WCED Examination Preparation Learning Resource 2016: Calculus - Differentiation Mathematics Grade 12
Welcome to the Western Cape Education Department (WCED) Examination Preparation Learning Resource for Grade 12 Mathematics, focusing on Calculus - Differentiation. This resource is designed to help Grade 12 learners prepare effectively for the National Senior Certificate (NSC) Mathematics examination, with particular emphasis on the differentiation section of Calculus.
What is Calculus?
Calculus is the branch of mathematics that deals with the study of change. In Grade 12 Mathematics, Calculus focuses mainly on two areas:
- Differentiation - the study of rates of change and gradients of curves
- Integration - the process of finding the area under curves
This resource specifically addresses differentiation, a fundamental concept in Calculus that helps us understand how things change and how mathematical relationships behave.
Importance of Differentiation
Differentiation is a vital tool in mathematics with far-reaching applications in various fields including:
- Physics (motion, forces, energy)
- Economics (marginal costs and revenues)
- Engineering
- Biology (population growth)
- Chemistry (reaction rates)
Basic Differentiation Rules
The Power Rule
The power rule is one of the most basic differentiation rules. If y = x^n, then:
Constant Multiple Rule
If y = k f(x), where k is a constant:
Sum and Difference Rules
If y = f(x) g(x):
Product Rule
If y = u v, where both u and v are functions of x:
Quotient Rule
If y = u/v, where both u and v are functions of x and v 0:
Chain Rule
If y = f(g(x)):
Differentiation of Specific Functions
Trigonometric Functions
d/dx(cos x) = -sin x
d/dx(tan x) = sec x
Exponential Functions
d/dx(e^kx) = ke^kx
d/dx(a^x) = a^x ln a
Logarithmic Functions
d/dx(log_a x) = 1/(x ln a)
Applications of Differentiation
In the Grade 12 Mathematics curriculum, differentiation is applied to solve various problems:
1. Determining Equations of Tangents
Given a function f(x), the equation of the tangent at a point (a, f(a)) is:
Example: Find the equation of the tangent to y = x at the point (2, 4).
Solution:
First, find the derivative: dy/dx = 2x
At x = 2, the gradient of the tangent is: dy/dx = 2(2) = 4
Using the point-gradient form: y - 4 = 4(x - 2)
Simplifying: y - 4 = 4x - 8
y = 4x - 4
2. Finding Stationary Points
Stationary points occur where dy/dx = 0. These can be:
- Local maxima
- Local minima
- Points of inflection
Tip: To determine the nature of a stationary point:
- Find the second derivative: dy/dx
- Substitute the x-coordinate of the stationary point
- If dy/dx > 0, it's a local minimum
- If dy/dx < 0, it's a local maximum
- If dy/dx = 0, use the first derivative test
Example: Find and classify the stationary points of y = x - 3x + 2x.
Solution:
First derivative: dy/dx = 3x - 6x + 2
Setting dy/dx = 0: 3x - 6x + 2 = 0
Solving using the quadratic formula: x = (6 (36 - 24))/6 = (6 12)/6
x = (6 + 23)/6 = (3 + 3)/3 and x = (6 - 23)/6 = (3 - 3)/3
To classify these points, we find the second derivative: dy/dx = 6x - 6
For x, dy/dx = 6((3 + 3)/3) - 6 = 6 + 23 - 6 = 23 > 0, so x is a local minimum
For x, dy/dx = 6((3 - 3)/3) - 6 = 6 - 23 - 6 = -23 < 0, so x is a local maximum
3. Optimization Problems
Differentiation helps us find maximum and minimum values, which is useful in optimization problems.
Example: Find the dimensions of a rectangle with perimeter 20 cm that has the maximum possible area.
Solution:
Let the length be x and the width be y.
Perimeter: 2x + 2y = 20 x + y = 10 y = 10 - x
Area: A = xy = x(10 - x) = 10x - x
To maximize the area, find dA/dx: dA/dx = 10 - 2x
Setting dA/dx = 0: 10 - 2x = 0 x = 5
Therefore, y = 10 - 5 = 5
The rectangle with maximum area is a square with dimensions 5 cm 5 cm.
4. Rate of Change Problems
Differentiation helps us understand how quantities change in relation to each other.
Example: The height of a ball thrown vertically upward is given by h(t) = -4.9t + 15t + 2, where t is in seconds and h in meters. Find the maximum height reached by the ball.
Solution:
The velocity is the derivative of height with respect to time: v(t) = dh/dt = -9.8t + 15
At maximum height, the velocity is zero: -9.8t + 15 = 0 t = 15/9.8 1.53 seconds
Maximum height = h(15/9.8) = -4.9(15/9.8) + 15(15/9.8) + 2 13.5 meters
Graph Sketching Using Calculus
One of the key applications of differentiation in Grade 12 Mathematics is graph sketching. The following steps are used:
- Determine the domain of the function
- Find the x- and y-intercepts
- Find the stationary points by setting dy/dx = 0
- Determine the nature of the stationary points
- Find points of inflection
- Identify asymptotes (if any)
- Determine the end behavior
- Sketch the graph
Example: Sketch the graph of y = x - 3x + 2.
Solution:
Domain: All real numbers
y-intercept: y(0) = 2
x-intercepts: x - 3x + 2 = 0
When x = 1: 1 - 3 + 2 = 0, so x = 1 is a solution
Using polynomial division: (x - 3x + 2)/(x - 1) = x - 2x - 2
Therefore, x - 3x + 2 = (x - 1)(x - 2x - 2)
Setting x - 2x - 2 = 0 and solving using the quadratic formula:
x = 1 3, so x-intercepts are at (1, 0), (1 + 3, 0), and (1 - 3, 0)
Stationary points: dy/dx = 3x - 6x = 3x(x - 2)
Setting dy/dx = 0: 3x(x - 2) = 0 x = 0 or x = 2
These are stationary points at (0, 2) and (2, -2)
Second derivative: dy/dx = 6x - 6
At x = 0: dy/dx = -6 < 0, so (0, 2) is a local maximum
At x = 2: dy/dx = 6 > 0, so (2, -2) is a local minimum
Point of inflection: dy/dx = 0 6x - 6 = 0 x = 1
At x = 1: y = 1 - 3 + 2 = 0, so the point of inflection is (1, 0)
End behavior: As x , y and as x -, y -
Using all this information, we can sketch the graph.
Examination Tips
When approaching differentiation questions in your examination:
- Always show your working clearly
- Start by identifying what the question is asking you to find
- Write down the relevant differentiation rules
- Check your answers by differentiating back (when appropriate)
- For graph sketching, ensure you label all key features
- Be careful with signs, especially when working with trigonometric functions
- Practice different types of problems before the examination
- Manage your time effectively during the examination
Common Mistakes to Avoid
- Forgetting to apply the chain rule when differentiating composite functions
- Misapplying the product and quotient rules
- Errors with trigonometric derivatives, especially signs
- Confusing the notation dy/dx with y'
- Forgetting to find both x and y coordinates when determining stationary points
- Incorrectly evaluating the second derivative to determine the nature of stationary points
- Misidentifying points of inflection
- Forgetting to verify solutions in applied problems
Practice Problems
Regular practice is crucial for mastering differentiation. Try these problems to test your understanding:
- Find the derivative of y = 2x - 5x + 3x - 7
- Differentiate y = sin(2x) + cos(3x)
- Find dy/dx when y = (3x + 5)(2x - 1)
- Find the derivative of y = e^2x ln(x)
- Determine the equation of the tangent to y = x - 4x + 3 at the point where x = 1
- Find and classify the stationary points of y = 2x - 9x + 12x + 1
- A rectangular box with no lid is to be made from a 24 cm by 12 cm sheet of cardboard by cutting equal squares from each corner and folding up the edges. Find the dimensions of the box with maximum volume.
Summary
Differentiation is a fundamental concept in Calculus that provides powerful tools for analyzing functions and solving real-world problems. By understanding and applying the differentiation rules and techniques covered in this resource, you will be well-prepared to tackle differentiation questions in your Grade 12 Mathematics examination.
Remember that consistent practice is key to mastering differentiation. Work through various problems, including past examination papers, and seek help when needed. Good luck with your examination preparation!
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