Mathematics Methods SCSA Revision 2020 PDF

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Year 12 ATAR Methods Semester 2 revision

Year 12 ATAR METHODS – Semester 2 Revision

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Year 12 ATAR METHODS – Semester 2 Revision

1. Differentiation Topic 3.1: Further differentiation and applications Exponential functions 3.1.1 estimate the limit of

฀฀ ℎ−1 ℎ

as ℎ → 0, using technology, for various values of ฀฀ > 0

3.1.2 identify that ฀฀ is the unique number ฀฀ for which the above limit is 1

3.1.3 establish and use the formula

฀฀ (฀฀ ฀ ฀ ) ฀฀฀฀

= ฀฀ ฀฀

3.1.4 use exponential functions of the form ฀฀฀฀฀฀฀฀ and their derivatives to solve practical problems

Trigonometric functions

3.1.5 establish the formulas

฀฀ (sin ฀฀) ฀฀฀฀

= cos ฀฀ and฀฀฀฀ (cos ฀฀) = − sin ฀฀ by graphical treatment, numerical ฀฀

estimations of the limits, and informal proofs based on geometric constructions 3.1.6 use trigonometric functions and their derivatives to solve practical problems Differentiation rules 3.1.7 examine and use the product and quotient rules

3.1.8 examine the notion of composition of functions and use the chain rule for determining the derivatives of composite functions

3.1.9 apply the product, quotient and chain rule to differentiate functions such as ฀฀฀฀ ฀฀ , tan ฀฀, ฀฀ ฀฀ , ฀ ฀ sin ฀฀, 1

฀฀ − ฀฀ sin ฀฀ and ฀฀(฀฀฀฀ − ฀฀)

The second derivative and applications of differentiation 3.1.10 use the increments formula: ฀฀฀฀ ≈

฀฀฀฀

฀฀฀฀

× ฀฀฀฀ to estimate the change in the dependent variable ฀฀

resulting from changes in the independent variable ฀฀

3.1.11 apply the concept of the second derivative as the rate of change of the first derivative function 3.1.12 identify acceleration as the second derivative of position with respect to time 3.1.13 examine the concepts of concavity and points of inflection and their relationship with the second derivative 3.1.14 apply the second derivative test for determining local maxima and minima 3.1.15 sketch the graph of a function using first and second derivatives to locate stationary points and points of inflection 3.1.16 solve optimisation problems from a wide variety of fields using first and second derivatives

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Year 12 ATAR METHODS – Semester 2 Revision

Exponential functions Euler’s number

฀ ฀ = 2.71828 (correct to 5 d.p) 1 ฀฀ � ฀฀

Consider the expression lim �1 +

฀฀→∞

฀฀

where ฀ ฀ > 0

1 ฀ ฀

∴ lim �1 + ฀฀�

� where ฀ ฀ > 0

฀฀ ฀฀ −1 ฀ ฀ ฀฀→0

lim �

฀฀

1 ฀฀ �1 + � ฀฀ 2 2.25 2.48832 2.59374 2.70481 2.71692 2.71827 2.71828

1 2 5 10 100 1000 10000 1000000 ฀฀→∞

Consider the expression ฀฀฀ ฀ − 1 � ฀฀ 0.69315 0.91629 0.99325 0.99695 1.00063 0.99990 1.00001 1.00000

�

2 2.5 2.7 2.71 2.72 2.718 2.7183 2.71828

= ฀฀

฀฀ ฀฀ −1

∴ lim � ฀฀→0

฀

฀

� = 1 when ฀ ฀ = ฀฀

Differentiating exponential functions ฀฀฀฀

For ฀ ฀ = ฀฀ ฀ ฀ , = lim� ฀฀฀฀ ℎ→0

ℎ

= ฀฀ ฀ ฀ lim� ℎ→0

฀฀฀฀ ฀฀฀฀

�

�

฀฀ ฀฀ �฀฀ ℎ−1�

= lim � ℎ→0

ℎ

฀฀ (฀฀+ℎ) −฀฀ ฀฀

= lim � ℎ→0

฀฀(฀฀+ℎ)−฀฀(฀฀)

ℎ

฀฀ ℎ−1

= ฀฀ ฀ ฀ × 1

ℎ

�

�

= ฀฀ ฀฀

I.e. ฀฀ ฀฀ differentiates to itself. ฀฀(฀฀) = ฀฀ ฀฀

As shown on page 1 ฀฀ ฀฀ −1

lim � ฀฀→0

฀ ฀ ฀฀ ℎ−1

∴ lim � ℎ→0

฀

฀

� = 1 when ฀ ฀ = ฀฀ �=1

฀฀′(฀฀) = ฀฀ ฀฀

More on growth and decay ฀฀฀฀

If ฀ ฀ = ฀฀0 ฀฀ ฀฀฀฀ then = ฀ ฀ × ฀฀0 ฀฀ ฀฀฀฀ ฀฀฀฀ ∴ if

฀฀฀฀

฀฀฀฀

= ฀ ฀ × ฀฀

= ฀ ฀ × ฀฀

Then ฀ ฀ = ฀฀0 ฀฀ ฀฀฀฀

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Year 12 ATAR METHODS – Semester 2 Revision

Trigonometric functions Establishing the formulas Geometrically

฀฀

฀฀฀฀

(sin ฀฀) = cos ฀฀ and (cos ฀฀) = − sin ฀฀ ฀฀฀฀ ฀฀

The diagram on the right shows a unit circle with ฀฀฀฀ tangent to the circle ฀฀ and ฀฀ measured in radians with 0 < ฀ ฀ < . 2

It follows that tan ฀฀

฀฀฀฀

∴ ฀฀฀฀

฀฀฀฀

= tan ฀฀

= ฀฀฀฀

= tan ฀ ฀ × 1 1

Area ∆฀฀฀฀฀฀ = × ฀฀฀฀ × ฀฀฀฀ × sin ฀฀ 2 = 2 × 1 × 1 × sin ฀฀ 1

=

sin ฀฀ 2

1

Area sector ฀฀฀฀฀฀ = × ฀฀ 2 × ฀฀ 2 1

= × 12 × ฀฀ 2

Area ∆฀฀฀฀฀฀

=

฀฀ 2

1 = 2 × ฀฀฀฀ 1

× ฀฀฀฀

= × tan ฀ ฀ × 1 2

=

tan ฀฀ 2

Comparing these three areas, we can see that Area ∆฀฀฀฀฀฀

<

sin ฀฀ 2

<

For:

sin ฀฀ 2

sin ฀฀

sin ฀฀ ฀ ฀

Area sector ฀฀฀฀฀฀

<

฀฀ 2

<

<

฀฀ 2

< <

฀฀

1

Area ∆฀฀฀฀฀฀ tan ฀฀ 2

Multiply both sides by 2 Divide both sides by ฀฀

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Year 12 ATAR METHODS – Semester 2 Revision ฀฀

For:

2

฀฀

2 ฀฀ tan

฀ ฀ cos ฀฀

<

cos ฀฀

Putting these together:

<

We know that as ฀฀ → 0, cos ฀฀ → 1

We can see that

sin ฀฀

Multiply both sides by cos ฀฀

sin ฀฀

<

cos ฀฀

Multiply both sides by 2 and replace tan ฀฀

sin ฀฀ cos ฀฀

< <

Divide both sides by ฀฀

sin ฀฀ ฀ ฀

sin ฀฀ ฀ ฀

<

1 sin ฀฀ ฀

is “sandwiched” between cos ฀฀ and 1, then as ฀฀ → 0, ฀

฀ ฀

tends towards 1.

Using limits

From first principal’s if ฀฀(฀฀ ) = sin ฀฀ ฀฀(฀฀+ℎ)−฀฀(฀฀) ℎ

฀฀ ′ (฀฀) = lim ℎ→0

ℎ→0

= lim ℎ→0

= lim ℎ→0

= lim

sin(฀฀+ℎ)−sin ฀฀

Use the identity sin(฀ ฀ ± ฀฀) = sin ฀ ฀ cos ฀ ฀ ±

ℎ

sin ฀ ฀ cos ฀฀

sin ฀ ฀ cos ℎ+sin ℎ cos ฀฀−sin ฀฀ sin ℎ cos ฀฀ ℎ

ℎ

− lim

sin ℎ ℎ→0 ℎ

= cos ฀ ฀ lim

ℎ→0

sin ฀฀−sin ฀ ฀ cos ℎ

Split into two limits

ℎ

− sin ฀ ฀ lim

1−cos ℎ ℎ ℎ→0 sin ℎ ℎ→0 ℎ

Now we need to find out what lim

Factorise out cos ฀฀ and sin ฀฀ 1−cos ℎ ℎ ℎ→0

and lim

ℎ

are.

Using your calculator in radians getting ℎ as close to zero as you can from both the negative and the positive sides. The table suggests that: ℎ→0

lim

sin ℎ ℎ

ℎ→0

= 1 and lim

Returning to ฀฀’(฀฀)

∴

฀฀

฀฀฀฀

(sin ฀฀)

1−cos ℎ ℎ

=0 sin ℎ

= cos ฀ ฀ lim ℎ→0

ℎ

1−cos ℎ ℎ

− sin฀ ฀ lim ℎ→0

= cos ฀ ฀ × 1 − sin ฀ ฀ × 0

-0.02 -0.01 -0.001 -0.0001 0 0.0001 0.001 0.01 0.02

Substitute in

sin ℎ ℎ

sin ℎ

1−cos ℎ

0.9999 0.99998 0.999998 0.999999998 undefined 0.999999998 0.999998 0.99998 0.9999

-0.0100 -0.0050 -0.0005 -0.00005 undefined 0.00005 0.0005 0.0050 0.0100

ℎ

= 1 and

1−cos ℎ ℎ

ℎ

=0

= cos ฀฀

= cos ฀฀

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Year 12 ATAR METHODS – Semester 2 Revision

From first principals if ฀฀(฀฀) = cos ฀฀

ℎ ฀฀ ′ (฀฀) = lim฀฀(฀฀+ℎ)−฀฀(฀฀) ℎ→0

฀฀ = lim cos(฀฀+ℎ)−cos ℎ ℎ→0

ℎ→0

= lim

cos ฀ ฀ cos ℎ−sin ฀ ฀ sin ℎ−cos ฀฀

= − lim

ℎ→0

ℎ

cos ฀฀−cos ฀ ฀ cos ℎ sin ฀ ฀ sin ℎ − lim ℎ ℎ ℎ→0 sin ℎ

= − sin ฀ ฀ lim

∴

฀฀

฀฀฀฀

Use the identity cos(฀ ฀ ± ฀฀) = cos ฀ ฀ cos ฀฀ ∓ sin ฀ ฀ sin ฀฀

ℎ→0

ℎ

1−cos ℎ ℎ

− cos ฀ ฀ lim ℎ→0

= − sin ฀ ฀ × 1 − cos ฀ ฀ × 0

Split into two limits and factorise out -1 Factorise out sin ฀฀ and cos ฀฀ Substitute in

sin ℎ ℎ

= 1 and

1−cos ℎ ℎ

=0

(cos ฀฀) = − sin ฀฀

Graphically

Angle ฀฀ (radians)

sin ฀฀

cos ฀฀

0

0

1

฀฀ 6

1 2

√3 2

฀฀ 4

฀฀ 3

฀฀

1

√2

= √3 2 1

√2 2

1

√2

=

tan ฀฀ 0 1

√2 2

√3

=

√3 3

1

1 2

√3

0

undefined

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Year 12 ATAR METHODS – Semester 2 Revision

Drawing ฀ ฀ = sin ฀฀ and then sketching the tangent at key points helps us determine the function of the gradient.

By sketching the points of the tangents above, as well as others against ฀ ฀ = sin ฀฀ we can see that the ฀฀฀฀ ฀฀฀฀

= cos ฀฀.

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Year 12 ATAR METHODS – Semester 2 Revision

Drawing gradient. ฀ ฀ = cos ฀฀ and then sketching the tangent at key points helps us determine the function of the

By sketching the points of the tangents above, as well as others against ฀ ฀ = cos ฀฀ we can see that the ฀฀฀฀ = −sin ฀฀. ฀฀฀฀

฀฀(฀฀ ) = sin ฀฀

฀฀(฀฀) = cos ฀฀

฀฀′(฀฀) = cos ฀฀

฀฀ ′ (฀฀) = − sin ฀฀

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Year 12 ATAR METHODS – Semester 2 Revision

Differentiation rules Product rule ฀ ฀ = ฀฀(฀฀) × ฀฀(฀฀) Quotient rule

฀฀฀฀ = ฀฀′(฀฀) × ฀฀(฀฀) + ฀฀(฀฀) × ฀฀′(฀฀) ฀฀฀฀

฀฀(฀฀) ฀฀= ฀฀(฀฀)

฀฀฀฀ ฀฀′(฀฀) × ฀฀(฀฀) − ฀฀(฀฀) × ฀฀′(฀฀) = [฀฀( ฀฀)]2 ฀฀฀฀

฀ ฀ = [฀฀(฀฀)]฀฀

฀฀฀฀ = ฀฀[฀฀(฀฀)]฀฀−1 × ฀฀′(฀฀) ฀฀฀฀

฀ ฀ = ฀฀ ฀฀฀฀+฀฀

฀฀′ = ฀฀฀฀ ฀฀฀฀+฀฀

฀ ฀ = ฀฀(฀฀), ฀ ฀ = ฀฀(฀฀) ฀ ฀ = sin(฀฀฀฀ + ฀฀)

฀ ฀ = cos(฀฀฀฀ + ฀฀)

Chain rule

฀฀฀฀ ฀฀฀฀ ฀฀฀฀ = × ฀฀฀฀ ฀฀฀฀ ฀฀฀฀

฀฀′ = ฀ ฀ cos(฀฀฀฀ + ฀฀)

฀฀′ = − ฀฀sin(฀฀฀฀ + ฀฀)

The second derivative and applications of differentiation Incremental change

Using the notation ฀฀฀฀ as the small change in ฀฀ and ฀฀฀฀ as the small change in ฀฀, then

฀฀฀฀ ฀฀฀฀

฀฀฀฀

= lim ฀฀฀฀ ฀฀฀฀→0

If ฀฀฀฀ is small then

฀฀฀฀

฀฀฀฀

฀฀฀฀

≈ ฀฀฀฀

If finding a percentage change, divide both sides by ฀฀ and write the percentage change of ฀฀ as Marginal rates of change

฀฀฀฀ ฀฀

.

In economic situations there are three main functions Cost, ฀฀(฀฀), the cost of producing ฀฀ items.

Revenue, ฀฀(฀฀), the income from selling ฀฀ units.

Profit, ฀฀(฀฀ ) = ฀฀(฀฀ ) − ฀฀(฀฀)

The marginal cost, ฀฀′(฀฀) gives the approximate cost of producing one more item.

Similarly, ฀฀′(฀฀) (and ฀฀′(฀฀)) are the approximate extra revenue (and profit) from the sale of one more item.

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Year 12 ATAR METHODS – Semester 2 Revision

Application of the second derivative Second derivative Differentiating a function twice give the gradient function of the gradient. Notation includes Displacement, Velocity and Acceleration

฀฀ 2 ฀฀

฀฀฀฀ 2

, ฀฀" or ฀฀′′(฀฀)

Displacement ฀฀(฀฀) or ฀฀(฀฀) is the distance a particle is from the origin. Velocity is the rate of change of displacement ∴ ฀฀ = ฀฀฀฀

฀฀฀฀

฀฀2 ฀฀

Acceleration is the rate of change of velocity. ∴ ฀฀ = ฀฀฀฀ = ฀฀฀฀2 ฀฀฀฀

Curve sketching When given a curve to sketch look for the following features. 1. What is the basic shape of the graph, (cubic, parabola, hyperbola etc) 2. Find any intercepts 3. Find limits at infinity 4. Check for asymptotes 5. Identify any turning points 6. Find points of inflection Stationary and inflection points

For minimum (concave up) at ฀ ฀ = ฀฀: ฀฀ ′ = 0, ฀฀ ′′ > 0 ฀฀

฀฀ ′

฀฀− − \

฀฀

0 −

฀฀+ + /

For maximum (concave down) at ฀ ฀ = ฀฀: ฀฀ ′ = 0, ฀฀ ′′ < 0 ฀฀

฀฀ ′

฀฀− + /

฀฀

0 −

฀฀+ − \

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Year 12 ATAR METHODS – Semester 2 Revision

For horizontal inflection point at ฀ ฀ = ฀฀: ฀฀ ′ = ฀฀′′ = 0, ฀฀ ′′(฀฀+ ) and ฀฀ ′′(฀฀−) have opposite signs. ฀฀

฀฀ ′

฀฀"

฀฀− ±

฀฀

±

0

0

฀฀+ ±

∓

Note: if the first and second derivative are both 0 at the same point, it may not be a horizontal point of inflection. For example, ฀ ฀ = ฀฀ 4 . You will need to do the sign test to check. For oblique inflection point at ฀ ฀ = ฀฀:

฀฀′ ≠ 0, ฀฀ ′′ = 0. ฀฀′′(฀฀+ ) and ฀฀ ′′(฀฀− ) have opposite signs.

Optimisation 1. Draw a diagram if possible. 2. Define the variables 3. Define the function to be optimised 4. Use the relationship between the variables to express the function in terms of one variable 5. Determine stationary points 6. Determine nature of turning points 7. Check end points of the domain for global maximum or minimum 8. State conclusion in words.

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Year 12 ATAR METHODS – Semester 2 Revision

Examples 1.

A botanist has found that the rate of the loss of leaves from a particular rainforest tree is proportional

a)

The botanist wishes to study lichens that grow on the leaves of this tree. From past observations, he has found that of 100 leaves tagged, 80 remain after three years. Use this information to determine the value of the constant ฀฀ correct to 3 decimal places.

to the number of leaves on the tree.If the initial number of leaves on the tree is ฀฀0 , then the number of these leaves that remain after time t years is ฀฀(฀฀) = ฀฀0 ฀฀ −฀฀฀฀ where k is a positive constant.

80 = 100 × ฀฀ −3฀฀

฀฀ −3฀฀ =

4

5

฀฀ ≈ 0.074

Using your result from part a), calculate the number of leaves that the botanist should initially tag if he needs to have 200 leaves to study in eight years’ time.

b)

When ฀ ฀ = 8 the botanist requires 200 leaves so the initial number ฀฀0 must satisfy 200 = ฀฀0 ฀฀ −8฀฀ So ฀฀0 = 200฀฀ 8×0.074 = 362.6 The botanist should tag approximately 363 leaves

Determine the rate of change of the amount of leaves in the year found in part b). and interpret your result.

c)

Rate of change is

฀฀฀฀ ฀฀฀฀

= ฀ ฀ × ฀฀

= −0.074 × 362.6 −26.8 In the 8th year the leaves are decreasing by 27 leaves/year.

d) ฀฀0 2 1 2

If initially leaves are tagged, calculate when only half of them will remain on the tree. Give your answer correct to two significant figures.

= ฀฀0 ฀฀ ...