Precalculus · Spring 2nd Semester

Exam Study Guide

46 problems across 5 days. Trig, vectors, complex, parametric, polar, conics, matrices, sequences.

Day 0

Overview

Five days, eight topics, one exam. The unit circle below is the foundation for half of these problems — drag around it to watch the angle, sine, cosine, and tangent move together.

The Unit Circle, live

Drag anywhere on the circle to see live values · tap or click a marked angle to lock it

Angleπ/6

QuadrantI

cos θ√3/2

sin θ1/2

tan θ√3/3

Referenceπ/6

Your 5-day plan

Day 1 · Trig + Vectors

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Day 2 · Complex + Parametric

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Day 3 · Polar

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Day 4 · Conics + Matrices

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Day 5 · Sequences + Series

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Quick formula reference

Unit circle & trig identities

For a point on the unit circle: $\cos\theta = x$, $\sin\theta = y$, $\tan\theta = y/x$. Reciprocals: $\sec = 1/\cos$, $\csc = 1/\sin$, $\cot = 1/\tan$.

Pythagorean identity$$\sin^2\theta + \cos^2\theta = 1$$

Vectors

For $\vec u = a\hat{\mathbf{i}} + b\hat{\mathbf{j}}$ and $\vec v = c\hat{\mathbf{i}} + d\hat{\mathbf{j}}$:

Operations Sum: $\vec u + \vec v = (a+c)\hat{\mathbf{i}} + (b+d)\hat{\mathbf{j}}$
Scalar: $k\vec u = (ka)\hat{\mathbf{i}} + (kb)\hat{\mathbf{j}}$
Dot product: $\vec u \cdot \vec v = ac + bd$
Magnitude: $\|\vec u\| = \sqrt{a^2 + b^2}$
Position vector $\vec{PQ}$: $Q - P$

For example Let $\vec u = 3\hat{\mathbf{i}} + 4\hat{\mathbf{j}}$ and $\vec v = -2\hat{\mathbf{i}} + 5\hat{\mathbf{j}}$:
Sum: $\vec u + \vec v = (3{-}2)\hat{\mathbf{i}} + (4{+}5)\hat{\mathbf{j}} = \hat{\mathbf{i}} + 9\hat{\mathbf{j}}$
Scalar: $2\vec u = 6\hat{\mathbf{i}} + 8\hat{\mathbf{j}}$
Dot product: $\vec u \cdot \vec v = (3)(-2) + (4)(5) = -6 + 20 = 14$
Magnitude: $\|\vec u\| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$
Position vector for $P=(1, 2)$, $Q=(4, 7)$: $\vec{PQ} = 3\hat{\mathbf{i}} + 5\hat{\mathbf{j}}$

Complex numbers in polar form

For $z = r\,\text{cis}\,\theta$ and $w = s\,\text{cis}\,\phi$:

$z \cdot w = (rs)\,\text{cis}(\theta + \phi)$
$\dfrac{z}{w} = \dfrac{r}{s}\,\text{cis}(\theta - \phi)$
Keep angle in $[0°, 360°)$. If magnitude turns negative, flip sign and add $180°$.

For example Let $z = 6\,\text{cis}\,40°$ and $w = 2\,\text{cis}\,70°$:
$z \cdot w = (6 \cdot 2)\,\text{cis}(40°{+}70°) = 12\,\text{cis}\,110°$
$z / w = (6/2)\,\text{cis}(40°{-}70°) = 3\,\text{cis}(-30°) = 3\,\text{cis}\,330°$ (adding $360°$ to bring into range)

Polar ↔ rectangular

Polar to rectangular: $x = r\cos\theta,\quad y = r\sin\theta$
Rectangular to polar: $r = \sqrt{x^2 + y^2},\quad \tan\theta = y/x$ (watch quadrant)

For example Polar to rectangular: $(4, \pi/3)$ → $x = 4\cos(\pi/3) = 4 \cdot \tfrac{1}{2} = 2$, $\ y = 4\sin(\pi/3) = 4 \cdot \tfrac{\sqrt{3}}{2} = 2\sqrt{3}$. Point: $(2, 2\sqrt{3})$.
Rectangular to polar: $(3, 3)$ → $r = \sqrt{9+9} = 3\sqrt{2}$. $\tan\theta = 3/3 = 1$, and $(3,3)$ is in Q1, so $\theta = \pi/4$. Point: $(3\sqrt{2}, \pi/4)$.

Rose curves $r = a\cos(n\theta)$ or $r = a\sin(n\theta)$: if $n$ even → $2n$ petals; if $n$ odd → $n$ petals.

Conics — standard forms

Circle: $(x-h)^2 + (y-k)^2 = r^2$
Ellipse: $\dfrac{(x-h)^2}{a^2} + \dfrac{(y-k)^2}{b^2} = 1$
Hyperbola: $\dfrac{(x-h)^2}{a^2} - \dfrac{(y-k)^2}{b^2} = 1$ (opens horizontally)
Parabola: $(y-k)^2 = 4p(x-h)$ (opens horizontally)

For example Circle $(x-2)^2 + (y+3)^2 = 16$ → center $(2, -3)$, radius $r = \sqrt{16} = 4$.
Ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$ → center $(0, 0)$, $a = 5$ (horizontal, since under $x$ and bigger), $b = 3$. Vertices at $(\pm 5, 0)$, co-vertices at $(0, \pm 3)$.
Hyperbola $\dfrac{x^2}{4} - \dfrac{y^2}{9} = 1$ → center $(0, 0)$, opens horizontally, $a = 2$, $b = 3$. Vertices at $(\pm 2, 0)$, asymptote slopes $\pm b/a = \pm 3/2$.
Parabola $(y-1)^2 = 8(x-2)$ → vertex $(2, 1)$, $4p = 8 \Rightarrow p = 2$, opens right. Focus at $(4, 1)$, directrix $x = 0$.

Sequences & series

Arithmetic $n$th term: $a_n = a_1 + (n-1)d$
Arithmetic sum: $S_n = \dfrac{n}{2}(a_1 + a_n)$
Geometric $n$th term: $a_n = a_1 r^{n-1}$
Geometric sum: $S_n = a_1 \dfrac{1 - r^n}{1 - r}$
Infinite geometric (when $|r|<1$): $S = \dfrac{a_1}{1 - r}$

For example Arithmetic sequence $5, 8, 11, 14, \ldots$ has $a_1 = 5$, $d = 3$. So $a_{10} = 5 + 9(3) = 32$ and $S_{10} = \dfrac{10}{2}(5 + 32) = 185$.
Geometric sequence $2, 6, 18, 54, \ldots$ has $a_1 = 2$, $r = 3$. So $a_5 = 2 \cdot 3^4 = 162$ and $S_5 = 2 \cdot \dfrac{1 - 3^5}{1 - 3} = 242$.
Infinite geometric: $a_1 = 10$, $r = 1/2$ converges to $S = \dfrac{10}{1 - 1/2} = 20$.

Limits: The last 8 problems on the exam are limits — use your separate limits study guide for those, they're not covered here.

Day 1

Trig & Vectors

Exact values from the unit circle, quadrant signs, terminal-side problems, trig equations in $[0, 2\pi)$, plus vector basics.

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Video tutorials for Day 1

▶ The Unit Circle, Basic Introduction Organic Chem Tutor ▶ Solving Trigonometric Equations — factoring, identities, multiple angles Organic Chem Tutor ▶ Vectors — Precalculus (components, magnitude, direction) Organic Chem Tutor ▶ Dot Product of Two Vectors Organic Chem Tutor

The unit circle & reference angles

Sixteen special angles drive every problem on Day 1. You need cosine, sine, and tangent at each of them cold.

Memory trick — the sine row

Read sine across the first quadrant as $\dfrac{\sqrt{0}}{2},\ \dfrac{\sqrt{1}}{2},\ \dfrac{\sqrt{2}}{2},\ \dfrac{\sqrt{3}}{2},\ \dfrac{\sqrt{4}}{2}$ — i.e., $0,\ \tfrac{1}{2},\ \tfrac{\sqrt{2}}{2},\ \tfrac{\sqrt{3}}{2},\ 1$ for $0,\ \pi/6,\ \pi/4,\ \pi/3,\ \pi/2$. Cosine is the same row reversed. Tangent is sine over cosine.

Don't confuse $\sqrt{2}/2$ and $\sqrt{3}/2$. $\sqrt{3}/2 \approx 0.866$ (the bigger one) is for $30°$ and $60°$. $\sqrt{2}/2 \approx 0.707$ (the smaller one) is exclusively for $45°$.

Reference angle by quadrant Q1: $\theta_{\text{ref}} = \theta$ · Q2: $\theta_{\text{ref}} = \pi - \theta$
Q3: $\theta_{\text{ref}} = \theta - \pi$ · Q4: $\theta_{\text{ref}} = 2\pi - \theta$

ASTC — "All Students Take Calculus" Q1: All positive · Q2: only Sin positive
Q3: only Tan positive · Q4: only Cos positive

For deeper trig-equation practice (interactive equation solver, full quiz, calculator-mode problems), your existing Trig Equations Guide covers it.

Find the exact value (problems 1–7)

1 $\cos\dfrac{\pi}{6}$ Exact value ▼

Locate the angle. $\pi/6 = 30°$ — sits in Quadrant I.

Read $\cos$ off the unit circle. At $\pi/6$, the point on the unit circle is $\left(\dfrac{\sqrt{3}}{2},\ \dfrac{1}{2}\right)$. Cosine is the $x$-coordinate.

State the value. $\cos(\pi/6) = \sqrt{3}/2$. (Positive — Q1.)

Answer$\cos\dfrac{\pi}{6} = \dfrac{\sqrt{3}}{2}$

cos(π/6):

2 $\sin\dfrac{3\pi}{4}$ Exact value ▼

Locate the angle. $3\pi/4 = 135°$ — Quadrant II.

Find the reference angle. $\theta_{\text{ref}} = \pi - 3\pi/4 = \pi/4$.

Sign + value. In Q2, sine is positive. $\sin(\pi/4) = \sqrt{2}/2$, so $\sin(3\pi/4) = +\sqrt{2}/2$.

Answer$\sin\dfrac{3\pi}{4} = \dfrac{\sqrt{2}}{2}$

sin(3π/4):

3 $\tan\dfrac{\pi}{2}$ Exact value ▼

Locate the angle. $\pi/2 = 90°$ — straight up the $y$-axis.

Read $\sin$ and $\cos$. At $\pi/2$, the point is $(0, 1)$. So $\cos(\pi/2) = 0$ and $\sin(\pi/2) = 1$.

Compute $\tan = \sin/\cos$. $\tan(\pi/2) = 1/0$, which is undefined. (Vertical line — infinite slope.)

Answer$\tan\dfrac{\pi}{2}$ is undefined

4 $\sin\dfrac{5\pi}{3}$ Exact value ▼

Locate the angle. $5\pi/3 = 300°$ — Quadrant IV.

Find the reference angle. $\theta_{\text{ref}} = 2\pi - 5\pi/3 = \pi/3$.

Sign + value. In Q4, sine is negative. $\sin(\pi/3) = \sqrt{3}/2$, so $\sin(5\pi/3) = -\sqrt{3}/2$.

Answer$\sin\dfrac{5\pi}{3} = -\dfrac{\sqrt{3}}{2}$

sin(5π/3):

5 $\tan 210°$ Exact value ▼

Locate the angle. $210°$ — Quadrant III ($180° < 210° < 270°$).

Find the reference angle. $\theta_{\text{ref}} = 210° - 180° = 30°$.

Sign + value. In Q3, both $\sin$ and $\cos$ are negative, so $\tan = \sin/\cos$ is positive. $\tan 30° = \sqrt{3}/3$, so $\tan 210° = +\sqrt{3}/3$.

Answer$\tan 210° = \dfrac{\sqrt{3}}{3}$

tan(210°):

6 $\csc 150°$ Exact value ▼

Rewrite as $1/\sin$. $\csc\theta = 1/\sin\theta$.

Find $\sin 150°$. $150°$ is in Q2; reference angle is $180° - 150° = 30°$. In Q2 sine is positive, so $\sin 150° = +\sin 30° = 1/2$.

Take the reciprocal. $\csc 150° = 1/(1/2) = 2$.

Answer$\csc 150° = 2$

csc(150°):

7 $\sec 30°$ Exact value ▼

Rewrite as $1/\cos$. $\sec\theta = 1/\cos\theta$.

Find $\cos 30°$. $30°$ is in Q1, $\cos 30° = \sqrt{3}/2$.

Take the reciprocal and rationalize. $\sec 30° = 1/(\sqrt{3}/2) = 2/\sqrt{3} = \dfrac{2\sqrt{3}}{3}$.

Answer$\sec 30° = \dfrac{2\sqrt{3}}{3}$

sec(30°):

Identify the quadrant (problem 8)

8 Name the quadrant: (a) $\cos\theta > 0$ and $\tan\theta < 0$ (b) $\sin\theta > 0$ and $\cos\theta > 0$ Quadrants ▼

Part (a)

Where is $\cos > 0$? Q1 and Q4 (positive $x$-coordinate).

Where is $\tan < 0$? Q2 and Q4 ($\sin$ and $\cos$ have opposite signs).

Intersection. Both true in Q4.

Part (b)

Where are both positive? $\sin > 0$ in Q1, Q2. $\cos > 0$ in Q1, Q4. Both: Q1.

Answer(a) Quadrant IV · (b) Quadrant I

Terminal side & reference triangle (problems 9–10)

When a point $(x, y)$ lies on the terminal side of $\theta$, drop a perpendicular to the $x$-axis to form a right triangle. Then:

$r = \sqrt{x^2 + y^2}$ · $\sin\theta = y/r$ · $\cos\theta = x/r$ · $\tan\theta = y/x$

For example If the terminal side of $\theta$ passes through $(3, -4)$:
$r = \sqrt{3^2 + (-4)^2} = \sqrt{25} = 5$
$\sin\theta = -4/5$, $\cos\theta = 3/5$, $\tan\theta = -4/3$.
Sanity check: the point is in Q4 (positive $x$, negative $y$). In Q4, sine and tangent are negative, cosine is positive — matches.

$r$ is always positive (it's a length). The signs of $\sin$, $\cos$, $\tan$ come from the signs of $x$ and $y$ — which is just ASTC.

9 The point $(5, -12)$ lies on the terminal side of $\theta$. Find $\sin\theta$. Terminal side ▼

Find $r$. $r = \sqrt{x^2 + y^2} = \sqrt{25 + 144} = \sqrt{169} = 13$. (Always positive.)

Apply the definition. $\sin\theta = y/r = -12/13$.

Sanity check the sign. The point $(5, -12)$ is in Q4 (positive $x$, negative $y$). In Q4 sine is negative — checks out.

Answer$\sin\theta = -\dfrac{12}{13}$

-12/13:

10 $\cos\theta = 3/5$ and $\theta$ lies in Quadrant IV. Find $\tan\theta$. Terminal side ▼

Set up the reference triangle. $\cos\theta = x/r = 3/5$, so $x = 3$ and $r = 5$.

Solve for $y$ via Pythagorean. $x^2 + y^2 = r^2 \Rightarrow 9 + y^2 = 25 \Rightarrow y^2 = 16 \Rightarrow y = \pm 4$.

Pick the right sign for Q4. In Q4, $y$ is negative. So $y = -4$.

Compute $\tan\theta$. $\tan\theta = y/x = -4/3$.

Answer$\tan\theta = -\dfrac{4}{3}$

-4/3:

Solve in $[0, 2\pi)$ (problems 11–14)

The 6-step strategy

Isolate the trig function. Use algebra; treat $\sin x$ or $\tan x$ like a variable.
Reference angle. Ask: what Q1 angle gives this value?
ASTC. Is the value positive or negative? That picks the two quadrants.
Apply formulas: Q1: $\alpha$ · Q2: $\pi - \alpha$ · Q3: $\pi + \alpha$ · Q4: $2\pi - \alpha$.
Multiple angles? For $\sin(n\theta)$, work in $[0, 2n\pi)$ first, then divide by $n$ at the end. (Tangent has period $\pi$, not $2\pi$.)
Verify all answers land in $[0, 2\pi)$.

Square-root rule. When you take a square root (like $\tan^2 x = 3 \Rightarrow \tan x = \pm\sqrt{3}$ in #14), always include $\pm$. Forgetting it cuts your solutions in half.

Don't divide out a common factor like $\cos x$ in #12 — you'll lose its zeros. Factor and set each piece to zero instead.

Reciprocals first. If you see $\csc$, $\sec$, or $\cot$, rewrite as $1/\sin$, $1/\cos$, $1/\tan$ before doing anything else. The reference angles are the same.

11 $\sqrt{2}\,\sin x = 1$ Trig equation ▼

Isolate $\sin x$. $\sin x = 1/\sqrt{2} = \sqrt{2}/2$.

Reference angle. $\sin(\pi/4) = \sqrt{2}/2$, so $\theta_{\text{ref}} = \pi/4$.

Find all quadrants where sine is positive. Sine $> 0$ in Q1 and Q2.

Write both solutions. Q1: $x = \pi/4$. Q2: $x = \pi - \pi/4 = 3\pi/4$.

Answer$x = \dfrac{\pi}{4},\ \dfrac{3\pi}{4}$

12 $2\cos x \sin x = \cos x$ Trig equation ▼

Move everything to one side. $2\cos x \sin x - \cos x = 0$.

Factor. $\cos x\,(2\sin x - 1) = 0$. (Don't divide by $\cos x$ — you'd lose its zeros.)

First factor: $\cos x = 0$. $x = \pi/2,\ 3\pi/2$.

Second factor: $2\sin x - 1 = 0 \Rightarrow \sin x = 1/2$. Reference $\pi/6$. Sine $> 0$ in Q1, Q2. So $x = \pi/6,\ 5\pi/6$.

Combine all four. $x = \pi/6,\ \pi/2,\ 5\pi/6,\ 3\pi/2$.

Answer$x = \dfrac{\pi}{6},\ \dfrac{\pi}{2},\ \dfrac{5\pi}{6},\ \dfrac{3\pi}{2}$

13 $1 - 2\cos x = 0$ Trig equation ▼

Isolate $\cos x$. $2\cos x = 1 \Rightarrow \cos x = 1/2$.

Reference angle. $\cos(\pi/3) = 1/2$, so $\theta_{\text{ref}} = \pi/3$.

Where is $\cos > 0$? Q1 and Q4.

Solutions. Q1: $x = \pi/3$. Q4: $x = 2\pi - \pi/3 = 5\pi/3$.

Answer$x = \dfrac{\pi}{3},\ \dfrac{5\pi}{3}$

14 $\tan^2 x = 3$ Trig equation ▼

Take square root (both signs). $\tan x = \pm\sqrt{3}$.

Reference angle. $\tan(\pi/3) = \sqrt{3}$, so $\theta_{\text{ref}} = \pi/3$.

$\tan x = \sqrt{3}$ — Q1 & Q3. $x = \pi/3$ (Q1) and $x = \pi + \pi/3 = 4\pi/3$ (Q3).

$\tan x = -\sqrt{3}$ — Q2 & Q4. $x = \pi - \pi/3 = 2\pi/3$ (Q2) and $x = 2\pi - \pi/3 = 5\pi/3$ (Q4).

All four. $x = \pi/3,\ 2\pi/3,\ 4\pi/3,\ 5\pi/3$.

Answer$x = \dfrac{\pi}{3},\ \dfrac{2\pi}{3},\ \dfrac{4\pi}{3},\ \dfrac{5\pi}{3}$

Top trig-equation pitfalls

Forgetting the $\pm$ when taking a square root — half your solutions vanish.
Stopping at one quadrant. Most equations have 2 or 4 solutions. Always ask: "which other quadrant matches the sign?"
Dividing out a trig factor like $\cos x$ instead of factoring — you lose its zeros.
For multiple angles, not extending the range to $[0, 2n\pi)$ before dividing by $n$ at the end.
Using $2\pi$ as the tangent period — it's $\pi$. For $\tan(n\theta)$ you'll find $2n$ solutions, not $n$.

Vectors (problems 15–18)

A vector $\vec u = a\hat{\mathbf{i}} + b\hat{\mathbf{j}}$ has horizontal component $a$ and vertical component $b$. Likewise $\vec v = c\hat{\mathbf{i}} + d\hat{\mathbf{j}}$.

Add: $\vec u + \vec v = (a+c)\hat{\mathbf{i}} + (b+d)\hat{\mathbf{j}}$ · Scalar: $k\vec u = (ka)\hat{\mathbf{i}} + (kb)\hat{\mathbf{j}}$
Dot product: $\vec u \cdot \vec v = ac + bd$ (a scalar) · Magnitude: $\|\vec u\| = \sqrt{a^2 + b^2}$

For example $\vec u = 3\hat{\mathbf{i}} + 4\hat{\mathbf{j}}$, $\vec v = -2\hat{\mathbf{i}} + 5\hat{\mathbf{j}}$, $k = 2$:
$\vec u + \vec v = \hat{\mathbf{i}} + 9\hat{\mathbf{j}}$ · $2\vec u = 6\hat{\mathbf{i}} + 8\hat{\mathbf{j}}$
$\vec u \cdot \vec v = -6 + 20 = 14$ · $\|\vec u\| = \sqrt{9+16} = 5$

Component-by-component is the rule. Add components, scale components, dot products multiply matching components and add. Magnitude is just the Pythagorean theorem on $(a, b)$.

15 If $\vec u = 2\hat{\mathbf{i}} - 4\hat{\mathbf{j}}$ and $\vec v = 3\hat{\mathbf{i}} + 6\hat{\mathbf{j}}$, find $2\vec u - \vec v$. Vector ops ▼

Compute $2\vec u$. $2\vec u = 2(2\hat{\mathbf{i}} - 4\hat{\mathbf{j}}) = 4\hat{\mathbf{i}} - 8\hat{\mathbf{j}}$.

Subtract $\vec v$ component by component. $\hat{\mathbf{i}}$: $4 - 3 = 1$. $\hat{\mathbf{j}}$: $-8 - 6 = -14$.

Write the result. $2\vec u - \vec v = \hat{\mathbf{i}} - 14\hat{\mathbf{j}}$.

Answer$2\vec u - \vec v = \hat{\mathbf{i}} - 14\hat{\mathbf{j}}$

16 If $\vec v = 3\hat{\mathbf{i}} - 4\hat{\mathbf{j}}$, find $-2\vec v$. Vector ops ▼

Multiply each component by $-2$. $\hat{\mathbf{i}}$: $(-2)(3) = -6$. $\hat{\mathbf{j}}$: $(-2)(-4) = +8$.

Write the result. $-2\vec v = -6\hat{\mathbf{i}} + 8\hat{\mathbf{j}}$.

Answer$-2\vec v = -6\hat{\mathbf{i}} + 8\hat{\mathbf{j}}$

17 If $\vec u = 2\hat{\mathbf{i}} + 3\hat{\mathbf{j}}$ and $\vec v = -8\hat{\mathbf{i}} + 4\hat{\mathbf{j}}$, find $\vec u \cdot \vec v$. Dot product ▼

Multiply matching components. $\hat{\mathbf{i}}$: $(2)(-8) = -16$. $\hat{\mathbf{j}}$: $(3)(4) = 12$.

Add. (Dot product is a single number.) $-16 + 12 = -4$.

Answer$\vec u \cdot \vec v = -4$

-4:

18 Find the magnitude of $\vec v = -6\hat{\mathbf{i}} + 8\hat{\mathbf{j}}$. Magnitude ▼

Apply the formula $\|\vec v\| = \sqrt{a^2 + b^2}$. $\|\vec v\| = \sqrt{(-6)^2 + 8^2}$.

Compute the squares. $36 + 64 = 100$.

Take the square root. $\sqrt{100} = 10$. (Magnitude is always non-negative.)

Answer$\|\vec v\| = 10$

10:

Day 2

Complex & Parametric

Position vectors, multiplying and dividing complex numbers in polar form, and converting parametric equations to rectangular form.

0 / 5 attempted

Video tutorials for Day 2

▶ Complex Numbers in Polar Form & De Moivre's Theorem Organic Chem Tutor ▶ Parametric Equations — Eliminating the Parameter $t$ Organic Chem Tutor ▶ Vectors — Precalculus (review for the position-vector problem) Organic Chem Tutor

Position vector

A position vector $\vec{PQ}$ from $P$ to $Q$ is just $Q - P$ (subtract tail from head, component by component).

$\vec{PQ} = (Q_x - P_x)\hat{\mathbf{i}} + (Q_y - P_y)\hat{\mathbf{j}}$

For example $P = (1, 2)$ and $Q = (5, 8)$:
$\vec{PQ} = (5{-}1)\hat{\mathbf{i}} + (8{-}2)\hat{\mathbf{j}} = 4\hat{\mathbf{i}} + 6\hat{\mathbf{j}}$
$\vec{QP} = (1{-}5)\hat{\mathbf{i}} + (2{-}8)\hat{\mathbf{j}} = -4\hat{\mathbf{i}} - 6\hat{\mathbf{j}}$ (opposite direction)

Direction matters: $\vec{PQ} = -\vec{QP}$. Always read which point is the tail and which is the head — $P$ is the tail (start), $Q$ is the head (end).

19 Given $P(-2, 3)$ and $Q(1, -3)$, find the position vector $\vec{PQ}$. Position vector ▼

Identify tail and head. $P$ is the tail (start), $Q$ is the head (end). So $\vec{PQ} = Q - P$.

Subtract component by component. $x$: $1 - (-2) = 3$. $y$: $-3 - 3 = -6$.

Write the vector. $\vec{PQ} = 3\hat{\mathbf{i}} - 6\hat{\mathbf{j}}$. (Watch the sign on the $\hat{\mathbf{i}}$ component — subtracting a negative.)

Answer$\vec{PQ} = 3\hat{\mathbf{i}} - 6\hat{\mathbf{j}}$

Complex numbers in polar form (problems 20–21)

$z = r\,\text{cis}\,\theta$ is shorthand for $r(\cos\theta + i\sin\theta)$, where $r$ is the magnitude and $\theta$ is the argument.

Memory hook

Multiply: magnitudes multiply, angles add. Divide: magnitudes divide, angles subtract. (It mirrors the rules for $r e^{i\theta}$ — exponents add when you multiply, subtract when you divide.)

Multiply: $z \cdot w = (r s)\,\text{cis}(\theta + \phi)$
Divide: $z / w = (r/s)\,\text{cis}(\theta - \phi)$

For example $z = 6\,\text{cis}\,40°$ and $w = 2\,\text{cis}\,70°$:
$zw = 12\,\text{cis}\,110°$ · $z/w = 3\,\text{cis}(-30°) = 3\,\text{cis}\,330°$

If a magnitude comes out negative (like $-3\,\text{cis}\,\phi$), flip the sign to positive and add $180°$ to the angle: $-r\,\text{cis}\,\phi = r\,\text{cis}(\phi + 180°)$.

20 $z = 4\,\text{cis}\,50°$ and $w = -3\,\text{cis}\,60°$. Find $zw$ in polar form. Complex polar ▼

Multiply the magnitudes, add the angles. $zw = (4 \cdot -3)\,\text{cis}(50° + 60°) = -12\,\text{cis}\,110°$.

Magnitude is negative — fix it. Convention: magnitude must be positive. Flip the sign and add $180°$ to the angle.

Apply the fix. $-12\,\text{cis}\,110° = 12\,\text{cis}(110° + 180°) = 12\,\text{cis}\,290°$.

Check the angle is in $[0°, 360°)$. $290°$ is in range. Done.

Answer$zw = 12\,\text{cis}\,290°$

21 $z = 8\,\text{cis}\,120°$ and $w = 4\,\text{cis}\,40°$. Find $z/w$ in polar form. Complex polar ▼

Divide magnitudes, subtract angles. $z/w = (8/4)\,\text{cis}(120° - 40°) = 2\,\text{cis}\,80°$.

Verify the angle range. $80°$ is in $[0°, 360°)$. No adjustment needed.

Answer$z/w = 2\,\text{cis}\,80°$

Parametric to rectangular (problems 22–23)

A parametric equation expresses both $x$ and $y$ as functions of a third variable $t$ (the parameter) — typically time. The point $(x(t), y(t))$ traces out a curve as $t$ varies.

Strategy To eliminate the parameter: solve one equation for $t$ in terms of $x$ (or $y$), then substitute into the other equation.

For example $x = t + 1$, $y = t^2$:
Solve the simpler equation: $t = x - 1$.
Substitute: $y = (x-1)^2 = x^2 - 2x + 1$.
So the rectangular equation is $y = x^2 - 2x + 1$ — a parabola.

Watch for restrictions on the domain. For example, $x = \sqrt{t-2}$ forces $x \geq 0$ — your rectangular equation only describes part of the curve.

22 Convert to rectangular form: $x = t - 5,\ y = t^2 + 2$. Parametric ▼

Solve the simpler equation for $t$. $x = t - 5 \Rightarrow t = x + 5$.

Substitute into the $y$ equation. $y = (x + 5)^2 + 2$.

Expand carefully. $(x+5)^2 = x^2 + 10x + 25$. So $y = x^2 + 10x + 25 + 2$.

Simplify. $y = x^2 + 10x + 27$. (Don't forget the $+2$ at the end.)

Answer$y = x^2 + 10x + 27$

23 Convert to rectangular form: $x = \sqrt{t - 2},\ y = t + 2$. Parametric ▼

Square $x$ to clear the radical. $x^2 = t - 2 \Rightarrow t = x^2 + 2$.

Substitute into the $y$ equation. $y = (x^2 + 2) + 2 = x^2 + 4$.

Note the domain restriction. Original $x = \sqrt{t-2} \geq 0$, so the curve is only the right half of the parabola ($x \geq 0$).

Answer$y = x^2 + 4$ (for $x \geq 0$)

Day 3

Polar Coordinates & Curves

Converting between polar and rectangular, plotting points, and recognizing the four polar-curve families: roses, limaçons, lemniscates, circles.

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Video tutorials for Day 3

▶ Polar Coordinates — Conversion, Plotting, Negative $r$ Organic Chem Tutor ▶ Graphing Polar Equations — Roses, Limaçons, Cardioids, Lemniscates Organic Chem Tutor

Polar ↔ rectangular (problems 24–26)

Polar to rectangular: $x = r\cos\theta,\ y = r\sin\theta$
Rectangular to polar: $r = \sqrt{x^2 + y^2}$, $\tan\theta = y/x$ (use the quadrant of $(x, y)$ to get the right angle)

A negative $r$ means "go in the opposite direction." $(-r, \theta)$ is the same point as $(r, \theta + \pi)$.

24 Convert $\left(3,\ \dfrac{\pi}{6}\right)$ to rectangular coordinates. Polar → rect ▼

Apply $x = r\cos\theta$. $x = 3\cos(\pi/6) = 3 \cdot \dfrac{\sqrt{3}}{2} = \dfrac{3\sqrt{3}}{2}$.

Apply $y = r\sin\theta$. $y = 3\sin(\pi/6) = 3 \cdot \dfrac{1}{2} = \dfrac{3}{2}$.

Answer$\left(\dfrac{3\sqrt{3}}{2},\ \dfrac{3}{2}\right)$

25 Convert $(-1, 1)$ to polar coordinates. Rect → polar ▼

Find $r$. $r = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$.

Find the reference angle. $|\tan\theta_{\text{ref}}| = |y/x| = |1/(-1)| = 1$, so $\theta_{\text{ref}} = \pi/4$.

Pick the correct quadrant. $(-1, 1)$ has negative $x$, positive $y$ → Quadrant II.

Apply Q2 formula. $\theta = \pi - \pi/4 = 3\pi/4$.

Answer$\left(\sqrt{2},\ \dfrac{3\pi}{4}\right)$

26 Plot $\left(-4,\ \dfrac{\pi}{3}\right)$ and give an equivalent polar pair with $r > 0$. Polar plot ▼

Interpret a negative $r$. A negative radius means rotate to the angle, then point the opposite way. So $(-4, \pi/3)$ is at the same place as $(4, \pi/3 + \pi) = (4, 4\pi/3)$.

Compute the rectangular coordinates as a check. $x = -4\cos(\pi/3) = -4 \cdot 1/2 = -2$. $y = -4\sin(\pi/3) = -4 \cdot \sqrt{3}/2 = -2\sqrt{3}$.

Confirm the quadrant. $(-2, -2\sqrt{3})$ — both negative, Quadrant III. Matches the $4\pi/3$ direction (which is $240°$ — into Q3). Good.

AnswerEquivalent point with $r > 0$: $\left(4,\ \dfrac{4\pi}{3}\right)$. Rectangular: $(-2,\ -2\sqrt{3})$.

Polar curves (problems 27–30)

The four families you need to recognize:

Rose: $r = a\cos(n\theta)$ or $r = a\sin(n\theta)$. Petals: $n$ if $n$ odd, $2n$ if $n$ even.
Limaçon: $r = a \pm b\sin\theta$ or $r = a \pm b\cos\theta$. Compare $|a|$ vs $|b|$ for shape.
Lemniscate (figure-8): $r^2 = a^2\cos(2\theta)$ along $x$-axis, $r^2 = a^2\sin(2\theta)$ tilted.
Circle: $r = a\cos\theta$ or $r = a\sin\theta$ — circle of radius $|a|/2$ tangent to origin.

Snap recognition — read the equation

$\cos(n\theta)$ or $\sin(n\theta)$ with $n \geq 2$? Rose. · Constant $\pm$ trig? Limaçon. · $r^2 = $ something? Lemniscate. · Just $a\cos\theta$ or $a\sin\theta$? Circle.

Limaçon shapes by ratio: $|a/b| > 1$ is dimpled; $|a/b| = 1$ is a cardioid (heart, with a sharp point at origin); $|a/b| < 1$ has an inner loop.

$\cos$ vs $\sin$ orients the curve: $\cos$ symmetric about the $x$-axis; $\sin$ symmetric about the $y$-axis.

27 How many petals does $r = 6\cos(4\theta)$ have? Rose curve ▼

Identify the form. $r = a\cos(n\theta)$ with $a = 6,\ n = 4$.

Apply the petal rule. $n = 4$ is even, so the rose has $2n = 8$ petals.

Answer8 petals

28 Sketch $r = 3 - 2\sin\theta$. Limaçon ▼

Identify the form. $r = a - b\sin\theta$ with $a = 3,\ b = 2$.

Compare $|a|$ vs $|b|$. $|a| = 3 > |b| = 2$, so this is a dimpled limaçon (no inner loop, has an indent).

Find max and min $r$. Max when $\sin\theta = -1$ (at $\theta = 3\pi/2$): $r = 3 - 2(-1) = 5$. Min when $\sin\theta = 1$ (at $\theta = \pi/2$): $r = 3 - 2 = 1$.

Symmetry. Replacing $\theta$ with $\pi - \theta$ leaves $\sin\theta$ unchanged → symmetric about the $y$-axis (vertical axis).

AnswerDimpled limaçon, vertical axis of symmetry. Max $r = 5$ at $\theta = 3\pi/2$ (downward), min $r = 1$ at $\theta = \pi/2$ (small dimple at top).

29 Sketch $r^2 = 49\cos(2\theta)$. Lemniscate ▼

Identify the form. $r^2 = a^2\cos(2\theta)$ with $a^2 = 49$, so $a = 7$.

Recognize the family. This is a lemniscate (figure-8) along the $x$-axis (since $\cos$, not $\sin$).

Find the tips. Max $r^2$ when $\cos(2\theta) = 1$ (at $\theta = 0$ and $\theta = \pi$): $r^2 = 49 \Rightarrow r = \pm 7$. So the figure-8 reaches $(\pm 7, 0)$.

Domain. $r^2 \geq 0$, so we need $\cos(2\theta) \geq 0$, i.e. $\theta \in [-\pi/4, \pi/4] \cup [3\pi/4, 5\pi/4]$.

AnswerFigure-8 lemniscate along the $x$-axis, tips at $(\pm 7, 0)$.

30 Sketch $r = 3\cos\theta$. Polar circle ▼

Identify the form. $r = a\cos\theta$ with $a = 3$.

Recognize the family. This is a circle tangent to the origin, lying along the $x$-axis (because $\cos$).

Find center and radius. $r = 2(R)\cos\theta$ describes a circle of radius $R$ centered at $(R, 0)$. Here $2R = 3$, so $R = 3/2$.

Sanity check. At $\theta = 0$: $r = 3$ (rightmost point at $(3, 0)$). At $\theta = \pi/2$: $r = 0$ (passes through origin). Matches a circle of radius $3/2$ centered at $(3/2, 0)$.

AnswerCircle of radius $3/2$ centered at $(3/2,\ 0)$, passing through the origin.

Day 4

Conics & Matrices

Graphing conics, writing equations from key features, plus matrix inverses and determinants (2×2 and 3×3).

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Video tutorials for Day 4

▶ Conic Sections — Basic Introduction (circles, ellipses, hyperbolas, parabolas) Organic Chem Tutor ▶ Conic Sections Quiz — Graph & Identify in Standard Form Organic Chem Tutor ▶ Determinant of 2×2 and 3×3 Matrices Organic Chem Tutor ▶ Inverse of a 2×2 Matrix Organic Chem Tutor

Identify the conic at a glance

Look at the equation form before doing anything else:

Both squared, same coefficient, plus
e.g. $(x-h)^2 + (y-k)^2 = r^2$ → Circle

Both squared, different coefficients, plus
e.g. $\dfrac{(x-h)^2}{a^2} + \dfrac{(y-k)^2}{b^2} = 1$ → Ellipse

Both squared, minus sign
e.g. $\dfrac{(x-h)^2}{a^2} - \dfrac{(y-k)^2}{b^2} = 1$ → Hyperbola

Only one variable squared
e.g. $(y-k)^2 = 4p(x-h)$ → Parabola

Memory hooks

Ellipse: $a$ is always the larger denominator (under the major axis). Hyperbola: $a$ is under the positive term. Asymptote slope $= b/a$ for horizontal opening, $a/b$ for vertical opening. Parabola: the variable that's not squared is the direction it opens.

Graph the conic (problems 31–34)

31 $\dfrac{(x-2)^2}{16} + (y+3)^2 = 1$ Ellipse ▼

Identify the form. Both variables squared, both positive, different denominators → ellipse.

Read the center. Standard form is $\dfrac{(x-h)^2}{a^2} + \dfrac{(y-k)^2}{b^2} = 1$. Match: $h = 2,\ k = -3$. Center: $(2, -3)$. (Sign flips: $x-2$ means $h = +2$; $y+3$ means $k = -3$.)

Find $a$ and $b$. $a^2 = 16,\ b^2 = 1 \Rightarrow a = 4,\ b = 1$.

Determine orientation. The bigger denominator (16) is under $x$, so the major axis is horizontal.

Vertices and co-vertices. Vertices: center $\pm a$ along the major axis $\Rightarrow (2 \pm 4, -3) = (-2, -3)$ and $(6, -3)$. Co-vertices: $(2, -3 \pm 1) = (2, -2)$ and $(2, -4)$.

AnswerEllipse. Center $(2, -3)$. $a=4,\ b=1$, horizontal major axis. Vertices: $(-2, -3)$ and $(6, -3)$. Co-vertices: $(2, -2)$ and $(2, -4)$.

32 $\dfrac{(x+3)^2}{9} - \dfrac{y^2}{25} = 1$ Hyperbola ▼

Identify the form. Two squared terms with a minus sign between them → hyperbola.

Read the center. Match standard form $\dfrac{(x-h)^2}{a^2} - \dfrac{(y-k)^2}{b^2} = 1$. Center: $(-3, 0)$.

Find $a$ and $b$. $a^2 = 9,\ b^2 = 25 \Rightarrow a = 3,\ b = 5$.

Orientation. The positive term contains $x$, so the hyperbola opens horizontally (left and right).

Vertices. Vertices are along the opening axis at distance $a$ from center: $(-3 \pm 3, 0) = (-6, 0)$ and $(0, 0)$.

Asymptote slopes. For a horizontal hyperbola: slopes $= \pm b/a = \pm 5/3$. Asymptotes pass through the center.

AnswerHyperbola, opens horizontally. Center $(-3, 0)$. $a=3,\ b=5$. Vertices: $(-6, 0)$ and $(0, 0)$. Asymptote slopes: $\pm 5/3$.

33 $(y-2)^2 = 4x$ Parabola ▼

Identify the form. Only $y$ is squared → parabola opening horizontally (left or right).

Match standard form. Standard form is $(y-k)^2 = 4p(x-h)$. Compare to $(y-2)^2 = 4(x-0)$: $h = 0,\ k = 2$. Vertex: $(0, 2)$.

Solve for $p$. $4p = 4 \Rightarrow p = 1$.

Direction. $p > 0$ with $y$-squared → opens to the right.

Focus and directrix. Focus is $p$ units from the vertex in the opening direction: $(0+1, 2) = (1, 2)$. Directrix is the perpendicular line $p$ units the other way: $x = -1$.

AnswerParabola, opens right (since $y$ is squared). Vertex $(0, 2)$. $4p = 4 \Rightarrow p = 1$. Focus $(1, 2)$. Directrix $x = -1$.

34 $(x-3)^2 + (y+2)^2 = 9$ Circle ▼

Identify the form. Both squared, same coefficient (both 1), with a $+$ sign → circle.

Read the center. Standard form $(x-h)^2 + (y-k)^2 = r^2$. Match: $h = 3,\ k = -2$. Center: $(3, -2)$.

Find the radius. $r^2 = 9 \Rightarrow r = 3$.

AnswerCircle. Center $(3, -2)$, radius $r = 3$.

Write the equation (problems 35–36)

35 Write the equation of a hyperbola with vertices at $(0, \pm 4)$ and asymptote slope $\dfrac{4}{3}$. Write hyperbola ▼

Locate the vertices. Vertices are $(0, 4)$ and $(0, -4)$ — both on the $y$-axis. So the hyperbola opens vertically, and the center is the midpoint: $(0, 0)$.

Pick the correct standard form. Vertical hyperbola: $\dfrac{(y-k)^2}{a^2} - \dfrac{(x-h)^2}{b^2} = 1$. With center at origin: $\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1$.

Find $a$. $a$ = distance from center to vertex = 4. So $a^2 = 16$.

Find $b$ from the slope. For a vertical hyperbola, asymptote slope $= a/b$. Given slope $= 4/3$: $\dfrac{a}{b} = \dfrac{4}{3} \Rightarrow \dfrac{4}{b} = \dfrac{4}{3} \Rightarrow b = 3$. So $b^2 = 9$.

Write the equation. $\dfrac{y^2}{16} - \dfrac{x^2}{9} = 1$.

Answer$\dfrac{y^2}{16} - \dfrac{x^2}{9} = 1$ (vertical hyperbola; $a=4$ from vertices, $b=3$ from slope $a/b = 4/3$)

36 Write the equation of an ellipse with center $(2, -1)$, vertical major axis of length 8, minor axis length 4. Write ellipse ▼

Pick the correct form. Major axis is vertical, so the bigger denominator goes under $y$: $\dfrac{(x-h)^2}{b^2} + \dfrac{(y-k)^2}{a^2} = 1$.

Center. $(h, k) = (2, -1)$.

Find $a$ from the major axis. Major axis length $= 2a = 8 \Rightarrow a = 4$. So $a^2 = 16$.

Find $b$ from the minor axis. Minor axis length $= 2b = 4 \Rightarrow b = 2$. So $b^2 = 4$.

Assemble. $\dfrac{(x-2)^2}{4} + \dfrac{(y+1)^2}{16} = 1$.

Answer$\dfrac{(x-2)^2}{4} + \dfrac{(y+1)^2}{16} = 1$ (vertical major axis $\Rightarrow a^2$ under $y$; $a=4,\ b=2$)

Matrices (problems 37–39)

2×2 inverse: $\begin{bmatrix}a&b\\c&d\end{bmatrix}^{-1} = \dfrac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$
2×2 determinant: $\det = ad - bc$
3×3 determinant: expand along any row or column. Pick the one with the most zeros.

For example 2×2 determinant: $\begin{bmatrix}3 & 2 \\ 1 & 4\end{bmatrix}$ → $\det = (3)(4) - (2)(1) = 10$.
2×2 inverse of the same: swap diagonal, negate off-diagonal, divide by det → $\dfrac{1}{10}\begin{bmatrix}4 & -2 \\ -1 & 3\end{bmatrix} = \begin{bmatrix}2/5 & -1/5 \\ -1/10 & 3/10\end{bmatrix}$.
Singular check: if a matrix had $\det = 0$, no inverse exists.

Memory hooks for the 2×2 inverse

Swap the diagonal, negate the off-diagonal, divide by the determinant. That is: $a$ and $d$ trade places, $b$ and $c$ get a minus sign in front, then everything gets scaled by $1/\det$.

3×3 cofactor expansion along the first row: $\det = a_{11}(M_{11}) - a_{12}(M_{12}) + a_{13}(M_{13})$, where $M_{ij}$ is the $2 \times 2$ determinant of the matrix with row $i$ and column $j$ deleted. Watch the alternating signs.

37 Find the inverse of $\begin{bmatrix} -2 & -3 \\ -2 & 4 \end{bmatrix}$. 2×2 inverse ▼

Compute the determinant. $\det = ad - bc = (-2)(4) - (-3)(-2) = -8 - 6 = -14$.

Check it's nonzero. $-14 \neq 0$, so the inverse exists.

Apply the formula: swap diagonal, negate off-diagonal, divide by det. Swap $a$ and $d$ (both are $-2$ and $4$); negate $b$ and $c$. Result before scaling: $\begin{bmatrix}4 & 3 \\ 2 & -2\end{bmatrix}$.

Divide by the determinant. $A^{-1} = \dfrac{1}{-14}\begin{bmatrix}4 & 3 \\ 2 & -2\end{bmatrix} = \begin{bmatrix}-2/7 & -3/14 \\ -1/7 & 1/7\end{bmatrix}$.

Answer$\det = (-2)(4) - (-3)(-2) = -8 - 6 = -14$.
Inverse: $\dfrac{1}{-14}\begin{bmatrix} 4 & 3 \\ 2 & -2 \end{bmatrix} = \begin{bmatrix} -2/7 & -3/14 \\ -1/7 & 1/7 \end{bmatrix}$

37 (det) Find the determinant of $\begin{bmatrix} 2 & -3 \\ 4 & 4 \end{bmatrix}$. 2×2 det ▼

Apply $\det = ad - bc$. $a = 2,\ b = -3,\ c = 4,\ d = 4$.

Compute. $\det = (2)(4) - (-3)(4) = 8 - (-12) = 8 + 12 = 20$.

Answer$\det = (2)(4) - (-3)(4) = 8 + 12 = 20$

38 Find the determinant of $\begin{bmatrix} -2 & 5 \\ 4 & -3 \end{bmatrix}$. 2×2 det ▼

Apply $\det = ad - bc$. $a = -2,\ b = 5,\ c = 4,\ d = -3$.

Compute. $\det = (-2)(-3) - (5)(4) = 6 - 20 = -14$.

Answer$\det = (-2)(-3) - (5)(4) = 6 - 20 = -14$

39 Find the determinant of $\begin{bmatrix} 1 & 1 & -2 \\ 3 & 4 & 1 \\ -1 & 0 & 3 \end{bmatrix}$. 3×3 det ▼

Pick the row or column with the most zeros. Row 3 contains a $0$: $[-1,\ 0,\ 3]$. Expand along it — the zero kills one of the three minor calculations.

Write the cofactor expansion. $\det = (-1)\cdot C_{31} + (0)\cdot C_{32} + (3)\cdot C_{33}$, where $C_{ij} = (-1)^{i+j} M_{ij}$ and $M_{ij}$ is the $2\times 2$ minor with row $i$ and column $j$ deleted. Sign pattern for row 3: $+, -, +$.

Compute $M_{31}$ (delete row 3, column 1). Remaining: $\begin{bmatrix}1 & -2 \\ 4 & 1\end{bmatrix}$. $M_{31} = (1)(1) - (-2)(4) = 1 + 8 = 9$.

Compute $M_{33}$ (delete row 3, column 3). Remaining: $\begin{bmatrix}1 & 1 \\ 3 & 4\end{bmatrix}$. $M_{33} = (1)(4) - (1)(3) = 1$.

Combine with the signs. $\det = (-1)(+9) + 0 + (3)(+1) = -9 + 3 = -6$.

AnswerExpand along the bottom row (has a zero): $\det = (-1)\begin{vmatrix}1&-2\\4&1\end{vmatrix} - 0 + 3\begin{vmatrix}1&1\\3&4\end{vmatrix} = (-1)(9) + 3(1) = -6$

Day 5

Sequences & Series

Arithmetic and geometric sequences: nth-term formulas, specific terms, finite sums, and infinite geometric sums.

0 / 7 attempted

Video tutorials for Day 5

▶ Arithmetic Sequences & Series — Basic Introduction Organic Chem Tutor ▶ Geometric Sequences & Series — including infinite sums Organic Chem Tutor

Arithmetic or geometric? Quick test

Given a sequence, the first thing to do is figure out which type:

Arithmetic — constant difference. Subtract consecutive terms; if you always get the same number, that's $d$.
Geometric — constant ratio. Divide consecutive terms; if you always get the same number, that's $r$.

Memory hooks

Arithmetic adds, geometric multiplies. Arithmetic terms grow linearly; geometric terms grow (or shrink) exponentially.

Infinite geometric sum $S = \dfrac{a_1}{1-r}$ only converges when $|r| < 1$. Otherwise the series diverges and there's no sum to find.

40 Write the $n$th term of the arithmetic sequence: $3,\ 7,\ 11,\ 15,\ \ldots$ Arithmetic nth ▼

Identify the type. Differences: $7-3 = 4$, $11-7 = 4$, $15-11 = 4$. Constant difference → arithmetic with $d = 4$.

First term. $a_1 = 3$.

Apply $a_n = a_1 + (n-1)d$. $a_n = 3 + (n-1)(4)$.

Simplify. $a_n = 3 + 4n - 4 = 4n - 1$.

Answer$a_1 = 3,\ d = 4$. $a_n = 3 + (n-1)\cdot 4 = 4n - 1$.

41 Find $a_{25}$ of the arithmetic sequence: $35,\ 33,\ 31,\ 29,\ \ldots$ (worksheet shows '32' — almost certainly a typo for '31'; solving the most likely intended version) Arithmetic term ▼

Identify the type (using the corrected sequence). Differences: $33-35=-2,\ 31-33=-2,\ 29-31=-2$. Arithmetic with $d = -2$.

First term. $a_1 = 35$.

Apply $a_n = a_1 + (n-1)d$. $a_{25} = 35 + (25-1)(-2) = 35 + 24(-2)$.

Compute. $a_{25} = 35 - 48 = -13$.

Answer$a_1 = 35,\ d = -2$. $a_{25} = 35 + 24\cdot(-2) = 35 - 48 = -13$.

42 Write the $n$th term of the geometric sequence: $3,\ -12,\ 48,\ -192,\ \ldots$ Geometric nth ▼

Identify the type. Ratios: $-12/3 = -4,\ 48/(-12) = -4,\ -192/48 = -4$. Constant ratio → geometric with $r = -4$.

First term. $a_1 = 3$.

Apply $a_n = a_1 \, r^{n-1}$. $a_n = 3 \cdot (-4)^{n-1}$.

Sanity check the alternating signs. At $n=1$: $3 \cdot (-4)^0 = 3$ ✓. At $n=2$: $3 \cdot (-4)^1 = -12$ ✓. At $n=3$: $3 \cdot (-4)^2 = 48$ ✓.

Answer$a_1 = 3,\ r = -4$. $a_n = 3\cdot(-4)^{n-1}$.

43 Find the 13th term of the sequence: $-1,\ 2,\ -4,\ 8,\ \ldots$ Geometric term ▼

Identify the type. Ratios: $2/(-1) = -2,\ -4/2 = -2,\ 8/(-4) = -2$. Geometric, $r = -2$.

First term. $a_1 = -1$.

Apply $a_n = a_1\, r^{n-1}$. $a_{13} = -1 \cdot (-2)^{12}$.

Evaluate $(-2)^{12}$. Even exponent → positive. $(-2)^{12} = 2^{12} = 4096$.

Combine. $a_{13} = -1 \cdot 4096 = -4096$.

AnswerGeometric: $a_1 = -1,\ r = -2$. $a_{13} = -1\cdot(-2)^{12} = -1\cdot 4096 = -4096$.

44 Find the sum: $15 + 22 + 29 + 36 + \ldots + 106$. Arithmetic sum ▼

Identify the type. Differences: $22-15=7,\ 29-22=7$. Arithmetic, $d = 7$. First term $a_1 = 15$, last term $a_n = 106$.

Find $n$. From $a_n = a_1 + (n-1)d$: $106 = 15 + (n-1)(7) \Rightarrow 91 = 7(n-1) \Rightarrow n - 1 = 13 \Rightarrow n = 14$.

Apply the arithmetic-sum formula. $S_n = \dfrac{n}{2}(a_1 + a_n)$.

Compute. $S_{14} = \dfrac{14}{2}(15 + 106) = 7 \cdot 121 = 847$.

AnswerArithmetic, $a_1 = 15,\ d = 7$. Find $n$: $106 = 15 + (n-1)\cdot 7 \Rightarrow n = 14$.
$S_{14} = \dfrac{14}{2}(15 + 106) = 7 \cdot 121 = 847$.

45 Find the sum of the first 12 terms: $1 + 5 + 25 + 125 + \ldots$ Geometric sum ▼

Identify the type. Ratios: $5/1 = 5,\ 25/5 = 5$. Geometric, $r = 5$, $a_1 = 1$. Want $S_{12}$.

Apply $S_n = a_1\,\dfrac{1 - r^n}{1 - r}$. $S_{12} = 1 \cdot \dfrac{1 - 5^{12}}{1 - 5} = \dfrac{1 - 5^{12}}{-4}$.

Compute $5^{12}$. $5^{12} = 244{,}140{,}625$.

Plug in and simplify. $S_{12} = \dfrac{1 - 244{,}140{,}625}{-4} = \dfrac{-244{,}140{,}624}{-4} = 61{,}035{,}156$.

AnswerGeometric, $a_1 = 1,\ r = 5,\ n = 12$. $S_{12} = \dfrac{1 - 5^{12}}{1 - 5} = \dfrac{1 - 244{,}140{,}625}{-4} = 61{,}035{,}156$.

46 Find the sum of the infinite geometric series: $16,\ 8,\ 4,\ 2,\ \ldots$ Infinite sum ▼

Identify the type. Ratios: $8/16 = 1/2,\ 4/8 = 1/2$. Geometric, $r = 1/2$, $a_1 = 16$.

Check convergence. $|r| = 1/2 < 1$, so the infinite series converges and the formula applies.

Apply $S = \dfrac{a_1}{1 - r}$. $S = \dfrac{16}{1 - 1/2} = \dfrac{16}{1/2}$.

Compute. $S = 32$.

Answer$a_1 = 16,\ r = 1/2$ (and $|r| < 1$, so the sum converges). $S = \dfrac{a_1}{1-r} = \dfrac{16}{1 - 1/2} = \dfrac{16}{1/2} = 32$.