Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Analyze past examination papers.
• Apply differentiation techniques to solve problems.
• Use integration to calculate areas.
• Solve trigonometric equations.
• Apply the sine, cosine and area rules.

Lesson summary

This week marks a crucial stage in your Grade 12 Mathematics journey: intense revision and focused preparation for your final examinations. Mastering mathematical concepts isn't just about acing the exam; it's about developing critical thinking, problem-solving skills, and logical reasoning – skills highly valued in various fields across South Africa, from engineering and finance to technology and even everyday life. Think about budgeting your finances, analyzing data related to crime statistics, or even understanding the spread of diseases – all of these scenarios involve mathematical principles.

Lesson notes

This week, we will revisit the following key areas with a focus on exam-oriented problem-solving: Calculus (Differentiation and Integration), Trigonometry (Trigonometric equations and Rules), and Data Handling/Probability.

A. Calculus (Differentiation and Integration)

Differentiation: Differentiation is the process of finding the derivative of a function. The derivative represents the instantaneous rate of change of the function.

Rules of Differentiation: Power Rule: If f(x) = x n , then f'(x) = nx n-1 *.

Constant Rule: If f(x) = c (where c is a constant), then f'(x) =

0. Constant Multiple Rule: If f(x) = cf(x) (where c is a constant), then f'(x) = cf'(x)*.

Sum/Difference Rule: If f(x) = u(x) ± v(x), then f'(x) = u'(x) ± v'(x)*.

Product Rule: If f(x) = u(x)v(x), then f'(x) = u'(x)v(x) + u(x)v'(x)*.

Quotient Rule: If f(x) = u(x) / v(x), then f'(x) = [u'(x)v(x) - u(x)v'(x)] / [v(x)] 2 *.

Chain Rule: If f(x) = g(h(x)), then f'(x) = g'(h(x)) h'(x).

Applications of Differentiation: Finding stationary points (maxima, minima, points of inflection), determining intervals where a function is increasing or decreasing, solving optimization problems (e.g., maximizing area, minimizing cost).

Example 1 (Optimization): A farmer in KwaZulu-Natal wants to fence off a rectangular vegetable garden next to a long brick wall. He has 40 meters of fencing. What are the dimensions of the garden that maximize the area enclosed?

Solution: Let the length of the garden perpendicular to the wall be x, and the length parallel to the wall be y. Since the farmer only needs to fence three sides, 2x + y =

4

0. The area A is given by A = xy. We want to maximize

A. Express y in terms of x: y = 40 - 2x.

Substitute into the area equation: A = x(40 - 2x) = 40x - 2x 2 . Find the derivative of A with respect to x: dA/dx = 40 - 4x. Set the derivative equal to zero and solve for x: 40 - 4x = 0 => x =

1

0. Find the second derivative: d 2 A/dx 2 = -

4. Since the second derivative is negative, we have a maximum at x =

1

0. Find y: y = 40 - 2(10) =

2

0. Therefore, the dimensions that maximize the area are 10 meters by 20 meters.

Integration: Integration is the reverse process of differentiation. It's finding the anti-derivative of a function.

Rules of Integration: Power Rule: ∫x n dx = ( x n+1 / (n+1)) + C, where n ≠ -1 and C is the constant of integration.

Constant Multiple Rule: ∫cf(x) dx = c∫f(x) dx.

Sum/Difference Rule: ∫[u(x) ± v(x)] dx = ∫u(x) dx ± ∫v(x) dx.

Integration by Substitution: Used when the integrand contains a function and its derivative (or a multiple thereof).

Integration by Parts: Used when the integrand is a product of two functions. ∫u dv = uv - ∫v du.

Applications of Integration: Finding the area under a curve, finding the area between two curves, calculating volumes of solids of revolution.

Example 2 (Area Under a Curve): Calculate the area bounded by the curve y = x 2 , the x-axis, and the lines x = 1 and x =

3. Solution: Set up the definite integral: Area = ∫ 1 3 x 2 dx.

Find the antiderivative of x 2 : ( x 3 / 3).

Evaluate the definite integral: [(3 3 / 3) - (1 3 / 3)] = [9 - (1/3)] = 26/

3. Therefore, the area is 26/3 square units.

B. Trigonometry (Trigonometric equations and Rules)

Solving Trigonometric Equations: Remember to find the general solution and then apply the given interval to find specific solutions within that interval.

Example 3: Solve the equation 2sin(x) - 1 = 0 for x ∈ [0°; 360°].

Solution: Isolate sin(x): sin(x) = 1/

2. Find the reference angle: x = arcsin(1/2) = 30°. Determine the quadrants where sin(x) is positive (Quadrants I and II).

Find the solutions in the given interval: Quadrant I: x = 30°*.

Quadrant II: x = 180° - 30° = 150°*.

Therefore, the solutions are x = 30° and x = 150°. Sine, Cosine and Area Rules Sine Rule: a/sinA = b/sinB = c/sinC Cosine Rule: a 2 = b 2 + c 2 - 2bc.cosA Area Rule: Area = 1/2 ab * sinC Example 4: In triangle ABC, AB = 10cm, BC = 8cm, and angle BAC = 60°. Calculate the size of angle AC

B. Solution: Use the Sine Rule: 8/sin60 = 10/sinC sinC = (10 * sin60)/8 sinC = (10 * √3/2)/8 sinC = (5√3)/8 C = arcsin((5√3)/8) C ≈ 81.79 degrees

C. Data Handling and Probability Measures of Central Tendency: Mean, median, and mode.

Measures of Dispersion: Range, interquartile range (IQR), standard deviation.

Probability: Permutations, combinations, Venn diagrams.

Example 5 (Probability): A bag contains 5 red balls and 3 blue balls. Two balls are drawn at random without replacement. What is the probability that both balls are red?

Solution: Probability of the first ball being red: 5/

8. Probability of the second ball being red, given the first was red: 4/

7. Probability of both balls being red: (5/8) * (4/7) = 20/56 = 5/

1

4. Guided Practice (With Solutions)

Question 1: Differentiate the function f(x) = 3x 4 - 2x 2 + 5x -

7. Solution: f'(x) = 12x 3 - 4x + 5. (Applying the power rule to each term). Remember that the derivative of a constant is zero.