Further Mathematics Lesson Note SS2 Third Term

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Further Maths Enotes SS2 – Edudelight.com

FURTHER MATHEMATICS S S 2

SCHEME FOR 3^RD TERM

Review of second term’s work and introduction of binomial expression using Pascal triangle
Binomial expression of (a+b)ⁿ
Differentiation, limits and function
Differentiation from first principle
Differentiation of transcendent
Rules of differentiation
Rate of change Maximum and minimum problem
Higher derivatives of implement functions mechanics
Mechanics
Vector and dimension
Operational research
revision and examination

WEEK 1

REVISION TO SECOND TERM WORK AND INTRODUCTION TO BINOMIAL EXPRESSION USING PASCAL TRIANGLES

Revision

The distance between the points; P(x, 7) and Q(6, 19) is 13 units find the value of x

distance btw PQ =√(x₂–x¹)² + (y₂–y₁)²

√(6 x)² + (19 – 7)²

√(6 – x)² + 12² = 13

(6 – x)(6 – x) + 144 = 13²

36 – 6x – 6x + x² + 144 – 169 = 0

36 – 12x + x² – 25 = 0

x² – 12x + x² – 25 = 0

(x² – 11x)( – x + 11) = 0

x (x – 11) – 1 (x – 11) = 0

(x – 11) ( x – 1) = 0-

x = 11 or x = 1

Find the equation of a circle with centre (5, -1) and radius 3 units

(x – a) ²+(y – b)² = r²

(x – 5)² + (y + 1)² = 9

x² + y² – 10x +2y+ 1= 9

X² + y² – 6x + 2y =8

x² + y² – 6x + 2y -8 = 0

Find the center and radius of the circle whose equation

X² + y² – 2x – 4y + 1 = 0

Soln:

X² – 2x + y² – 4y + 1 = 0

X² – 2x + (- 1)² + y² – 4y + (- 2) = – 1 + 1 + 4

(x – 1)² + (y – 2)² = 4

r^{2 = 4}

r = √4 = 2 Center=(1,2)

If two people each toss a coin. What is the the probability of both coins showing tails.

P(TT) = p(T)×P(T)=1/2×1/2= ¼

Two dice are toss once. What is the probability of obtaining 9 points

1 2 3 4 5 6

1 2 3 4 5 6 7 = 4 = 1

2 3 4 5 6 7 8 369 9

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12

What force will you give a mass of 5kg at velocity 5ms^-1 in 3 mins?

f = ma

f = mv

where m = 5kg, v = 4.5, t = 3min = 180S

f = 5×4.5 = 0.1215N

180

f = ma

32 = 8m

m = 4

s = ut + 1 at²

45 = 40t + 1 x 10 t²

45 = 40t + 5t²

5t² + 40t – 45 = 0

(5t² + 45t) (-5t–45) = 0

5t(t+9) – 5 (t–9) = 0

(t+9) (5t–5) = 0

(t =-9) or 5t = 5

t = – 9 or t = 1

In how many ways can 10 people be seated on a bench if only 3 places are available.

solution

10p3 = 10 ! 10 ! = 10x9x8x7

10-3 ! 7 ! 7 !

= 720 ways

f = m(v–u) = 3(v–45)

t 0.3

10 = 3(v–5)

0.3

3 = 3 v – 15

18 = 3v

v = 6 m/s

INTRODUCTION TO BINOMIAL EXPRESSION USING PASCAL TRIANGLE

Consider the expansion of the following

(x+y)⁰ = 1

(x+y)¹ = 1x +1y

(x+y)² = 1x² +2xy +1y²

(x+y)³ = 1x³+ 3x²y+3xy²+1y³

Pascal Triangle

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

It is used in determining the coefficient term of the powers of the binomial expression.

Ex (1). Expand (x + y)⁵ using Pascal’s triangle

Solution

(x + y)⁵ the coefficient are 1 5 10 10 5 1

1x + x⁴y + x³y2 +x²y³ + xy⁴ + y⁵

X⁵ + x⁴y + x³y² + x³y² + xy⁴ + y+ y⁵

Notice that

There are 6 terms
In each of the term involved in the expression the power of x and y put together is 5. We say that the expression is homogenous
While the power of x is decreasing, the power of y is increasing

Ex (2). Using Pascal’s Triangle, simplify correct to 5 decimal places (1.01)⁴

Solution

(1.04)⁴ = (1+0.01)⁴ = 1⁴ + (0.01) + 6(0.01)² + 4(0.01)³ + 1 + (0.01)³

= 1 + 0.04 + 0.0006 + 0.000004 + 0.00000001 =1.04060401

= 1.04060

Exercises

Using Pascal’s triangle, expand and simplify completely: (2x + 3y)⁴

Solution

= (2x)⁴ + 3y⁰ + 4(2x)+3(3y)¹ + 6 (2x)² (3y)¹+ 4(2x(3y)³ + 1 (2x)⁰ (3y)⁴

= 2x⁴ + 96x³ y + 216x²y² + 216xy³ + 81y⁴

sing Pascal’s triangle, expand and simplify completely: (x – 2y)⁵

Solution

(x – 2y)⁵ 1 (x)⁵ + (2y)⁰ + 5(x)⁴ (2y) + 10(x)³ (2y)² + 10(x)² (2y)³ + 5(x) (-2y) + 1(x)⁰ (-2y)⁵

= x⁵ + -(10 x⁴ y) + 40 x³ y² + (- 80x² y³) + 80xy⁴ + (32y²)

= x⁵ – 10x⁴y + 40x³y²+ 80x²y³ + 80xy⁴ – 32y⁵

Assignment

1.Using Pascal’s triangle, expand and simplify (1 – 5x)⁵

2. Find the coefficient of x³ in the expansion of (x³-1/x³)^1989 (©Onyedelmagnifico)

WEEK 2

BINOMIAL EXPANSION OF (A + B)^N

(a + b)ⁿ = aⁿ+ nc, a^n-1 b + nc₂, a^n-2 b² +

_ _ _ nc_r a^n-r b^r + _ _ _ _ _ + bⁿ

where n is a positive integer

but nc₁ = n ! =n(n–1)! = n

(n–1)! n(n–1)!

nc₂ = n ! = n(n –1) (n–2)! = n(n–1)

(n–2)!2! (n–2)! 2! 2 !

nc₃ = n ! = n(n -1)(n–2)(n-3) = n(n-1)(n-2)

(n-3)! (n-3)! 3! 3!

Hence,

(a+b)ⁿ + aⁿ + na^n-1b + n(n-1) a^n-1b² + n(n-1)(n-2) —(n+r+1)

2 ! r !

xa^n-rb^r + _ _ _ _ _ _ _ _ bⁿ

Ex.1a) Write down the binomial expansion of 1+1_x⁶ Simplify all the terms

b) Use the expansion in (a) to evaluate (1.0025)⁶ correct to five significant figures

Solution

(a) 1+1 ⁶ = (1)⁶ 6c, (1⁵) 1 ¹ + 6c₂(1)⁴ 1²

4x 4x b4x

+ 1 + 6c₃(1)³ 1 _x ³ + 6c₄ (1)² 1 ⁴ + 6c₅ (1)ⁿ 1 ⁵ + 6c₃(1)³ 1 ⁶

4x 4x 4x 4x

= 1 + 6³ 1 + 15 1 ² + 20 1 ³ + 15 1 ⁴ + 6 1 ⁵ + 1 ⁶

4x 16x 64x 256x 1024x 4096x

= 1 + 3x + 15x +² + 5x³ + 15x⁴ + 3x⁵ + 1x⁶

2 16 16 256 512 4096

b) (1.0025)⁶ = (1+0.0025)⁶ = 1 + 25 ⁶ = 1 + 1 ⁶

10000 400

put 1x = 1

4 400

400x = 4

x = 4 = 1 = 0.01

400 100

.:. (1.0025)⁶ = 1 + 3 (0.01) + 15 (0.01)² + 5 (0.01)³ + 15 (0.01)⁴ + 3 (0.01)⁵ + 1 (0.01)⁶ 2 16 16 256 512 4096

+ 1 + 0.015 + 0.00009 + 0.0000003125

= 1.0150940625

= 1.0151

Ex 2.i. Using the binomial theorem, expand (it 2x)⁵, simplifying all the terms.

ii. Use you expansion to calculate of value of 1.02⁵ correct to six significant figures.

If the first three terms of the expansion of (1+px)ⁿ in ascending power of x are 1 + 20x

+160x. Find the value of n and p

Solution

(1+2x)⁵ = 1 + 5c, (2x) + 5c₂ (2x)² + 5C₃ (2x)³ + 5c₄ (2x)⁴ + 5C₅(2x)⁵

= 1 + 5(2x) + 10(2x)² + 10(2x)³ + 5(2x)⁴ + (2x)⁵

= 1 + 10x + 40x² + 80x³ + 80x⁴ + 32x⁵

(1 + 0.02)⁵ = (1.02)⁵

Put it 0.02 = 1 + 2x

0.02 = 2x

2 2

x = 0.01

= (1.02)⁵ = 1 + 10(0.01) + 40(0.01)² + 80 (0.01)³ + 80 (0.01)⁴ + 32(0.01)⁵

= 1 + 0.1 + 0.004 + 0.00008 + 0.00000008

= 1.10408(6 s.f)

(b) (1+px)ⁿ = 1 + 20x + 160x² + _ _ _ _ _ _ _ _ _

(1+px)ⁿ = 1 + nc, (px) + nc₂ (px) + _ _ _ _ _ _ _ _

1 + npx + n(n-1)² p²x²

= 1 + 20x + 160x²

By comparing coeff.

np = 20 _ _ _ _ _ _ _ _ _ (1)

n(n-r)p² = 160 _ _ _ _ _ (2)

from eqn (1)

p = 20 _ _ _ _ _ _ _(3)

Subst. for p = 20 in eqn (2)

n n-1 x 20 ² = 160

2 n

2000

n(n-1) x 400 = 160

2 n

200n – 200 = 160n

200n – 160n = 200

40n = 200

n = 200

n = 5

Subst. for in eqn. (3)

p = 20 = 4

Ex 3. a) Obtain the first four term of the expansion of 2 + 1 ⁸ in ascending power of x.

Hence find the value of (2.005)⁸ 2x

Solution

2 +1 ⁸ =

= 2⁸ + 8c₁ 1 x + 8c₂ 1 ² + 8c₃ 1 ³

4 4x 4x

2⁸ (1 + 8² 1 ¹ + 28 1 ² + 56 1 ³

4x 4x 4x

2⁸ (1+2x + 7 x² + 7 x³ + _ _ _ _ _ _)

4 8

b) Write (2.005) = 2 + 0.005

Put 2 + 1 x = 2 + 0.005

x = 0.005 x 2

x = 0.01

Hence, (2.005)⁸ =2⁸ 1+2(0.01) + 7(0.0001) + 7 (0.000001)

4 8

= 2⁸ + 2⁹ +(0.01) + 2⁶ (7) (0.0001 + 2⁵ (7) 0.000001 + _ _ _ _ _ _ _ _ _ _

= 256 + 5.12 + 0.0448 + 0.000224

= 261.17

BINOMIAL THEOREM FOR NEGATIVE INDEX

Ex 4. a) Write and simplify the first four terms of the expansion of 1- 1 ^-4

Solution

BINOMIAL THEOREM

(a+b)ⁿ = aⁿ + nc, a^n-1b + nc₂ a^n-2 b +_ _ _ _ _ _

nc_r a^n-r b^r + _ _ _ _ _ _ bⁿ

1- 1 ^-4 = 1 + (-4) -1 + 10 – 1 ² + – 20 1^s

3x 3x 3x 3x

= 1 + 4 x + 10 x + 20x³

3 9 27

Exercises

1^.a) Using the binomial theorem, write down and simplify all the term of the expansion of (1+1//2a ⁶

b) Given that 3.15⁶ = 3⁶ x b, use your result to estimate the value of b, correct to 4 decimal places.

2. Obtain the first five term of the expansion of (1+3x)^-1/2 x < 1

3. Write down the first three terms of the expression (1+ax)ⁿ in ascending powers of x. If the coefficient of x and x² are 2 and 3/2 respectively. Find the value of a and n.

Solution

a). 1+1 ⁶ = 1 + 6c₁ 1 + 6c₂ 1 ² + 6c₃ 1 ³ + 6c₄ 1 ⁴ + 6c₅ 1 ⁵ + 6c₆ 1

2a 2a 2a 2a 2a 2a 2a

= 1 + 6 1a + 15 1a ² 20 1 ³ + 15 1 +⁴ + 6 1 ⁵ + 1a ⁶

2 2 2a 2a 2a 2a

= 1 + 3a + 15 a² + 20 a³+ 15 a⁴ + 6 + a⁵ + 1 a

4 8 16 32 64

= 1 + 3a + 15 a² + 5 a³ + 15 a⁴ + 3 a⁵ + 1 a⁶

4 2 16 16 64

b). (3.15) = (3+0.15)

3(1+0.05)

.:. (3.15)⁶ = 3⁶ (1+0.05)⁶

Put 1 + 0.05 = 1 + 1 a

0.05 = a

a = 2 x 0.05

a = 0.1

(3.15)⁶ = 3⁶(1+3(0.1) + 15 (0.01) + 5 (0.001) + 15 (0.0001) + 3 (0.00001) + 1 (0.0000001)

4 2 16 16 16

= 3⁶(1+0.3 + 0.0375 + 0.0025 + 0.0009375 + 0.000001875

= 3⁶ (1.34009)

If 3.15⁶ = 3⁶ x b

3⁶(1.34009) = 3⁶b

3⁶ 3⁶

b = 1.3401 (4 d.p)

2. (1+3x)^-1/2 = 1 + (-1/2)3x +

+ (3x)⁴

= 1 – x + x² + = x³ + x⁴

3. (1 + ax)ⁿ = 1 + nC₁(ax) + nC₂ (ax)² + nC₃ (ax)³+ ………

=1 + n(ax) + (ax)² + (ax)³

But coeff of x is 2

And “ of x² is ³/₂

i.e an = 2 ………………………………..(1)

a²= ³/₂

= ³/₂

= ⁶/₂

a²(n² – n) = 3

a²n² – a²n – 3……………………………(2)

From equ (1)

A = ………………………………..(3)

Subst for a = in equ 1 (2)

( )² n² – ( )² n = 3

(n²) – (n) = 3

4 – = 3

4- 3 =

1 =

n = 4

subst for n in equ (3)

a = =

ASSIGNMENT:

FURTHER MATHEMATICS FOR SENIOR SECONDARY SCHOOLS (E. Egbe, G.A. Odili)

EXERCISE 36.2; NO 1, 2, 3, and 5 PAGE 519

2. Given that the coefficients of x² and y² in the expansion of

Are in the ratio 2:1 (© Onyedelmagnifico)

Week 3

DIFFERENTIATION

Differentiation in mathematics has to do with the difference between two values in a scale

Limit of a function (Properties of Limits)

Lim = K

x a

Lim F(x) + F₂(x) + F₃ (x) + ……………….F_n (x)]

a x a x a x a x

I.e. the limit of the sum of a finite no of functions is equal to the sum of their respective limits.

Lim [F₁ (x) – F₂ (x) = Lim F₁ (x) – Lim F₂ (x)
Lim [F₁(x) F₂(x). F₃ (x) + …………………Fn (x)]

= Lim F₁(x). Lim F₂ (x). Lim F₃ ………………… Lim F_n (x)

a x a x a x a x

x a

= Provide Lim F₂ (x) 0

x a

Lim K f(x) = K Lim f(x)

Lim = 1

x a

Ex 1: Evaluate Lim

Solution

= =

x a

Evaluate Lim

Solution

x a

Lim

x a

Lim (x + 5)

5 + 5 =10

x a

Evaluate Lim

Solution

x Infinityinfint

We know that Lim = 0

Lim

x a

Lim

x a

Lim

= = 3

Ex 1: Differentiate 3x² – x + 5 wrt x from first principle

Solution

Write Y = 3x² – x + 5

Y + = 3(x + x)² – (x + x) + 5

6 + y = 3(x² + x x + x dx + ( x)² – x – dx + 5)

= 3(x² + 2x x + ( x)²) – x – x + 5

y + y = 3x² + 6x x + 3( x)² – x – x + 5

But y = 3x² – x + 5

= 3x² + 6x x + 3 ( x)² – x – x + 5 – (3x² – x + 5)

² _ 6x x + 3( x)² – x – x + 5 – 3x² + x – 5

x x + 3( x)² –

Divide through by x

= + –

Lim 0

= 6x – 1

2. Find the derivative of y =x³ from first principle.

Solution

Y = x³

y + y = (x + x)³

= x³ + 3x² x + 3x x² + ( x)³

y = x³ + 3x² x + 3x x² + ( x)³ _ (x³)

y = 3x² + 3x ( x)² + 3x( x)² + ( x)³

y =3x² x + 3x ( x)² + ( x)³

= + +

= 3x² + 3x x + ( x)²

= 3x² = 3x²

Differentiation of polynomials

If y = xⁿ ; if y = axⁿ

= nx^n-1 = anx^n-1

The derivative of x⁷is 7x⁶

The Derivative of x^1/2 is ½ x^-1/2 =

The derivative of a constant is Zero

Examples:

Find the derivatives of the following
2x³ – 5x²+2

Let 6 = 2x³ – 5x² + 2

= 6x² – 10x

Let y = x² + 2x + x^-1

= 2x + 2 + -x^-2

2x + 2-

+ – 3

= + – 3

= x^-1/2 + – x^-3/2

–

Exercises

Find the limiting values of g(x) = (x-2) (x+2) as x approaches 3

Solution

g(x) = (x – 2) (x + 2)

= x² + 2x – 2x – 4

= x² – 4

Lin g (x) = 9 – 4

x 3 =5

If f(x) = 3x², find

x a

Lim

Solution

F¹(x) =

F¹(x) = 6x

Find the gradient of the tangent to the curve

y=2x² + 5x – 3 at the point where x = 1

Solution

is the gradient function of the curve

= 2x² + 5x – 3

= 4x + 5

= m = 9

Find the first principle, the derivative of with respect to x

Solution

Let 6 =

6 + =

y = –

y =

= 3x – 3x – 3 x

x(x + x)

= -3 x

x(x + x)

y = – 3 x

x x(x + x)

= – 3

x(x + x)

Lim dy = -3

dx x (x + 0)

-3 = -3. 1 = -3 x^-2 = -3x^-2

x² x²

Assignment

FURTHER MATHEMATICS FOR SENIOR SECONDARY SCHOOLS (E. Egbe, G.A. Odili)

EXERCISE 20.6; NO 1, 2, 3, 9 and 12 PAGE 295

Further Maths Enotes SS2 – Edudelight.com

WEEK 4

Differentiation of transcendent (Trig function)

Derivative of y = sinx

If y = sinx

y + y = sin (x + x)

y = sin (x + x) – sinx

Apply trig. formula

Sin A – Sin B = 2Cos A + B sin A – B

2 2

Where A = x + x, B = x

.:. y = 2Cos (x + x)+x Sin+ x –x

2 2

= 2Cos 2x + x . Sin x

2 2

= 2Cos x + x . – Sin x

2 2

dy = 2Cos x + x . Sin x

dx 2 2

Multiply numerator and denominator by 2

= Cos x + x .Sin x

2 2

= Cos x + x . x Sin x

2 2

when x = 0

dy = Cos x

= Cos x

2. The derivative of Cos x

if y = Co sx

dy = – Sinx

3. The derivative of tanx

If y = ta nx

dy = sec²x

4. The derivative of secx

if y = secx

dy = secx tanx

5. The derivative of cosecx

If y = cosec x

dy = – cosecx cot x

6. If y = e^x

dy = e^x

7. If y = Log_ax

dy = 1

dx x

8. If y = a^x

dy = a^x Log_ea

Ex. 1. Find the derivative of cos2x

Solution

If y = cos2x

let u = 2x

du = 2

y = cosU

dy = -sinU

dy = 2x – sin u

= 2x – sin2x

-2 sin 2x

2. If y = sin 1/3 x find x

Solution

Let u = 1/3 x

du = 1

dx = 3

y = sinU

dy = cosU

dy = 1 cosU

du 3

= 1 cos 1 x

3 3

3. Find the derivative of e^2x

Solution

If y = e^2x

Let u = 2x

du = 2

y = e^u

dy = e^u

dy = 2e^u

= 2e^2x

Exercises

Find the derivative of the following
Tan5x
Cos²x
Find the derivative of Log_a √1 + x
Find the derivative of a^2x
If y = e^sin4x, find dy

Solution

1a. if y = tan 5x

Let u = 5x

du= 5

y = tanU

dy = sec²u

.:. dy = 5.sec²u

= 5sec² (5x)

1b. If y = cos²x

let u = cosx

du = – sinx

y = u²

dy = 2u

dy = – 2u . sin x

= – 2cosx.sinx

2a. if y = Log_a 1 + x

Let u = 1 + x = (1 + x)^1/2

du = 1 (1 + x)^-1/2= 1 . 1 = 1

dx 2 2 (1 + x)^1/2 2 1 + x

y = Log_au

dy = 1

du u

dy = 1 1

dx u . 2 1 + x

1 . 1

1 + x 2 1 + x

2(1 + x)

3. If y = a^2x

Introduce the log of both sides

Log_ey = 2x Log_ea

Diff wrt x

dy = 2log_ea

y dx

multiply thru by y

dy = 2y log_ea

2a^2x log_ea

4. If y = e^sin4x find dy

let u = sin4x

du = 4cos4x

y = e^u

dy = e^u

dy = e^u – 4cos4x

= e^sin4x 4cos4x

4e^sin4x cos4x

ASSIGNMENT: FURTHER MATHS PROJECT, PAGE 197, 29C, NO. 4C PAGE 198, NO. 8B, C

WEEK 5

RULES OF DIFFERENTIATION

1. Derivation of a sum

If y = u + v [where u and v are function of x]

dy = du + dv

dx dx dx

2. Function of a function

If y = f(u)

and u = h(x)

chain rule for differentiation

dy = du x dy

dx dx du

Ex. 1. Find the derivative of each of the following

y = (3x² – 2)³
y = 1

(1 + x²)

Solution

Let u = 3x² – 2

du = 6x

y = u³

dy = 3u²

.:. dy = 6x.3u² = 6x X 3(3x² – 2)

= 18x (3x² – 2)

Let u = 1 + x²

dy = 2x

y = 1 = 1 = u^-1/2

u u^1/2

dy = – ½ u^-3/2

dy = 2x . -1 u^-3/2

dx 2

-xu^3/2

= – x . 1

u^-3/2

= – x = -x

u³ (1 + x²)³

3. Derivative of a product

If y =uv, where u and v are functions of x then dy = dv + du

0 dx dx dx

Ex. 1. Find the derivative of each of the following

y = (3 + 2x) (1 – x)
y = x (1 + 2x)²

Solution

Let u = 3 + 2x, v = 1- x

du = 2, dv = -1

dx dx

dy = dv du

dx dx dx

(3 + 2x) x – 1 + 1 – x(2)

-1(3+ 2x) + 2(1 – x)

= -1 -4x

b. Let u = x, v = (1 + 2x)²

du = 1 , dv = 4 (1 + 2x)

dx 2 x dx

dy = x x 4(1 + 2x) + ( 1 + 2x)²X1

dx = 2 x

= 4 x (1 + 2x) + (1 + 2x)²

2 x

4. Derivative of a Quotient

If y = u, where u and v are functions of x and = 0

dy = du dy

dx dx dx

V²

Ex. 3. Find the derivative 2 + x

x² + 2x + 7

Solution

Let u = 2 + x, v = x² + 2x + 7

du = 1 dv = 2x + 2

dx dx

dy = 1(x² + 2x + 7) – (2 + x) ( 2x + 2)

(x² + 2x + 7)²

x² + 2X + 7 – (4X + 4 + 2x² + 2x)

(x² + 2x + 7)²

x² + 2x + 7 – (6x + 2x²+ 4)

(x² + 2x + 7)²

x² + 2x + 7 – 6x – 2x² – 4

(x² + 2x + 7)²

= – x²– 4x + 3

(x² + 2x + 7)²

5. Implicit differentiation

Consider x²y + xy³ + 3x = 0, here the relationship between y and x is set to be implicit, it is not expressed in a direct way. It is defined in term of x and y.

Ex 4. Differentiate each of the following implicitly.

x² + y² = 25

Differentiate wrt x

2x + 2y dy = 0

2y dy = – 2x

dy = -2x = – x

dx 2x y

4y²x – 5x²y³ + 4y = 0

Differentiate wrt x

8yx dy + 4y² – 10xy³ – 15y²x² dy + 4dy

dx dx dx

= 0

Collect like terms

8xy dy – 15x²y² dy + 4dy = 10xy³-4y²

dx dx dx

dy (8xy – 15x²y² + 4) = 10xy³ – 4y²

dy = 10xy³ – 4y²

dx 8xy – 15x²y² + 4

Exercise

Find the derivation of x + 1 – 3

If y = 5 find dy

(6 – x²) dx

If f(x) = (1 _ 2x + 3x²) (4 – 5x²), Find f¹(x)
Find the derivative of 3 + 2x –x²

1 + x

Differentiate x²y + y² + 4x + 1, Implicitly.

Solution

Let y = x + 1 – 3

dy = 1 + – 1

dx 2 x 2 x³

= 1 – 1

2 x 2 x³

y = 5

(6 – x²)³

Let u = 6 – x²

du = -2x

y = 5 = 5.1 = 5.u^-3

u³ u³

dy = -15u^-4

dy = -2x(-15u^-4)

+30 x u^-4

+ 30 x . 1 = + 30x

u⁴ u⁴

dy = -30x

dx (6 – x²)⁴

If f(x) = (1 – 2x + 3x²) (4 – 5x²),

f¹(x) = dv + du

dx dx

Let u = 1-2x + 3x², v = 4 – 5x²

du = -2 +6x dv = -10x

dx dx

dy = 1 – 2x + 3x² (-10x) + (4 – 5x²) (-2 + 6x)

– 10x + 20x² – 30x² + [-8 + 24x + 10x² – 30x³]

– 60x³ + 30x² + 14x – 8

Let u = 3 + 2x – x², v = 1 + x

du = 2 -2x v = (1 + x)^1/2

dx dv = 1 (1 + x) ^-1/2

dx 2

dy = du – dv

dx dx dx

V²

(1 + x)^1/2 2(1 – 1x) – (3 + 2x – x²) x 1

2(1 + x)^1/2

1 + x

Multiply through by 2(1 + x)^1/2

4(1 + x) (1 – x) – (3 + 2x -x²) = 4(1 – x²) -3 -2x + x²

2(1 + x)^3/2 2(1 + x)^3/2

= 4 – 4x² -3 -2x + x²

2(1 + x)^3/2

= 1 – 2x – 3x²

2(1 + x)^3/2

x²y + y² + 4x = 1

2xy + x²dy + 2y dy + 4 = 0

dx dx

x²dy + 2y dy + 2xy + 4 = 0

dx dx

dy (x² + 2y) = -2xy – 4

dy = -2(xy + 2)

dx x² – 2y

ASSIGNMENT

FURTHER MATHS PROJECT, PAGE 187, EXERCISE 11A, NO. 16, 18, 19B, 23E.

WEEK 6

APPLICATION OF DIFFERENCES

Rate of Change

If y = f(x), dy can be interpreted as the rate at which y is changing respect to x.

If y increase as x increase dy > 0 and if y decreases as x increases, dy < 0

dx dx

Ex. 1. The radius of a circle is increase at the rate of 0.01cm/s. Find the rate at which the area is increasing when the radius of the circle is 15cm.

Solution

A = r²

dA = 2 r, dr = 0.01cm/s

dr dt

dA = dA . dr

dt dr dt

= 2 r. 0.01

= 2 x 5 x 0.01

= 10 x 0.01

= 0.1 cm/s

= 0.1 x 3.142 = 0.3142cm/s

Tangents and Normal to curve

For any curve, dy is the gradient function

Recall that equation of the line of gradient m is y-y₁ = m(x – x₁). From this equation, we can obtain the equation of the tangent.

The normal to a curve is the straight line perpendicular to the tangent at the point of contact.

Equation of normal is

y – y, = m¹(x – x₁)

Where m¹ = – 1

Ex.1. Find the equation of the tangent and the normal to the curve y = 2x² – x² + 3x + 1 at the point x = 1

y = 2x³ – x² + 3x + 1

dy = 6x² – 2x + 3

at x = 1

dy = 6 – 2 + 3 = 7

.:. gradient of the tangent at x = 1 is 7

at point x = 1

y=2-1+3+1=5

Equation of the tangent at x =1

y – y₁ = m(x – x₁)

y – 5 = 7 (x – 1)

y – 5 = 7x -7

y = 7x – 2 equation of the tangent

Equation of the normal at x = 1

y = y₁ = m¹ (x – x₁)

y – y₁ = -1 (x – x₁)

y – 5 = -1 (x – 1)

y – 5 = -x + 1

7(y – 5) = 1 –x

7y – 35 = 1 – x

x + 7x – 36 = 0

Ex. 2. Find the equation of the tangent to the curve x²y + y³x + 3x – 13 = 0 at the point ( 1, 2)

Solution

x²y + y³x + 3x – 13 = 0

2xy + x² dy + 3y² x dy + y³ + 3 = 0

dx dx

x² dy + 3y²x dy = -2xy – y³ -3

dx dx

dy (x² + 3xy²) = -2xy –y³-3

dy = -2xy –y³-3

dx x² + 3xy²

m = dy

at X = 1, Y = 2

m = -2(1)(2) -2³ -3

1 + 3(4)

= – 4 – 8 – 3 = -15

13 13

equation of tangent is

y – y₁ = m(x – x₁)

y – 2 = – 15 (x – 1)

13(y – 2) = -15x + 15

13y – 26 = – 15x + 15

15x + 13y – 26 – 15 = 0

15x + 13y – 41 = 0

Maximum and Minimum Value

The necessary and sufficient conditions for a maximum point at x = a on the curve y = f(x) are:
f¹(a) = 0 or dy = 0

f¹¹ (a) < 0 d²y =0

The necessary and sufficient conditions for a maximum point at x = a on the curve y = f(x) are:
f¹(a) = 0
f¹¹ (a) > 0
The curve y = f(x) has a point of inflexion at x = a if;
f¹¹(a) = 0 f¹¹9a) = 0

f¹¹(a^–) < 0 or f¹¹(a^–) > 0

f¹¹(a⁺) > 0 f¹¹(a⁺) , 0

Both maximum and minimum point are called turning points.

Ex. 1. Find the stationary point of a curve whose equation is y = x³ + x² -3x + 4

Solution

y = 1 x³ + x² – 3x + 4

dy = x² + 2x – 3

-3x²

(x²+3x) – (x-3

X(x+3) -1(x+3)

(x+3)(x-1)

At stationary points dy = 0

(x + 3) (x + 1) = 0

x = -3 or x = 1

2. Find the turning point on the curve

y = x⁴+ 5 x³ – 2x²–3x+ 1 and distinguish between them (b) sketch the graph of the following

2 3

Solution

y = x⁴+ 5 x³ – 2x+2– 3x+ 1

2 3

dy = 2x³ + 5x² – 4x -3, d²y = 6x² + 10x -4

at stationary points, dy = 0

solve

2x³ + 5x² – 4x – 3

(x + 1) is a factor

2x² + 7x + 3

x – 1 2x³ + 5x² -4x -3

2x³– 2x²

+ 7x² – 4x – 3

7x² – 7x

+ 3x – 3

3x – 3

(x – 1) (2x²+ 7x + 3)

(x – 1) (2x + 1) (x + 3) = 0

x = 1, x = – ½, x = -3 are the stationary points.

For minimum points

f¹(a) = 0

f¹¹(a) > 0

Subt. for x = 1, x = -0.5, x =-3 into d²y²

at x = 1, d²y² = 12

dx²

AT X = -0.5, d²y = – 7.5

dx²

at x = -3, d²y = 20

Hence there is minimum point at x = 1 and x = -3

There is maximum point at x = – 0.5

Y_min= ½ + 5/3- 2 – 3 + 1 = 0.5 + 1.7 – 2 – 3 + 1 = -1.8

Y_min =-40.5 – 45 – 18 + 9 + 1 + – 93.5

Y_max= -0.3125 + -0.21 -0.5 + 1.5 + 1

= 1.76

-1/2

-20

-40

-60

-80

-100

-3

Exercises

Find the maximum and the minimum points of curve (a). y = 2x² – 3² – 1x + 4 (b). Sketch the curve. (WAEC 2013)
Find the equation of the tangent to the curve x²+ y² = 1 at point 1, 3

4 2

The equation of a curve is y = x(3 – x²)

Find the equation of its normal when x = 2. (WAEC 2012)

Solution

y = 2x² – 3x² – 12x + 4

dy = 6x² – 6x – 12

= 6(x² – x – 2)

at stationary pt dy = 0

i.e x² – x – 2 = 0

(x + 1) (x + 2) = 0

X= -1 or x = 2

d²y = 12x – 6

dx²

at x = -1, d²y = – 12 – 6 = – 18 < 0

d²y

There is a maximum point at x = -1

at x = 2, d²y = 12(2) -6 = 18>0 (minimum point)

.:. There is a minimum point at x = 2

b) y_max = 2(-1)³ – 3(-1)² – 12( – 1) + 4

-2 -3 + 12 + 4 = 11

y_min 16 – 12 – 24 + 4

= – 16

x + 2y dy = 0

2 dx

2y dy = – x

dx 2

2y 2y

dy = -x

dx 4y

at (1,3/2), dy = -1

dx 6
m =-1

eqn of tangent is y –y, = m(x-x)

y- 3/2 = -1 (x-1)

y – 3 = -x + 1

1 2 6

2y – 3 = x + 1

2 6

6(2y-3) = 2(-x+1)

12y-18 = -2x+2

12y + 2x – 18 – 2 = 0

12y + 2x – 20 = 0 or 2x +12y – 20

3. y = x(3 – x²)

y = 3x – x³

dy = 3 – 3x²

at x = 2

dy = m = 3 – 3(4) = 3 – 12 = -9

at x = 2 y = 2(3 – 4)

2(-1) = – 2

eqn of its normal = y-y₁ = 1 (x-x₁)

y- -2 = -1 (x-2)

-9

y + 2 = x – 2

9( y + 2) = x – 2

9y + 18 = x – 2

9y – x + 18 +2 = 0 .: 9y –x + 20 = 0

– x + 9y +20 = 0

x – 9y – 20 =0

Assignment further maths project, page 227 Exercise 12, no 8, 12 and 21

WEEK 7

REVIEW OF FIRST HALF TERM`S WORK AND PERIODIC TEST.

Further Maths Enotes SS2 – Edudelight.com

WEEK 8

HIGHER DERIVATIVES

Given y = f(x), dy is also a function of x

The derivative of dy wrt x is d²y (second derivative)

dx dx²

The derivative of d²y is d³y (3^rd derivative)

dxdy³

The derivate of d³y + d⁴y (4^th derivative)

dy³ dy⁴

Ex: find the first, second and third derivate of each of the fill

a) 3x⁵ – 2x + x² -1

b) 1n x

c) sin 3x²

Solutions

a) y = 3x ⁵ – 2x⁴ + x²-1

dy = 15x⁴ – 8x³ + 2x

d²y = 60x³ – 24x² +2

dx²

d³y = 180x² – 48x

dy³

b) y + in x

dy = 1

dx x

d²y = -x^-2 = -1

dx² x²

d³y = 2x^-3 = 2

dy³ x3

c) sin 3x²

y = sin 3x²

dy = 6x cos 3x²

Let u = 3x²

du = 6x

y = 6x cos u

dy = 6x (-sin u) + 6(cos u)

= -6x sin u + 6 cos u

= – 6x (sin 3x²) + 6 cos 3x²

d²y = 6x – 6x sin 3x² + 6 cos 3x²

dx²

d²y = 6 cos 3x² – 36x² sin 3x²

dx²

d³y = -6 sin 3x² d (3x²) – 72x sin 3x² + -36x² cos 3x² d(3x²) ^6x

dy³ dx dx

= -36x sin 3x² – 72x sin 3x² – 216x³ cos3x²

= -108x sin 3x² – 216x³ cos3x²

= – 108x (x sin 3x² – 2x² cos 3x²)

Exercises

Find d³y given that

dy³

y = sinx²
y = 3x . e^-x
y = log_e (1+2x)²

Solutions

y = sin x

dy = 2x cos x²

d²y = 2 cos x² + 2x – sin x² d (x²)^2x

dx² dx

= 2cos x² – 2x sin x² (2x)

= 2cos x² – 4x² sinx²

d³y = 2 – sin x²d (x²)^2x– 8x sinx² + (-4x) cos x² d(x²)^2x

dx³dx dx

= -4x sinx² – 8x sinx – 8x³ cos x²

= -4x (sin x² + 2 sinx² sinx² + 2x²cosx²)

y = 3x e^-x

dx = 3. e^-x + 3. e^-x

= 3e^-x – 3. e^-x

d²y = 3. -e^-x – 3. e^-x +- e^-x. 3x

dx²

=- 3. e^-x – 3. e^-x – e^-x.3x

d³y = -3. e^-x – 3. e^-x + e^-x.3x + e^-x.3x + – e^-x.3

dx³

= 3. e^-x + 3. e^-x + e^-x3x – 3. e^-x

Y = log_e (1+2x)²

dy = 1 .4(1+2x)

dx (1+2x)²

= 4 + 8x = 4(1+2x) = 4

(1+2x)² (1+2x)(1+2x) 1+2x

d²y = -2(1+2x)^-2

dx²

d³y = -4(1+2x)^-3 – 4

dx² (1+2x)³

ASSIGNMENT: FURTHER MATHS PROJECT, PAGE 198 EXERCISE 11B, NO. 15A, B, C, D

WEEK 9

MECHANICS

Scalar products of two vectors in the Dimensions

The scalar or dot product of two vectors a and b is when as a.b and defined as

a.b = a b Cos

where is the angle between the vectors a and b

If a = a₁ i +a₂j + a₃j

and b = b₁ i + b₂j + b₃k

then a-b = (a₁ i + a₂j + a₃k).(b₁ i +b₂j + b₃k)

= a₁ b₁ i.i + a₁ b₂ i.j + a₁ b₃ i.k + a₂ b₁ j.i + a₂ b₂ j.j + a₂ b₃ j.k)

Note that i.j and k are mutually perpendicular unit vector

.:. i.i = 1

j.j = 1, k.k =1

i.j = 0, j.k = 0, k.i = 1

.:. a.b = a₁ b₁, + a₂ b₂ + a₃ b₃

Angles between two vectors

CoS = a.b

a b

Ex (1). find the scalar product of this pair of vector.

a = 2i + 3j + 4k, b = 5i – 2j + k

Solution

a.b = (2i + 3j + 4k) (5i – 2j + k)

10 – 6 + 4 = 8

(2). Find the cosine of angles between the vectors 2i + 3j – k and 3i – 5j + 2k

Solution

a.b = 6 – 15 – 2 = -11

a = 4 + 9 + 1 = 14

b = 9 + 25 + 4 = 38

Cos = -11 = -11 = – 0.4769

14 x 38 532

= Cos^-1 0.4769

Projectionshu

Projection of the vector a on the vector b

= b.a

where b = b

b hu

Basic properties of dot products

Commutatively

a.b = a.b

Scalar multiplication

(k.a).b = a.(k.b)

where k is a constant.

Let a = a₁i + a+2+j

a.a = a ²

Distributivity

a*(b=c) = a*b + a*c

Given that a – b = 0 (a and b are perpendicular then a b Cos = 0

then a b Cos = 0

Exercises

find the scalar product of 9i – 2j + and i– 3j – 4k
Find the angle between the vectors

3i – 2j + k and 4i + 3j – 2k

Find the projection on the vector a on the vector b if a = 2i + 3j – 7k and b = 4i – 2j + 3k
Find the value of for the vectors 4i + 5 j and 3i – j are perpendicular

Solution

a.b = 9 + 6 – 4 = 11
Cos a-b but a.b = 12-6-2-4

a b a = 9 + 4 + 1 = 14

b = 16 + 9 + 4 29

.:. Cos = 4 =4 = 0.1985 .:. = 78.5⁰

14 x 29 406

b.a where b = b = 4i – 2j + 3k = 4i-2j+3k

b 16+4+9 29

1 4i – 2j +3k.2i+3j-7k

1 8 – 6 – 21

1 (-19) = -19

29 29

The vectors are perpendicular when a.b = 0

(4i+5 j).(3i-j) = 0

12 – 5 = 0

12 = 5

5 5

.:. = 12

ASSIGNMENT: FURTHER MATHS PROJECT, PAGE 280, NO. 10 AND 2C

Further Maths Enotes SS2 – Edudelight.com

WEEK 10

VECTOR OR CROSS PRODUCT

The vector produce (cross product) of a and b is written as axb

if a = a₁ i + a₂ j + a₃k

and b = b₁i + b₂j + b₃k

a x b = i j k

a₁ a₂ a₃

b₁ b₂ b₃

=(a₂ b₃ – a₃ b₂) i – (a₁ b₃ a₃ b₁) j + (a₁ b₂ – a₂ b₁) k

Ex 1. If P = 2i + 4j +3k and q = i + 5j -2k. Find the vector product of p and q.

Solution

p x q = i j k

2 4 3

1 5 -2

= (-8-15)i – (-4-3)j + (10-4)k

-23i + 7j + 6k

2. If a = 3i – j + 2k, b = i + 3j – 2k, Determine the magnitude and direction cosine of the product vector (axb)

Solution

a x b = i j k

3 -1 2 =(2-6)i – (-6-2)j + (9+1)k

1 3 -2 -4i + 8j + 10k

a x b = 16+64+100 = 180

a = 9+1+4 = 13 b = 1 + 9 + 4 = 13

.:. c = a = 3 m = 1 n= – 2

p 13 13 13

l¹ = 1 m¹ = 3 n¹ = -2

13 13 13

Cos = ll¹ + mm¹ + nn¹

3 x 1 + 1 x 3 + 2 -2

13 13 13 13 13 13

3 + 3 -4 = 2 = 0.1538

13 13 13 13

= 81.2⁰

Exercises

Find the vector product of axb when

a. a = I + 2j – k = 2i + 3J + k

b. a = 2i + 3j + 4k, b = 5i – 2j + k

c. a = 2i + 3j – k, b = I – 2j + 3k

i j k 1 2 -1 2 3 1

solution

a x b = = (2 + 3) i – (1 + 2) j + (3 – 4) k

= 5i – 3j – k

i j k 2 3 4 5 -2 1

a x b = = (3 + 8) I – (2 – 20) j + (-4 -15)k

= 11i +18j =19k

i J k 2 3 -1 1 -2 3

a x b = = (9 – 2) I – (6 + i) j +(-4-3) k

=7i – 7j – 7k

ASSIGNMENT

Given A B = and BC = find the angle between the vector AB and AC (WASSCE 2015)
Given that r = 2i – j, s = 3i +5j and t= 6i – 2j, find the magnitude of 2r + s – t (WASSCE 2015 obj)
Given that x = 3i – j, y = 2i + kj
Find the Vector product X x y

If the cosine of angle between x and y is find the value of constant k (WASSCE 2014).

33 minutes read