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Various topics in mathematics including binomial series expansion, rational functions, partial fractions, implicit differentiation, parametric equations, curves, exponential growth and decay, differential equations, and vector equations of lines. It includes formulas, examples, and problem-solving strategies.
What you will learn
Typology: Lecture notes
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Examination Board: AQA
Chapter 1 โ Binomial series expansion
Chapter 2 โ Rational functions and division of polynomials
Chapter 3 โ Partial fractions and applications
Chapter 4 โ Implicit differentiation and applications
Chapter 5 โ Parametric equations for curves and differentiation
Chapter 6 โ Further trigonometry with integration
Chapter 7 โ Exponential growth and decay
Chapter 8 โ Differential equations
Chapter 9 โ Vector equations of lines
Chapter 2 โ Rational functions and division of polynomials
To simplify rational expressions:
Eg: ๐ฅ^2 + ๐ฅ โ 6 3๐ฅ^2 โ 12
To multiply rational expressions:
To divide rational expressions:
Eg: 2๐ฅ + 3 ๐ฅ^2 โ ๐ฅ
To add or subtract rational expressions:
Eg: 1 ๐ฅ + 2
๐ท(๐) is a polynomial (degree ๐). (๐๐ + ๐) is the divisor (degree 1 ). ๐ธ(๐) is the quotient (degree ๐ โ 1). ๐น is the remainder (degree 0 ).
Eg: 4๐ฅ^3 โ 3๐ฅ^2 + 3๐ฅ + 3 = (2๐ฅ + 1)(2๐ฅ^2 โ 2.5๐ฅ + 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 ๐ฅ, including any missing powers of ๐ฅ with a 0 coefficient if necessary (as place-holders).
The factor theorem can be extended for factors of the form (๐๐ฅ + ๐):
More generally, the remainder theorem states that:
Eg:
๐โ๐๐ 6๐ฅ^3 โ 5๐ฅ^2 + 2๐ฅ + 5 ๐๐ ๐๐๐ฃ๐๐๐๐ ๐๐ฆ (2๐ฅ + 1) ๐กโ๐ ๐๐๐๐๐๐๐๐๐ ๐๐ :
3 โ 5 (โ
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|๐๐ฅ + ๐| + ๐ถ
Eg:
(Note: result obtained by partial fractions โ 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 + ๐ฆ)โ1^ = 1 โ ๐ฆ + ๐ฆ^2 โ ๐ฆ^3 + ๐ฆ^3 โ โฏ ๐ฃ๐๐๐๐ ๐๐๐ |๐ฆ| < 1
(1 + ๐ฆ)โ2^ = 1 โ 2๐ฆ + 3๐ฆ^2 โ 4๐ฆ^3 + โฏ ๐ฃ๐๐๐๐ ๐๐๐ |๐ฆ| < 1
Eg:
Chapter 4 โ Implicit differentiation and applications
Most functions we have dealt with are explicitly defined (๐ฆ in terms of ๐ฅ, for instance). Sometimes a function is difficult or impossible to write as ๐ฆ = ๐(๐ฅ), and so they are written implicitly.
Eg: ๐ฅ๐ฆ โ ๐ฆ^3 = ๐ฅ^2 + 3
(ln ๐ฆ) =
(sin ๐ฆ) = cos ๐ฆ
Note: Frequently techniques such as chain rule or product rule are required in conjunction with these rules.
Eg: ๐ฅ๐ฆ โ ๐ฆ^3 = ๐ฅ^2 + 3
If necessary, this can be rearranged to give ๐๐ฆ ๐๐ฅ:
๐ ๐ ๐ ๐
Chapter 6 โ 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(๐ด ยฑ ๐ต) = sin ๐ด cos ๐ต ยฑ cos ๐ด sin ๐ต
cos(๐ด ยฑ ๐ต) = cos ๐ด cos ๐ต โ sin ๐ด sin ๐ต
tan(๐ด ยฑ ๐ต) =
tan ๐ด ยฑ tan ๐ต 1 โ tan ๐ด tan ๐ต
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 2๐ด = 2 sin ๐ด cos ๐ด
cos 2๐ด = cos^2 ๐ด โ sin^2 ๐ด = 1 โ 2 sin^2 ๐ด = 2 cos^2 ๐ด โ 1
tan 2๐ด =
2 tan ๐ด 1 โ tan^2 ๐ด
Note: These can readily be derived from the ๐ด ยฑ ๐ต formulae.
Eg:
sin(๐ด ยฑ ๐ต) = sin ๐ด cos ๐ต ยฑ cos ๐ด sin ๐ต
๐ฟ๐๐ก ๐ด = ๐ต โน sin(๐ด + ๐ด) = sin ๐ด cos ๐ด + cos ๐ด sin ๐ด
โน sin 2๐ด = 2 sin ๐ด cos ๐ด
To integrate either sin^2 ๐ฅ or cos^2 ๐ฅ write in terms of cos 2๐ฅ.
Note: The second and third versions of the cos 2๐ด formula given above are derived from the first by using the identity sin^2 ๐ด + cos^2 ๐ด = 1.
Eg:
โซ cos^2 ๐ฅ ๐๐ฅ = โซ
cos 2๐ฅ + 1 2
Functions of the form ๐ sin ๐ + ๐ cos ๐ can be written in the form ๐ sin(๐ ยฑ ๐ผ)^ or ๐ cos(๐ ยฑ ๐ผ).
Note: There are formulae for calculating ๐ผ directly from ๐ and ๐, but since a change of sign on either the sin ๐ part, cos ๐ 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 ๐ using ๐ ^2 = ๐^2 + ๐^2. Step 2: Choose a function to convert into: (๐ sin(๐ + ๐ผ), ๐ sin(๐ โ ๐ผ), ๐ cos(๐ + ๐ผ) or ๐ cos(๐ โ ๐ผ). Step 3: Expand this using the appropriate sin(๐ด ยฑ ๐ต) or cos(๐ด ยฑ ๐ต) formulae (in formula book). Step 4: Compare coefficients on the left and right sides of the identity and produce equations for ๐ผ. Step 5: Solve for ๐ผ, using non-primary solutions if necessary to ensure the value fits both equations.
Eg:
Write 3 sin ๐ โ 2 cos ๐ in the form ๐ sin(๐ โ ๐ผ)
sin(๐ด โ ๐ต) = sin ๐ด cos ๐ต โ cos ๐ด sin ๐ต โน sin(๐ โ ๐ผ) = sin ๐ cos ๐ผ โ cos ๐ sin ๐ผ
โน โ13 sin(๐ โ ๐ผ) = โ13(sin ๐ cos ๐ผ โ cos ๐ sin ๐ผ) = 3 sin ๐ โ 2 cos ๐
โน โ13 cos ๐ผ sin ๐ โ โ13 sin ๐ผ cos ๐ = 3 sin ๐ โ 2 cos ๐
โน cos ๐ผ =
๐๐๐ sin ๐ผ =
โน 3 sin ๐ โ 2 cos ๐ = โ๐๐ ๐ฌ๐ข๐ง(๐ฝ โ ๐๐. ๐ยฐ)
Note: If the primary solutions to the two equations for ๐ผ 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 ๐ผ that fits both equations, although it may be negative or obtuse.
To solve an equation of the form ๐ sin ๐ฅ + ๐ cos ๐ฅ = ๐, first write ๐ sin ๐ฅ + ๐ cos ๐ฅ in the form ๐ cos(๐ฅ ยฑ ๐ผ) or ๐ sin(๐ฅ ยฑ ๐ผ), then rearrange and solve for ๐ฅ.
Chapter 8 โ Differential equations
A differential equation is an equation which involves at least one derivative of a variable with respect to another variable.
Eg: ๐๐ฆ ๐๐ฅ
= ๐๐ก^ sin ๐ฅ ๐๐
A first order differential equation is one in which the highest order of derivative is the first.
Eg: ๐๐ฆ ๐๐ฅ
= ln ๐ก
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 ๐ฅ is proportional to ๐ฅ:
๐๐ฅ ๐๐ก = ๐๐ฅ^ ๐ > 0
The rate of decrease of ๐ฅ is proportional to ๐ฅ: ๐๐ฅ ๐๐ก = โ๐๐ฅ^ ๐ > 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.
๐๐ ๐๐ก
The general solution of the first order differential equation ๐๐ฆ ๐๐ฅ = ๐(๐ฅ)โ(๐ฆ) (^) is given by:
โซ
1 โ(๐ฆ) ๐๐ฆ = โซ ๐(๐ฅ)๐๐ฅ + ๐ถ^ where^ ๐ถ^ is an arbitrary constant.
Note: This method is known as the โSeparation of Variablesโ method, since it involves using division or multiplication to rearrange the differential equation so as to have all ๐ฅ terms on the side of the ๐๐ฅ and all the ๐ฆ terms on the side with the ๐๐ฆ. 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 ๐ = 4 at ๐ก = 0), and solving to find the value of the arbitrary constant.
Chapter 9 โ Vector equations of lines
A vector is a quantity with a direction. is represented by bold type such as ๐, with a line
underneath, ๐ฃ (easier for hand-written work), or, for a vector between two points, ๐ด๐ตโโโโโ.
The magnitude (or size) of the vector ๐ด๐ตโโโโโ is written |๐ด๐ตโโโโโ | or occasionally just ๐ด๐ต. The magnitude of ๐ is written as |๐| or occasionally ๐ฃ.
In two dimensions, the vector ๐ = [
๐]^ has component^ ๐^ acting in the^ ๐ฅ^ direction and component^ ๐ acting in the ๐ฆ direction. For 3 dimensions, the third component acts in the ๐ง direction.
The vector between points ๐ด and ๐ต is denoted ๐ด๐ตโโโโโ and can be calculated from the position vectors
of ๐ด and ๐ต (usually written as ๐๐ดโโโโโ and ๐๐ตโโโโโ respectively) as follows:
Eg:
๐๐๐๐๐ก ๐ด: (2,3, โ1)^ ๐๐๐๐๐ก ๐ต: (โ5,0,2)^ โน ๐๐ดโโโโโ = [
The magnitude of the 2D vector [
] and the 3D vector [
] are given by:
Note: This is simply an application of Pythagorasโ theorem, and gives the length of the vector.
Eg:
|[
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 (๐ฅ 1 , ๐ฆ 1 , ๐ง 1 )^ and (๐ฅ 2 , ๐ฆ 2 , ๐ง 2 )^ is:
|[
Provided ๐ and ๐ are non-zero vectors (zero vectors donโt have a specific direction):
๐ โ ๐ = 0 โบ ๐ ๐๐๐ ๐ ๐๐๐ ๐๐๐๐๐๐๐๐๐๐ข๐๐๐
Note: This is because perpendicular vectors are at 90ยฐ, giving cos ๐ = 0.
The vector equation of a line is given in the form ๐ = ๐ + ๐๐ where ๐ is the position vector (any point on the line) and ๐ is the direction vector (any vector with the same direction as the line).
An example might be:
๐ = [
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 ๐ or ๐ will determine the point.
Eg:
๐ฟ 1 : ๐ = [
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:
๐๐ค๐ ๐๐๐๐๐ โ๐๐ฃ๐ ๐๐๐๐๐๐ก๐๐๐ ๐ฃ๐๐๐ก๐๐๐ [
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.