Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Apply differentiation to solve optimisation problems.
• Solve integration problems using substitution.
• Apply trigonometry to solve 2D and 3D problems.
• Determine the equation of tangents to circles.
• Apply principles to solve conditional probability problems.

Lesson summary

This week focuses on consolidating our understanding of key Grade 12 Mathematics topics in preparation for the preliminary examinations. This is a crucial stage as prelims serve as a vital indicator of preparedness for the final NSC examinations and provide an opportunity to identify areas needing further attention. Mastering these concepts is not just about passing exams; it equips you with analytical and problem-solving skills essential for various careers and navigating real-world challenges in South Africa, from engineering and finance to data analysis and technology.

Lesson notes

Differentiation Differentiation is the process of finding the derivative of a function, which represents the instantaneous rate of change of the function. This concept is fundamental in optimisation problems.

Rules of Differentiation: Power Rule: If f(x) = x n , then f'(x) = nx n-1 Constant Multiple Rule: If f(x) = kg(x), then f'(x) = k*g'(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)

Example 1: Optimisation A farmer in Limpopo wants to fence off a rectangular vegetable garden next to a long wall. He has 80 meters of fencing. What are the dimensions of the garden that maximize the area? Let the width of the garden be x meters and the length be y meters. Since the wall acts as one side, the fencing required is x + 2y =

8

0. We want to maximize the area A = xy.

Express y in terms of x: 2y = 80 - x => y = 40 - x/2 Substitute into the area equation: A = x(40 - x/2) = 40x - x 2 /2 Differentiate A with respect to x: dA/dx = 40 - x Set dA/dx = 0 to find the critical points: 40 - x = 0 => x = 40 Find the second derivative: d 2 A/dx 2 = -

1. Since the second derivative is negative, the area is maximized when x =

4

0. Calculate y: y = 40 - 40/2 = 20 Therefore, the dimensions that maximize the area are width x = 40 meters and length y = 20 meters. Integration Integration is the reverse process of differentiation. It's used to find the area under a curve, among other applications.

Substitution Rule: This rule is used when the integrand contains a function and its derivative. ∫f(g(x))g'(x) dx = F(g(x)) + C, where F'(x) = f(x)

Example 2: Integration by Substitution Evaluate ∫2x(x 2 + 1) 3 dx Let u = x 2 +

1. Then du/dx = 2x, so du = 2x dx Substitute: ∫u 3 du Integrate: (u 4 / 4) + C Substitute back: ((x 2 + 1) 4 / 4) + C Trigonometry Trigonometry deals with the relationships between the sides and angles of triangles.

Sine Rule: a/sin A = b/sin B = c/sin C Cosine Rule: a 2 = b 2 + c 2 - 2bc cos A Area Rule: Area = (1/2)ab sin C Example 3: Solving a 3D Trigonometry Problem A tower is standing on level ground. From a point A, the angle of elevation of the top of the tower is 30°. From a point B, which is 50 m away from A and in the same horizontal line as the foot of the tower, the angle of elevation is 45°. Calculate the height of the tower. Let the height of the tower be h. Let the distance from A to the foot of the tower be x. tan 30° = h/x => x = h/tan 30° = h√3 Let the distance from B to the foot of the tower be y. y = x - 50 = h√3 - 50 tan 45° = h/y => h = y = h√3 - 50 h - h√3 = -50 => h(1-√3) = -50 h = -50 / (1 - √3) = -50 / (1 - 1.732) = -50 / -0.732 ≈ 68.3 m Analytical Geometry Analytical geometry involves using coordinate systems to study geometric shapes.

Equation of a Circle: (x - a) 2 + (y - b) 2 = r 2 , where (a, b) is the center and r is the radius.

Tangent to a Circle: The tangent at a point on the circle is perpendicular to the radius at that point.

Example 4: Finding the Equation of a Tangent Find the equation of the tangent to the circle x 2 + y 2 = 25 at the point (3, 4). The center of the circle is (0, 0). The radius at (3, 4) has a gradient of 4/

3. The tangent is perpendicular to the radius, so its gradient is -3/

4. Using the point-slope form of a line, y - y 1 = m(x - x 1 ), the equation of the tangent is y - 4 = (-3/4)(x - 3) Simplifying, y = (-3/4)x + 9/4 + 4 = (-3/4)x + 25/4 Therefore, the equation of the tangent is 3x + 4y =

2

5. Probability Probability is the measure of the likelihood of an event occurring.

Conditional Probability: P(A|B) = P(A ∩ B) / P(B), where P(A|B) is the probability of A given that B has occurred.

Example 5: Conditional Probability In a class of 30 students, 18 take Mathematics, 12 take Physics, and 5 take both. What is the probability that a student takes Physics given that they take Mathematics? Let M be the event that a student takes Mathematics, and P be the event that a student takes Physics. P(M) = 18/30, P(P) = 12/30, P(M ∩ P) = 5/30 P(P|M) = P(M ∩ P) / P(M) = (5/30) / (18/30) = 5/18 Guided Practice (With Solutions)

Question 1: A company wants to minimize the cost of producing cylindrical cans. The volume of each can must be 500 cm 3 . Find the radius and height of the can that minimizes the surface area. (Surface Area = 2πr 2 + 2πrh, Volume = πr 2 h)

Solution: Express h in terms of r using the volume equation: h = 500 / (πr 2 ) Substitute h into the surface area equation: A = 2πr 2 + 2πr(500 / (πr 2 )) = 2πr 2 + 1000/r Differentiate A with respect to r: dA/dr = 4πr - 1000/r 2 Set dA/dr = 0: 4πr = 1000/r 2 => r 3 = 1000/(4π) => r = (250/π) 1/3 ≈ 4.30 cm Calculate h: h = 500 / (π(4.30) 2 ) ≈ 8.60 cm

Commentary: This question combines volume, surface area, and differentiation to solve a practical optimization problem.