Lesson Notes By Weeks and Term v5 - Grade 12
Differential calculus – Week 7 focus
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Differential calculus – Week 7 focus
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Subject: Mathematics
Class: Grade 12
Term: 2nd Term
Week: 7
Theme: General lesson support
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Differential calculus is a cornerstone of advanced mathematics, providing the tools to understand and model change. In Grade 12, we delve deeper into its applications and theoretical underpinnings. This week, our focus is on applications of derivatives, specifically rate of change, optimisation problems, and curve sketching. These skills aren't just theoretical; they are crucial for understanding real-world phenomena, from modelling the spread of a disease to optimizing business operations. In South Africa, with its dynamic economic landscape and diverse environmental challenges, understanding these concepts is vital for informed decision-making and problem-solving.
2.1 Rate of Change The derivative, denoted as dy/dx or f'(x), represents the instantaneous rate of change of y with respect to x. It tells us how much y changes for a tiny change in x. This is fundamental to understanding how quantities interact and influence each other. Related rates problems involve finding the rate of change of one quantity in terms of the rate of change of another, when both quantities are related by an equation. Key Steps for Solving Related Rates Problems: Identify Variables: Assign symbols to all quantities involved.
Identify Given Rates: Determine which rates of change are known (e.g., dx/dt).
Identify the Unknown Rate: Determine the rate of change you need to find (e.g., dy/dt).
Establish a Relationship: Find an equation that relates the variables involved. This might involve geometry (e.g., Pythagoras' theorem, volume formulas), or other physical principles.
Differentiate with Respect to Time (t): Apply the chain rule to differentiate the equation with respect to time. Remember that if x and y are functions of t, then d/dt(x^2) = 2x(dx/dt).
Substitute Known Values: Plug in the known rates and values of the variables at the instant you are considering.
Solve for the Unknown Rate: Solve the resulting equation for the unknown rate of change.
Interpret the Result: State your answer with the correct units.
Example 1 (Related Rates): Water is being pumped into a conical tank at a rate of 2 m³/min. The tank has a height of 10 m and a radius of 4 m. How fast is the water level rising when the water is 5 m deep?
Solution: Variables: Let V be the volume of water, h be the height of the water, and r be the radius of the water surface.
Given Rate: dV/dt = 2 m³/min Unknown Rate: dh/dt when h = 5 m Relationship: V = (1/3)πr²h. We need to eliminate 'r' because we only have information about 'h'. Using similar triangles, r/h = 4/10 => r = (2/5)h.
Therefore, V = (1/3)π((2/5)h)²h = (4/75)πh³.
Differentiate: dV/dt = (4/25)πh² (dh/dt)
Substitute: 2 = (4/25)π(5)² (dh/dt)
Solve: 2 = 4π (dh/dt) => dh/dt = 1/(2π) m/min.
Interpretation: The water level is rising at a rate of 1/(2π) meters per minute when the water is 5 m deep. 2.2 Optimisation Problems Optimisation problems involve finding the maximum or minimum value of a function subject to certain constraints. This is crucial for efficiently allocating resources, maximising profit, or minimising costs. Key Steps for Solving Optimisation Problems: Understand the Problem: Read the problem carefully and identify what quantity needs to be maximised or minimised.
Define Variables: Assign variables to all relevant quantities.
Formulate the Objective Function: Write an equation for the quantity to be optimised (e.g., area, volume, cost). This is the function you will differentiate.
Identify Constraints: Write down any constraints or relationships between the variables. These constraints may need to be used to eliminate one of the variables from the objective function, leaving it as a function of a single variable. Express the Objective Function in Terms of One Variable: Use the constraints to eliminate variables until the objective function is expressed in terms of a single variable.
Find Critical Points: Differentiate the objective function and set the derivative equal to zero to find the critical points. Also, consider endpoints of the domain.
Determine Maximum or Minimum: Use the first derivative test (sign change around the critical point) or the second derivative test (concavity at the critical point) to determine whether each critical point corresponds to a maximum or minimum.
Answer the Question: State your answer in the context of the original problem, including units.
Example 2 (Optimisation): A farmer in KwaZulu-Natal wants to fence off a rectangular grazing area along a straight river. He has 400 meters of fencing. What are the dimensions of the grazing area that will maximise the enclosed area? (
Note: The river acts as one side of the rectangle, so no fencing is needed there.)
Solution: Understand: Maximise the area of a rectangular grazing area.
Variables: Let x be the length of the fence perpendicular to the river and y be the length of the fence parallel to the river. Area = A, Perimeter = P Objective Function: A = xy Constraint: P = 2x + y = 400 (fencing constraint).
One Variable: From the constraint, y = 400 - 2x. Substituting into the objective function, A = x(400 - 2x) = 400x - 2x².
Critical Points: dA/dx = 400 - 4x. Setting dA/dx = 0, we get 400 - 4x = 0 => x =
1
0
0. Maximum/Minimum: d²A/dx² = -4, which is negative, indicating a maximum at x =
1
0
0. Answer: x = 100 meters, y = 400 - 2(100) = 200 meters. The dimensions that maximize the area are 100 m by 200 m. The maximum area is 20,000 m². 2.3 Curve Sketching Curve sketching involves using calculus to accurately sketch the graph of a function, identifying key features such as intercepts, turning points (local maxima and minima), and points of inflection.