Pre-Calculus Study Notebook

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VECTORS

Unit 1 · Problems 1–5

MAGNITUDE = length of arrow = \(\sqrt{x^2+y^2}\)

DIRECTION = angle \(\theta = \tan^{-1}(y/x)\)

DOT PRODUCT = \(\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta\)

PERPENDICULAR ⟹ dot product = 0

Given \(\vec{v} = \langle 3, 4 \rangle\), find its magnitude:

Step 1: \(|\vec{v}| = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25}\)

Step 2: \(|\vec{v}| = \mathbf{5}\) ✔

💡 3-4-5 is a Pythagorean triple — memorize it!

VECTORS — Magnitude

Find the magnitude of the vector \(\vec{u} = \langle -5,\; 12 \rangle\).

💡 Hint: Remember — negative signs disappear when you square!

💡 EXPLANATION

\(|\vec{u}| = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = \mathbf{13}\)
Key: 5-12-13 is a Pythagorean triple! Add it to your memory bank. 🧠

VECTORS — Dot Product

Are vectors \(\vec{a} = \langle 3,\; -4 \rangle\) and \(\vec{b} = \langle 8,\; 6 \rangle\) perpendicular?

💡 Hint: Compute \(\vec{a} \cdot \vec{b}\) first. If it equals 0, they're perpendicular!

💡 EXPLANATION

\(\vec{a}\cdot\vec{b} = (3)(8) + (-4)(6) = 24 - 24 = \mathbf{0}\)
Since the dot product is 0, the vectors ARE perpendicular! ✔
Trick: "Perpendicular = Product is ZERO" — P.P.Z! 🎯

VECTORS — Component Form

A vector has magnitude \(10\) and direction angle \(60°\). What are its component form coordinates \(\langle x,\; y \rangle\)?

💡 Hint: \(x = r\cos\theta\), \(y = r\sin\theta\)

💡 EXPLANATION

\(x = 10\cos 60° = 10 \cdot \tfrac{1}{2} = 5\)
\(y = 10\sin 60° = 10 \cdot \tfrac{\sqrt{3}}{2} = 5\sqrt{3}\)
Memory: cos60° = ½, sin60° = √3/2. Think: "cosine comes first, sine is bigger at 60°" 📐

VECTORS — Angle Between

Find the angle between \(\vec{p} = \langle 1,\; 0 \rangle\) and \(\vec{q} = \langle 0,\; 1 \rangle\) using the dot product formula.

💡 EXPLANATION

\(\vec{p}\cdot\vec{q} = (1)(0)+(0)(1) = 0\)
\(\cos\theta = \dfrac{0}{|\vec{p}||\vec{q}|} = 0 \Rightarrow \theta = \cos^{-1}(0) = 90°\)
These are just the x and y unit vectors — obviously perpendicular! 😄

VECTORS — Vector Addition

If \(\vec{u} = \langle 2,\; -3 \rangle\) and \(\vec{v} = \langle -5,\; 7 \rangle\), what is \(2\vec{u} - \vec{v}\)?

💡 Hint: Scale first, THEN subtract component by component.

💡 EXPLANATION

\(2\vec{u} = \langle 4, -6 \rangle\)
\(2\vec{u} - \vec{v} = \langle 4-(-5),\; -6-7 \rangle = \langle 9,\; -13 \rangle\) ✔
Watch out! Subtracting a negative: \(4 - (-5) = 4 + 5 = 9\) — don't drop the sign! ⚠️

🔲

MATRICES

Unit 2 · Problems 6–10

ROW × COLUMN = rule for multiplication

DETERMINANT 2×2: \(ad - bc\)

IDENTITY = \(I\), like multiplying by 1

INVERSE EXISTS only if \(\det \neq 0\)

Find det of \(A = \begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix}\):

det(A) = \((3)(4) - (2)(1) = 12 - 2 = \mathbf{10}\)

💡 "ad minus bc" — diagonal down MINUS diagonal up!

MATRIX — Determinant

Calculate \(\det\begin{pmatrix} 5 & -2 \\ 3 & 4 \end{pmatrix}\).

💡 EXPLANATION

\(\det = (5)(4) - (-2)(3) = 20 - (-6) = 20 + 6 = \mathbf{26}\)
Trap alert! \((-2)(3) = -6\), and then \(-(-6) = +6\). Double negatives trip everyone up! ⚠️

MATRIX — Multiplication

Find \(AB\) where \(A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\) and \(B = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\).

💡 Hint: \(B\) is a "swap matrix" — it swaps the columns of \(A\)!

💡 EXPLANATION

Row 1: \([1·0+2·1,\; 1·1+2·0] = [2,\; 1]\)
Row 2: \([3·0+4·1,\; 3·1+4·0] = [4,\; 3]\)
Result: \(\begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}\) ✔ — the columns of A are swapped!

MATRIX — Inverse

Find the inverse of \(M = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}\).

💡 Formula: \(M^{-1} = \dfrac{1}{\det M}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}\)

💡 EXPLANATION

\(\det M = (2)(2)-(3)(1) = 4-3 = 1\)
\(M^{-1} = \frac{1}{1}\begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix}\) ✔
Memory trick: Swap main diagonal, negate off-diagonal, divide by det. "Swap-Negate-Divide!" 🔄

MATRIX — System of Equations

Use matrices to solve: \(\begin{cases} 2x + y = 5 \\ x + 3y = 10 \end{cases}\)

💡 Set up \(AX = B\), then \(X = A^{-1}B\). Or just substitute!

💡 EXPLANATION

From eq1: \(y = 5 - 2x\). Sub into eq2:
\(x + 3(5-2x) = 10 \Rightarrow x + 15 - 6x = 10 \Rightarrow -5x = -5 \Rightarrow x = 1\)
Then \(y = 5 - 2(1) = 3\). So \(\mathbf{(1,\; 3)}\) ✔

MATRIX — Properties

Which of the following is TRUE about matrix multiplication?

💡 EXPLANATION

✅ D is correct: Matrix multiplication IS associative: \((AB)C = A(BC)\)
❌ A: \(AB \neq BA\) in general — order matters!
❌ B: The correct identity rule is \(A \cdot I = A\), not \(I\)
❌ C: You can only multiply if columns of A = rows of B
Remember: "MATRICES ARE NOT COMMUTATIVE!" — This trips up almost everyone! 🚨

🌀

POLAR GRAPHS

Unit 3 · Problems 11–14

POLAR → RECT \(x = r\cos\theta\), \(y = r\sin\theta\)

RECT → POLAR \(r = \sqrt{x^2+y^2}\), \(\theta = \tan^{-1}(y/x)\)

ROSE \(r = a\cos(n\theta)\): n even → 2n petals, n odd → n petals

CARDIOID \(r = a(1 \pm \cos\theta)\) — heart shape ❤️

Convert polar \((r,\theta) = (4, 30°)\) to rectangular:

\(x = 4\cos30° = 4 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3}\)

\(y = 4\sin30° = 4 \cdot \frac{1}{2} = 2\)

Rectangular: \(\mathbf{(2\sqrt{3},\; 2)}\) ✔

POLAR — Conversion

Convert the polar point \(\left(6,\; \dfrac{\pi}{4}\right)\) to rectangular coordinates.

💡 \(\frac{\pi}{4}\) radians = 45°, and \(\cos45° = \sin45° = \frac{\sqrt{2}}{2}\)

💡 EXPLANATION

\(x = 6\cos\!\frac{\pi}{4} = 6 \cdot \frac{\sqrt{2}}{2} = 3\sqrt{2}\)
\(y = 6\sin\!\frac{\pi}{4} = 6 \cdot \frac{\sqrt{2}}{2} = 3\sqrt{2}\)
Since \(\frac{\pi}{4} = 45°\), both components are equal — the point lies on the line \(y = x\)! 🎯

POLAR — Graph Identification

The polar equation \(r = 3\cos(2\theta)\) produces which type of graph?

💡 Hint: Look at the coefficient of \(\theta\). Is it \(n\) even or odd?

💡 EXPLANATION

\(r = a\cos(n\theta)\) is a rose curve.
Here \(n = 2\) (even), so petals = \(2n = 4\). ✔
RULE: "Even n → 2n petals, Odd n → n petals." Write this on your wrist! ✍️

POLAR — Rectangular to Polar

Convert the rectangular point \((-\sqrt{3},\; 1)\) to polar form \((r,\;\theta)\) where \(r > 0\) and \(0 \le \theta < 2\pi\).

💡 Careful! The point is in Quadrant II. Adjust your angle!

💡 EXPLANATION

\(r = \sqrt{3+1} = 2\)
Reference angle: \(\tan^{-1}\!\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} = 30°\)
But point is in Q2 (x<0, y>0): \(\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\) ✔
Always identify the quadrant FIRST before finding the angle! 🧭

POLAR — Equation Recognition

Which polar equation represents a circle centered at the origin with radius 5?

💡 EXPLANATION

\(r = 5\) means: every point is exactly 5 units from the origin, at ANY angle. That's a circle! ✔
— \(r = 5\sin\theta\) is a circle shifted up
— \(\theta = 5\) is a straight ray (line through origin)
— \(r = 5\cos\theta + 5\) is a cardioid-like shape
Simple rule: "\(r = \text{const}\) = CIRCLE centered at origin" 🔵

📏

TRIGONOMETRY

Unit 4 · Problems 15–18

SOH-CAH-TOA sin=Opp/Hyp, cos=Adj/Hyp, tan=Opp/Adj

ALL STUDENTS TAKE CALCULUS Q1:All+, Q2:Sin+, Q3:Tan+, Q4:Cos+

PYTHAGOREAN ID \(\sin^2\theta + \cos^2\theta = 1\)

DOUBLE ANGLE \(\sin2\theta = 2\sin\theta\cos\theta\)

If \(\sin\theta = \frac{3}{5}\) and \(\theta\) is in Q1, find \(\cos\theta\):

Step 1: \(\cos^2\theta = 1 - \sin^2\theta = 1 - \frac{9}{25} = \frac{16}{25}\)

Step 2: \(\cos\theta = \frac{4}{5}\) (positive in Q1) ✔

TRIG — Exact Values

What is the exact value of \(\tan\!\left(\dfrac{2\pi}{3}\right)\)?

💡 \(\frac{2\pi}{3} = 120°\) — reference angle is \(60°\), located in Q2!

💡 EXPLANATION

Reference angle = 60°, \(\tan60° = \sqrt{3}\)
In Q2: sine is +, cosine is −, so tangent is negative.
\(\tan\!\frac{2\pi}{3} = -\sqrt{3}\) ✔
"All Students Take Calculus" — In Q2, only Sin is positive, so Tan is negative! 🎓

TRIG — Identity

Simplify: \(\dfrac{\sin^2\theta + \cos^2\theta}{\cos\theta}\)

💡 EXPLANATION

\(\sin^2\theta + \cos^2\theta = 1\) (Pythagorean identity!)
So the expression becomes: \(\dfrac{1}{\cos\theta} = \sec\theta\) ✔
Step 1 always: Look for \(\sin^2+\cos^2\) — replace it with 1 instantly! 🔥

TRIG — Equation Solving

Solve for \(\theta\) in \([0, 2\pi)\): \(2\sin\theta - \sqrt{3} = 0\)

💡 Isolate \(\sin\theta\), find reference angle, then check which quadrants have positive sine!

💡 EXPLANATION

\(2\sin\theta = \sqrt{3} \Rightarrow \sin\theta = \dfrac{\sqrt{3}}{2}\)
Reference angle: \(\sin^{-1}\!\frac{\sqrt{3}}{2} = \frac{\pi}{3}\)
Sine is positive in Q1 and Q2: \(\theta = \frac{\pi}{3}\) and \(\pi - \frac{\pi}{3} = \frac{2\pi}{3}\) ✔
Always find 2 answers when solving trig equations! (unless restricted) 📐

TRIG — Double Angle

If \(\sin\theta = \dfrac{1}{3}\) and \(\theta\) is in Q1, find \(\sin(2\theta)\).

💡 Use \(\sin2\theta = 2\sin\theta\cos\theta\). You need to find \(\cos\theta\) first!

💡 EXPLANATION

\(\cos\theta = \sqrt{1-\frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}\) (Q1: positive)
\(\sin2\theta = 2 \cdot \frac{1}{3} \cdot \frac{2\sqrt{2}}{3} = \frac{4\sqrt{2}}{9}\) ✔
Two-step process: Find missing trig value → Apply double angle formula 🔁

🔗

MIXED — Connections

Unit 5 · Problems 19–20

These final problems connect multiple topics. Think carefully about which tool to use! 🛠️

MIXED — Vectors + Trig

A force vector \(\vec{F}\) has magnitude \(20\) N and makes an angle of \(150°\) with the positive x-axis. What is the x-component of \(\vec{F}\)?

💡 Use components: \(F_x = |\vec{F}|\cos\theta\). Remember: \(150° = 180° - 30°\) is in Q2!

💡 EXPLANATION

\(F_x = 20\cos150° = 20 \cdot (-\frac{\sqrt{3}}{2}) = -10\sqrt{3}\) N ✔
In Q2: cosine is NEGATIVE, so the x-component is negative (pointing left).
Always check the sign! A negative x-component means the force points leftward. 🎯

MIXED — Polar + Matrix

A rotation matrix is given by \(R = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}\). If you rotate the point \((1, 0)\) by \(\theta = 90°\), what is the new point?

💡 Multiply the matrix by the column vector \(\begin{pmatrix}1\\0\end{pmatrix}\). What are \(\cos90°\) and \(\sin90°\)?

💡 EXPLANATION

\(\cos90° = 0,\; \sin90° = 1\)
\(R = \begin{pmatrix}0 & -1\\1 & 0\end{pmatrix}\)
\(R\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}0\cdot1+(-1)\cdot0\\1\cdot1+0\cdot0\end{pmatrix} = \begin{pmatrix}0\\1\end{pmatrix}\) ✔
Rotating the x-axis point 90° gives the y-axis point — beautiful geometry! 🌟

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