Lesson Note on Differentiation: Applications to Rate, Gradient and Optimization for SSS 2

Class: Senior Secondary School 2 (SS2 / SSS2)
Term: 2nd Term
Week: 4
Age: 16 years
Duration: 45 minutes
Subject: Further Mathematics
Curriculum Theme: N/A
Previous Lesson: Differentiation: Rules, Product and Quotient Rule.
Topic: DIFFERENTIATION
Subject Matter: Application of differentiation (rate of change, gradient, maximum and minimum values, equation of motion)

Specific Objectives

By the end of the lesson, pupils should be able to:

Cognitive Domain:

• Define rate of change and gradient in the context of differentiation.
• State the conditions for maximum and minimum values of a function.
• Recall the relationship between displacement, velocity, and acceleration using differentiation.

Affective Domain:

• Appreciate the practical applications of differentiation in real-world problems.
• Develop an interest in solving optimization problems.
• Show confidence in handling differentiation problems involving rates and motion.

Psychomotor Domain:

• Calculate the rate of change of various quantities.
• Determine the gradient of a curve at a specific point.
• Find maximum and minimum values of given functions.
• Solve problems related to the equation of motion using differentiation.

Social Domain:

• Collaborate with peers to solve application problems of differentiation.
• Communicate their solutions clearly to the class.
• Participate actively in class discussions on differentiation applications.

Reference Materials

The following resources were used in planning this lesson:

• Senior Secondary Schools Education Curriculum
• State Unified Scheme of Work
• P.N. Chikezie, et al. (2018). Further Mathematics for Senior Secondary Schools 2. Learn Africa Publications.
• https://www.mathsisfun.com/calculus/derivatives-applications.html
• https://www.khanacademy.org/math/ap-calculus-ab/ab-applications-of-differentiation

Instructional Materials

The teacher will teach this lesson with the aid of:

• A chart showing examples of rate of change, gradient, and optimization problems.
• Whiteboard and markers.

Rationale for the Lesson

This lesson helps pupils understand how differentiation is used to solve practical problems in various fields. It enables pupils to apply mathematical concepts to real-world situations such as calculating speed, finding maximum efficiency, and determining the slope of a curve.

Prerequisite/Previous Knowledge

Pupils should have prior knowledge of basic differentiation rules for algebraic functions, including power rule, product rule, and quotient rule.

Lesson Content/Board Summary

DIFFERENTIATION: APPLICATIONS

1. Rate of Change

The rate of change describes how one quantity changes in relation to another. If a quantity y depends on another quantity x, the rate of change of y with respect to x is given by the derivative `dy/dx`.

If the quantities change with respect to time (t), then the rate of change is `dy/dt`.

Example 1: The radius of a circle is increasing at a rate of 2 cm/s. Find the rate at which the area is increasing when the radius is 5 cm. (Given: Area `A = πr²`)

Solution:

• Step 1: Write down the given information and formula.
  Given `dr/dt = 2 cm/s`, `r = 5 cm`. Find `dA/dt`.
  Formula: `A = πr²`
• Step 2: Differentiate A with respect to r.
  `dA/dr = d/dr(πr²) = 2πr`
• Step 3: Use the chain rule to find `dA/dt`.
  `dA/dt = (dA/dr) * (dr/dt)`
  `dA/dt = (2πr) * (2)`
  `dA/dt = 4πr`
• Step 4: Substitute the value of r.
  When `r = 5 cm`, `dA/dt = 4π(5) = 20π cm²/s`

Example 2: The volume `V` of a sphere is given by `V = (4/3)πr³`. Find the rate of change of volume with respect to its radius `r` when `r = 3 cm`.

Solution:

• Step 1: Write down the formula.
  `V = (4/3)πr³`
• Step 2: Differentiate V with respect to r.
  `dV/dr = d/dr((4/3)πr³) = (4/3)π * 3r² = 4πr²`
• Step 3: Substitute the value of r.
  When `r = 3 cm`, `dV/dr = 4π(3)² = 4π(9) = 36π cm³/cm`

2. Gradient of a Function (Tangent)

The derivative `dy/dx` of a function `y = f(x)` gives the gradient (slope) of the tangent to the curve at any point `(x, y)`.

Example 1: Find the gradient of the curve `y = x³ – 2x + 5` at the point where `x = 2`.

Solution:

• Step 1: Differentiate the function `y` with respect to `x`.
  `dy/dx = d/dx(x³ – 2x + 5) = 3x² – 2`
• Step 2: Substitute the given `x` value into the derivative.
  At `x = 2`, `dy/dx = 3(2)² – 2 = 3(4) – 2 = 12 – 2 = 10`
• Step 3: State the gradient.
  The gradient of the curve at `x = 2` is 10.

Example 2: Find the equation of the tangent to the curve `y = x² + 3x – 1` at the point `(1, 3)`.

Solution:

• Step 1: Differentiate the function `y` with respect to `x` to find the gradient.
  `dy/dx = d/dx(x² + 3x – 1) = 2x + 3`
• Step 2: Substitute the x-coordinate of the given point into the derivative to find the gradient `m`.
  At `x = 1`, `m = 2(1) + 3 = 5`
• Step 3: Use the point-slope form of a line `(y – y₁) = m(x – x₁)` with `(x₁, y₁) = (1, 3)` and `m = 5`.
  `y – 3 = 5(x – 1)`
  `y – 3 = 5x – 5`
• Step 4: Rearrange the equation to the standard form.
  `y = 5x – 2` or `5x – y – 2 = 0`

3. Maximum and Minimum Values (Optimization)

Differentiation can be used to find the maximum or minimum values of a function, which are often referred to as stationary points or turning points.

Conditions for stationary points:

• First Derivative Test: `dy/dx = 0` at a stationary point.
• Second Derivative Test:
  • If `d²y/dx² < 0`, the point is a maximum.
  • If `d²y/dx² > 0`, the point is a minimum.
  • If `d²y/dx² = 0`, the test is inconclusive (it could be a point of inflection).

Example 1: Find the stationary points of the function `y = x³ – 3x² + 2` and determine their nature.

Solution:

• Step 1: Find the first derivative and set it to zero.
  `dy/dx = 3x² – 6x`
  `3x² – 6x = 0`
  `3x(x – 2) = 0`
  So, `x = 0` or `x = 2`.
• Step 2: Find the corresponding y-values.
  When `x = 0`, `y = (0)³ – 3(0)² + 2 = 2`. Point: `(0, 2)`
  When `x = 2`, `y = (2)³ – 3(2)² + 2 = 8 – 12 + 2 = -2`. Point: `(2, -2)`
• Step 3: Find the second derivative.
  `d²y/dx² = d/dx(3x² – 6x) = 6x – 6`
• Step 4: Test the nature of each stationary point using the second derivative.
  At `x = 0`: `d²y/dx² = 6(0) – 6 = -6`. Since `-6 < 0`, `(0, 2)` is a maximum point.
  At `x = 2`: `d²y/dx² = 6(2) – 6 = 12 – 6 = 6`. Since `6 > 0`, `(2, -2)` is a minimum point.

Example 2: A rectangular field is to be fenced off with 100 meters of fencing. What dimensions will maximize the area of the field?

Solution:

• Step 1: Define variables and set up equations.
  Let the length be `L` and width be `W`.
  Perimeter `P = 2L + 2W = 100` => `L + W = 50` => `W = 50 – L`
  Area `A = L * W = L(50 – L) = 50L – L²`
• Step 2: Differentiate the area function with respect to L and set to zero.
  `dA/dL = 50 – 2L`
  `50 – 2L = 0` => `2L = 50` => `L = 25` meters.
• Step 3: Find the corresponding width.
  `W = 50 – L = 50 – 25 = 25` meters.
• Step 4: Use the second derivative test to confirm it’s a maximum.
  `d²A/dL² = d/dL(50 – 2L) = -2`.
  Since `-2 < 0`, the area is maximized when `L = 25` meters.
• Step 5: State the dimensions.
  The dimensions that maximize the area are `L = 25 m` and `W = 25 m` (a square).

4. Equation of Motion

Differentiation is fundamental in kinematics to describe motion. If `s` is displacement, `v` is velocity, and `a` is acceleration, all as functions of time `t`:

• Velocity: `v = ds/dt` (Rate of change of displacement with respect to time).
• Acceleration: `a = dv/dt = d²s/dt²` (Rate of change of velocity with respect to time, or second derivative of displacement).

Example 1: The displacement `s` (in meters) of a particle moving in a straight line is given by `s(t) = t³ – 6t² + 9t + 5`, where `t` is time in seconds. Find:

a) The velocity of the particle at `t = 2` seconds.

b) The acceleration of the particle at `t = 3` seconds.

Solution:

• Step 1: Find the velocity function `v(t)`.
  `v(t) = ds/dt = d/dt(t³ – 6t² + 9t + 5) = 3t² – 12t + 9`
• Step 2: Calculate velocity at `t = 2`.
  `v(2) = 3(2)² – 12(2) + 9 = 3(4) – 24 + 9 = 12 – 24 + 9 = -3 m/s`
• Step 3: Find the acceleration function `a(t)`.
  `a(t) = dv/dt = d/dt(3t² – 12t + 9) = 6t – 12`
• Step 4: Calculate acceleration at `t = 3`.
  `a(3) = 6(3) – 12 = 18 – 12 = 6 m/s²`

Example 2: A particle’s displacement `s` (in meters) is given by `s(t) = 2t² – 8t + 10`. Find the time `t` when the particle is momentarily at rest.

Solution:

• Step 1: Find the velocity function `v(t)`.
  `v(t) = ds/dt = d/dt(2t² – 8t + 10) = 4t – 8`
• Step 2: Set velocity to zero for “momentarily at rest”.
  `4t – 8 = 0`
• Step 3: Solve for `t`.
  `4t = 8`
  `t = 2` seconds.

Teaching Methods/Instructional Techniques

Discussion, Lecture, Demonstration, Question and Answer, Visual Aids

Instructional Procedures

Step 1: Introduction

Time: 3 minutes
Teaching Skill: Set Induction
Teacher’s Activity: The teacher greets the pupils and asks them what they remember about differentiation. The teacher then introduces the topic by explaining that differentiation is not just about finding derivatives but also has many practical uses.
Pupils’ Activity: Pupils respond to the greetings and recall basic differentiation rules. They listen attentively to the introduction.
Learning Point: Pupils recall prior knowledge and are introduced to the lesson’s focus on applications.

Step 2: Rate of Change

Time: 8 minutes
Teaching Skill: Explanation/Demonstration
Teacher’s Activity: The teacher explains the concept of rate of change using `dy/dx` or `dA/dt` and demonstrates Example 1 and Example 2 from the board summary, explaining each step clearly.
Pupils’ Activity: Pupils listen, take notes, and ask questions for clarification. They observe how the examples are solved.
Learning Point: Pupils understand how to calculate rates of change using differentiation.

Step 3: Gradient of a Function (Tangent)

Time: 7 minutes
Teaching Skill: Explanation/Demonstration
Teacher’s Activity: The teacher explains that the derivative represents the gradient of the tangent to a curve. The teacher demonstrates Example 1 and Example 2 from the board summary, showing how to find the gradient and the equation of the tangent.
Pupils’ Activity: Pupils follow the explanations, copy the examples, and try to grasp the method.
Learning Point: Pupils learn to find the gradient of a curve and the equation of its tangent at a given point.

Step 4: Maximum and Minimum Values (Optimization)

Time: 10 minutes
Teaching Skill: Explanation/Demonstration
Teacher’s Activity: The teacher explains the conditions for maximum and minimum points using the first and second derivative tests. The teacher demonstrates Example 1 and Example 2 from the board summary, emphasizing the steps to identify and classify stationary points.
Pupils’ Activity: Pupils pay close attention to the conditions and solution steps, asking questions about the second derivative test.
Learning Point: Pupils can identify stationary points and determine their nature (maximum or minimum).

Step 5: Equation of Motion

Time: 7 minutes
Teaching Skill: Explanation/Demonstration
Teacher’s Activity: The teacher explains the relationship between displacement, velocity, and acceleration using differentiation (`v = ds/dt`, `a = dv/dt`). The teacher demonstrates Example 1 and Example 2 from the board summary.
Pupils’ Activity: Pupils follow the derivations and calculations, understanding how differentiation links these kinematic quantities.
Learning Point: Pupils apply differentiation to solve problems involving motion.

Step 6: Evaluation/Review

Time: 5 minutes
Teaching Skill: Questioning/Assessment
Teacher’s Activity: The teacher evaluates the learning by asking the following questions:

1. Define rate of change and state its formula.
2. How do you find the gradient of a curve at a specific point?
3. State the conditions for a maximum point and a minimum point using the second derivative test.
4. If the displacement `s` of a particle is given by `s(t) = 3t² – 2t + 1`, find its velocity function `v(t)`.

Pupils’ Activity: Pupils answer orally and in writing.
Learning Point: Pupils demonstrate understanding of the lesson.

Step 7: Conclusion

Time: 5 minutes
Teaching Skill: Summarization
Teacher’s Activity: The teacher summarizes the key applications of differentiation covered: rate of change, gradient, maximum/minimum values, and equations of motion. The teacher assigns homework, which includes practice problems from the textbook on these applications.
Pupils’ Activity: Pupils listen to the summary, ask any final questions, and copy the homework assignment.
Learning Point: Pupils consolidate their understanding of the various applications of differentiation.

Lesson Keywords

• Differentiation – The process of finding the derivative of a function.
• Rate of Change – How one quantity changes in response to another.
• Gradient – The slope of the tangent to a curve at a given point.
• Maximum Value – The highest point a function reaches in a given interval, where the first derivative is zero and the second derivative is negative.
• Minimum Value – The lowest point a function reaches in a given interval, where the first derivative is zero and the second derivative is positive.
• Displacement – The change in position of an object.
• Velocity – The rate of change of displacement with respect to time.
• Acceleration – The rate of change of velocity with respect to time.

Differentiation

For pupils who grasp the concepts quickly, the teacher can provide more complex problems involving implicit differentiation or related rates with multiple variables. For pupils who are struggling, the teacher will provide simpler functions and guide them through step-by-step solutions, focusing on one application at a time before combining them.

Note for teachers using this lesson plan

Teachers should ensure pupils have a strong foundation in basic differentiation rules before introducing applications. Emphasize the real-world relevance of each application to sustain pupil interest. Encourage pupils to draw diagrams for optimization problems where applicable. Provide ample practice problems for each application to reinforce understanding.