Mathematics - Additional (0606) | Calculus

Welcome to Calculus: The Math of Change!

Hello future Mathematicians! Calculus might sound scary, but it is one of the most powerful and exciting areas of maths you will learn. At its heart, Calculus is simply the study of how things change—whether it's the speed of a car, the slope of a curve, or the volume of water flowing into a tank.

You won't be given any formulas for Calculus in your exam paper, so learning these rules and concepts well is absolutely crucial. Don't worry, we'll break it down step-by-step!

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SECTION 1: Differentiation (Finding the Rate of Change)

1.1 The Concept of the Derived Function

The core idea of differentiation is finding the instantaneous rate of change of a function. Think of it like a car's speedometer: while average speed is easy to calculate, differentiation tells you the exact speed at a precise moment in time.

The result of differentiating a function \(y = f(x)\) is the derived function or derivative, which gives the gradient (slope) of the curve at any point \(x\).

Key Notations:

• The derivative of \(y\) with respect to \(x\) is written as: \(\frac{dy}{dx}\)
• If the function is \(f(x)\), the derivative is written as: \(f'(x)\)
• The second derivative (differentiating twice) is written as: \(\frac{d^2y}{dx^2}\) or \(f''(x)\)

Did you know? Differentiation is formally based on the idea of a limit, where we look at how the change in \(y\) (\(\delta y\)) over the change in \(x\) (\(\delta x\)) behaves as \(\delta x\) approaches zero (\(\delta x \to 0\)). However, in Add Maths, you only need an informal understanding of this idea—no need to differentiate from first principles!

1.2 Standard Derivatives and Basic Rules (Syllabus 14.3)

These are the building blocks. You must know these rules by heart!

The Power Rule

If \(y = ax^n\), then \(\frac{dy}{dx} = n a x^{n-1}\).

Memory Aid (Power Rule): Bring the power down and reduce the power by one.

Example:

• If \(y = 5x^3\), then \(\frac{dy}{dx} = 3 \times 5x^{3-1} = 15x^2\).
• If \(y = \frac{1}{x} = x^{-1}\), then \(\frac{dy}{dx} = (-1)x^{-2} = -\frac{1}{x^2}\).
• If \(y = 7\) (a constant), then \(\frac{dy}{dx} = 0\).

Standard Functions (MUST be in Radians!)

For trigonometric functions, all angles must be measured in radians.

1. Exponential:
If \(y = e^x\), then \(\frac{dy}{dx} = e^x\). (The easiest one!)

2. Logarithmic:
If \(y = \ln x\), then \(\frac{dy}{dx} = \frac{1}{x}\).

3. Trigonometric:

• If \(y = \sin x\), then \(\frac{dy}{dx} = \cos x\).
• If \(y = \cos x\), then \(\frac{dy}{dx} = -\sin x\).
• If \(y = \tan x\), then \(\frac{dy}{dx} = \sec^2 x\).

The Sum and Constant Multiple Rules

Differentiation works easily across sums and differences, and constants just multiply along:
If \(y = 3x^2 + 5e^x - 2\), then \(\frac{dy}{dx} = 6x + 5e^x - 0\).

SECTION 2: Advanced Differentiation Techniques

2.1 The Chain Rule (Composite Functions)

This rule is used when you have a function inside another function, like \(y = (3x^2 + 4)^5\).

If \(y\) is a function of \(u\), and \(u\) is a function of \(x\), then:
\[ \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \]

Analogy: Peeling an Onion. Differentiate the outside layer first, treat the inside as \(u\), and then multiply by the derivative of the inside layer (\(\frac{du}{dx}\)).

Example: Differentiate \(y = (3x^2 + 4)^5\).

1. Outside (Power Rule): \(5(3x^2 + 4)^4\)
2. Inside (Derivative of \(3x^2 + 4\)): \(6x\)
3. Chain Rule: \(\frac{dy}{dx} = 5(3x^2 + 4)^4 \times (6x) = 30x(3x^2 + 4)^4\)

The Chain Rule is essential for standard functions involving \((ax+b)\):

• If \(y = \sin(2x+1)\), then \(\frac{dy}{dx} = \cos(2x+1) \times 2 = 2\cos(2x+1)\).
• If \(y = e^{4x}\), then \(\frac{dy}{dx} = e^{4x} \times 4 = 4e^{4x}\).

2.2 The Product Rule (Multiplying Functions) (Syllabus 14.4)

If \(y\) is the product of two functions, \(u\) and \(v\), where \(y = uv\), then:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]

Example: Differentiate \(y = x^2 e^x\). Let \(u = x^2\) and \(v = e^x\).

• \(\frac{du}{dx} = 2x\)
• \(\frac{dv}{dx} = e^x\)
• \(\frac{dy}{dx} = (x^2)(e^x) + (e^x)(2x) = x e^x (x + 2)\)

2.3 The Quotient Rule (Dividing Functions) (Syllabus 14.4)

If \(y\) is the quotient of two functions, \(u\) and \(v\), where \(y = \frac{u}{v}\), then:
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

Memory Aid (Quotient Rule): "Low d-high minus high d-low, square the bottom and away we go!" (Where 'low' is \(v\) and 'high' is \(u\)).

Important Note: While you can use the Quotient Rule for division, often it is easier to rewrite the expression and use the Product Rule or Chain Rule instead. For example, rewriting \(\frac{x}{e^x}\) as \(x e^{-x}\) makes it a product rule question.

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SECTION 3: Applications of Differentiation

3.1 Gradients, Tangents, and Normals (Syllabus 14.5)

The Tangent

The derivative, \(\frac{dy}{dx}\), evaluated at a specific point \(x_1\), gives the gradient of the tangent line at that point.

To find the equation of the tangent, use the point-slope form: \(y - y_1 = m_{tan} (x - x_1)\).

The Normal

The normal is the line perpendicular to the tangent at the point of contact.

If the gradient of the tangent is \(m_{tan}\), the gradient of the normal is \(m_{norm}\):
\[ m_{norm} = -\frac{1}{m_{tan}} \]

3.2 Stationary Points (Maxima and Minima) (Syllabus 14.6, 14.8, 14.9)

A stationary point (or turning point) is a point on the curve where the gradient is zero. At these points, the curve is momentarily flat.

To find stationary points:
Step 1: Find the first derivative, \(\frac{dy}{dx}\).
Step 2: Set \(\frac{dy}{dx} = 0\) and solve for \(x\).
Step 3: Substitute the \(x\) values back into the original equation \(y=f(x)\) to find the corresponding \(y\) coordinates.

Note: You are only expected to deal with maxima and minima. Points of inflexion are NOT included in this syllabus.

Distinguishing between Maxima and Minima (Syllabus 14.9)

We need to determine if the stationary point is a peak (maximum) or a valley (minimum). There are two standard tests:

A) The Second Derivative Test (Preferred method)

This uses the second derivative, \(\frac{d^2y}{dx^2}\):

• If \(\frac{d^2y}{dx^2} > 0\) (Positive), it is a Minimum point (happy face, curving up).
• If \(\frac{d^2y}{dx^2} < 0\) (Negative), it is a Maximum point (sad face, curving down).

B) The First Derivative Test (Checking the gradient either side)

If the second derivative test gives zero (or is too complicated), check the sign of \(\frac{dy}{dx}\) just before and just after the stationary point \(x=a\):

• Maximum: Gradient changes from \((+)\) to \((0)\) to \((-)\).
• Minimum: Gradient changes from \((-)\) to \((0)\) to \((+)\).

Justification is Key: In an exam, you must show the values you obtain for \(\frac{d^2y}{dx^2}\) or the changes in \(\frac{dy}{dx}\) to fully justify your conclusion.

3.3 Connected Rates of Change and Small Increments (Syllabus 14.7)

Connected Rates of Change

This is where variables are related by time, \(t\). For example, if we know how fast the volume of a balloon is changing (\(\frac{dV}{dt}\)) and how the volume relates to the radius (\(V\) and \(r\)), we can find how fast the radius is changing (\(\frac{dr}{dt}\)).

We use the Chain Rule, usually in terms of time \(t\):
\[ \frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt} \]

Small Increments and Approximations

When \(x\) changes by a small amount \(\delta x\), the resulting small change in \(y\), \(\delta y\), can be approximated using the derivative:
\[ \delta y \approx \frac{dy}{dx} \times \delta x \]

This formula basically says: Small change in \(y\) is approximately equal to the gradient multiplied by the small change in \(x\).

To find the new value of \(y\): \(y_{new} \approx y_{original} + \delta y\).

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SECTION 4: Integration (The Reverse Process)

4.1 Indefinite Integration (The Anti-Derivative) (Syllabus 14.10)

Integration is the reverse process of differentiation. If differentiation gives us the rate of change, integration finds the original function.

When you integrate an expression, you perform indefinite integration.

The Rule for Integration (The Reverse Power Rule)

If you integrate \(x^n\):
\[ \int x^n dx = \frac{x^{n+1}}{n+1} + C, \quad \text{where } n \neq -1 \]

Memory Aid: Increase the power by one, then divide by the new power.

The Arbitrary Constant, \(+C\): Since the derivative of any constant is zero, when we reverse differentiation, we lose information about the original constant. Therefore, you must always include the arbitrary constant \(+C\) for indefinite integrals.

4.2 Standard Integrals (Syllabus 14.11, 14.12)

These are the reverse rules of our standard derivatives, usually generalized for the form \((ax+b)\). Remember, if you integrate a function of \((ax+b)\), you must divide by the derivative of the inside, which is \(a\).

I. Power Rule (General Form)

\[ \int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C, \quad \text{for rational } n \neq -1 \]

II. The Special Case \(n = -1\)

If \(n=-1\), we cannot divide by zero. Since \(\frac{d}{dx} (\ln x) = \frac{1}{x}\), the integral is:
\[ \int \frac{1}{x} dx = \ln|x| + C \]
\[ \int \frac{1}{ax+b} dx = \frac{1}{a} \ln|ax+b| + C \]

III. Exponentials and Trigonometry (Angles always in Radians!)

• \(\int e^{ax+b} dx = \frac{1}{a} e^{ax+b} + C\)
• \(\int \sin(ax+b) dx = -\frac{1}{a} \cos(ax+b) + C\)
• \(\int \cos(ax+b) dx = \frac{1}{a} \sin(ax+b) + C\)
• \(\int \sec^2(ax+b) dx = \frac{1}{a} \tan(ax+b) + C\)

4.3 Definite Integrals and Area (Syllabus 14.13)

A definite integral has upper and lower limits (\(a\) and \(b\)). It represents the net area under the curve \(y=f(x)\) between those limits.

\[ \int_{a}^{b} f(x) dx = [F(x)]_a^b = F(b) - F(a) \]

(Where \(F(x)\) is the integrated form of \(f(x)\)).

Evaluating Plane Areas

The definite integral calculates area. You must be very careful if the curve goes below the x-axis, as the integral result will be negative. Area must always be positive!

Area between a curve and the x-axis:

• If the curve is entirely above the x-axis, Area = \(\int_{a}^{b} y dx\).
• If the curve is entirely below the x-axis, Area = \(|\int_{a}^{b} y dx|\) (take the positive value).
• If the curve crosses the x-axis, you must find the x-intercepts, split the integral into separate regions, and treat the negative areas as positive before summing them up.

Area between two curves:
Area = \(\int_{a}^{b} (y_{upper} - y_{lower}) dx\)
You must first find the points of intersection (\(a\) and \(b\)) by setting the two equations equal to each other.

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SECTION 5: Calculus in Kinematics (Motion)

Kinematics applies calculus to the motion of objects moving in a straight line. The relationships between displacement (\(s\)), velocity (\(v\)), and acceleration (\(a\)) are fundamental.

5.1 Differentiation in Kinematics (Syllabus 14.14)

Velocity is the rate of change of displacement. Acceleration is the rate of change of velocity.

1. Displacement (\(s\)) \(\to\) Velocity (\(v\))
   \[ v = \frac{ds}{dt} \]
2. Velocity (\(v\)) \(\to\) Acceleration (\(a\))
   \[ a = \frac{dv}{dt} \]
3. Therefore: \(a = \frac{d^2s}{dt^2}\)

Example: If displacement \(s = 3t^3 - 10t^2 + 4t + 8\):
Velocity \(v = 9t^2 - 20t + 4\)
Acceleration \(a = 18t - 20\)

Important Distinction:

• Displacement (\(s\)): Position relative to a fixed origin (can be positive or negative).
• Distance: Total length travelled (always positive). To find distance, you must identify where the velocity is zero (where the object changes direction) and sum the magnitude of the displacement in each segment.
• Speed: Magnitude of velocity (always positive).

5.2 Integration in Kinematics (Syllabus 14.14)

Integration is used to reverse the process, finding the original functions for speed or displacement.

1. Acceleration (\(a\)) \(\to\) Velocity (\(v\))
   \[ v = \int a dt \]
2. Velocity (\(v\)) \(\to\) Displacement (\(s\))
   \[ s = \int v dt \]

When integrating, you will need to use the arbitrary constant \(+C\). To find the value of \(C\), you must be given initial conditions (e.g., \(v=5\) when \(t=0\), or \(s=2\) when \(t=1\)).

5.3 Kinematic Graphs (Syllabus 14.15)

The key relationships derived from calculus also apply to the graphs:

1. Velocity-Time Graph:

• The gradient gives the acceleration (\(\frac{dv}{dt}\)).
• The area under the graph gives the displacement (\(\int v dt\)).

2. Displacement-Time Graph:

• The gradient gives the velocity (\(\frac{ds}{dt}\)).

3. Acceleration-Time Graph:

• The area under the graph gives the change in velocity (\(\int a dt\)).