Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Solve optimisation problems using calculus.
• Apply trigonometric identities to solve equations.
• Calculate probabilities of compound events.
• Analyze statistical data.
• Solve complex Euclidean Geometry problems.

Lesson summary

This week marks a critical stage in your Grade 12 Mathematics journey: focused revision and preparation for the final examinations. We'll concentrate on areas commonly encountered in the final exam, addressing potential weaknesses and solidifying understanding. This targeted revision will boost your confidence and equip you with the problem-solving skills needed to succeed. Mathematics plays a vital role in everyday life, from budgeting your monthly income and understanding interest rates on loans (especially relevant in the current South African economic climate) to making informed decisions about investments and interpreting data related to socio-economic issues in our country.

Lesson notes

2.1 Calculus: Optimisation Problems Optimisation problems involve finding the maximum or minimum value of a function, subject to certain constraints.

The key steps are: Identify the objective function: This is the function you want to maximise or minimise (e.g., area, volume, profit).

Identify the constraint(s): These are the conditions that limit the possible values of the variables (e.g., fixed perimeter, limited resources). Express the objective function in terms of a single variable: Use the constraint(s) to eliminate one or more variables from the objective function.

Find the critical points: Differentiate the objective function with respect to the remaining variable and set the derivative equal to zero. Solve for the critical points. Determine the nature of the critical points: Use the first or second derivative test to determine whether each critical point is a maximum, a minimum, or a point of inflection.

Answer the question: Make sure to provide the values of the variables that optimise the objective function, as well as the optimised value itself.

Example 1: A farmer in KwaZulu-Natal wants to fence off a rectangular plot of land alongside a straight river. He has 800 meters of fencing. What are the dimensions of the plot that will maximise the area enclosed? (

Note: No fencing is required along the river).

Solution: Objective function: Area, A = l w (where l is the length and w is the width)

Constraint: Perimeter of the fence, 2w + l = 800 Express A in terms of a single variable: From the constraint, l = 800 - 2w. Substituting into the area equation, A = (800 - 2w) w = 800w - 2w 2 Find critical points: dA/dw = 800 - 4*w = 0 => w = 200 Determine the nature of the critical point: d 2 A/dw 2 = -4 (negative), so w = 200 is a maximum.

Answer: If w = 200, then l = 800 - 2(200) =

4

0

0. Therefore, the dimensions that maximise the area are width = 200 meters and length = 400 meters. The maximum area is 200 * 400 = 80000 m 2 . 2.2 Trigonometry: Identities and Equations Mastering trigonometric identities is crucial for solving trigonometric equations and simplifying complex expressions.

Some important identities include: sin 2 θ + cos 2 θ = 1 tan θ = sin θ / cos θ sin 2θ = 2 sin θ cos θ cos 2θ = cos 2 θ - sin 2 θ = 2 cos 2 θ - 1 = 1 - 2 sin 2 θ Solving trigonometric equations often involves using these identities to simplify the equation and express it in terms of a single trigonometric function. Remember to find all solutions within the specified interval.

Example 2: Solve the equation 2cos 2 x - sin x - 1 = 0 for x ∈ [0°; 360°] Solution: Use the identity cos 2 x = 1 - sin 2 x: 2(1 - sin 2 x) - sin x - 1 = 0 Simplify: 2 - 2sin 2 x - sin x - 1 = 0 => -2sin 2 x - sin x + 1 = 0 Multiply by -1: 2sin 2 x + sin x - 1 = 0 Factorise: (2sin x - 1)(sin x + 1) = 0 Solve for sin x: 2sin x - 1 = 0 => sin x = 1/2 OR sin x + 1 = 0 => sin x = -1 Find the solutions for x: sin x = 1/2 => x = 30° or x = 150° (using the CAST diagram or the reference angle) sin x = -1 => x = 270° Therefore, the solutions are x = 30°, 150°, and 270°. 2.3 Probability: Compound Events Compound events involve the occurrence of two or more events. We use tree diagrams, Venn diagrams, and contingency tables to represent and calculate probabilities of these events.

Key concepts include: Independent events: The occurrence of one event does not affect the probability of the other event. P(A and B) = P(A)

P(B)

Dependent events: The occurrence of one event affects the probability of the other event. P(A and B) = P(A) P(B|A) (where P(B|A) is the probability of B given that A has occurred)

Mutually exclusive events: The events cannot occur at the same time. P(A or B) = P(A) + P(B)

Non-mutually exclusive events: The events can occur at the same time. P(A or B) = P(A) + P(B) - P(A and B)

Example 3: In a class of 30 learners, 18 take Mathematics, 12 take Physical Sciences, and 5 take both. What is the probability that a randomly selected learner takes either Mathematics or Physical Sciences?

Solution: Let M be the event that a learner takes Mathematics and S be the event that a learner takes Physical Sciences. P(M) = 18/30, P(S) = 12/30, P(M and S) = 5/30 Since learners can take both subjects, the events are not mutually exclusive. P(M or S) = P(M) + P(S) - P(M and S) = 18/30 + 12/30 - 5/30 = 25/30 = 5/6 Therefore, the probability that a randomly selected learner takes either Mathematics or Physical Sciences is 5/6. 2.4 Statistics: Measures of Central Tendency and Dispersion Measures of central tendency (mean, median, mode) describe the 'average' value of a dataset. Measures of dispersion (range, variance, standard deviation) describe the spread or variability of the data. The standard deviation is particularly important as it indicates how closely the data points cluster around the mean.