HL Applications & Interpretation Revision Notes

Worked Examples: Error & Bounds Example 1: Calculating Percentage Error
The exact area of a rectangle is \(125.4 \text{ cm}^2\). A student measures it and estimates the area to be \(120 \text{ cm}^2\). Calculate the percentage error.
\(v_A = 125.4\) and \(v_E = 120\).
\(\epsilon = \left| \frac{125.4 - 120}{120} \right| \times 100\% = \left| \frac{5.4}{120} \right| \times 100\% = 0.045 \times 100\% = 4.5\%\).
Example 2: Establishing Bounds
The length of a fence is given as \(45\) m, correct to the nearest metre. Write down the upper and lower bounds.
Since it is to the nearest \(1\) m, the maximum error is \(\pm 0.5\) m.
Lower Bound = \(44.5\) m, Upper Bound = \(45.5\) m. (\(44.5 \le L < 45.5\)).

Worked Examples: Advanced Exponents Example 1: Numerical Evaluation
Evaluate \(27^{\frac{2}{3}}\) without a calculator.
\(27^{\frac{2}{3}} = (\sqrt[1]{27})^2 = (3)^2 = 9\).
Example 2: Solving Exponential Equations
Solve the equation \(x^{-\frac{1}{2}} = \frac{1}{5}\) for \(x\).
Using the negative exponent rule: \(\frac{1}{x^{1/2}} = \frac{1}{5} \implies \frac{1}{\sqrt{x}} = \frac{1}{5}\).
Therefore, \(\sqrt{x} = 5\). Squaring both sides gives \(x = 25\).

Worked Examples: Logarithm Laws Example 1: Expanding and Condensing
Express \(\ln\left(\frac{\sqrt{x}}{y^3}\right)\) in terms of \(\ln x\) and \(\ln y\).
Using the Quotient Law: \(\ln(\sqrt{x}) - \ln(y^3)\).
Using rational exponents and the Power Law: \(\frac{1}{2}\ln x - 3\ln y\).
Example 2: Solving Exponential Equations with Logarithms
Solve the equation \(5 \times 3^x = 40\) exactly.
Isolate the base: \(3^x = 8\).
Apply the natural logarithm to both sides: \(\ln(3^x) = \ln(8)\).
Use the Power Law: \(x \ln(3) = \ln(8) \implies x = \frac{\ln(8)}{\ln(3)}\).

Worked Examples: Finite Sequences Example 1: Arithmetic Sum
An arithmetic sequence has \(u_1 = 4\) and a common difference \(d = 5\). Find the sum of the first \(15\) terms.
Using the sum formula: \(S_{15} = \frac{15}{2}(2(4) + (15-1)(5))\).
\(S_{15} = 7.5(8 + 14(5)) = 7.5(8 + 70) = 7.5(78) = 585\).
Example 2: Geometric Term
A geometric sequence has \(u_1 = 3\) and \(u_2 = 12\). Find the 5th term (\(u_5\)).
Find \(r\): \(r = \frac{12}{3} = 4\).
Find \(u_5\): \(u_5 = 3(4)^{5-1} = 3(4)^4 = 3(256) = 768\).

Worked Examples: Infinite Series Example 1: Finding the Sum to Infinity
An infinite geometric series is given by \(18 + 6 + 2 + \frac{2}{3} + \dots\) Find its exact sum.
Find the common ratio (\(r\)): \(r = \frac{6}{18} = \frac{1}{3}\).
Check convergence: Since \(|1/3| < 1\), the series converges.
Calculate \(S_\infty\): \(S_\infty = \frac{18}{1 - 1/3} = \frac{18}{2/3} = 18 \times \frac{3}{2} = 27\).
Example 2: Evaluating Sigma Notation to Infinity
Evaluate exactly: \(\sum_{k=1}^{\infty} 5(0.2)^{k-1}\)
This is an infinite geometric series. The first term (\(k=1\)) is \(u_1 = 5(0.2)^0 = 5\).
The common ratio is the base of the exponent: \(r = 0.2\).
Since \(|0.2| < 1\), \(S_\infty = \frac{5}{1 - 0.2} = \frac{5}{0.8} = 6.25\).

Worked Examples: Financial Mathematics Example 1: Basic Compound Interest
$5000 is invested at \(4\%\) p.a., compounded quarterly. Find the value after 5 years.
Using the formula: \(FV = 5000 \times \left(1 + \frac{4}{100 \times 4}\right)^{4 \times 5} = 5000(1.01)^{20} \approx \$6100.95\).
Using GDC: N=20, I%=4, PV=-5000, PMT=0, P/Y=4, C/Y=4 \(\rightarrow\) Solve FV.
Example 2: Loan Repayment (Amortization)
A $20,000 car loan is charged \(6\%\) p.a. compounded monthly. The buyer wants to pay it off in 3 years with monthly payments. Calculate the monthly payment.
Using GDC: N=36 (months), I%=6, PV=20000 (loan received), FV=0 (paid off), P/Y=12, C/Y=12. Solve for PMT.
PMT \(\approx -608.44\). The monthly payment is $608.44.

Worked Examples: Linear Systems Example 1: Solving a 3x3 System
Solve the system:
\(x + 2y - z = 5\)
\(3x - y + 2z = 8\)
\(2x + y + z = 7\)
Using GDC: MENU \(\rightarrow\) Equation \(\rightarrow\) Simultaneous (F1) \(\rightarrow\) Unknowns: 3 (F2).
Enter the matrix coefficients: \([1, 2, -1, 5]\), \([3, -1, 2, 8]\), \([2-4]\).
Solve: \(x = 1, y = 3, z = 2\).

Worked Examples: Complex Forms and Conversions Example 1: Cartesian to Polar/Euler Conversion
Convert \(z = 1 + i\) into Euler form.
Find \(r\): \(r = \sqrt{1^2 + 1^2} = \sqrt{2}\).
Find \(\theta\): \(\tan\theta = \frac{1}{1} = 1 \implies \theta = \frac{\pi}{4}\).
Write in Euler form: \(z = \sqrt{2}e^{i\frac{\pi}{4}}\).
Example 2: Polar to Cartesian Conversion
Convert \(w = 4\operatorname{cis}\left(\frac{5\pi}{6}\right)\) into Cartesian form \(a+bi\).
Expand using sine and cosine: \(w = 4\left( \cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6} \right)\).
Substitute exact trig values: \(w = 4\left( -\frac{\sqrt{3}}{2} + i\frac{1}{2} \right) = -2\sqrt{3} + 2i\).

Worked Examples: Exponential Arithmetic Example 1: Multiplying and Dividing
Given \(z_1 = 4e^{i\frac{\pi}{2}}\) and \(z_2 = 2e^{i\frac{\pi}{6}}\), find \(z_1 z_2\) and \(\frac{z_1}{z_2}\).
Product: \(z_1 z_2 = (4 \times 2) e^{i(\frac{\pi}{2} + \frac{\pi}{6})} = 8e^{i\frac{4\pi}{6}} = 8e^{i\frac{2\pi}{3}}\).
Quotient: \(\frac{z_1}{z_2} = \left(\frac{4}{2}\right) e^{i(\frac{\pi}{2} - \frac{\pi}{6})} = 2e^{i\frac{2\pi}{6}} = 2e^{i\frac{\pi}{3}}\).

Worked Examples: Matrix Algebra Example 1: Addition and Scalar Multiplication
Let \(A = \begin{pmatrix} 4 & 1 \\ 2 & 0 \end{pmatrix}\) and \(B = \begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix}\). Find \(2A - B\).
Scalar multiply \(A\): \(2A = \begin{pmatrix} 8 & 2 \\ 4 & 0 \end{pmatrix}\).
Subtract \(B\) element-by-element: \(\begin{pmatrix} 8-1 & 2-(-2) \\ 4-3 & 0-5 \end{pmatrix} = \begin{pmatrix} 7 & 4 \\ 1 & -5 \end{pmatrix}\).
Example 2: Matrix Multiplication
Let \(A = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}\) and \(B = \begin{pmatrix} 5 \\ -1 \end{pmatrix}\). Find \(AB\).
Top Row \(\times\) Column: \((2 \times 5) + (3 \times -1) = 10 - 3 = 7\).
Bottom Row \(\times\) Column: \((1 \times 5) + (4 \times -1) = 5 - 4 = 1\).
Resulting Matrix: \(AB = \begin{pmatrix} 7 \\ 1 \end{pmatrix}\).

Worked Examples: Eigenvalues & Diagonalization Example 1: Finding Eigenvalues
Find the exact eigenvalues of \(A = \begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix}\).
Set up characteristic equation: \(\begin{vmatrix} 4-\lambda & 2 \\ 1 & 3-\lambda \end{vmatrix} = 0\).
Calculate determinant: \((4-\lambda)(3-\lambda) - (2 \times 1) = 0\).
Expand and Factorize: \(\lambda^2 - 7\lambda + 10 = 0 \implies (\lambda - 5)(\lambda - 2) = 0\).
The exact eigenvalues are \(\lambda_1 = 5\) and \(\lambda_2 = 2\).
Example 2: Finding Eigenvectors and Diagonalizing
Find the eigenvector for \(\lambda_1 = 5\) and construct \(P\) and \(D\). (Assume \(\lambda_2=2\) gives \(\vec{v}_2 = \begin{pmatrix} -1 \\ 1 \end{pmatrix}\)).
Substitute \(\lambda = 5\): \(\begin{pmatrix} 4-5 & 2 \\ 1 & 3-5 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies \begin{pmatrix} -1 & 2 \\ 1 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\).
Create equation from top row: \(-1x + 2y = 0 \implies x = 2y\). Let \(y = 1 \implies x = 2\).
Eigenvector \(\vec{v}_1 = \begin{pmatrix} 2 \\ 1 \end{pmatrix}\).
Construct \(P\) and \(D\): \(P = \begin{pmatrix} 2 & -1 \\ 1 & 1 \end{pmatrix}\), \(D = \begin{pmatrix} 5 & 0 \\ 0 & 2 \end{pmatrix}\).
Diagonalized Form: \(A = \begin{pmatrix} 2 & -1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 5 & 0 \\ 0 & 2 \end{pmatrix} P^{-1}\).

Worked Examples: Linear Equations Example 1: Finding an Equation from Two Points
A straight line passes through \(A(2, 5)\) and \(B(6, 17)\). Find its equation in the form \(y = mx + c\).
Find the gradient \(m\): \(m = \frac{17 - 5}{6 - 2} = \frac{12}{4} = 3\).
Use point-gradient form with point \(A\): \(y - 5 = 3(x - 2)\).
Expand and solve for \(y\): \(y - 5 = 3x - 6 \implies y = 3x - 1\).
Example 2: Perpendicular Lines
Line \(L_1\) has equation \(y = -\frac{1}{4}x + 2\). Line \(L_2\) is perpendicular to \(L_1\) and passes through \((1, 3)\). Find the equation of \(L_2\) in general form \(ax + by + d = 0\).
The gradient of \(L_1\) is \(-\frac{1}{4}\). The perpendicular gradient is the negative reciprocal: \(m_2 = 4\).
Use point \((1, 3)\): \(y - 3 = 4(x - 1) \implies y - 3 = 4x - 4 \implies y = 4x - 1\).
Rearrange to general form (integers): \(4x - y - 1 = 0\).

Worked Examples: Domain and Range Restrictions Example 1: Denominator Restriction
Find the maximal domain of \(f(x) = \frac{3}{2x - 8}\).
The denominator cannot be zero: \(2x - 8 \neq 0 \implies 2x \neq 8 \implies x \neq 4\).
Domain: \(x \in \mathbb{R}, x \neq 4\).
Example 2: Square Root Restriction
Find the domain of \(g(x) = \sqrt{10 - 2x}\).
The inside of the root must be \(\ge 0\): \(10 - 2x \ge 0 \implies 10 \ge 2x \implies 5 \ge x\).
Domain: \(x \le 5\).
Example 3: Finding Range Algebraically
Find the range of the quadratic function \(h(x) = x^2 - 4x + 7\).
Find the \(x\)-coordinate of the vertex: \(x = -\frac{b}{2a} = -\frac{-4}{2(1)} = 2\).
Find the \(y\)-coordinate: \(h(2) = (2)^2 - 4(2) + 7 = 4 - 8 + 7 = 3\).
Since \(a > 0\), the parabola opens upwards, making the vertex a minimum.
Range: \(y \ge 3\).

Worked Examples: Composite Functions Example 1: Algebraic Composition
Let \(f(x) = 2x - 3\) and \(g(x) = x^2 + 1\). Find an expression for \((f \circ g)(x)\).
\((f \circ g)(x) = f(g(x)) = f(x^2 + 1)\).
Substitute \((x^2+1)\) into \(f(x)\): \(2(x^2 + 1) - 3 = 2x^2 + 2 - 3 = 2x^2 - 1\).
Example 2: Evaluating Composites
Using the same functions, evaluate \((g \circ f)(4)\).
First find \(f(4)\): \(f(4) = 2(4) - 3 = 8 - 3 = 5\).
Substitute the result into \(g(x)\): \(g(5) = 5^2 + 1 = 25 + 1 = 26\).

Worked Examples: Inverse Functions Example 1: Finding an Inverse Function Algebraically
Let \(f(x) = \frac{3x + 1}{x - 2}\). Find \(f^{-1}(x)\).
Step 1: Set \(y = f(x) \implies y = \frac{3x + 1}{x - 2}\).
Step 2: Swap \(x\) and \(y\): \(x = \frac{3y + 1}{y - 2}\).
Step 3: Multiply out the denominator: \(x(y - 2) = 3y + 1 \implies xy - 2x = 3y + 1\).
Step 4: Collect all \(y\) terms on one side: \(xy - 3y = 2x + 1\).
Step 5: Factor out \(y\): \(y(x - 3) = 2x + 1\).
Step 6: Divide to isolate \(y\): \(y = \frac{2x + 1}{x - 3}\).
Answer: \(f^{-1}(x) = \frac{2x + 1}{x - 3}\).
Example 2: Restricting the Domain
Explain why \(h(x) = x^2 - 6x\) does not have an inverse on \(x \in \mathbb{R}\), and state a restricted domain that would allow an inverse to exist.
\(h(x)\) is a parabola, which fails the horizontal line test (it is a many-to-one function).
To make it one-to-one, we restrict it to one half of the parabola by finding the vertex:
Vertex \(x = -b/2a = -(-6)/2 = 3\).
A valid restricted domain is \(x \ge 3\) (or \(x \le 3\)).

Worked Examples: Graph Transformations Example 1: Identifying Transformations
Describe the full sequence of geometric transformations that maps \(f(x) = x^2\) onto the graph \(g(x) = -3(x - 4)^2 + 5\).
1. Translation: \(4\) units RIGHT and \(5\) units UP.
2. Stretch: A vertical stretch by a scale factor of \(3\).
3. Reflection: A reflection in the \(x\)-axis (due to the negative sign in front of the \(3\)).
Example 2: Composite Coordinate Shifts
The point \((2, 6)\) lies on the curve \(y = f(x)\). Find the coordinates of the corresponding point on the curve \(y = f(2x) - 1\).
The \(f(2x)\) implies a horizontal stretch by scale factor \(\frac{1}{2}\). The \(x\)-coordinate halves: \(2 \times \frac{1}{2} = 1\).
The \(-1\) implies a vertical translation DOWN \(1\). The \(y\)-coordinate drops: \(6 - 1 = 5\).
New coordinate: \((1, 5)\).

Worked Examples: Logistic Modelling Example 1: Interpreting the Logistic Formula
The population of fish in a lake \(t\) years after introduction is modelled by \(P(t) = \frac{5000}{1 + 24e^{-0.3t}}\).
(a) Find the initial population.
Substitute \(t = 0\): \(P(0) = \frac{5000}{1 + 24e^0} = \frac{5000}{1 + 24(1)} = \frac{5000}{25} = 200\) fish.
(b) State the carrying capacity.
The numerator dictates the maximum limit. As \(t \to \infty\), the term \(e^{-0.3t} \to 0\), leaving \(\frac{5000}{1+0}\).
The carrying capacity is exactly \(5000\) fish.

Worked Examples: Piecewise Functions Example 1: Evaluating Piecewise Boundaries
A taxi fare \(C(x)\) for travelling \(x\) kilometres is given by:
\(C(x) = \begin{cases} 5 & \text{for } 0 \le x \le 2 \\ 5 + 1.5(x - 2) & \text{for } x > 2 \end{cases}\)
Calculate the cost of a \(10\) km journey.
Because \(10 > 2\), we must use the second equation in the piecewise definition:
\(C(10) = 5 + 1.5(10 - 2) = 5 + 1.5(8) = 5 + 12 = \$17\).

Worked Examples: Linearising Data Example 1: Extracting Exponential Parameters from a Linear Graph
The mass of bacteria \(M\) after \(t\) hours is modelled by \(M = a \times b^t\).
A scientist plots a graph of \(\ln M\) against \(t\). The graph forms a perfect straight line with a \(y\)-intercept of \(4\) and a gradient of \(1.5\). Find the exact values of the constants \(a\) and \(b\).
Step 1: Write out the linearised formula.
\(\ln M = \ln a + t\ln b\).
Step 2: Match the graph features to the formula.
The \(y\)-intercept is \(\ln a\). Therefore: \(\ln a = 4 \implies a = e^4\).
The gradient is \(\ln b\). Therefore: \(\ln b = 1.5 \implies b = e^{1.5}\).
Step 3: State the final model.
\(M = e^4 \times (e^{1.5})^t = e^4 \times e^{1.5t}\).
Example 2: Formulating a Power Model
Variables \(P\) and \(Q\) obey a power law \(P = k Q^n\). When \(\log_{10} P\) is plotted on the vertical axis against \(\log_{10} Q\) on the horizontal axis, the line passes through \((2, 7)\) and \((5, 13)\). Find \(n\).
Step 1: Linearise.
\(\log P = \log k + n \log Q\).
Here, the gradient of the line is exactly \(n\).
Step 2: Calculate the gradient.
\(n = m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{13 - 7}{5 - 2} = \frac{6}{3} = 2\).
The power \(n\) is exactly \(2\).

Worked Examples: 3D Coordinates Example 1: Finding 3D Distance and Midpoint
Given points \(A(1, 2, 3)\) and \(B(4, -2, 15)\), find the exact distance \(AB\) and the midpoint \(M\).
Distance: \(AB = \sqrt{(4 - 1)^2 + (-2 - 2)^2 + (15 - 3)^2} = \sqrt{3^2 + (-4)^2 + 12^2}\).
\(AB = \sqrt{9 + 16 + 144} = \sqrt{169} = 13\) units.
Midpoint: \(M = \left( \frac{1 + 4}{2}, \frac{2 - 2}{2}, \frac{3 + 15}{2} \right) = (2.5, 0, 9)\).

Worked Examples: SOH CAH TOA Example 1: Angle of Elevation
A person stands \(50\) m from the base of a building. The angle of elevation to the top of the building is \(31^\circ\). Calculate the height of the building.
We have the Adjacent (\(50\)) and want the Opposite (\(h\)). Use Tangent:
\(\tan(31^\circ) = \frac{h}{50} \implies h = 50 \tan(31^\circ) \approx 30.0\) m.

Worked Examples: Advanced Triangle Rules Example 1: Finding an Angle with the Cosine Rule
A triangle has sides \(a=5\), \(b=7\), and \(c=8\). Find the largest angle.
The largest angle (\(C\)) is opposite the largest side (\(c=8\)).
Rearranged Cosine Rule: \(\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{5^2 + 7^2 - 8^2}{2(5)(7)} = \frac{25 + 49 - 64}{70} = \frac{10}{70} = \frac{1}{7}\).
\(C = \cos^{-1}\left(\frac{1}{7}\right) \approx 81.8^\circ\).
Example 2: Finding Area
Using the triangle above, calculate its exact area using \(\sin C\).
Since \(\cos C = \frac{1}{7}\), we use \(\sin^2 C + \cos^2 C = 1 \implies \sin C = \sqrt{1 - (1/7)^2} = \sqrt{\frac{48}{49}} = \frac{4\sqrt{3}}{7}\).
Area \(= \frac{1}{2}ab \sin C = \frac{1}{2}(5)(7)\left(\frac{4\sqrt{3}}{7}\right) = 10\sqrt{3}\).

Worked Examples: Radians Example 1: Conversions and Sector Area
A circle has a radius of \(12\) cm. A sector is formed by an angle of \(150^\circ\).
(a) Convert the angle to radians exactly.
\(150^\circ \times \frac{\pi}{180^\circ} = \frac{150\pi}{180} = \frac{5\pi}{6}\) radians.
(b) Calculate the exact area of the sector.
\(A = \frac{1}{2}r^2\theta = \frac{1}{2}(12^2)\left(\frac{5\pi}{6}\right) = \frac{1}{2}(144)\left(\frac{5\pi}{6}\right) = 72 \left(\frac{5\pi}{6}\right) = 60\pi \text{ cm}^2\).

Worked Examples: The Unit Circle Example 1: Using the Pythagorean Identity
Given that \(\sin \theta = \frac{3}{5}\) and \(\frac{\pi}{2} < \theta < \pi\) (obtuse angle), find the exact value of \(\cos \theta\).
Use \(\cos^2 \theta + \sin^2 \theta = 1 \implies \cos^2 \theta + \left(\frac{3}{5}\right)^2 = 1 \implies \cos^2 \theta = 1 - \frac{9}{25} = \frac{16}{25}\).
\(\cos \theta = \pm \sqrt{\frac{16}{25}} = \pm \frac{4}{5}\).
Because \(\theta\) is in the second quadrant (\(\frac{\pi}{2} < \theta < \pi\)), cosine is negative. \(\cos \theta = -\frac{4}{5}\).

Worked Examples: Trig Functions Example 1: Extracting Graph Parameters
The height of water in a harbour is given by \(h(t) = 4.2 \sin\left(\frac{\pi}{6}t\right) + 6.8\) metres, where \(t\) is hours after midnight.
(a) Find the maximum and minimum heights.
Max \(= d + a = 6.8 + 4.2 = 11.0\) m.
Min \(= d - a = 6.8 - 4.2 = 2.6\) m.
(b) Find the period of the tides.
\(b = \frac{\pi}{6}\). Period \(= \frac{2\pi}{b} = \frac{2\pi}{\pi/6} = 12\) hours.

Worked Examples: Voronoi Boundaries Example 1: Finding the Equation of a Boundary
Sites \(A(2, 5)\) and \(B(6, 1)\) dictate two adjacent Voronoi regions. Find the exact equation of the boundary separating them.
1. Midpoint: \(M = \left(\frac{2+6}{2}, \frac{5+1}{2}\right) = (4, 3)\).
2. Gradient \(AB\): \(m = \frac{1 - 5}{6 - 2} = \frac{-4}{4} = -1\).
3. Perpendicular Gradient: \(m_\perp = \frac{-1}{-1} = 1\).
4. Equation: \(y - 3 = 1(x - 4) \implies y = x - 1\).

Worked Examples: Vector Mathematics Example 1: Dot Product and Angles
Let \(p = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}\) and \(q = \begin{pmatrix} 4 \\ 5 \\ -1 \end{pmatrix}\). Find the angle between \(p\) and \(q\).
1. Dot Product: \(p \cdot q = (2)(4) + (-1)(5) + (3)(-1) = 8 - 5 - 3 = 0\).
2. Conclusion: Since \(p \cdot q = 0\), \(\cos \theta = 0\). The angle is exactly \(90^\circ\) (perpendicular).
Example 2: Unit Vectors
Find a unit vector in the same direction as \(v = \begin{pmatrix} 3 \\ 0 \\ -4 \end{pmatrix}\).
Magnitude \(|v| = \sqrt{3^2 + 0^2 + (-4)^2} = \sqrt{9 + 16} = 5\).
Unit Vector \(= \frac{1}{5} \begin{pmatrix} 3 \\ 0 \\ -4 \end{pmatrix} = \begin{pmatrix} 0.6 \\ 0 \\ -0.8 \end{pmatrix}\).

Worked Examples: Vector Lines Example 1: Intersecting Lines
Line \(L_1\) has equation \(r_1 = \begin{pmatrix} 2 \\ 5 \end{pmatrix} + t \begin{pmatrix} 1 \\ -2 \end{pmatrix}\). Line \(L_2\) has equation \(r_2 = \begin{pmatrix} -4 \\ 2 \end{pmatrix} + s \begin{pmatrix} 3 \\ 1 \end{pmatrix}\). Find their point of intersection.
Equate the \(x\) and \(y\) components to form simultaneous equations:
\(x\)-eq: \(2 + t = -4 + 3s \implies t - 3s = -6\)
\(y\)-eq: \(5 - 2t = 2 + s \implies -2t - s = -3\)
Solve via GDC or substitution. Let’s substitute \(t = 3s - 6\) into the \(y\)-equation:
\(-2(3s - 6) - s = -3 \implies -6s + 12 - s = -3 \implies -7s = -15 \implies s = \frac{15}{7}\).
Substitute \(s\) into \(r_2\) to find the intersection coordinate:
\(x = -4 + 3(15/7) = 17/7\), and \(y = 2 + (15/7) = 29/7\). Point is \((17/7, 29/7)\).

Worked Examples: Kinematics Example 1: Position, Velocity, and Speed
A ship leaves a port at \(O(0, 0)\) and moves with a constant velocity vector \(v = \begin{pmatrix} 12 \\ -5 \end{pmatrix}\) km/h.
(a) Find the speed of the ship.
Speed \(= |v| = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13\) km/h.
(b) Find the position vector of the ship after 3 hours.
\(r(3) = r_0 + 3v = \begin{pmatrix} 0 \\ 0 \end{pmatrix} + 3 \begin{pmatrix} 12 \\ -5 \end{pmatrix} = \begin{pmatrix} 36 \\ -15 \end{pmatrix}\).
Example 2: Closest Distance Between Two Moving Objects
If Object A is at \(r_A(t) = \begin{pmatrix} 0 \\ 10 \end{pmatrix} + t\begin{pmatrix} 4 \\ 0 \end{pmatrix}\) and Object B is at \(r_B(t) = \begin{pmatrix} 20 \\ 0 \end{pmatrix} + t\begin{pmatrix} 0 \\ 5 \end{pmatrix}\), find the distance between them at \(t=2\).
Find position of A at \(t=2\): \(r_A(2) = (8, 10)\).
Find position of B at \(t=2\): \(r_B(2) = (20, 10)\).
Distance \(= \sqrt{(20 - 8)^2 + (10 - 10)^2} = \sqrt{12^2 + 0} = 12\) units.

Worked Examples: Outliers & Sampling Example 1: Identifying Outliers
A dataset of test scores has a lower quartile of \(45\) and an upper quartile of \(71\). A student scored \(105\). Is this score a mathematical outlier?
1. Find the IQR: \(IQR = Q_3 - Q_1 = 71 - 45 = 26\).
2. Calculate Upper Boundary: \(Q_3 + 1.5(IQR) = 71 + 1.5(26) = 71 + 39 = 110\).
3. Conclusion: Since \(105 < 110\), the score is strictly within the boundary and is not an outlier.
Example 2: Stratified Sampling
A school has \(400\) juniors and \(600\) seniors. A sample of \(50\) students is needed. How many juniors should be sampled using stratified sampling?
1. Total Population: \(400 + 600 = 1000\).
2. Junior Proportion: \(\frac{400}{1000} \times 50 = 20\) juniors.

Worked Examples: Advanced Probability Example 1: Using Conditional Probability Formulas
Given that \(P(A) = 0.6\), \(P(B) = 0.5\), and \(P(A \cup B) = 0.8\), find \(P(A|B)\).
1. Find \(P(A \cap B)\): Use \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).
\(0.8 = 0.6 + 0.5 - P(A \cap B) \implies P(A \cap B) = 1.1 - 0.8 = 0.3\).
2. Find Conditional: \(P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.3}{0.5} = 0.6\).
*(Note: Because \(P(A|B) = P(A)\), events A and B are mathematically independent!)*

Worked Examples: The Binomial Distribution Example 1: Binomial Probabilities and Expected Value
A biased coin has a \(0.6\) probability of landing on heads. It is flipped \(20\) times.
(a) Find the probability of getting exactly \(14\) heads.
Let \(X \sim B(20, 0.6)\). We need \(P(X = 14)\).
Using GDC (Bpd): \(x=14, \text{Numtrial}=20, p=0.6 \implies P(X=14) \approx 0.124\).
(b) Find the expected number of heads and the variance.
\(E(X) = np = 20 \times 0.6 = 12\).
\(Var(X) = np(1-p) = 20 \times 0.6 \times 0.4 = 4.8\).

Worked Examples: Linear Transformations Example 1: Transforming Mean and Variance
A factory produces bags of flour. The weight of a bag, \(X\), has an expected value \(E(X) = 500\) g and a variance \(Var(X) = 16\) g\(^2\). The factory changes the packaging process. The new weight \(Y\) is modelled by the transformation \(Y = 1.2X - 5\).
(a) Find the new expected weight, \(E(Y)\).
\(E(Y) = E(1.2X - 5) = 1.2E(X) - 5\).
\(E(Y) = 1.2(500) - 5 = 600 - 5 = 595\) g.
(b) Find the new variance and standard deviation of \(Y\).
\(Var(Y) = Var(1.2X - 5) = (1.2)^2 Var(X)\). (The \(-5\) is ignored for spread!)
\(Var(Y) = 1.44 \times 16 = 23.04\) g\(^2\).
The standard deviation is \(\sqrt{23.04} = 4.8\) g.

Worked Examples: Regression Analysis Example 1: Spearman’s Rank Calculations
Two judges rank 5 competitors (A to E). Judge 1 ranks them: 1, 2, 3, 4, 5. Judge 2 ranks them: 2, 1, 4, 3, 5. Find Spearman’s Rank manually.
1. Calculate differences squared (\(d^2\)):
A: \(1-2 = -1 \implies d^2 = 1\)
B: \(2-1 = 1 \implies d^2 = 1\)
C: \(3-4 = -1 \implies d^2 = 1\)
D: \(4-3 = 1 \implies d^2 = 1\)
E: \(5-5 = 0 \implies d^2 = 0\)
Total \(\sum d^2 = 4\).
2. Formula: \(r_s = 1 - \frac{6(4)}{5(25 - 1)} = 1 - \frac{24}{120} = 1 - 0.2 = 0.8\).
Example 2: Evaluating \(SS_{res}\)
A quadratic model is proposed for a dataset yielding an \(SS_{res}\) of 2.45. An exponential model is tested on the exact same dataset yielding an \(SS_{res}\) of 0.82. Which model is a better fit?
The exponential model is a better fit because its Sum of Square Residuals is much smaller, indicating the predicted curve passes much closer to the true data points.

Worked Examples: Distributions Example 1: Normal Probabilities
Bicycle stopping distances are modelled by \(N(6.76, 0.12^2)\) metres. Find the probability a randomly chosen bicycle stops in less than \(6.5\) m.
Using GDC (Norm CD): Lower = \(-1\times 10^{99}\), Upper = \(6.5\), \(\sigma=0.12\), \(\mu=6.76\).
\(P(X < 6.5) \approx 0.0151\).
Example 2: Poisson Distribution Sums
The number of emails received per hour follows \(X \sim Po(4)\). The number of texts follows \(Y \sim Po(6)\). Find the probability of receiving exactly \(12\) messages in total in one hour.
Let \(T = X + Y\). Therefore \(T \sim Po(4 + 6) \implies T \sim Po(10)\).
We need \(P(T = 12)\). Using GDC (Poisson PD): \(\lambda = 10\), \(x=12\).
\(P(T = 12) \approx 0.0948\).

Worked Examples: Hypothesis Conclusions Example 1: Interpreting Test Output
A \(\chi^2\) test for independence evaluates gender and career choice at a \(5\%\) significance level. The GDC gives \(\chi^2 = 8.54\) and \(p = 0.0736\). The critical value is \(9.49\). State the conclusion.
Method 1 (\(p\)-value): Since \(p = 0.0736 > 0.05\), there is insufficient evidence to reject \(H_0\).
Method 2 (Critical Value): Since the test statistic \(8.54\) is less than the critical value \(9.49\), it is not in the critical region. Do not reject \(H_0\).

Worked Examples: Markov Chains Example 1: Setting up and finding the Steady State
Customers buy either Brand A or Brand B. If they buy A, the probability they buy A next week is \(0.8\). If they buy B, the probability they switch to A is \(0.3\).
1. Set up the Transition Matrix \(T\) (Columns = "From", Rows = "To"):
\(T = \begin{pmatrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{pmatrix}\) (Notice columns sum to 1).
2. Find the steady state algebraically:
Let \(s = \begin{pmatrix} a \\ b \end{pmatrix}\). We know \(Ts = s\) and \(a + b = 1 \implies b = 1 - a\).
\(\begin{pmatrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{pmatrix} \begin{pmatrix} a \\ 1-a \end{pmatrix} = \begin{pmatrix} a \\ 1-a \end{pmatrix}\).
Top row equation: \(0.8a + 0.3(1-a) = a\).
\(0.8a + 0.3 - 0.3a = a \implies 0.5a + 0.3 = a \implies 0.3 = 0.5a \implies a = 0.6\).
Since \(a = 0.6\), \(b = 0.4\). The long-term steady state is \(60\%\) Brand A and \(40\%\) Brand B.

Worked Examples: Tangents and Normals Example 1: Using the Power Rule
Find the derivative of \(f(x) = 4x^3 - \frac{2}{x^2} + 5x\).
First, rewrite with negative exponents: \(f(x) = 4x^3 - 2x^{-2} + 5x^1\).
Apply the power rule: \(f'(x) = 12x^2 - (-2)(2)x^{-3} + 5 = 12x^2 + \frac{4}{x^3} + 5\).
Example 2: Equation of a Normal Line
Find the equation of the normal to the curve \(y = x^2 - 4x\) at the point where \(x = 3\).
1. Find the \(y\)-coordinate: \(y = (3)^2 - 4(3) = 9 - 12 = -3\). Point is \((3, -3)\).
2. Find the tangent gradient (\(m_T\)): \(f'(x) = 2x - 4\). At \(x = 3\), \(m_T = 2(3) - 4 = 2\).
3. Find the normal gradient (\(m_N\)): \(m_N = -\frac{1}{2}\).
4. Equation: \(y - (-3) = -\frac{1}{2}(x - 3) \implies y + 3 = -0.5x + 1.5 \implies y = -0.5x - 1.5\).

Worked Examples: Advanced Rules Example 1: The Chain Rule
Differentiate \(y = \cos(x^3 + 2x)\).
Let \(u = x^3 + 2x \implies u' = 3x^2 + 2\).
The derivative of \(\cos(u)\) is \(-\sin(u)\).
Using Chain Rule: \(y' = -(3x^2 + 2)\sin(x^3 + 2x)\).
Example 2: The Product Rule
Differentiate \(f(x) = x^2 e^{3x}\).
Let \(u = x^2 \implies u' = 2x\).
Let \(v = e^{3x} \implies v' = 3e^{3x}\) (chain rule required here!).
Product Rule: \(f'(x) = u'v + uv' = (2x)(e^{3x}) + (x^2)(3e^{3x}) = e^{3x}(2x + 3x^2)\).
Example 3: The Quotient Rule
Differentiate \(g(x) = \frac{\ln x}{x}\).
Let \(u = \ln x \implies u' = \frac{1}{x}\).
Let \(v = x \implies v' = 1\).
Quotient Rule: \(g'(x) = \frac{u'v - uv'}{v^2} = \frac{(\frac{1}{x})(x) - (\ln x)(1)}{x^2} = \frac{1 - \ln x}{x^2}\).

Worked Examples: Optimization & Concavity Example 1: Classifying Turning Points
Given \(f(x) = 2x^3 - 9x^2 + 12x\), find the coordinates of the turning points and classify them using the second derivative test.
1. Set first derivative to zero: \(f'(x) = 6x^2 - 18x + 12 = 0\).
Divide by 6: \(x^2 - 3x + 2 = 0 \implies (x-1)(x-2) = 0\).
Turning points are at \(x = 1\) and \(x = 2\).
2. Find \(y\)-coordinates: \(f(1) = 2 - 9 + 12 = 5 \implies (1, 5)\).
\(f(2) = 16 - 36 + 24 = 4 \implies (2, 4)\).
3. Find second derivative: \(f''(x) = 12x - 18\).
4. Classify:
At \(x=1\), \(f''(1) = 12 - 18 = -6\). Since \(f''(x) < 0\), \((1, 5)\) is a Local Maximum.
At \(x=2\), \(f''(2) = 24 - 18 = 6\). Since \(f''(x) > 0\), \((2, 4)\) is a Local Minimum.
Example 2: Finding Points of Inflexion
Find the coordinates of the point of inflexion for \(f(x)\) from Example 1.
1. Set second derivative to zero: \(f''(x) = 12x - 18 = 0\).
\(12x = 18 \implies x = 1.5\).
2. Find \(y\)-coordinate: \(f(1.5) = 2(1.5)^3 - 9(1.5)^2 + 12(1.5) = 4.5\).
The point of inflexion is at \((1.5, 4.5)\).

Worked Examples: Integration Example 1: Definite Integrals & Area
Find the exact area enclosed by the curve \(y = 3x^2 - 2\) and the \(x\)-axis between \(x=1\) and \(x=3\).
Area \(= \int_1^3 (3x^2 - 2) dx\).
Find the antiderivative: \([x^3 - 2x]_1^3\).
Evaluate at bounds: \(( (3)^3 - 2(3) ) - ( (1)^3 - 2(1) ) = (27 - 6) - (1 - 2) = 21 - (-1) = 22\).
Example 2: Integration by Substitution
Evaluate the indefinite integral \(\int 2x e^{x^2} dx\).
Let \(u = x^2\). Then \(\frac{du}{dx} = 2x \implies dx = \frac{du}{2x}\).
Substitute \(u\) and \(dx\): \(\int 2x e^u \frac{du}{2x}\).
The \(2x\) terms cancel: \(\int e^u du = e^u + C\).
Substitute back for \(x\): \(e^{x^2} + C\).

Worked Examples: Kinematics Example 1: Integrating Acceleration
A particle starts from rest (\(v=0, t=0\)). Its acceleration is \(a(t) = 6t\). Find its velocity \(v(t)\).
\(v(t) = \int 6t dt = \frac{6t^2}{2} + C = 3t^2 + C\).
Use initial condition \(v(0) = 0 \implies 0 = 3(0)^2 + C \implies C = 0\).
Velocity is \(v(t) = 3t^2\).
Example 2: Total Distance Travelled (With Direction Change)
A particle has velocity \(v(t) = 2t - 4\). Find the total distance travelled from \(t=0\) to \(t=4\).
The particle stops when \(v(t) = 0 \implies 2t - 4 = 0 \implies t = 2\).
We must split the integral: Distance \(= |\int_0^2 (2t - 4) dt| + |\int_2^4 (2t - 4) dt|\).
\(s(t) = [t^2 - 4t]\).
\(s(0) = 0\). \(s(2) = 4 - 8 = -4\). \(s(4) = 16 - 16 = 0\).
Distance \(= |-4 - 0| + |0 - (-4)| = 4 + 4 = 8\) metres.

Worked Examples: Solving DEs Example 1: Separation of Variables
Solve \(\frac{dy}{dx} = 2xy\) given \(y(0) = 3\).
Separate: \(\frac{1}{y} dy = 2x dx\).
Integrate: \(\int \frac{1}{y} dy = \int 2x dx \implies \ln|y| = x^2 + C\).
Convert to exponential: \(y = e^{x^2 + C} = A e^{x^2}\).
Use initial condition \(y(0) = 3 \implies 3 = A e^0 \implies A = 3\).
Solution: \(y = 3e^{x^2}\).
Example 2: Euler’s Method (AHL 5.16)
Given \(\frac{dy}{dx} = x + y\), \(y(0) = 1\), use Euler’s method with step \(h=0.1\) to find \(y(0.1)\).
Formula: \(y_1 = y_0 + h \times f(x_0, y_0)\).
\(y_1 = 1 + 0.1 \times (0 + 1) = 1 + 0.1 = 1.1\).

Worked Examples: Maclaurin Series Example 1: Manual Generation
Find the first three non-zero terms of the Maclaurin series for \(f(x) = e^{2x}\).
\(f(x) = e^{2x} \implies f(0) = 1\).
\(f'(x) = 2e^{2x} \implies f'(0) = 2\).
\(f''(x) = 4e^{2x} \implies f''(0) = 4\).
Substitute into formula: \(f(x) = 1 + x(2) + \frac{x^2}{2}(4) = 1 + 2x + 2x^2\).
Example 2: Substitution Method
The formula booklet gives \(\sin x \approx x - \frac{x^3}{3!} + \dots\)
Find the series for \(\sin(x^2)\).
Substitute \(x^2\) for \(x\): \(\sin(x^2) \approx (x^2) - \frac{(x^2)^3}{6} = x^2 - \frac{x^6}{6}\).