Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Solve problems involving financial mathematics.
• Apply differentiation to solve optimization problems.
• Calculate and interpret measures for statistical data.
• Solve Euclidean Geometry problems.
• Effectively manage your time during examinations.

Lesson summary

This week marks a crucial stage in your Grade 12 Mathematics journey: final examination preparation. We are dedicating Week 10 to intensive revision and strategic exam techniques. This is not just about memorizing formulas; it's about deeply understanding the concepts and applying them confidently. Mathematics is a foundational subject that unlocks opportunities in various fields, from engineering and finance to data science and education. Your success in mathematics will significantly impact your future career prospects and your ability to contribute meaningfully to South Africa's economic and social development.

Lesson notes

2.1 Financial Mathematics Simple Interest: Interest earned only on the principal amount.

Formula: A = P(1 + in)

Where: A = Accumulated amount P = Principal amount i = Interest rate (per annum as a decimal) n = Number of years Compound Interest: Interest earned on the principal amount and previously accumulated interest.

Formula: A = P(1 + i)^n Where: A = Accumulated amount P = Principal amount i = Interest rate (per annum as a decimal) n = Number of compounding periods Annuities: A series of equal payments made at regular intervals.

Future Value Annuity (FV): The total value of the annuity at the end of the term. Used for savings plans, retirement funds.

Formula:* FV = x[(1+i)^n - 1]/i Where: FV = Future value x = Payment amount i = Interest rate per period n = Number of periods Present Value Annuity (PV): The amount of money needed now to fund a series of future payments. Used for loans, mortgages.

Formula:* PV = x[1 - (1+i)^-n]/i Where: PV = Present value x = Payment amount i = Interest rate per period n = Number of periods Sinking Funds: An annuity where payments are made to accumulate a specific sum of money over time. (Essentially the same formula as a future value annuity, but we solve for x).

Example 1: Simple Interest Sipho invests R5000 in a savings account that pays simple interest at a rate of 8% per annum. How much will he have after 3 years?

Solution: P = R5000, i = 0.08, n = 3 A = 5000(1 + 0.08 * 3) = 5000(1 + 0.24) = 5000(1.24) = R6200 Example 2: Compound Interest Zanele invests R8000 in an account that pays compound interest at a rate of 10% per annum, compounded quarterly. How much will she have after 5 years?

Solution: P = R8000, i = 0.10/4 = 0.025 (quarterly rate), n = 5 * 4 = 20 (number of quarters) A = 8000(1 + 0.025)^20 = 8000(1.025)^20 = 8000 * 1.6386 = R13108.80 Example 3: Future Value Annuity Thabo invests R500 per month into a retirement annuity for 25 years. The annuity earns interest at a rate of 9% per annum, compounded monthly. What will be the future value of his investment?

Solution: x = R500, i = 0.09/12 = 0.0075, n = 25 * 12 = 300 FV = 500[(1 + 0.0075)^300 - 1] / 0.0075 = 500[(1.0075)^300 - 1] / 0.0075 = 500[9.3817 - 1] / 0.0075 = 500 * 8.3817 / 0.0075 = R558,780 2.2 Calculus: Optimization Optimization problems involve finding the maximum or minimum value of a function subject to certain constraints.

The key steps are: Formulate the objective function: Define the function you want to maximize or minimize (e.g., area, volume, profit).

Identify constraints: Determine any limitations or conditions that must be satisfied (e.g., fixed perimeter, limited resources). Express the objective function in terms of a single variable: Use the constraints to eliminate variables from the objective function.

Differentiate the objective function: Find the derivative of the function. Set the derivative equal to zero and solve for the variable: Find the critical points where the function has a maximum or minimum. Verify that the critical point is a maximum or minimum: Use the second derivative test or consider the endpoints of the interval.

Answer the question: State the maximum or minimum value and the corresponding values of the variables.

Example 4: Optimization A farmer has 200 meters of fencing to enclose a rectangular field. What dimensions will maximize the area of the field?

Solution: Objective function: A = lw (area)

Constraint: 2l + 2w = 200 (perimeter) => l + w = 100 => w = 100 - l Substitute: A = l(100 - l) = 100l - l^2 Differentiate: dA/dl = 100 - 2l Set derivative to zero: 100 - 2l = 0 => 2l = 100 => l = 50 Verify: d^2A/dl^2 = -2 (negative, so it's a maximum)

Answer: l = 50m, w = 100 - 50 = 50m. The dimensions are 50m x 50m (a square), and the maximum area is 2500 m^2. 2.3 Data Handling Measures of Central Tendency: Mean: Average of the data set (sum of values divided by the number of values).

Median: Middle value when the data set is ordered.

Mode: The value that appears most frequently in the data set.

Measures of Dispersion: Range: Difference between the largest and smallest values in the data set.

Interquartile Range (IQR): Difference between the upper quartile (Q3) and the lower quartile (Q1). IQR = Q3 - Q

1. The IQR contains the middle 50% of the data.

Standard Deviation: A measure of how spread out the data is from the mean. A low standard deviation indicates that the data points are clustered closely around the mean, while a high standard deviation indicates that the data points are more spread out.

Example 5: Data Handling Consider the following data set representing the test scores of 10 students: 60, 70, 75, 80, 80, 85, 90, 90, 90,

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5. Solution: Mean: (60+70+75+80+80+85+90+90+90+95)/10 = 815/10 = 81.5 Median: Since there are 10 values (even), the median is the average of the 5th and 6th values: (80+85)/2 = 82.5 Mode: 90 (appears three times)

Range: 95 - 60 = 35 Q1 (Lower Quartile): The median of the lower half (60, 70, 75, 80, 80) is 75.