Math Course: Binomial Series, Rational Functions, Parametric Equations, Exponential Growth, Lecture notes of Trigonometry

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NotNotNotNot

The Not-Formula Book for C

Everything you need to know for Core 4 that won’t be in the formula book

Examination Board: AQA

Brief

This document is intended as an aid for revision. Although it includes some examples

and explanation, it is primarily not for learning content, but for becoming familiar

with the requirements of the course as regards formulae and results. It cannot

replace the use of a text book, and nothing produces competence and familiarity with

mathematical techniques like practice. This document was produced as an addition

to classroom teaching and textbook questions, to provide a summary of key points

and, in particular, any formulae or results you are expected to know and use in this

module.

Contents

Chapter 1 – Binomial series expansion

Chapter 2 – Rational functions and division of polynomials

Chapter 3 – Partial fractions and applications

Chapter 4 – Implicit differentiation and applications

Chapter 5 – Parametric equations for curves and differentiation

Chapter 6 – Further trigonometry with integration

Chapter 7 – Exponential growth and decay

Chapter 8 – Differential equations

Chapter 9 – Vector equations of lines

Chapter 2 – Rational functions and division of polynomials

To simplify rational expressions:

1. Factorise all algebraic expressions.
2. Cancel any factors that are common to the numerator and denominator.

Eg: 𝑥^2 + 𝑥 − 6 3𝑥^2 − 12

To multiply rational expressions:

1. Factorise all algebraic expressions.
2. Write as a single fraction.
3. Cancel any factors that are common to the numerator and the denominator.

To divide rational expressions:

1. Convert the division to a multiplication of the reciprocal.
2. Follow the instructions above.

Eg: 2𝑥 + 3 𝑥^2 − 𝑥

÷

4𝑥^2 − 9

(𝑥 − 1)^2

𝑥^2 − 𝑥

×

(𝑥 − 1)^2

4𝑥^2 − 9

(2𝑥 + 3)(𝑥 − 1)^2

(𝑥^2 − 𝑥)(4𝑥^2 − 9)

(2𝑥 + 3)(𝑥 − 1)^2

To add or subtract rational expressions:

1. Factorise all algebraic expressions.
2. Write each rational expression with the same denominator.
3. Add or subtract to get a single rational expression.
4. Simplify and factorise the numerator.
5. Cancel any factors common to the denominator and the numerator.

Eg: 1 𝑥 + 2

(𝑥 + 2)^2

(𝑥 + 2)^2

𝑷(𝒙) is a polynomial (degree 𝑛). (𝒂𝒙 + 𝒃) is the divisor (degree 1 ). 𝑸(𝒙) is the quotient (degree 𝑛 − 1). 𝑹 is the remainder (degree 0 ).

Eg: 4𝑥^3 − 3𝑥^2 + 3𝑥 + 3 = (2𝑥 + 1)(2𝑥^2 − 2.5𝑥 + 2.75) + 0.

When dividing a polynomial by a linear expression the remainder will be a constant and the quotient will always be one degree lower than the polynomial.

To apply algebraic long division , first write the polynomial and divisor in descending powers of 𝑥, including any missing powers of 𝑥 with a 0 coefficient if necessary (as place-holders).

The factor theorem can be extended for factors of the form (𝑎𝑥 + 𝑏):

More generally, the remainder theorem states that:

Eg:

𝑊ℎ𝑒𝑛 6𝑥^3 − 5𝑥^2 + 2𝑥 + 5 𝑖𝑠 𝑑𝑖𝑣𝑖𝑑𝑒𝑑 𝑏𝑦 (2𝑥 + 1) 𝑡ℎ𝑒 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑖𝑠:

3 − 5 (−

2

• 2 (−

After splitting a rational function into partial fractions, it is often necessary to integrate. The following two results are the most commonly used in this case:

ln|𝑎𝑥 + 𝑏| + 𝐶

(𝑎𝑥 + 𝑏)^2

Eg:

4𝑥^2 + 3𝑥 + 3

(3𝑥 + 2)(𝑥 − 1)^2

(𝑥 − 1)^2

(Note: result obtained by partial fractions – working omitted for the purposes of this example)

Another common application of partial fractions is binomial expansion. The following two common results are easily derived from the binomial formula (in the formula book), but may be useful to memorise in their own right:

(1 + 𝑦)−1^ = 1 − 𝑦 + 𝑦^2 − 𝑦^3 + 𝑦^3 − ⋯ 𝑣𝑎𝑙𝑖𝑑 𝑓𝑜𝑟 |𝑦| < 1

(1 + 𝑦)−2^ = 1 − 2𝑦 + 3𝑦^2 − 4𝑦^3 + ⋯ 𝑣𝑎𝑙𝑖𝑑 𝑓𝑜𝑟 |𝑦| < 1

Eg:

= (1 + 𝑥)−1^ + (1 + 2𝑥)−1^ = 1 − 𝑥 + 𝑥^2 − ⋯ + 1 − (2𝑥) + (2𝑥)^2 − ⋯

Chapter 4 – Implicit differentiation and applications

Most functions we have dealt with are explicitly defined (𝑦 in terms of 𝑥, for instance). Sometimes a function is difficult or impossible to write as 𝑦 = 𝑓(𝑥), and so they are written implicitly.

Eg: 𝑥𝑦 − 𝑦^3 = 𝑥^2 + 3

(𝑦𝑛) = 𝑛𝑦𝑛−^

(ln 𝑦) =

(sin 𝑦) = cos 𝑦

[𝑓(𝑦)] = 𝑓′(𝑦)

Note: Frequently techniques such as chain rule or product rule are required in conjunction with these rules.

Eg: 𝑥𝑦 − 𝑦^3 = 𝑥^2 + 3

+ 𝑦 − 3𝑦^2

If necessary, this can be rearranged to give 𝑑𝑦 𝑑𝑥:

𝒅𝒚 𝒅𝒙

Chapter 6 – Further trigonometry with integration

Compound angle identities can be used to manipulate trigonometric functions with different input values.

The six identities can be summarised as shown:

sin(𝐴 ± 𝐵) = sin 𝐴 cos 𝐵 ± cos 𝐴 sin 𝐵

cos(𝐴 ± 𝐵) = cos 𝐴 cos 𝐵 ∓ sin 𝐴 sin 𝐵

tan(𝐴 ± 𝐵) =

tan 𝐴 ± tan 𝐵 1 ∓ tan 𝐴 tan 𝐵

, [𝐴 ± 𝐵 ≠ (𝑘 +

) 𝜋]

Note: These results are in the formula book , so you are not required to learn them. They are included here for the sake of completeness.

You will need to know (or be able to derive from the formulae above) the double angle formulae:

sin 2𝐴 = 2 sin 𝐴 cos 𝐴

cos 2𝐴 = cos^2 𝐴 − sin^2 𝐴 = 1 − 2 sin^2 𝐴 = 2 cos^2 𝐴 − 1

tan 2𝐴 =

2 tan 𝐴 1 − tan^2 𝐴

Note: These can readily be derived from the 𝐴 ± 𝐵 formulae.

Eg:

sin(𝐴 ± 𝐵) = sin 𝐴 cos 𝐵 ± cos 𝐴 sin 𝐵

𝐿𝑒𝑡 𝐴 = 𝐵 ⟹ sin(𝐴 + 𝐴) = sin 𝐴 cos 𝐴 + cos 𝐴 sin 𝐴

⟹ sin 2𝐴 = 2 sin 𝐴 cos 𝐴

To integrate either sin^2 𝑥 or cos^2 𝑥 write in terms of cos 2𝑥.

Note: The second and third versions of the cos 2𝐴 formula given above are derived from the first by using the identity sin^2 𝐴 + cos^2 𝐴 = 1.

Eg:

∫ cos^2 𝑥 𝑑𝑥 = ∫

cos 2𝑥 + 1 2

Functions of the form 𝑎 sin 𝜃 + 𝑏 cos 𝜃 can be written in the form 𝑅 sin(𝜃 ± 𝛼)^ or 𝑅 cos(𝜃 ± 𝛼).

𝑅 = √𝑎^2 + 𝑏^2

Note: There are formulae for calculating 𝛼 directly from 𝑎 and 𝑏, but since a change of sign on either the sin 𝜃 part, cos 𝜃 part or both will cause equivalent changes in the formulae, they are not included here. The method described below allows for any variation of form, converting into any of the four varieties of simplified function.

Step 1: Calculate 𝑅 using 𝑅^2 = 𝑎^2 + 𝑏^2. Step 2: Choose a function to convert into: (𝑅 sin(𝜃 + 𝛼), 𝑅 sin(𝜃 − 𝛼), 𝑅 cos(𝜃 + 𝛼) or 𝑅 cos(𝜃 − 𝛼). Step 3: Expand this using the appropriate sin(𝐴 ± 𝐵) or cos(𝐴 ± 𝐵) formulae (in formula book). Step 4: Compare coefficients on the left and right sides of the identity and produce equations for 𝛼. Step 5: Solve for 𝛼, using non-primary solutions if necessary to ensure the value fits both equations.

Eg:

Write 3 sin 𝜃 − 2 cos 𝜃 in the form 𝑅 sin(𝜃 − 𝛼)

𝑅 = √3^2 + (−2)^2 = √

sin(𝐴 − 𝐵) = sin 𝐴 cos 𝐵 − cos 𝐴 sin 𝐵 ⟹ sin(𝜃 − 𝛼) = sin 𝜃 cos 𝛼 − cos 𝜃 sin 𝛼

⟹ √13 sin(𝜃 − 𝛼) = √13(sin 𝜃 cos 𝛼 − cos 𝜃 sin 𝛼) = 3 sin 𝜃 − 2 cos 𝜃

⟹ √13 cos 𝛼 sin 𝜃 − √13 sin 𝛼 cos 𝜃 = 3 sin 𝜃 − 2 cos 𝜃

⟹ cos 𝛼 =

𝑎𝑛𝑑 sin 𝛼 =

⟹ 3 sin 𝜃 − 2 cos 𝜃 = √𝟏𝟑 𝐬𝐢𝐧(𝜽 − 𝟑𝟑. 𝟕°)

Note: If the primary solutions to the two equations for 𝛼 do not match, you will need to look at secondary solutions. Sketch the graph and look further afield. There will always be a value for 𝛼 that fits both equations, although it may be negative or obtuse.

To solve an equation of the form 𝑎 sin 𝑥 + 𝑏 cos 𝑥 = 𝑐, first write 𝑎 sin 𝑥 + 𝑏 cos 𝑥 in the form 𝑅 cos(𝑥 ± 𝛼) or 𝑅 sin(𝑥 ± 𝛼), then rearrange and solve for 𝑥.

Chapter 8 – Differential equations

A differential equation is an equation which involves at least one derivative of a variable with respect to another variable.

Eg: 𝑑𝑦 𝑑𝑥

= 𝑒𝑡^ sin 𝑥 𝑜𝑟

𝑑^2 𝑚

𝑑𝑡^2

A first order differential equation is one in which the highest order of derivative is the first.

Eg: 𝑑𝑦 𝑑𝑥

= 2𝑥^4 + 3 𝑜𝑟 𝑥^2

= ln 𝑡

Note: We will be dealing only with first order differential equations in this module.

Certain common statements involving derivatives need to be interpreted in the form of a differential equation:

The rate of increase of 𝑥 is proportional to 𝑥:

𝑑𝑥 𝑑𝑡 = 𝑘𝑥^ 𝑘 > 0

The rate of decrease of 𝑥 is proportional to 𝑥: 𝑑𝑥 𝑑𝑡 = −𝑘𝑥^ 𝑘 > 0

Note: These statements can be given in a variety of ways, so you will need to be able to recognise a number of different types.

Eg: The volume of a snowball decreases at a rate proportional to its volume.

𝑑𝑉 𝑑𝑡

The general solution of the first order differential equation 𝑑𝑦 𝑑𝑥 = 𝑔(𝑥)ℎ(𝑦) (^) is given by:

∫

1 ℎ(𝑦) 𝑑𝑦 = ∫ 𝑔(𝑥)𝑑𝑥 + 𝐶^ where^ 𝐶^ is an arbitrary constant.

Note: This method is known as the ‘Separation of Variables’ method, since it involves using division or multiplication to rearrange the differential equation so as to have all 𝑥 terms on the side of the 𝑑𝑥 and all the 𝑦 terms on the side with the 𝑑𝑦. To solve then it is necessary to be able to integrate each side (not necessarily straightforward; recall substitution, inspection and integration by parts).

A particular solution is obtainable from the general solution by substituting in the values of a specific condition (eg 𝑉 = 4 at 𝑡 = 0), and solving to find the value of the arbitrary constant.

Chapter 9 – Vector equations of lines

A vector is a quantity with a direction. is represented by bold type such as 𝒗, with a line

underneath, 𝑣 (easier for hand-written work), or, for a vector between two points, 𝐴𝐵⃗⃗⃗⃗⃗.

The magnitude (or size) of the vector 𝐴𝐵⃗⃗⃗⃗⃗ is written |𝐴𝐵⃗⃗⃗⃗⃗ | or occasionally just 𝐴𝐵. The magnitude of 𝒗 is written as |𝒗| or occasionally 𝑣.

In two dimensions, the vector 𝒗 = [

𝑏]^ has component^ 𝑎^ acting in the^ 𝑥^ direction and component^ 𝑏 acting in the 𝑦 direction. For 3 dimensions, the third component acts in the 𝑧 direction.

The vector between points 𝐴 and 𝐵 is denoted 𝐴𝐵⃗⃗⃗⃗⃗ and can be calculated from the position vectors

of 𝐴 and 𝐵 (usually written as 𝑂𝐴⃗⃗⃗⃗⃗ and 𝑂𝐵⃗⃗⃗⃗⃗ respectively) as follows:

Eg:

𝑃𝑜𝑖𝑛𝑡 𝐴: (2,3, −1)^ 𝑃𝑜𝑖𝑛𝑡 𝐵: (−5,0,2)^ ⟹ 𝑂𝐴⃗⃗⃗⃗⃗ = [

] 𝑂𝐵⃗⃗⃗⃗⃗ = [

]

𝐴𝐵⃗⃗⃗⃗⃗ = 𝑂𝐵⃗⃗⃗⃗⃗ − 𝑂𝐴⃗⃗⃗⃗⃗ = [

] − [

] = [

]

The magnitude of the 2D vector [

] and the 3D vector [

] are given by:

|[

]| = √𝑎^2 + 𝑏^2 |[

]| = √𝑎^2 + 𝑏^2 + 𝑐^2

Note: This is simply an application of Pythagoras’ theorem, and gives the length of the vector.

Eg:

|[

]| = √2^2 + (−2)^2 + 1^2 = √9 = 𝟑

The results above can be combined to calculate the distance between two points (although this is easy enough to do in two separate steps): The distance between (𝑥 1 , 𝑦 1 , 𝑧 1 )^ and (𝑥 2 , 𝑦 2 , 𝑧 2 )^ is:

|[

]| = √(𝑥 2 − 𝑥 1 )^2 + (𝑦 2 − 𝑦 1 )^2 + (𝑧 2 − 𝑧 1 )^2

Provided 𝒂 and 𝒃 are non-zero vectors (zero vectors don’t have a specific direction):

𝒂 ∙ 𝒃 = 0 ⟺ 𝒂 𝑎𝑛𝑑 𝒃 𝑎𝑟𝑒 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟

Note: This is because perpendicular vectors are at 90°, giving cos 𝜃 = 0.

The vector equation of a line is given in the form 𝒓 = 𝒂 + 𝜆𝒃 where 𝒂 is the position vector (any point on the line) and 𝒃 is the direction vector (any vector with the same direction as the line).

An example might be:

𝒓 = [

] + 𝜆 [

]

This goes through the point (3,4,1) and points in the direction of [

]

Note: Since any point on the line will suffice as a position vector and any scalar multiple of the direction vector will also point in the same direction, there is no limit to the number of different equations that could be produced for the same line.

To find the point of intersection of two lines (if it exists), generate a general point for each, and set them equal to each other. The solutions (if consistent) of the resulting three simultaneous equations for 𝜆 or 𝜇 will determine the point.

Eg:

𝐿 1 : 𝑟 = [

] + 𝜆 [

] 𝐿 2 : 𝑟 = [

] + 𝜇 [

]

𝐿 1 𝐺𝑒𝑛𝑒𝑟𝑎𝑙 𝑃𝑜𝑖𝑛𝑡: [

] 𝐿 2 𝐺𝑒𝑛𝑒𝑟𝑎𝑙 𝑃𝑜𝑖𝑛𝑡: [

]

[

] = [

] ⟹

𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝜆 = 1 𝑖𝑛𝑡𝑜 𝑡ℎ𝑒 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑝𝑜𝑖𝑛𝑡 𝑓𝑜𝑟 𝐿 1 : [

]

The angle between two lines is defined to be the angle between their direction vectors.

Eg:

For the lines given above, the angle between would be found by applying the dot product formula to

the vectors [

] and [

].

If two lines are parallel, their direction vectors will be scalar multiples of one another. That is, one can be produced by multiplying each element of the other by a particular number.

If two lines are neither parallel nor do they intersect , they are said to be skew.

Note: It is, of course, not possible for a line in two dimensions to be skew. To prove skewness, first show that the lines are not parallel (introduce a scalar variable to multiply by the direction vector and demonstrate that the resulting equations show a contradiction), then show that the lines do not intersect (by constructing simultaneous equations from the general points, as shown previously, and demonstrating that they produce a contradiction).

Eg:

𝑇𝑤𝑜 𝑙𝑖𝑛𝑒𝑠 ℎ𝑎𝑣𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 [

] 𝑎𝑛𝑑 [

]. 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒𝑦 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙.

𝐼𝑓 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙: [

] = 𝑎 [

] 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑠𝑐𝑎𝑙𝑎𝑟 𝑎.

If it has already been demonstrated that these two lines do not intersect, these two results are sufficient to conclude the lines must be skew.