5 =2–6+5 =1 S(3) = =4–9+5 =0 Assume S(k) = 2k −1 − 3k + 5 Prove S(k+1) = 2k +1−1 − 3(k + 1) + 5 = 2k − 3k − 3 + 5 = 2k − 3k + 2 LHS = 4U k −1 − 5U k − 2 + 2U k − 3 = 4 2k −1 − 3k + 5 − 5 2k − 2 − 3(k − 1) + 5 + 2 2 k − 3 − 3(k − 2) + 5 = 2k +1 − 12k + 20 − 5.2k − 2 + 15k − 40 + 2k − 2 − 6 − 22 = 2 k 2 − 5.2−2 + 2 −2 − 3k + 2 ( ) ( ( ) ( RHS = 2k − 3k + 2 ) ) = 2 − 3k + 2 k Hence if n = k True for n = k + 1 True for n = 1, 2, 3, … ∴must be true for all positive integer n http://fatmuscle.cjb.net 5