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Precalculusthe reference sheet

Every topic in the course, boiled down to its rule, its formula, and one worked example. For review and quick lookups, not for learning a topic the first time.

12 units, 43 sections, 296 topics. Press / to search.

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Unit 1 18 topics

Sequences and Series

Arithmetic Series, Finite Geometric Series, and The Binomial Theorem.

1.1 Arithmetic Series

11011291138114711561165117411831192111011110 columns1 + 2 + 3 + … + 1010 + 9 + 8 + … + 1Every column totals 11.10 columns × 11 = 110,so the sum is 55S = n(a1 + an) / 2= 10(1 + 10) / 2 = 55
Write the series forward and backward and add: every pair has the same total. Half of n pairs gives the sum.
  1. 1.1.1

    Sigma Notation

    ∑\sum means add. Plug each integer from the lower limit to the upper limit into the expression and add the results. There are (upper minus lower plus 1) terms.

    e.g. ∑k=14(2k+1)=3+5+7+9=24\sum_{k=1}^{4} (2k + 1) = 3 + 5 + 7 + 9 = 24

    ∑i=1nai=a1+a2+⋯+an\displaystyle \sum_{i=1}^{n} a_i = a_1 + a_2 + \cdots + a_n
  2. 1.1.2

    Properties of Finite Series

    A sum can be split across addition, and a constant factor can be pulled out. Summing a constant nn times gives nn times the constant.

    e.g. ∑i=110(3i+2)=3⋅55+2⋅10=185\sum_{i=1}^{10} (3i + 2) = 3 \cdot 55 + 2 \cdot 10 = 185

    ∑(ai+bi)=∑ai+∑bi\displaystyle \sum (a_i + b_i) = \sum a_i + \sum b_i
    ∑c ai=c∑ai\displaystyle \sum c\,a_i = c \sum a_i
    ∑i=1nc=cn\displaystyle \sum_{i=1}^{n} c = cn
    ∑i=1ni=n(n+1)2\displaystyle \sum_{i=1}^{n} i = \frac{n(n+1)}{2}
  3. 1.1.3

    Expressing an Arithmetic Series in Sigma Notation

    Write the general term a1+(k−1)da_1 + (k - 1)d, simplify it, and sum from k=1k = 1 to the number of terms.

    e.g. 4+7+10+⋯+31=∑k=110(3k+1)4 + 7 + 10 + \cdots + 31 = \sum_{k=1}^{10} (3k + 1)

    ∑k=1n(a1+(k−1)d)\displaystyle \sum_{k=1}^{n} \big(a_1 + (k - 1)d\big)
  4. 1.1.4

    Finding the Sum of an Arithmetic Series

    Average the first and last terms, then multiply by the number of terms. Use the second form when the last term is unknown.

    e.g. 2+5+8+⋯+292 + 5 + 8 + \cdots + 29: n=10n = 10, S=10(2+29)2=155S = \tfrac{10(2 + 29)}{2} = 155

    Sn=n(a1+an)2\displaystyle S_n = \frac{n(a_1 + a_n)}{2}
    Sn=n2(2a1+(n−1)d)\displaystyle S_n = \frac{n}{2}\big(2a_1 + (n - 1)d\big)
  5. 1.1.5

    Finding the First Term of an Arithmetic Series

    Substitute the known values into a sum formula and solve for a1a_1.

    e.g. S8=100S_8 = 100, d=3d = 3: 100=82(2a1+21)⇒25=2a1+21⇒a1=2100 = \tfrac82(2a_1 + 21) \Rightarrow 25 = 2a_1 + 21 \Rightarrow a_1 = 2

    a1=2Snn−an\displaystyle a_1 = \frac{2S_n}{n} - a_n
  6. 1.1.6

    Calculating the Number of Terms in an Arithmetic Series

    If the last term is known, use the formula below. If only the sum is known, set up SnS_n, solve the quadratic in nn, and keep the positive whole-number answer.

    e.g. 5+9+13+⋯+815 + 9 + 13 + \cdots + 81: n=81−54+1=20n = \tfrac{81 - 5}{4} + 1 = 20

    n=an−a1d+1\displaystyle n = \frac{a_n - a_1}{d} + 1
  7. 1.1.7

    Modeling With Arithmetic Series

    Use when a quantity changes by the same amount each step and you need the total. Identify a1a_1, dd, and nn, then apply the sum formula.

    e.g. 15 rows, 20 seats in row 1, 2 more per row: last row 20+14⋅2=4820 + 14 \cdot 2 = 48, total 15(20+48)2=510\tfrac{15(20 + 48)}{2} = 510

1.2 Finite Geometric Series

  1. 1.2.1

    The Sum of a Finite Geometric Series

    A geometric series adds terms that each get multiplied by the same ratio rr. With first term aa, ratio r≠1r \ne 1, and nn terms, use the sum formula.

    e.g. 2+6+18+542 + 6 + 18 + 54: a=2a = 2, r=3r = 3, n=4n = 4, so S=2(1−81)1−3=80S = \frac{2(1 - 81)}{1 - 3} = 80

    Sn=a(1−rn)1−r\displaystyle S_n = \frac{a(1 - r^n)}{1 - r}
  2. 1.2.2

    The Sum of the First N Terms of a Geometric Series

    Identify the first term a1a_1, the ratio rr, and the number of terms nn, then substitute. If r=1r = 1, every term is the same and the sum is na1n a_1.

    e.g. First 6 terms of 1,2,4,…1, 2, 4, \dots: S6=1(1−64)1−2=63S_6 = \frac{1(1 - 64)}{1 - 2} = 63

    Sn=a1(1−rn)1−r\displaystyle S_n = \frac{a_1(1 - r^n)}{1 - r}
    Sn=na1  if  r=1\displaystyle S_n = n a_1 \;\text{if}\; r = 1
  3. 1.2.3

    Writing Geometric Series in Sigma Notation

    Write the general term a1rk−1a_1 r^{k-1} and let kk run from 1 to nn. Find nn by solving a1rn−1=a_1 r^{n-1} = last term.

    e.g. 3+6+12+⋯+96=∑k=163⋅2k−13 + 6 + 12 + \dots + 96 = \sum_{k=1}^{6} 3 \cdot 2^{k-1}, since 3⋅25=963 \cdot 2^5 = 96

    ∑k=1na1rk−1\displaystyle \sum_{k=1}^{n} a_1 r^{k-1}
  4. 1.2.4

    Sums of Finite Geometric Series Given in Sigma Notation

    Plug the starting index into the term to get the first term aa. The ratio rr is the base of the exponent. The number of terms is top index minus bottom index plus 1.

    e.g. ∑k=042k\sum_{k=0}^{4} 2^k: a=1a = 1, r=2r = 2, n=5n = 5, so S=1−321−2=31S = \frac{1 - 32}{1 - 2} = 31

    n=top−bottom+1\displaystyle n = \text{top} - \text{bottom} + 1
    S=a(1−rn)1−r\displaystyle S = \frac{a(1 - r^n)}{1 - r}
  5. 1.2.5

    Solving Geometric Series Problems Using Exponential Equations and Inequalities

    To find how many terms reach a target, set the sum formula equal to (or greater than) the target, isolate rnr^n, and take logs. For "at least" questions, round nn up. Dividing by the log of a number less than 1 (a negative number) flips the inequality.

    e.g. 1+2+4+⋯>10001 + 2 + 4 + \dots > 1000: 2n−1>1000⇒n>log⁡21001≈9.972^n - 1 > 1000 \Rightarrow n > \log_2 1001 \approx 9.97, so n=10n = 10

    a(1−rn)1−r≥T\displaystyle \frac{a(1 - r^n)}{1 - r} \ge T
    ⇒  isolate rn, take logs\displaystyle \Rightarrow\; \text{isolate } r^n \text{, take logs}
  6. 1.2.6

    Modeling With Geometric Series

    Use a geometric series when an amount changes by the same percent each step and you need the running total. The first amount is aa; the ratio is 1 plus the growth rate, or 1 minus the decay rate.

    e.g. A ball falls 10 ft, then each fall is half the last. First 4 falls: 10(1−0.54)1−0.5=18.75\frac{10(1 - 0.5^4)}{1 - 0.5} = 18.75 ft

    r=1±rate\displaystyle r = 1 \pm \text{rate}
    Sn=a(1−rn)1−r\displaystyle S_n = \frac{a(1 - r^n)}{1 - r}
  7. 1.2.7

    Modeling Financial Problems Using Geometric Series

    Equal deposits PP made at the end of each period, earning rate ii per period, form a geometric series with ratio 1+i1 + i. Its sum is the future value after nn deposits.

    e.g. $100 at the end of each year for 3 years at 10% per year: 100+110+121=100⋅1.13−10.1=$331100 + 110 + 121 = 100 \cdot \frac{1.1^3 - 1}{0.1} = \$331

    FV=P⋅(1+i)n−1i\displaystyle FV = P \cdot \frac{(1 + i)^n - 1}{i}

1.3 The Binomial Theorem

11112113311464115101051row 0row 1row 2row 3row 4row 5(a + b)4 =a4 + 4a3b + 6a2b2 + 4ab3 + b4Row n holds the coefficientsof (a + b)n.Each entry is the sum of thetwo above it: 3 + 3 = 6.
Each number in Pascal’s triangle is the sum of the two above it. Row 4 gives the coefficients 1, 4, 6, 4, 1 of (a + b)⁴.
  1. 1.3.1

    Pascal's Triangle and the Binomial Coefficients

    Each entry is the sum of the two entries above it. Row nn (counting the top as row 0) lists (n0)\binom{n}{0} through (nn)\binom{n}{n}.

    e.g. Row 4 is 1,4,6,4,11, 4, 6, 4, 1, so (42)=6\binom42 = 6

    (nr)=n!r! (n−r)!\displaystyle \binom{n}{r} = \frac{n!}{r!\,(n - r)!}
    (nr)=(n−1r−1)+(n−1r)\displaystyle \binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}
  2. 1.3.2

    Expanding a Binomial Using Binomial Coefficients

    Powers of aa fall from nn to 0 while powers of bb rise from 0 to nn. Coefficients come from row nn of Pascal's triangle. Keep any negative sign with bb.

    e.g. (x−2)3=x3−6x2+12x−8(x - 2)^3 = x^3 - 6x^2 + 12x - 8

    (a+b)n=∑k=0n(nk)an−kbk\displaystyle (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
  3. 1.3.3

    The Special Case of the Binomial Theorem

    With a=1a = 1, the expansion of (1+x)n(1 + x)^n uses the binomial coefficients directly. Setting x=1x = 1 shows that row nn of Pascal's triangle sums to 2n2^n.

    e.g. (1+x)4=1+4x+6x2+4x3+x4(1 + x)^4 = 1 + 4x + 6x^2 + 4x^3 + x^4

    (1+x)n=∑k=0n(nk)xk\displaystyle (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k
    =1+nx+(n2)x2+⋯\displaystyle = 1 + nx + \binom{n}{2}x^2 + \cdots
  4. 1.3.4

    Approximating Values Using the Binomial Theorem

    Write the number as (1+x)n(1 + x)^n with xx small and keep the first few terms. Later terms are tiny.

    e.g. 1.025≈1+5(0.02)+10(0.0004)=1.1041.02^5 \approx 1 + 5(0.02) + 10(0.0004) = 1.104

    (1+x)n≈1+nx+(n2)x2\displaystyle (1 + x)^n \approx 1 + nx + \binom{n}{2}x^2

Unit 2 16 topics

Rational Functions

Rational Expressions and Equations, and Rational Functions.

2.4 Rational Expressions and Equations

abcd=⟹ad = bcThen check that no denominator is zero.
Cross-multiplying is multiplying both sides by both denominators at once. It only works when each side is a single fraction.
  1. 2.4.1

    Closure Properties of Polynomials

    Polynomials are closed under addition, subtraction, and multiplication: the result is always a polynomial. They are not closed under division, since a quotient like 1x\frac{1}{x} is not a polynomial.

    e.g. (x+1)(x−2)=x2−x−2(x + 1)(x - 2) = x^2 - x - 2 is a polynomial; xx+1\frac{x}{x + 1} is not

  2. 2.4.2

    Closure Properties of Rational Expressions

    Rational expressions are closed under addition, subtraction, multiplication, and division by a nonzero rational expression: the result is always a ratio of polynomials.

    e.g. 1x+1x+1=2x+1x(x+1)\frac{1}{x} + \frac{1}{x + 1} = \frac{2x + 1}{x(x + 1)}, still rational

    ab+cd=ad+bcbd\displaystyle \frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}
    ab⋅cd=acbd\displaystyle \frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}
  3. 2.4.3

    Rational Equations With Three Terms

    Multiply every term by the LCD to clear the fractions, then solve. Reject any answer that makes a denominator zero.

    e.g. 3x−12=1x\frac{3}{x} - \frac{1}{2} = \frac{1}{x}: multiply by 2x2x: 6−x=26 - x = 2, so x=4x = 4

  4. 2.4.4

    Advanced Rational Equations

    Factor every denominator first to find the LCD and the excluded values. Multiply through, solve the resulting equation (often a quadratic), and discard extraneous roots.

    e.g. xx−2−1x+2=8x2−4\frac{x}{x - 2} - \frac{1}{x + 2} = \frac{8}{x^2 - 4} gives x2+x−6=0x^2 + x - 6 = 0, so x=−3x = -3 (x=2x = 2 is extraneous)

  5. 2.4.5

    Further Advanced Rational Equations

    Use the same method with more terms or harder denominators. List the excluded values before solving and check every answer against them. If every answer is excluded, there is no solution.

    e.g. 2x−1+1x+1=4x2−1\frac{2}{x - 1} + \frac{1}{x + 1} = \frac{4}{x^2 - 1}: 2(x+1)+(x−1)=4⇒x=12(x + 1) + (x - 1) = 4 \Rightarrow x = 1, which is excluded, so no solution

2.5 Rational Functions

xyVA x = −2HA y = 2hole (1, 4/3)y = (2x2 − 2) ÷ ((x − 1)(x + 2))= 2(x + 1)(x − 1) ÷ ((x − 1)(x + 2))= 2(x + 1) ÷ (x + 2), x ≠ 1Canceled factor (x − 1): holeLeftover zero of the bottom:vertical asymptote x = −2Same degree top and bottom:HA is y = 2 ÷ 1 = 2The graph never reaches eitherdashed line or the open circle.
A factor that cancels leaves a hole; a zero of the denominator that remains becomes a vertical asymptote. With equal degrees, the horizontal asymptote is the ratio of leading coefficients.
  1. 2.5.1

    Roots of Rational Functions

    The zeros are the xx-values that make the numerator 0 but not the denominator. Cancel common factors first: a cancelled factor gives a hole, not a root.

    e.g. f(x)=x2−4x−2f(x) = \frac{x^2 - 4}{x - 2}: root only at x=−2x = -2 (there is a hole at x=2x = 2)

    p(x)q(x)=0\displaystyle \frac{p(x)}{q(x)} = 0
    ⇒  p(x)=0 and q(x)≠0\displaystyle \Rightarrow\; p(x) = 0 \text{ and } q(x) \ne 0
  2. 2.5.2

    Vertical Asymptotes of Rational Functions

    Cancel common factors first. Each xx-value that still makes the denominator zero is a vertical asymptote.

    e.g. f(x)=x+1(x−3)(x+1)f(x) = \frac{x + 1}{(x - 3)(x + 1)}: vertical asymptote x=3x = 3 only

  3. 2.5.3

    Locating Holes in Rational Functions

    A factor that cancels completely from the numerator and denominator gives a hole. Its xx-value makes that factor zero; its yy-value comes from the simplified function. If the factor still remains in the denominator, it is an asymptote instead.

    e.g. f(x)=x2−1x−1=x+1f(x) = \frac{x^2 - 1}{x - 1} = x + 1 for x≠1x \ne 1: hole at (1,2)(1, 2)

  4. 2.5.4

    Horizontal Asymptotes of Rational Functions

    Compare the degree nn of the numerator with the degree mm of the denominator. Leading coefficients are ana_n and bmb_m.

    e.g. 3x2+12x2−x\frac{3x^2 + 1}{2x^2 - x}: horizontal asymptote y=32y = \tfrac32

    n<m:  y=0\displaystyle n < m: \; y = 0
    n=m:  y=anbm\displaystyle n = m: \; y = \frac{a_n}{b_m}
    n>m:  none\displaystyle n > m: \; \text{none}
  5. 2.5.5

    End Behavior of Rational Functions

    Divide to write ff as a polynomial q(x)q(x) plus a proper fraction. The fraction goes to 0 as x→±∞x \to \pm\infty, so ff behaves like q(x)q(x). If the numerator's degree is one more than the denominator's, qq is linear and gives a slant asymptote.

    e.g. x2+1x=x+1x\frac{x^2 + 1}{x} = x + \frac{1}{x}: slant asymptote y=xy = x

    f(x)=q(x)+r(x)d(x)\displaystyle f(x) = q(x) + \frac{r(x)}{d(x)}
    ⇒  f(x)≈q(x) for large ∣x∣\displaystyle \Rightarrow\; f(x) \approx q(x) \text{ for large } \lvert x \rvert
  6. 2.5.6

    Infinite Limits of Rational Functions

    Near a vertical asymptote x=ax = a, the function goes to ∞\infty or −∞-\infty. Find the sign by testing a value just left and just right of aa.

    e.g. f(x)=1x−2f(x) = \frac{1}{x - 2}: lim⁡x→2+f(x)=∞\lim_{x \to 2^+} f(x) = \infty, lim⁡x→2−f(x)=−∞\lim_{x \to 2^-} f(x) = -\infty

    lim⁡x→a−f(x)=±∞\displaystyle \lim_{x \to a^-} f(x) = \pm\infty
    lim⁡x→a+f(x)=±∞\displaystyle \lim_{x \to a^+} f(x) = \pm\infty
  7. 2.5.7

    Infinite Limits of Rational Functions: Advanced Cases

    Find the sign of each factor near aa. If (x−a)(x - a) has an even power in the denominator, both one-sided limits have the same sign. An odd power gives opposite signs.

    e.g. f(x)=x−3(x−1)2f(x) = \frac{x - 3}{(x - 1)^2} near x=1x = 1: numerator near −2-2, denominator small and positive, so f→−∞f \to -\infty from both sides

  8. 2.5.8

    The Domain and Range of a Rational Function

    The domain is all real numbers except the zeros of the original denominator, including holes. For the range, solve y=f(x)y = f(x) for xx and exclude the yy-values with no solution. Also exclude a hole's yy-value if the graph does not reach it elsewhere.

    e.g. f(x)=2x+1x−3f(x) = \frac{2x + 1}{x - 3}: x=3y+1y−2x = \frac{3y + 1}{y - 2}, so domain x≠3x \ne 3, range y≠2y \ne 2

  9. 2.5.9

    Identifying a Rational Function From a Graph

    Each vertical asymptote x=ax = a gives a factor (x−a)(x - a) in the denominator. Each xx-intercept rr gives a factor (x−r)(x - r) in the numerator. Use the horizontal asymptote or a known point to find the constant kk.

    e.g. Vertical asymptote x=1x = 1, xx-intercept −2-2, horizontal asymptote y=3y = 3: f(x)=3(x+2)x−1f(x) = \frac{3(x + 2)}{x - 1}

    f(x)=k(x−r)x−a\displaystyle f(x) = \frac{k(x - r)}{x - a}
  10. 2.5.10

    Identifying a Rational Function From a Graph Containing Holes

    A hole at x=cx = c means (x−c)(x - c) is a factor of both the numerator and the denominator. Build the function from the asymptotes and intercepts, then multiply top and bottom by (x−c)(x - c).

    e.g. The graph of y=2xy = \frac{2}{x} with a hole at (1,2)(1, 2): f(x)=2(x−1)x(x−1)f(x) = \frac{2(x - 1)}{x(x - 1)}

    f(x)=g(x) (x−c)h(x) (x−c)\displaystyle f(x) = \frac{g(x)\,(x - c)}{h(x)\,(x - c)}
  11. 2.5.11

    Identifying the Graph of a Rational Function

    Find the features from the equation and match them to a graph: zeros, vertical asymptotes, holes, the horizontal or slant asymptote, the yy-intercept, and the sign between key xx-values.

    e.g. f(x)=x−1x+2f(x) = \frac{x - 1}{x + 2}: zero at x=1x = 1, asymptotes x=−2x = -2 and y=1y = 1, yy-intercept −12-\tfrac12

Unit 3 21 topics

Inequalities

Quadratic Inequalities, Polynomial Inequalities, and Non-Polynomial Inequalities.

3.6 Quadratic Inequalities

xy14+−+sign(x − 1)(x − 4) > 0Roots: x = 1 and x = 4Test one point in each region:x = 0: (−1)(−4) = 4 > 0 ✓x = 2: (1)(−2) = −2 < 0 ✗x = 5: (4)(1) = 4 > 0 ✓x < 1 or x > 4where the graph is above the x-axis
The product is positive where the parabola is above the x-axis. The roots split the line into regions; test one point in each, and the answer is x < 1 or x > 4.
  1. 3.6.1

    Solving Elementary Quadratic Inequalities

    For c>0c > 0: x2<cx^2 < c gives the values between the roots, and x2>cx^2 > c gives the values outside them. If c<0c < 0, x2<cx^2 < c has no solution and x2>cx^2 > c holds for all xx.

    e.g. x2≤9⇒−3≤x≤3x^2 \le 9 \Rightarrow -3 \le x \le 3

    x2<c  ⇒  −c<x<c\displaystyle x^2 < c \;\Rightarrow\; -\sqrt{c} < x < \sqrt{c}
    x2>c  ⇒  x<−c\displaystyle x^2 > c \;\Rightarrow\; x < -\sqrt{c}
    or  x>c\displaystyle \text{or}\; x > \sqrt{c}
  2. 3.6.2

    Solving Quadratic Inequalities From Graphs

    f(x)>0f(x) > 0 where the graph is above the xx-axis, and f(x)<0f(x) < 0 where it is below. The xx-intercepts are the boundaries. Include them only for ≤\le or ≥\ge.

    e.g. y=(x−1)(x−5)y = (x - 1)(x - 5) opens up, so y<0y < 0 for 1<x<51 < x < 5.

  3. 3.6.3

    Solving Quadratic Inequalities Using the Graphical Method

    Move everything to one side, find the roots, and sketch the parabola. If it opens up, it is negative between the roots and positive outside them. If it opens down, the reverse.

    e.g. x2−2x−8>0⇒(x−4)(x+2)>0⇒x<−2x^2 - 2x - 8 > 0 \Rightarrow (x - 4)(x + 2) > 0 \Rightarrow x < -2 or x>4x > 4

    a>0:  ax2+bx+c<0\displaystyle a > 0:\; ax^2 + bx + c < 0
    between the roots\displaystyle \text{between the roots}
  4. 3.6.4

    Solving Quadratic Inequalities Using the Sign Table Method

    Move everything to one side and factor. Mark the roots on a number line, find the sign of each factor on each interval, and multiply the signs. Keep the intervals with the sign you need.

    e.g. −x2+x+6≥0⇒(x−3)(x+2)≤0⇒−2≤x≤3-x^2 + x + 6 \ge 0 \Rightarrow (x - 3)(x + 2) \le 0 \Rightarrow -2 \le x \le 3

  5. 3.6.5

    Determining Unknown Parameters in Quadratic Equations With Real Solutions

    ax2+bx+c=0ax^2 + bx + c = 0 has real solutions when the discriminant is at least 0, and two distinct real solutions when it is positive. Solve that inequality for the parameter.

    e.g. x2+kx+9=0x^2 + kx + 9 = 0: k2−36≥0⇒k≤−6k^2 - 36 \ge 0 \Rightarrow k \le -6 or k≥6k \ge 6

    b2−4ac≥0\displaystyle b^2 - 4ac \ge 0
  6. 3.6.6

    Determining Unknown Parameters in Quadratic Equations With No Real Solutions

    A quadratic has no real solutions when its discriminant is negative. Solve that inequality for the parameter.

    e.g. x2+4x+k=0x^2 + 4x + k = 0: 16−4k<0⇒k>416 - 4k < 0 \Rightarrow k > 4

    b2−4ac<0\displaystyle b^2 - 4ac < 0
  7. 3.6.7

    Determining Unknown Parameters in Quadratic Equations: Advanced Cases

    If the parameter is in the x2x^2 coefficient, the equation is only quadratic when that coefficient is not 0, so check that case separately. A quadratic is positive for all xx when it opens up and has no real roots.

    e.g. kx2+4x+1=0kx^2 + 4x + 1 = 0 has two real solutions when k≠0k \ne 0 and 16−4k>016 - 4k > 0, so k<4k < 4, k≠0k \ne 0.

    a≠0  and  b2−4ac>0\displaystyle a \ne 0 \;\text{and}\; b^2 - 4ac > 0
    ax2+bx+c>0  for all x\displaystyle ax^2 + bx + c > 0 \;\text{for all } x
      ⟺  a>0,  b2−4ac<0\displaystyle \iff a > 0,\; b^2 - 4ac < 0

3.7 Polynomial Inequalities

  1. 3.7.1

    Inequalities Involving Powers of Binomials

    An even power (x−a)n(x - a)^n is never negative and equals 0 only at x=ax = a. An odd power (x−a)n(x - a)^n has the same sign as x−ax - a.

    e.g. (x−3)4>0⇒x≠3(x - 3)^4 > 0 \Rightarrow x \ne 3; (x−2)5<0⇒x<2(x - 2)^5 < 0 \Rightarrow x < 2

    (x−a)even≥0\displaystyle (x - a)^{\text{even}} \ge 0
    (x−a)odd has the sign of x−a\displaystyle (x - a)^{\text{odd}} \text{ has the sign of } x - a
  2. 3.7.2

    Solving Polynomial Inequalities Using a Graphical Method

    Move everything to one side and factor. Plot the zeros, use the end behavior, and remember the graph crosses at odd-multiplicity zeros and bounces at even ones. Read off where the graph is above or below the axis.

    e.g. x(x−2)(x+3)>0x(x - 2)(x + 3) > 0: zeros −3,0,2-3, 0, 2, so −3<x<0-3 < x < 0 or x>2x > 2

  3. 3.7.3

    Solving Polynomial Inequalities Using Special Factoring Techniques and the Graphical

    Factor with grouping, difference of squares, or sum and difference of cubes, then use the graph's zeros and end behavior to read off the solution.

    e.g. x3−x2−4x+4≤0⇒(x−1)(x−2)(x+2)≤0⇒x≤−2x^3 - x^2 - 4x + 4 \le 0 \Rightarrow (x - 1)(x - 2)(x + 2) \le 0 \Rightarrow x \le -2 or 1≤x≤21 \le x \le 2

    a2−b2=(a−b)(a+b)\displaystyle a^2 - b^2 = (a - b)(a + b)
    a3−b3=(a−b)(a2+ab+b2)\displaystyle a^3 - b^3 = (a - b)(a^2 + ab + b^2)
    a3+b3=(a+b)(a2−ab+b2)\displaystyle a^3 + b^3 = (a + b)(a^2 - ab + b^2)
  4. 3.7.4

    Solving Polynomial Inequalities Using the Sign Table Method

    Factor completely and list the zeros. Make a sign table with one row per factor and multiply the signs on each interval. A factor with an even power never changes sign.

    e.g. (x−1)2(x+4)<0(x - 1)^2(x + 4) < 0: (x−1)2>0(x-1)^2 > 0 for x≠1x \ne 1, so x+4<0⇒x<−4x + 4 < 0 \Rightarrow x < -4

3.8 Non-Polynomial Inequalities

  1. 3.8.1

    Solving Radical Inequalities

    First require the radicand to be at least 0. When both sides are nonnegative, square both sides. Combine the conditions.

    e.g. x−3<2⇒0≤x−3<4⇒3≤x<7\sqrt{x - 3} < 2 \Rightarrow 0 \le x - 3 < 4 \Rightarrow 3 \le x < 7

    A<B,  B>0\displaystyle \sqrt{A} < B,\; B > 0
    ⇒  0≤A<B2\displaystyle \Rightarrow\; 0 \le A < B^2
  2. 3.8.2

    Solving Inequalities Involving Exponential Functions

    Write both sides with the same base, then compare exponents. If the base is greater than 1 keep the direction; if it is between 0 and 1, flip it. You can also take logs of both sides.

    e.g. (13)x≥9⇒3−x≥32⇒x≤−2\left(\tfrac13\right)^x \ge 9 \Rightarrow 3^{-x} \ge 3^2 \Rightarrow x \le -2

    b>1:  bu>bv  ⟺  u>v\displaystyle b > 1:\; b^u > b^v \iff u > v
    0<b<1:  bu>bv  ⟺  u<v\displaystyle 0 < b < 1:\; b^u > b^v \iff u < v
  3. 3.8.3

    Solving Inequalities Involving Logarithmic Functions

    The argument of a log must be positive, so include that condition. Rewrite in exponential form. Keep the direction for a base greater than 1; flip it for a base between 0 and 1.

    e.g. log⁡2(x−1)<3⇒0<x−1<8⇒1<x<9\log_2(x - 1) < 3 \Rightarrow 0 < x - 1 < 8 \Rightarrow 1 < x < 9

    b>1:  log⁡bu<c\displaystyle b > 1:\; \log_b u < c
      ⟺  0<u<bc\displaystyle \iff 0 < u < b^c
  4. 3.8.4

    Solving Inequalities Involving Exponential Functions and Polynomials

    An exponential like exe^x or 2x2^x is always positive, so you can divide by it without flipping the sign. For expressions with 4x4^x and 2x2^x, substitute u=2xu = 2^x, solve in uu, then convert back.

    e.g. ex(x2−4)<0⇒x2−4<0⇒−2<x<2e^x(x^2 - 4) < 0 \Rightarrow x^2 - 4 < 0 \Rightarrow -2 < x < 2

    bx>0  for all x\displaystyle b^x > 0 \;\text{for all } x
  5. 3.8.5

    Solving Inequalities Involving Positive and Negative Factors

    A product or quotient is positive when its factors have the same sign and negative when they have opposite signs. A factor that is always positive, like x2+1x^2 + 1 or exe^x, does not affect the sign.

    e.g. (x2+1)(x−5)≤0⇒x−5≤0⇒x≤5(x^2 + 1)(x - 5) \le 0 \Rightarrow x - 5 \le 0 \Rightarrow x \le 5

    AB>0  ⟺  A,B same sign\displaystyle AB > 0 \iff A, B \text{ same sign}
    AB<0  ⟺  A,B opposite signs\displaystyle AB < 0 \iff A, B \text{ opposite signs}
  6. 3.8.6

    Solving Inequalities Involving Geometric Sequences

    Write the term or sum formula, set up the inequality, and solve for nn using logs or by testing powers. nn must be a positive integer, so take the first one that works. Dividing by the log of a number less than 1 flips the sign.

    e.g. First term of 3,6,12,…3, 6, 12, \ldots over 1000: 3⋅2n−1>1000⇒2n−1>333.3‾⇒n=103 \cdot 2^{n-1} > 1000 \Rightarrow 2^{n-1} > 333.\overline{3} \Rightarrow n = 10

    an=a1rn−1\displaystyle a_n = a_1 r^{n-1}
    Sn=a1(1−rn)1−r\displaystyle S_n = \frac{a_1(1 - r^n)}{1 - r}
  7. 3.8.7

    Solving Rational Inequalities

    Get 0 on one side and a single fraction on the other. The zeros of the numerator and denominator split the number line. Test each interval. Never include a zero of the denominator, and do not multiply by a denominator whose sign is unknown.

    e.g. x−2x+3≥0⇒x<−3\dfrac{x - 2}{x + 3} \ge 0 \Rightarrow x < -3 or x≥2x \ge 2

  8. 3.8.8

    Further Solving of Rational Inequalities

    Subtract to get 0 on one side, combine over a common denominator, and simplify the numerator. Then use a sign table on the critical values.

    e.g. Critical values 1 and 3, so x<1x < 1 or x>3x > 3.

    2x−1<1\displaystyle \frac{2}{x - 1} < 1
    ⇒  3−xx−1<0\displaystyle \Rightarrow\; \frac{3 - x}{x - 1} < 0
  9. 3.8.9

    Solving Two-Variable Nonlinear Inequalities

    Graph the boundary curve: dashed for << or >>, solid for ≤\le or ≥\ge. Test a point not on the curve and shade the side where it makes the inequality true.

    e.g. y>x2−4y > x^2 - 4: dashed parabola; (0,0)(0,0) gives 0>−40 > -4, true, so shade above.

  10. 3.8.10

    Further Solving of Two-Variable Nonlinear Inequalities

    For a system, graph each inequality and shade the region where all of them hold. A point is a solution only if it satisfies every inequality.

    e.g. Is (1,3)(1, 3) in y≥x2y \ge x^2, x2+y2<16x^2 + y^2 < 16? 3≥13 \ge 1 and 1+9<161 + 9 < 16, so yes.

Unit 4 42 topics

Trigonometry

The Inverse Trigonometric Functions, Trigonometric Identities, The Sum and Difference Formulas, and more.

4.9 The Inverse Trigonometric Functions

−π/2π/2arcsin xrange [−π/2, π/2]input −1 ≤ x ≤ 10πarccos xrange [0, π]input −1 ≤ x ≤ 1−π/2π/2arctan xrange (−π/2, π/2)input: any real x
Each inverse trig function returns an angle from one fixed arc, so it gives exactly one answer. Arcsine and arctangent use the right half of the circle; arccosine uses the top half, and arctangent never reaches ±π/2.
  1. 4.9.1

    Graphing the Inverse Sine Function

    The graph of y=sin⁡−1xy = \sin^{-1} x (also written arcsin⁡x\arcsin x) is sine restricted to [−π2,π2][-\tfrac{\pi}{2}, \tfrac{\pi}{2}], reflected over y=xy = x. It increases from (−1,−π2)(-1, -\tfrac{\pi}{2}) through (0,0)(0, 0) to (1,π2)(1, \tfrac{\pi}{2}).

    e.g. sin⁡−11=π2\sin^{-1} 1 = \tfrac{\pi}{2}, sin⁡−1(−12)=−π6\sin^{-1}(-\tfrac12) = -\tfrac{\pi}{6}

    y=sin⁡−1x  ⟺  sin⁡y=x\displaystyle y = \sin^{-1} x \iff \sin y = x
    domain [−1,1]\displaystyle \text{domain } [-1, 1]
    range [−π2,π2]\displaystyle \text{range } [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]
  2. 4.9.2

    Graphing the Inverse Cosine Function

    The graph of y=cos⁡−1xy = \cos^{-1} x is cosine restricted to [0,π][0, \pi], reflected over y=xy = x. It decreases from (−1,π)(-1, \pi) through (0,π2)(0, \tfrac{\pi}{2}) to (1,0)(1, 0).

    e.g. cos⁡−10=π2\cos^{-1} 0 = \tfrac{\pi}{2}, cos⁡−1(−1)=π\cos^{-1}(-1) = \pi

    y=cos⁡−1x  ⟺  cos⁡y=x\displaystyle y = \cos^{-1} x \iff \cos y = x
    domain [−1,1]\displaystyle \text{domain } [-1, 1]
    range [0,π]\displaystyle \text{range } [0, \pi]
  3. 4.9.3

    Graphing the Inverse Tangent Function

    The graph of y=tan⁡−1xy = \tan^{-1} x is tangent restricted to (−π2,π2)(-\tfrac{\pi}{2}, \tfrac{\pi}{2}), reflected over y=xy = x. It increases through (0,0)(0, 0) with horizontal asymptotes y=±π2y = \pm\tfrac{\pi}{2}.

    e.g. tan⁡−11=π4\tan^{-1} 1 = \tfrac{\pi}{4}, tan⁡−1(−1)=−π4\tan^{-1}(-1) = -\tfrac{\pi}{4}

    y=tan⁡−1x  ⟺  tan⁡y=x\displaystyle y = \tan^{-1} x \iff \tan y = x
    domain (−∞,∞)\displaystyle \text{domain } (-\infty, \infty)
    range (−π2,π2)\displaystyle \text{range } (-\tfrac{\pi}{2}, \tfrac{\pi}{2})
  4. 4.9.4

    Evaluating Expressions Containing Inverse Trigonometric Functions

    Find the angle in the inverse function's range that has the given value. A trig function undoes its own inverse whenever the input is in the inverse's domain.

    e.g. cos⁡−1(−22)=3π4\cos^{-1}(-\tfrac{\sqrt2}{2}) = \tfrac{3\pi}{4}

    sin⁡(sin⁡−1x)=x,  −1≤x≤1\displaystyle \sin(\sin^{-1} x) = x,\; -1 \le x \le 1
    cos⁡(cos⁡−1x)=x,  −1≤x≤1\displaystyle \cos(\cos^{-1} x) = x,\; -1 \le x \le 1
    tan⁡(tan⁡−1x)=x for all x\displaystyle \tan(\tan^{-1} x) = x \text{ for all } x
  5. 4.9.5

    Further Evaluating Expressions Containing Inverse Trigonometric Functions

    For a trig function of an inverse trig value, draw a right triangle from the input and read off the ratio, choosing the sign by quadrant. For an inverse of a trig value, the answer must land in the inverse's range.

    e.g. cos⁡(sin⁡−135)=45\cos(\sin^{-1}\tfrac35) = \tfrac45 (3-4-5 triangle); sin⁡−1(sin⁡5π6)=π6\sin^{-1}(\sin\tfrac{5\pi}{6}) = \tfrac{\pi}{6}

    sin⁡−1(sin⁡θ)=θ\displaystyle \sin^{-1}(\sin\theta) = \theta
    only if −π2≤θ≤π2\displaystyle \text{only if } -\tfrac{\pi}{2} \le \theta \le \tfrac{\pi}{2}

4.10 Trigonometric Identities

xyθP (cos θ, sin θ)1cos θsin θI allII sinIII tanIV cosx-coordinate = cos θy-coordinate = sin θcos²θ + sin²θ = 1tan θ = sin θ ÷ cos θPositive in each quadrant:All, Sin, Tan, Cos
On the unit circle, the point at angle θ is (cos θ, sin θ), so the legs of the triangle are cos θ and sin θ. The quadrant labels show which ratios are positive there.
  1. 4.10.1

    Simplifying Expressions Using Basic Trigonometric Identities

    Rewrite everything in sine and cosine using the quotient and reciprocal identities, then cancel.

    e.g. tan⁡θcos⁡θ=sin⁡θcos⁡θ⋅cos⁡θ=sin⁡θ\tan\theta\cos\theta = \tfrac{\sin\theta}{\cos\theta}\cdot\cos\theta = \sin\theta

    tan⁡θ=sin⁡θcos⁡θ,  cot⁡θ=cos⁡θsin⁡θ\displaystyle \tan\theta = \frac{\sin\theta}{\cos\theta},\; \cot\theta = \frac{\cos\theta}{\sin\theta}
    sec⁡θ=1cos⁡θ,  csc⁡θ=1sin⁡θ\displaystyle \sec\theta = \frac{1}{\cos\theta},\; \csc\theta = \frac{1}{\sin\theta}
  2. 4.10.2

    Simplifying Expressions Using the Pythagorean Identity

    Holds for every angle. Look for sin⁡2θ+cos⁡2θ\sin^2\theta + \cos^2\theta and replace it with 1.

    e.g. 1−sin⁡2θcos⁡θ=cos⁡2θcos⁡θ=cos⁡θ\frac{1 - \sin^2\theta}{\cos\theta} = \frac{\cos^2\theta}{\cos\theta} = \cos\theta

    sin⁡2θ+cos⁡2θ=1\displaystyle \sin^2\theta + \cos^2\theta = 1
  3. 4.10.3

    Alternate Forms of the Pythagorean Identity

    Solve the identity for one squared term. When taking a square root, pick the sign from the angle's quadrant.

    e.g. sin⁡θ=35\sin\theta = \tfrac35 in Quadrant II: cos⁡θ=−1−925=−45\cos\theta = -\sqrt{1 - \tfrac{9}{25}} = -\tfrac45

    sin⁡2θ=1−cos⁡2θ\displaystyle \sin^2\theta = 1 - \cos^2\theta
    cos⁡2θ=1−sin⁡2θ\displaystyle \cos^2\theta = 1 - \sin^2\theta
    cos⁡θ=±1−sin⁡2θ\displaystyle \cos\theta = \pm\sqrt{1 - \sin^2\theta}
  4. 4.10.4

    Simplifying Expressions Using the Secant-Tangent Identity

    Divide the Pythagorean identity by cos⁡2θ\cos^2\theta to get it. Look for 1+tan⁡2θ1 + \tan^2\theta or sec⁡2θ−tan⁡2θ\sec^2\theta - \tan^2\theta.

    e.g. (sec⁡θ−tan⁡θ)(sec⁡θ+tan⁡θ)=sec⁡2θ−tan⁡2θ=1(\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = \sec^2\theta - \tan^2\theta = 1

    1+tan⁡2θ=sec⁡2θ\displaystyle 1 + \tan^2\theta = \sec^2\theta
  5. 4.10.5

    Alternate Forms of the Secant-Tangent Identity

    Rearrange 1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta as needed. Choose the sign of a square root from the quadrant.

    e.g. tan⁡θ=34\tan\theta = \tfrac34 in Quadrant III: sec⁡θ=−1+916=−54\sec\theta = -\sqrt{1 + \tfrac{9}{16}} = -\tfrac54

    tan⁡2θ=sec⁡2θ−1\displaystyle \tan^2\theta = \sec^2\theta - 1
    sec⁡2θ−tan⁡2θ=1\displaystyle \sec^2\theta - \tan^2\theta = 1
    sec⁡θ=±1+tan⁡2θ\displaystyle \sec\theta = \pm\sqrt{1 + \tan^2\theta}
  6. 4.10.6

    Simplifying Trigonometric Expressions Using the Cotangent-Cosecant Identity

    Divide the Pythagorean identity by sin⁡2θ\sin^2\theta to get it. Rearranged forms are cot⁡2θ=csc⁡2θ−1\cot^2\theta = \csc^2\theta - 1 and csc⁡2θ−cot⁡2θ=1\csc^2\theta - \cot^2\theta = 1.

    e.g. sin⁡2θ(1+cot⁡2θ)=sin⁡2θcsc⁡2θ=1\sin^2\theta(1 + \cot^2\theta) = \sin^2\theta\csc^2\theta = 1

    1+cot⁡2θ=csc⁡2θ\displaystyle 1 + \cot^2\theta = \csc^2\theta
  7. 4.10.7

    Simplifying Trigonometric Expressions Using Cofunction Identities

    A function of an angle equals its cofunction of the complementary angle. The pairs are sine and cosine, tangent and cotangent, secant and cosecant. In degrees, use 90∘−θ90^\circ - \theta.

    e.g. cos⁡(π2−θ)csc⁡θ=sin⁡θcsc⁡θ=1\cos(\tfrac{\pi}{2} - \theta)\csc\theta = \sin\theta\csc\theta = 1

    sin⁡(π2−θ)=cos⁡θ\displaystyle \sin(\tfrac{\pi}{2} - \theta) = \cos\theta
    cos⁡(π2−θ)=sin⁡θ\displaystyle \cos(\tfrac{\pi}{2} - \theta) = \sin\theta
    tan⁡(π2−θ)=cot⁡θ\displaystyle \tan(\tfrac{\pi}{2} - \theta) = \cot\theta
    sec⁡(π2−θ)=csc⁡θ\displaystyle \sec(\tfrac{\pi}{2} - \theta) = \csc\theta

4.11 The Sum and Difference Formulas

  1. 4.11.1

    The Sum and Difference Formulas for Sine

    The signs match: plus on the left gives plus on the right. Note sin⁡(A+B)≠sin⁡A+sin⁡B\sin(A + B) \ne \sin A + \sin B.

    e.g. sin⁡(x+π)=sin⁡xcos⁡π+cos⁡xsin⁡π=−sin⁡x\sin(x + \pi) = \sin x\cos\pi + \cos x\sin\pi = -\sin x

    sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\displaystyle \sin(A + B) = \sin A\cos B + \cos A\sin B
    sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\displaystyle \sin(A - B) = \sin A\cos B - \cos A\sin B
  2. 4.11.2

    The Sum and Difference Formulas for Cosine

    The sign flips: a sum on the left gives a minus on the right.

    e.g. cos⁡(x+π2)=cos⁡x⋅0−sin⁡x⋅1=−sin⁡x\cos(x + \tfrac{\pi}{2}) = \cos x\cdot 0 - \sin x\cdot 1 = -\sin x

    cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\displaystyle \cos(A + B) = \cos A\cos B - \sin A\sin B
    cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\displaystyle \cos(A - B) = \cos A\cos B + \sin A\sin B
  3. 4.11.3

    The Sum and Difference Formulas for Tangent

    The numerator keeps the sign and the denominator flips it. Undefined when the denominator is 0.

    e.g. tan⁡(x+π4)=tan⁡x+11−tan⁡x\tan(x + \tfrac{\pi}{4}) = \frac{\tan x + 1}{1 - \tan x}

    tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\displaystyle \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A\tan B}
    tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B\displaystyle \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A\tan B}
  4. 4.11.4

    Calculating Trigonometric Ratios Using the Sum Formula for Sine

    Write the angle as a sum or difference of special angles (30∘30^\circ, 45∘45^\circ, 60∘60^\circ, or their radian forms), then apply the formula. If ratios are given instead, find any missing ones with the Pythagorean identity.

    e.g. sin⁡75∘=sin⁡45∘cos⁡30∘+cos⁡45∘sin⁡30∘=6+24\sin 75^\circ = \sin 45^\circ\cos 30^\circ + \cos 45^\circ\sin 30^\circ = \frac{\sqrt6 + \sqrt2}{4}

    sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\displaystyle \sin(A + B) = \sin A\cos B + \cos A\sin B
  5. 4.11.5

    Calculating Trigonometric Ratios Using the Sum Formula for Cosine

    Split the angle into special angles, or find the missing ratios from given ones, then apply the formula. Watch the minus sign.

    e.g. cos⁡75∘=cos⁡45∘cos⁡30∘−sin⁡45∘sin⁡30∘=6−24\cos 75^\circ = \cos 45^\circ\cos 30^\circ - \sin 45^\circ\sin 30^\circ = \frac{\sqrt6 - \sqrt2}{4}

    cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\displaystyle \cos(A + B) = \cos A\cos B - \sin A\sin B
  6. 4.11.6

    Calculating Trigonometric Ratios Using the Sum Formula for Tangent

    Split the angle into special angles with known tangents (tan⁡30∘=33\tan 30^\circ = \tfrac{\sqrt3}{3}, tan⁡45∘=1\tan 45^\circ = 1, tan⁡60∘=3\tan 60^\circ = \sqrt3), apply the formula, and rationalize.

    e.g. tan⁡75∘=1+3/31−3/3=3+33−3=2+3\tan 75^\circ = \frac{1 + \sqrt3/3}{1 - \sqrt3/3} = \frac{3 + \sqrt3}{3 - \sqrt3} = 2 + \sqrt3

    tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\displaystyle \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A\tan B}
  7. 4.11.7

    Writing Sums of Trigonometric Functions in Amplitude-Phase Form

    A sum asin⁡x+bcos⁡xa\sin x + b\cos x is a single sine wave with amplitude RR. Find the phase φ\varphi from its cosine and sine, using both signs to fix the quadrant.

    e.g. sin⁡x+3cos⁡x\sin x + \sqrt3\cos x: R=2R = 2, φ=π3\varphi = \tfrac{\pi}{3}, so it equals 2sin⁡(x+π3)2\sin(x + \tfrac{\pi}{3})

    asin⁡x+bcos⁡x=Rsin⁡(x+φ)\displaystyle a\sin x + b\cos x = R\sin(x + \varphi)
    R=a2+b2\displaystyle R = \sqrt{a^2 + b^2}
    cos⁡φ=aR,  sin⁡φ=bR\displaystyle \cos\varphi = \tfrac{a}{R},\; \sin\varphi = \tfrac{b}{R}

4.12 The Double-Angle Formulas

  1. 4.12.1

    The Double-Angle Formula for Sine

    The sine of twice an angle. It comes from the sum formula with A=B=θA = B = \theta. Note sin⁡2θ≠2sin⁡θ\sin 2\theta \ne 2\sin\theta.

    e.g. 2sin⁡15∘cos⁡15∘=sin⁡30∘=122\sin 15^\circ\cos 15^\circ = \sin 30^\circ = \tfrac12

    sin⁡2θ=2sin⁡θcos⁡θ\displaystyle \sin 2\theta = 2\sin\theta\cos\theta
  2. 4.12.2

    Verifying Trigonometric Identities Using the Double-Angle Formula for Sine

    Start with the more complicated side and transform it into the other side. Replace sin⁡2θ\sin 2\theta with 2sin⁡θcos⁡θ2\sin\theta\cos\theta, or spot 2sin⁡θcos⁡θ2\sin\theta\cos\theta and collapse it.

    e.g. (sin⁡θ+cos⁡θ)2=sin⁡2θ+cos⁡2θ+2sin⁡θcos⁡θ=1+sin⁡2θ(\sin\theta + \cos\theta)^2 = \sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta = 1 + \sin 2\theta

    sin⁡2θ=2sin⁡θcos⁡θ\displaystyle \sin 2\theta = 2\sin\theta\cos\theta
  3. 4.12.3

    Using the Double-Angle Formula for Sine With the Pythagorean Theorem

    Given one ratio and the quadrant, draw a right triangle (or use the Pythagorean identity) to find the other ratio. Give it the sign from the quadrant, then double.

    e.g. cos⁡θ=513\cos\theta = \tfrac{5}{13} in Quadrant IV: sin⁡θ=−1213\sin\theta = -\tfrac{12}{13}, so sin⁡2θ=2(−1213)(513)=−120169\sin 2\theta = 2(-\tfrac{12}{13})(\tfrac{5}{13}) = -\tfrac{120}{169}

    sin⁡2θ=2sin⁡θcos⁡θ\displaystyle \sin 2\theta = 2\sin\theta\cos\theta
  4. 4.12.4

    The Double-Angle Formula for Cosine

    Three equivalent forms. Pick the one that uses the ratio you know.

    e.g. sin⁡θ=13\sin\theta = \tfrac13: cos⁡2θ=1−2⋅19=79\cos 2\theta = 1 - 2\cdot\tfrac19 = \tfrac79

    cos⁡2θ=cos⁡2θ−sin⁡2θ\displaystyle \cos 2\theta = \cos^2\theta - \sin^2\theta
    =2cos⁡2θ−1\displaystyle = 2\cos^2\theta - 1
    =1−2sin⁡2θ\displaystyle = 1 - 2\sin^2\theta
  5. 4.12.5

    Verifying Trigonometric Identities Using the Double-Angle Formulas for Cosine

    Choose the form that cancels the constant: with 1−cos⁡2θ1 - \cos 2\theta use 1−2sin⁡2θ1 - 2\sin^2\theta, and with 1+cos⁡2θ1 + \cos 2\theta use 2cos⁡2θ−12\cos^2\theta - 1.

    e.g. 1−cos⁡2θsin⁡2θ=2sin⁡2θ2sin⁡θcos⁡θ=tan⁡θ\frac{1 - \cos 2\theta}{\sin 2\theta} = \frac{2\sin^2\theta}{2\sin\theta\cos\theta} = \tan\theta

    1−cos⁡2θ=2sin⁡2θ\displaystyle 1 - \cos 2\theta = 2\sin^2\theta
    1+cos⁡2θ=2cos⁡2θ\displaystyle 1 + \cos 2\theta = 2\cos^2\theta
  6. 4.12.6

    Finding Exact Values of Trigonometric Expressions Using the Double-Angle Formulas for

    Spot a double-angle pattern and collapse it to one function of the doubled angle, then use a known exact value.

    e.g. cos⁡215∘−sin⁡215∘=cos⁡30∘=32\cos^2 15^\circ - \sin^2 15^\circ = \cos 30^\circ = \tfrac{\sqrt3}{2}

    cos⁡2θ−sin⁡2θ=cos⁡2θ\displaystyle \cos^2\theta - \sin^2\theta = \cos 2\theta
    2cos⁡2θ−1=cos⁡2θ\displaystyle 2\cos^2\theta - 1 = \cos 2\theta
    1−2sin⁡2θ=cos⁡2θ\displaystyle 1 - 2\sin^2\theta = \cos 2\theta
  7. 4.12.7

    Simplifying Expressions Using the Double-Angle Formula for Tangent

    Spot the pattern 2tan⁡θ1−tan⁡2θ\frac{2\tan\theta}{1 - \tan^2\theta} and collapse it to tan⁡2θ\tan 2\theta. Undefined when tan⁡2θ=1\tan^2\theta = 1.

    e.g. 2tan⁡15∘1−tan⁡215∘=tan⁡30∘=33\frac{2\tan 15^\circ}{1 - \tan^2 15^\circ} = \tan 30^\circ = \tfrac{\sqrt3}{3}

    tan⁡2θ=2tan⁡θ1−tan⁡2θ\displaystyle \tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}
  8. 4.12.8

    Verifying Trigonometric Identities Using the Double-Angle Formula for Tangent

    Replace tan⁡2θ\tan 2\theta with its formula, then simplify toward the other side. Multiplying top and bottom by the same expression often helps.

    e.g. tan⁡2θ=2tan⁡θ1−tan⁡2θ\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}; multiply top and bottom by cot⁡θ\cot\theta to get 2cot⁡θ−tan⁡θ\frac{2}{\cot\theta - \tan\theta}

    tan⁡2θ=2tan⁡θ1−tan⁡2θ\displaystyle \tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}

4.13 Elementary Trigonometric Equations

π/2π3π/22π1−1xy = ½y = sin xπ/6π/65π/65π/6sin x = ½ on [0, 2π):x = π/6 or x = 5π/6All solutions: x = π/6 + 2πn or x = 5π/6 + 2πn, for any integer n
The line y = ½ meets the unit circle at π/6 and 5π/6, and it meets the sine wave at the same two x-values. Adding 2πn to each gives every solution.
  1. 4.13.1

    Elementary Trigonometric Equations Containing Sine

    Isolate sin⁡x\sin x. If sin⁡x=k\sin x = k with −1≤k≤1-1 \le k \le 1, the solutions in [0,2π)[0, 2\pi) are α\alpha and π−α\pi - \alpha, where α=sin⁡−1k\alpha = \sin^{-1} k. Add 2π2\pi to any negative angle, and there is no solution if ∣k∣>1\lvert k \rvert > 1.

    e.g. sin⁡x=12⇒x=π6,5π6\sin x = \tfrac12 \Rightarrow x = \tfrac{\pi}{6}, \tfrac{5\pi}{6}

    sin⁡x=k⇒x=α,  π−α\displaystyle \sin x = k \Rightarrow x = \alpha,\; \pi - \alpha
    α=sin⁡−1k\displaystyle \alpha = \sin^{-1} k
  2. 4.13.2

    Elementary Trigonometric Equations Containing Cosine

    Isolate cos⁡x\cos x. If cos⁡x=k\cos x = k with −1≤k≤1-1 \le k \le 1, the solutions in [0,2π)[0, 2\pi) are α\alpha and 2π−α2\pi - \alpha, where α=cos⁡−1k\alpha = \cos^{-1} k. There is no solution if ∣k∣>1\lvert k \rvert > 1.

    e.g. cos⁡x=−12⇒x=2π3,4π3\cos x = -\tfrac12 \Rightarrow x = \tfrac{2\pi}{3}, \tfrac{4\pi}{3}

    cos⁡x=k⇒x=α,  2π−α\displaystyle \cos x = k \Rightarrow x = \alpha,\; 2\pi - \alpha
    α=cos⁡−1k\displaystyle \alpha = \cos^{-1} k
  3. 4.13.3

    Elementary Trigonometric Equations Containing Tangent

    tan⁡x=k\tan x = k has solutions for every real kk. In [0,2π)[0, 2\pi) they are α\alpha and α+π\alpha + \pi, where α=tan⁡−1k\alpha = \tan^{-1} k. If α\alpha is negative, use α+π\alpha + \pi and α+2π\alpha + 2\pi.

    e.g. tan⁡x=−1\tan x = -1: α=−π4\alpha = -\tfrac{\pi}{4}, so x=3π4,7π4x = \tfrac{3\pi}{4}, \tfrac{7\pi}{4}

    tan⁡x=k⇒x=α,  α+π\displaystyle \tan x = k \Rightarrow x = \alpha,\; \alpha + \pi
    α=tan⁡−1k\displaystyle \alpha = \tan^{-1} k
  4. 4.13.4

    Elementary Trigonometric Equations Containing Secant

    Rewrite as a cosine equation and solve that. There is no solution if ∣k∣<1\lvert k \rvert < 1.

    e.g. sec⁡x=2⇒cos⁡x=12⇒x=π3,5π3\sec x = 2 \Rightarrow \cos x = \tfrac12 \Rightarrow x = \tfrac{\pi}{3}, \tfrac{5\pi}{3}

    sec⁡x=k  ⟺  cos⁡x=1k\displaystyle \sec x = k \iff \cos x = \frac{1}{k}
  5. 4.13.5

    Elementary Trigonometric Equations Containing Cosecant

    Rewrite as a sine equation and solve that. There is no solution if ∣k∣<1\lvert k \rvert < 1.

    e.g. csc⁡x=−2⇒sin⁡x=−12⇒x=7π6,11π6\csc x = -2 \Rightarrow \sin x = -\tfrac12 \Rightarrow x = \tfrac{7\pi}{6}, \tfrac{11\pi}{6}

    csc⁡x=k  ⟺  sin⁡x=1k\displaystyle \csc x = k \iff \sin x = \frac{1}{k}
  6. 4.13.6

    Elementary Trigonometric Equations Containing Cotangent

    For k≠0k \ne 0, rewrite as a tangent equation. For k=0k = 0, use cos⁡x=0\cos x = 0, so x=π2,3π2x = \tfrac{\pi}{2}, \tfrac{3\pi}{2}.

    e.g. cot⁡x=3⇒tan⁡x=13⇒x=π6,7π6\cot x = \sqrt3 \Rightarrow \tan x = \tfrac{1}{\sqrt3} \Rightarrow x = \tfrac{\pi}{6}, \tfrac{7\pi}{6}

    cot⁡x=k  ⟺  tan⁡x=1k\displaystyle \cot x = k \iff \tan x = \frac{1}{k}
  7. 4.13.7

    General Solutions of Elementary Trigonometric Equations

    Find one solution α\alpha, then add every whole number of periods. Sine and cosine repeat every 2π2\pi, tangent every π\pi. Here nn is any integer.

    e.g. cos⁡x=12⇒x=±π3+2πn\cos x = \tfrac12 \Rightarrow x = \pm\tfrac{\pi}{3} + 2\pi n

    sin⁡x=k:  x=α+2πn\displaystyle \sin x = k:\; x = \alpha + 2\pi n
    or x=π−α+2πn\displaystyle \text{or } x = \pi - \alpha + 2\pi n
    cos⁡x=k:  x=±α+2πn\displaystyle \cos x = k:\; x = \pm\alpha + 2\pi n
    tan⁡x=k:  x=α+πn\displaystyle \tan x = k:\; x = \alpha + \pi n

4.14 Trigonometric Equations Containing Transformed Functions

π/2π3π/22π5π/2−1123xmidline y = 1amplitude|A| = 2period 2π/B = 2πy = A sin(B(x − C)) + DAamplitude |A|Bperiod 2π/BCphase shift CDmidline y = DHere A = 2, B = 1,C = 0, D = 1.
A sinusoid is the sine wave stretched, shifted, and lifted. Read the midline first; the amplitude is how far the wave rises above it, and the period is the length of one full cycle.
  1. 4.14.1

    General Solutions of Trigonometric Equations With Transformed Functions

    Let u=bx+cu = bx + c. Write the general solution for uu, then solve for xx. Dividing by bb also divides the period, so 2πn2\pi n becomes 2πnb\tfrac{2\pi n}{b}.

    e.g. sin⁡2x=12⇒2x=π6+2πn\sin 2x = \tfrac12 \Rightarrow 2x = \tfrac{\pi}{6} + 2\pi n or 5π6+2πn\tfrac{5\pi}{6} + 2\pi n, so x=π12+πnx = \tfrac{\pi}{12} + \pi n or 5π12+πn\tfrac{5\pi}{12} + \pi n

    sin⁡(bx+c)=k\displaystyle \sin(bx + c) = k
    ⇒  bx+c=α+2πn\displaystyle \Rightarrow\; bx + c = \alpha + 2\pi n
    or bx+c=π−α+2πn\displaystyle \text{or } bx + c = \pi - \alpha + 2\pi n
  2. 4.14.2

    Trigonometric Equations Containing Transformed Sine Functions

    Isolate sin⁡(bx+c)\sin(bx + c) and let u=bx+cu = bx + c. If 0≤x<2π0 \le x < 2\pi, then uu runs from cc to 2πb+c2\pi b + c, so list every solution for uu in that longer interval, then solve for xx.

    e.g. 2sin⁡2x=32\sin 2x = \sqrt3 on [0,2π)[0, 2\pi): 2x=π3,2π3,7π3,8π32x = \tfrac{\pi}{3}, \tfrac{2\pi}{3}, \tfrac{7\pi}{3}, \tfrac{8\pi}{3}, so x=π6,π3,7π6,4π3x = \tfrac{\pi}{6}, \tfrac{\pi}{3}, \tfrac{7\pi}{6}, \tfrac{4\pi}{3}

    asin⁡(bx+c)+d=k\displaystyle a\sin(bx + c) + d = k
    ⇒  sin⁡u=k−da\displaystyle \Rightarrow\; \sin u = \frac{k - d}{a}
  3. 4.14.3

    Trigonometric Equations Containing Transformed Cosine Functions

    Isolate cos⁡(bx+c)\cos(bx + c) and let u=bx+cu = bx + c. Shift the interval for xx to get the interval for uu, find every uu in it, then solve for xx.

    e.g. cos⁡(x−π4)=0\cos(x - \tfrac{\pi}{4}) = 0 on [0,2π)[0, 2\pi): u∈[−π4,7π4)u \in [-\tfrac{\pi}{4}, \tfrac{7\pi}{4}), u=π2,3π2u = \tfrac{\pi}{2}, \tfrac{3\pi}{2}, so x=3π4,7π4x = \tfrac{3\pi}{4}, \tfrac{7\pi}{4}

    acos⁡(bx+c)+d=k\displaystyle a\cos(bx + c) + d = k
    ⇒  cos⁡u=k−da\displaystyle \Rightarrow\; \cos u = \frac{k - d}{a}
  4. 4.14.4

    Trigonometric Equations Containing Transformed Tangent Functions

    Let u=bx+cu = bx + c and solve tan⁡u=k\tan u = k with period π\pi. Then solve for xx, so the period becomes πb\tfrac{\pi}{b}.

    e.g. tan⁡2x=1⇒2x=π4+πn⇒x=π8+πn2\tan 2x = 1 \Rightarrow 2x = \tfrac{\pi}{4} + \pi n \Rightarrow x = \tfrac{\pi}{8} + \tfrac{\pi n}{2}

    tan⁡(bx+c)=k\displaystyle \tan(bx + c) = k
    ⇒  bx+c=α+πn\displaystyle \Rightarrow\; bx + c = \alpha + \pi n

4.15 Advanced Trigonometric Equations

  1. 4.15.1

    Solving Trigonometric Equations Using the Sin-Cos-Tan Identity

    When an equation has the form sin⁡x=kcos⁡x\sin x = k\cos x, divide both sides by cos⁡x\cos x to get tan⁡x=k\tan x = k. First check that cos⁡x=0\cos x = 0 is not a solution, since those values would be lost.

    e.g. sin⁡x=3cos⁡x⇒tan⁡x=3⇒x=π3,4π3\sin x = \sqrt3\cos x \Rightarrow \tan x = \sqrt3 \Rightarrow x = \tfrac{\pi}{3}, \tfrac{4\pi}{3}

    tan⁡x=sin⁡xcos⁡x\displaystyle \tan x = \frac{\sin x}{\cos x}
  2. 4.15.2

    Solving Trigonometric Equations Using the Zero-Product Property

    Move everything to one side, factor, and set each factor equal to zero. Never divide both sides by a trig function, because that throws away solutions.

    e.g. 2sin⁡xcos⁡x=cos⁡x⇒cos⁡x(2sin⁡x−1)=02\sin x\cos x = \cos x \Rightarrow \cos x(2\sin x - 1) = 0, so x=π6,π2,5π6,3π2x = \tfrac{\pi}{6}, \tfrac{\pi}{2}, \tfrac{5\pi}{6}, \tfrac{3\pi}{2}

    AB=0⇒A=0  or  B=0\displaystyle AB = 0 \Rightarrow A = 0 \;\text{or}\; B = 0
  3. 4.15.3

    Quadratic Trigonometric Equations Containing Sine or Cosine

    Use sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1 to write everything in one function. Treat it as a quadratic in u=sin⁡xu = \sin x (or cos⁡x\cos x), factor or use the quadratic formula, and reject any root outside [−1,1][-1, 1].

    e.g. 2sin⁡2x−sin⁡x−1=0⇒(2sin⁡x+1)(sin⁡x−1)=02\sin^2 x - \sin x - 1 = 0 \Rightarrow (2\sin x + 1)(\sin x - 1) = 0, so x=π2,7π6,11π6x = \tfrac{\pi}{2}, \tfrac{7\pi}{6}, \tfrac{11\pi}{6}

    sin⁡2x+cos⁡2x=1\displaystyle \sin^2 x + \cos^2 x = 1
  4. 4.15.4

    Quadratic Trigonometric Equations Containing Tangent or Cotangent

    Rewrite in terms of tan⁡x\tan x alone, then solve the quadratic in u=tan⁡xu = \tan x. Tangent takes every real value, so no root is rejected for size, but check that no step divided by zero.

    e.g. tan⁡x+cot⁡x=2⇒tan⁡2x−2tan⁡x+1=0⇒tan⁡x=1\tan x + \cot x = 2 \Rightarrow \tan^2 x - 2\tan x + 1 = 0 \Rightarrow \tan x = 1, so x=π4,5π4x = \tfrac{\pi}{4}, \tfrac{5\pi}{4}

    cot⁡x=1tan⁡x\displaystyle \cot x = \frac{1}{\tan x}
    1+tan⁡2x=sec⁡2x\displaystyle 1 + \tan^2 x = \sec^2 x

Unit 5 30 topics

Vectors

Introduction to Vectors, Vectors in 2D Cartesian Coordinates, Vectors in 3D Cartesian Coordinates, and more.

5.16 Introduction to Vectors

xy3112uvu + vu = ⟨3, 1⟩v = ⟨1, 2⟩u + v = ⟨3 + 1, 1 + 2⟩= ⟨4, 3⟩Add the x parts, add the y parts.length |v| = √(x² + y²)|u + v| = √(4² + 3²) = 5
To add vectors, put the tail of v at the head of u; the sum runs from the first tail to the last head. It is also the diagonal of the parallelogram the two vectors make.
  1. 5.16.1

    Introduction to Vectors

    A vector has both magnitude and direction. A scalar has only size. A vector is drawn as an arrow from its tail to its head, and two vectors are equal when they have the same magnitude and direction, wherever they sit.

    e.g. A speed of 50 mph is a scalar; a velocity of 50 mph due north is a vector.

    BA→=−AB→\displaystyle \overrightarrow{BA} = -\overrightarrow{AB}
  2. 5.16.2

    The Triangle Law for the Addition of Two Vectors

    Place the tail of the second vector at the head of the first. The sum runs from the tail of the first to the head of the second. Order does not matter: a+b=b+a\mathbf{a} + \mathbf{b} = \mathbf{b} + \mathbf{a}.

    e.g. Walk 3 km east, then 4 km north: you end 32+42=5\sqrt{3^2 + 4^2} = 5 km from the start.

    AB→+BC→=AC→\displaystyle \overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}
  3. 5.16.3

    The Magnitude of a Vector

    The magnitude is the length of the vector. It is never negative, and only the zero vector has magnitude 00. Scaling by kk scales the length by ∣k∣\lvert k \rvert.

    e.g. ∣v∣=5⇒∣−3v∣=15\lvert \mathbf{v} \rvert = 5 \Rightarrow \lvert -3\mathbf{v} \rvert = 15

    ∣AB→∣=AB\displaystyle \lvert \overrightarrow{AB} \rvert = AB
    ∣kv∣=∣k∣ ∣v∣\displaystyle \lvert k\mathbf{v} \rvert = \lvert k \rvert\,\lvert \mathbf{v} \rvert
  4. 5.16.4

    Problem Solving Using Vector Diagrams

    Sketch the vectors head to tail to form a triangle. Use the Pythagorean theorem for a right angle, and the law of cosines or law of sines otherwise.

    e.g. A plane heads north at 120 mph with a 50 mph wind from the west: ground speed 1202+502=130\sqrt{120^2 + 50^2} = 130 mph.

    c2=a2+b2−2abcos⁡C\displaystyle c^2 = a^2 + b^2 - 2ab\cos C
    sin⁡Aa=sin⁡Bb\displaystyle \frac{\sin A}{a} = \frac{\sin B}{b}
  5. 5.16.5

    Parallel Vectors

    Two nonzero vectors are parallel exactly when one is a scalar multiple of the other. If k>0k > 0 they point the same way; if k<0k < 0 they point opposite ways.

    e.g. 2a−4b=2(a−2b)2\mathbf{a} - 4\mathbf{b} = 2(\mathbf{a} - 2\mathbf{b}), so it is parallel to a−2b\mathbf{a} - 2\mathbf{b}.

    u=kv,k≠0\displaystyle \mathbf{u} = k\mathbf{v}, \quad k \ne 0
  6. 5.16.6

    Unit Vectors

    A unit vector has magnitude 11. Divide any nonzero vector by its magnitude to get the unit vector in its direction. i\mathbf{i} and j\mathbf{j} are the unit vectors along the xx- and yy-axes.

    e.g. If ∣v∣=4\lvert \mathbf{v} \rvert = 4, the unit vector is 14v\tfrac14\mathbf{v}, and a vector of length 10 in that direction is 104v=2.5v\tfrac{10}{4}\mathbf{v} = 2.5\mathbf{v}.

    v^=v∣v∣\displaystyle \hat{\mathbf{v}} = \frac{\mathbf{v}}{\lvert \mathbf{v} \rvert}
  7. 5.16.7

    Linear Combinations of Vectors and Their Properties

    A linear combination is au+bva\mathbf{u} + b\mathbf{v} for scalars aa and bb. Vectors follow the usual commutative, associative, and distributive rules. If u\mathbf{u} and v\mathbf{v} are not parallel and au+bv=cu+dva\mathbf{u} + b\mathbf{v} = c\mathbf{u} + d\mathbf{v}, then a=ca = c and b=db = d.

    e.g. 2(a+3b)−(a−b)=a+7b2(\mathbf{a} + 3\mathbf{b}) - (\mathbf{a} - \mathbf{b}) = \mathbf{a} + 7\mathbf{b}

    u+v=v+u\displaystyle \mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}
    k(u+v)=ku+kv\displaystyle k(\mathbf{u} + \mathbf{v}) = k\mathbf{u} + k\mathbf{v}
    (a+b)v=av+bv\displaystyle (a + b)\mathbf{v} = a\mathbf{v} + b\mathbf{v}
  8. 5.16.8

    Describing the Position Vector of a Point Using Known Vectors

    The position vector of a point runs from the origin OO to that point. To reach any point, follow a path of known vectors and add them.

    e.g. PP is 13\tfrac13 of the way from AA to BB: OP→=a+13(b−a)=23a+13b\overrightarrow{OP} = \mathbf{a} + \tfrac13(\mathbf{b} - \mathbf{a}) = \tfrac23\mathbf{a} + \tfrac13\mathbf{b}

    AB→=b−a\displaystyle \overrightarrow{AB} = \mathbf{b} - \mathbf{a}
    OM→=12(a+b)  (midpoint)\displaystyle \overrightarrow{OM} = \tfrac12(\mathbf{a} + \mathbf{b}) \;\text{(midpoint)}

5.17 Vectors in 2D Cartesian Coordinates

  1. 5.17.1

    Two-Dimensional Vectors Expressed in Component Form

    A 2D vector is written by its horizontal and vertical changes. For the vector from AA to BB, subtract the coordinates of the start from the end.

    e.g. A(1,2)A(1, 2), B(4,−2)B(4, -2): AB→=⟨3,−4⟩=3i−4j\overrightarrow{AB} = \langle 3, -4 \rangle = 3\mathbf{i} - 4\mathbf{j}

    v=⟨a,b⟩=ai+bj\displaystyle \mathbf{v} = \langle a, b \rangle = a\mathbf{i} + b\mathbf{j}
    AB→=⟨x2−x1, y2−y1⟩\displaystyle \overrightarrow{AB} = \langle x_2 - x_1,\, y_2 - y_1 \rangle
  2. 5.17.2

    Addition and Scalar Multiplication of Cartesian Vectors in 2D

    Add or subtract matching components. A scalar multiplies every component.

    e.g. 2⟨1,−3⟩+⟨4,5⟩=⟨6,−1⟩2\langle 1, -3 \rangle + \langle 4, 5 \rangle = \langle 6, -1 \rangle

    ⟨a,b⟩+⟨c,d⟩\displaystyle \langle a, b \rangle + \langle c, d \rangle
    =⟨a+c, b+d⟩\displaystyle = \langle a + c,\, b + d \rangle
    k⟨a,b⟩=⟨ka,kb⟩\displaystyle k\langle a, b \rangle = \langle ka, kb \rangle
  3. 5.17.3

    Calculating the Magnitude of Cartesian Vectors in 2D

    Use the Pythagorean theorem on the components.

    e.g. ∣⟨−5,12⟩∣=25+144=13\lvert \langle -5, 12 \rangle \rvert = \sqrt{25 + 144} = 13

    ∣⟨a,b⟩∣=a2+b2\displaystyle \lvert \langle a, b \rangle \rvert = \sqrt{a^2 + b^2}
  4. 5.17.4

    Calculating the Direction of Cartesian Vectors in 2D

    The direction is the angle θ\theta measured counterclockwise from the positive xx-axis. Find the reference angle with inverse tangent, then place it in the vector's quadrant: add 180∘180^\circ if a<0a < 0, and add 360∘360^\circ if the vector is in Quadrant IV.

    e.g. ⟨−3,3⟩\langle -3, 3 \rangle: tan⁡−1(−1)=−45∘\tan^{-1}(-1) = -45^\circ, but the vector is in Quadrant II, so θ=135∘\theta = 135^\circ

    tan⁡θ=ba\displaystyle \tan\theta = \frac{b}{a}
  5. 5.17.5

    Calculating the Components of Cartesian Vectors in 2D

    Given magnitude and direction angle θ\theta, the horizontal component uses cosine and the vertical component uses sine.

    e.g. Magnitude 10 at 30∘30^\circ: ⟨10cos⁡30∘,10sin⁡30∘⟩=⟨53,5⟩\langle 10\cos 30^\circ, 10\sin 30^\circ \rangle = \langle 5\sqrt3, 5 \rangle

    v=⟨∣v∣cos⁡θ, ∣v∣sin⁡θ⟩\displaystyle \mathbf{v} = \langle \lvert\mathbf{v}\rvert\cos\theta,\, \lvert\mathbf{v}\rvert\sin\theta \rangle
  6. 5.17.6

    Velocity and Acceleration for Plane Motion

    Velocity and acceleration are vectors; work with each component separately. With constant acceleration a\mathbf{a} and initial velocity u\mathbf{u}, the velocity at time tt is u+at\mathbf{u} + \mathbf{a}t. Speed is the magnitude of velocity.

    e.g. u=⟨2,1⟩\mathbf{u} = \langle 2, 1 \rangle m/s, a=⟨1,−2⟩\mathbf{a} = \langle 1, -2 \rangle m/s2^2, t=3t = 3: v=⟨5,−5⟩\mathbf{v} = \langle 5, -5 \rangle, speed 525\sqrt2 m/s

    v=u+at\displaystyle \mathbf{v} = \mathbf{u} + \mathbf{a}t
    a=v−ut\displaystyle \mathbf{a} = \frac{\mathbf{v} - \mathbf{u}}{t}
    speed=∣v∣\displaystyle \text{speed} = \lvert \mathbf{v} \rvert
  7. 5.17.7

    Calculating Displacement for Plane Motion

    Displacement is the change in position. With constant velocity it is vt\mathbf{v}t. With constant acceleration use the formula below, then add the starting position to get the new position.

    e.g. u=⟨4,0⟩\mathbf{u} = \langle 4, 0 \rangle, a=⟨0,2⟩\mathbf{a} = \langle 0, 2 \rangle, t=3t = 3: s=⟨12,0⟩+⟨0,9⟩=⟨12,9⟩\mathbf{s} = \langle 12, 0 \rangle + \langle 0, 9 \rangle = \langle 12, 9 \rangle, a distance of 15

    s=ut+12at2\displaystyle \mathbf{s} = \mathbf{u}t + \tfrac12\mathbf{a}t^2
    r=r0+s\displaystyle \mathbf{r} = \mathbf{r}_0 + \mathbf{s}

5.18 Vectors in 3D Cartesian Coordinates

  1. 5.18.1

    Three-Dimensional Vectors in Component Form

    A 3D vector has xx-, yy-, and zz-components. k\mathbf{k} is the unit vector along the zz-axis. Subtract start coordinates from end coordinates.

    e.g. A(1,0,2)A(1, 0, 2), B(3,−1,5)B(3, -1, 5): AB→=⟨2,−1,3⟩\overrightarrow{AB} = \langle 2, -1, 3 \rangle

    ⟨a,b,c⟩=ai+bj+ck\displaystyle \langle a, b, c \rangle = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}
    AB→=⟨x2−x1,y2−y1,z2−z1⟩\displaystyle \overrightarrow{AB} = \langle x_2 - x_1, y_2 - y_1, z_2 - z_1 \rangle
  2. 5.18.2

    Addition and Scalar Multiplication of Cartesian Vectors in 3D

    Same as 2D: add matching components, and a scalar multiplies every component.

    e.g. ⟨1,2,−1⟩−2⟨0,3,1⟩=⟨1,−4,−3⟩\langle 1, 2, -1 \rangle - 2\langle 0, 3, 1 \rangle = \langle 1, -4, -3 \rangle

    ⟨a,b,c⟩+⟨d,e,f⟩\displaystyle \langle a, b, c \rangle + \langle d, e, f \rangle
    =⟨a+d, b+e, c+f⟩\displaystyle = \langle a + d,\, b + e,\, c + f \rangle
    k⟨a,b,c⟩=⟨ka,kb,kc⟩\displaystyle k\langle a, b, c \rangle = \langle ka, kb, kc \rangle
  3. 5.18.3

    Calculating the Magnitude of Cartesian Vectors in 3D

    Square all three components, add, and take the square root.

    e.g. ∣⟨2,3,6⟩∣=4+9+36=7\lvert \langle 2, 3, 6 \rangle \rvert = \sqrt{4 + 9 + 36} = 7

    ∣⟨a,b,c⟩∣=a2+b2+c2\displaystyle \lvert \langle a, b, c \rangle \rvert = \sqrt{a^2 + b^2 + c^2}

5.19 The Dot Product

θuv|u| cos θshadow of u on vu · v = |u||v| cos θ= u₁v₁ + u₂v₂u · v = 0 meansperpendicularu · v > 0: θ is acuteu · v < 0: θ is obtuse
The dot product is the length of v times the length of u’s shadow on v. When the vectors are perpendicular the shadow vanishes, so u · v = 0.
  1. 5.19.1

    Calculating the Dot Product Using Angle and Magnitude

    The dot product is a scalar. Use the angle θ\theta between the vectors placed tail to tail, with 0∘≤θ≤180∘0^\circ \le \theta \le 180^\circ. Perpendicular vectors have dot product 00.

    e.g. ∣u∣=4\lvert\mathbf{u}\rvert = 4, ∣v∣=5\lvert\mathbf{v}\rvert = 5, θ=60∘\theta = 60^\circ: u⋅v=20⋅12=10\mathbf{u}\cdot\mathbf{v} = 20 \cdot \tfrac12 = 10

    u⋅v=∣u∣ ∣v∣cos⁡θ\displaystyle \mathbf{u} \cdot \mathbf{v} = \lvert\mathbf{u}\rvert\,\lvert\mathbf{v}\rvert\cos\theta
  2. 5.19.2

    Calculating the Dot Product Using Components

    Multiply matching components and add the products. The result is a number, not a vector.

    e.g. ⟨2,−1,3⟩⋅⟨4,5,1⟩=8−5+3=6\langle 2, -1, 3 \rangle \cdot \langle 4, 5, 1 \rangle = 8 - 5 + 3 = 6

    ⟨a1,a2⟩⋅⟨b1,b2⟩\displaystyle \langle a_1, a_2 \rangle \cdot \langle b_1, b_2 \rangle
    =a1b1+a2b2\displaystyle = a_1 b_1 + a_2 b_2
    3D: add a3b3\displaystyle \text{3D: add } a_3 b_3
  3. 5.19.3

    The Angle Between Two Vectors

    Solve the dot product formula for cos⁡θ\cos\theta. A dot product of 00 means perpendicular, positive means acute, and negative means obtuse.

    e.g. ⟨1,0⟩\langle 1, 0 \rangle and ⟨1,1⟩\langle 1, 1 \rangle: cos⁡θ=11⋅2\cos\theta = \tfrac{1}{1 \cdot \sqrt2}, so θ=45∘\theta = 45^\circ

    cos⁡θ=u⋅v∣u∣ ∣v∣\displaystyle \cos\theta = \frac{\mathbf{u}\cdot\mathbf{v}}{\lvert\mathbf{u}\rvert\,\lvert\mathbf{v}\rvert}
  4. 5.19.4

    Calculating a Scalar Projection

    The scalar projection of u\mathbf{u} onto v\mathbf{v} is the signed length of the shadow of u\mathbf{u} along v\mathbf{v}. It is negative when the angle is obtuse.

    e.g. u=⟨2,1⟩\mathbf{u} = \langle 2, 1 \rangle onto v=⟨3,4⟩\mathbf{v} = \langle 3, 4 \rangle: 6+45=2\tfrac{6 + 4}{5} = 2

    compvu=u⋅v∣v∣\displaystyle \text{comp}_{\mathbf{v}}\mathbf{u} = \frac{\mathbf{u}\cdot\mathbf{v}}{\lvert\mathbf{v}\rvert}
    =∣u∣cos⁡θ\displaystyle = \lvert\mathbf{u}\rvert\cos\theta
  5. 5.19.5

    Calculating a Vector Projection

    The vector projection of u\mathbf{u} onto v\mathbf{v} is the vector along v\mathbf{v} whose length is the scalar projection. Note the squared magnitude in the denominator.

    e.g. u=⟨2,1⟩\mathbf{u} = \langle 2, 1 \rangle onto v=⟨3,4⟩\mathbf{v} = \langle 3, 4 \rangle: 1025⟨3,4⟩=⟨65,85⟩\tfrac{10}{25}\langle 3, 4 \rangle = \left\langle \tfrac65, \tfrac85 \right\rangle

    projvu=u⋅v∣v∣2 v\displaystyle \text{proj}_{\mathbf{v}}\mathbf{u} = \frac{\mathbf{u}\cdot\mathbf{v}}{\lvert\mathbf{v}\rvert^2}\,\mathbf{v}

5.20 The Cross Product

θuvu × vareau × v is perpendicularto both u and v|u × v| = |u||v| sin θ= parallelogram areaRight-hand rule: curl your fingersfrom u toward v; your thumb pointsalong u × v.Order matters: v × u = −(u × v)
The cross product of two 3D vectors is a new vector standing straight up from the plane they span. Its length equals the area of their parallelogram, and the right-hand rule gives its direction.
  1. 5.20.1

    The Cross Product of Two Vectors

    Defined only for vectors in 3D. u×v\mathbf u \times \mathbf v is a vector perpendicular to both u\mathbf u and v\mathbf v, pointing the way the right-hand rule gives. Its length is ∥u∥∥v∥sin⁡θ\lVert \mathbf u \rVert \lVert \mathbf v \rVert \sin\theta.

    e.g. ⟨1,0,0⟩×⟨0,1,0⟩=⟨0,0,1⟩\langle 1,0,0\rangle \times \langle 0,1,0\rangle = \langle 0,0,1\rangle, so i×j=k\mathbf i \times \mathbf j = \mathbf k

    u×v=⟨u2v3−u3v2,\displaystyle \mathbf u \times \mathbf v = \langle u_2v_3 - u_3v_2,
    u3v1−u1v3,  u1v2−u2v1⟩\displaystyle \quad u_3v_1 - u_1v_3,\; u_1v_2 - u_2v_1 \rangle
    ∥u×v∥=∥u∥∥v∥sin⁡θ\displaystyle \lVert \mathbf u \times \mathbf v \rVert = \lVert \mathbf u \rVert \lVert \mathbf v \rVert \sin\theta
  2. 5.20.2

    Properties of the Cross Product

    Swapping the order flips the sign. It distributes over addition and scalars pull out. u×v=0\mathbf u \times \mathbf v = \mathbf 0 exactly when the vectors are parallel, so u×u=0\mathbf u \times \mathbf u = \mathbf 0. It is not associative.

    e.g. i×j=k\mathbf i \times \mathbf j = \mathbf k, so j×i=−k\mathbf j \times \mathbf i = -\mathbf k

    v×u=−(u×v)\displaystyle \mathbf v \times \mathbf u = -(\mathbf u \times \mathbf v)
    u×(v+w)=u×v+u×w\displaystyle \mathbf u \times (\mathbf v + \mathbf w) = \mathbf u \times \mathbf v + \mathbf u \times \mathbf w
    (cu)×v=c(u×v)\displaystyle (c\mathbf u) \times \mathbf v = c(\mathbf u \times \mathbf v)
    (u×v)⋅u=0\displaystyle (\mathbf u \times \mathbf v) \cdot \mathbf u = 0
  3. 5.20.3

    Calculating the Cross Product Using Determinants

    Write a 3×33 \times 3 determinant with top row i,j,k\mathbf i, \mathbf j, \mathbf k, middle row the components of u\mathbf u, and bottom row the components of v\mathbf v. Expand along the top row: u×v\mathbf u \times \mathbf v is the sum of the three terms below. Keep the minus sign on j\mathbf j.

    e.g. ⟨1,2,3⟩×⟨4,5,6⟩=i(12−15)−j(6−12)+k(5−8)=⟨−3,6,−3⟩\langle 1,2,3\rangle \times \langle 4,5,6\rangle = \mathbf i(12-15) - \mathbf j(6-12) + \mathbf k(5-8) = \langle -3, 6, -3\rangle

    i∣u2u3v2v3∣\displaystyle \mathbf i \begin{vmatrix} u_2 & u_3 \\ v_2 & v_3 \end{vmatrix}
    −j∣u1u3v1v3∣\displaystyle - \mathbf j \begin{vmatrix} u_1 & u_3 \\ v_1 & v_3 \end{vmatrix}
    +k∣u1u2v1v2∣\displaystyle + \mathbf k \begin{vmatrix} u_1 & u_2 \\ v_1 & v_2 \end{vmatrix}
  4. 5.20.4

    Finding Areas Using the Cross Product

    The length of u×v\mathbf u \times \mathbf v is the area of the parallelogram with sides u\mathbf u and v\mathbf v. A triangle is half of that parallelogram. For a triangle PQRPQR, use two edge vectors from the same vertex.

    e.g. P(1,0,0)P(1,0,0), Q(2,2,0)Q(2,2,0), R(4,1,0)R(4,1,0): ⟨1,2,0⟩×⟨3,1,0⟩=⟨0,0,−5⟩\langle 1,2,0\rangle \times \langle 3,1,0\rangle = \langle 0,0,-5\rangle, so area =52= \tfrac52

    Apar=∥u×v∥\displaystyle A_{\text{par}} = \lVert \mathbf u \times \mathbf v \rVert
    Atri=12∥PQ→×PR→∥\displaystyle A_{\text{tri}} = \frac12 \lVert \overrightarrow{PQ} \times \overrightarrow{PR} \rVert
  5. 5.20.5

    The Scalar Triple Product

    u⋅(v×w)\mathbf u \cdot (\mathbf v \times \mathbf w) is a number, not a vector. It equals the determinant of the 3×33 \times 3 matrix whose rows are u\mathbf u, v\mathbf v, w\mathbf w. Cycling the order keeps the value; swapping two vectors flips the sign. It is 00 exactly when the three vectors are coplanar.

    e.g. u=⟨1,2,0⟩\mathbf u = \langle 1,2,0\rangle, v=⟨0,1,3⟩\mathbf v = \langle 0,1,3\rangle, w=⟨2,0,1⟩\mathbf w = \langle 2,0,1\rangle: v×w=⟨1,6,−2⟩\mathbf v \times \mathbf w = \langle 1,6,-2\rangle, so u⋅(v×w)=1+12+0=13\mathbf u \cdot (\mathbf v \times \mathbf w) = 1 + 12 + 0 = 13

    u⋅(v×w)\displaystyle \mathbf u \cdot (\mathbf v \times \mathbf w)
    =v⋅(w×u)\displaystyle = \mathbf v \cdot (\mathbf w \times \mathbf u)
    =w⋅(u×v)\displaystyle = \mathbf w \cdot (\mathbf u \times \mathbf v)
  6. 5.20.6

    Volumes of Parallelepipeds

    The volume of the parallelepiped with edges u\mathbf u, v\mathbf v, w\mathbf w from one corner is the absolute value of the scalar triple product. A volume of 00 means the edges are coplanar.

    e.g. Edges ⟨2,0,0⟩\langle 2,0,0\rangle, ⟨1,3,0⟩\langle 1,3,0\rangle, ⟨0,1,4⟩\langle 0,1,4\rangle: ⟨1,3,0⟩×⟨0,1,4⟩=⟨12,−4,1⟩\langle 1,3,0\rangle \times \langle 0,1,4\rangle = \langle 12,-4,1\rangle, so V=∣24+0+0∣=24V = \lvert 24 + 0 + 0 \rvert = 24

    V=∣u⋅(v×w)∣\displaystyle V = \lvert \mathbf u \cdot (\mathbf v \times \mathbf w) \rvert
  7. 5.20.7

    Finding Volumes of Tetrahedrons and Pyramids Using Vector Products

    Take three edge vectors from one vertex. A tetrahedron is 16\tfrac16 of the parallelepiped. A pyramid whose base is the parallelogram on u\mathbf u, v\mathbf v, with w\mathbf w running from that corner to the apex, is 13\tfrac13 of it.

    e.g. Vertices (0,0,0)(0,0,0), (2,0,0)(2,0,0), (0,3,0)(0,3,0), (0,0,4)(0,0,4): triple product =24= 24, so V=246=4V = \tfrac{24}{6} = 4

    Vtet=16∣u⋅(v×w)∣\displaystyle V_{\text{tet}} = \frac16 \lvert \mathbf u \cdot (\mathbf v \times \mathbf w) \rvert
    Vpyr=13∣u⋅(v×w)∣\displaystyle V_{\text{pyr}} = \frac13 \lvert \mathbf u \cdot (\mathbf v \times \mathbf w) \rvert

Unit 6 37 topics

Matrices

Introduction to Matrices, Matrix Multiplication, Determinants, and more.

6.21 Introduction to Matrices

  1. 6.21.1

    Introduction to Matrices

    A matrix is a rectangular array of numbers called entries. Its size is (rows) ×\times (columns), always rows first.

    e.g. (123456)\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix} is a 2×32 \times 3 matrix

    m×n=rows×columns\displaystyle m \times n = \text{rows} \times \text{columns}
  2. 6.21.2

    Index Notation for Matrices

    aija_{ij} is the entry of matrix AA in row ii and column jj, row first.

    e.g. For A=(123456)A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}: a12=2a_{12} = 2 and a23=6a_{23} = 6

    aij=entry in row i, column j\displaystyle a_{ij} = \text{entry in row } i,\ \text{column } j
  3. 6.21.3

    Adding and Subtracting Matrices

    Only matrices of the same size can be added or subtracted. Combine the entries in matching positions.

    e.g. (1234)+(5−102)=(6136)\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} + \begin{pmatrix} 5 & -1 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 6 & 1 \\ 3 & 6 \end{pmatrix}

    (A±B)ij=aij±bij\displaystyle (A \pm B)_{ij} = a_{ij} \pm b_{ij}
  4. 6.21.4

    Properties of Matrix Addition

    For matrices of the same size, addition is commutative and associative. The zero matrix OO is the additive identity, and −A-A undoes AA.

    e.g. A−B=A+(−B)A - B = A + (-B)

    A+B=B+A\displaystyle A + B = B + A
    (A+B)+C=A+(B+C)\displaystyle (A + B) + C = A + (B + C)
    A+O=A\displaystyle A + O = A
    A+(−A)=O\displaystyle A + (-A) = O
  5. 6.21.5

    Scalar Multiplication of Matrices

    Multiply every entry by the scalar. Scalars distribute over matrix addition.

    e.g. 3(1−204)=(3−6012)3\begin{pmatrix} 1 & -2 \\ 0 & 4 \end{pmatrix} = \begin{pmatrix} 3 & -6 \\ 0 & 12 \end{pmatrix}

    (kA)ij=k aij\displaystyle (kA)_{ij} = k\,a_{ij}
    k(A+B)=kA+kB\displaystyle k(A + B) = kA + kB
  6. 6.21.6

    Zero, Square, Diagonal and Identity Matrices

    A zero matrix OO has every entry 00. A square matrix is n×nn \times n. A diagonal matrix is square with 00 everywhere off the main diagonal (top-left to bottom-right). The identity InI_n is diagonal with every diagonal entry 11.

    e.g. (200−5)\begin{pmatrix} 2 & 0 \\ 0 & -5 \end{pmatrix} is diagonal but not the identity

    I2=(1001)\displaystyle I_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
  7. 6.21.7

    The Transpose of a Matrix

    Rows become columns: row ii of AA becomes column ii of ATA^T. An m×nm \times n matrix becomes n×mn \times m.

    e.g. (123456)T=(142536)\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}^T = \begin{pmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{pmatrix}

    (AT)ij=aji\displaystyle (A^T)_{ij} = a_{ji}
    (AT)T=A\displaystyle (A^T)^T = A
    (A+B)T=AT+BT\displaystyle (A + B)^T = A^T + B^T

6.22 Matrix Multiplication

  1. 6.22.1

    Multiplying a Matrix by a Column Vector

    Each entry of the result is a row of the matrix dotted with the vector. The number of columns must equal the number of entries in the vector.

    e.g. (213−1)(45)=(8+512−5)=(137)\begin{pmatrix} 2 & 1 \\ 3 & -1 \end{pmatrix}\begin{pmatrix} 4 \\ 5 \end{pmatrix} = \begin{pmatrix} 8 + 5 \\ 12 - 5 \end{pmatrix} = \begin{pmatrix} 13 \\ 7 \end{pmatrix}

    (abcd)(xy)\displaystyle \begin{pmatrix} a & b \\ c & d \end{pmatrix} \binom{x}{y}
    =(ax+bycx+dy)\displaystyle = \begin{pmatrix} ax + by \\ cx + dy \end{pmatrix}
  2. 6.22.2

    Multiplying Square Matrices

    The entry in row ii, column jj of ABAB is row ii of AA dotted with column jj of BB.

    e.g. (1234)(2013)=(2+20+66+40+12)=(461012)\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix} = \begin{pmatrix} 2 + 2 & 0 + 6 \\ 6 + 4 & 0 + 12 \end{pmatrix} = \begin{pmatrix} 4 & 6 \\ 10 & 12 \end{pmatrix}

    (AB)ij=(row i of A)\displaystyle (AB)_{ij} = (\text{row } i \text{ of } A)
    ⋅ (column j of B)\displaystyle \quad \cdot\, (\text{column } j \text{ of } B)
  3. 6.22.3

    Conformability for Matrix Multiplication

    ABAB is defined only when the number of columns of AA equals the number of rows of BB. The result takes the outer dimensions.

    e.g. (2×3)(3×4)(2 \times 3)(3 \times 4) gives 2×42 \times 4; (2×3)(2×3)(2 \times 3)(2 \times 3) is undefined

    (m×n)(n×p)→m×p\displaystyle (m \times n)(n \times p) \to m \times p
  4. 6.22.4

    Multiplying Matrices

    Check conformability first. Then each entry of ABAB is a row of AA dotted with a column of BB.

    e.g. (120013)(102104)=(52213)\begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 3 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 2 & 1 \\ 0 & 4 \end{pmatrix} = \begin{pmatrix} 5 & 2 \\ 2 & 13 \end{pmatrix}

    (AB)ij=ai1b1j+⋯+ainbnj\displaystyle (AB)_{ij} = a_{i1}b_{1j} + \cdots + a_{in}b_{nj}
  5. 6.22.5

    Powers of Matrices

    AnA^n means AA multiplied by itself nn times, so only square matrices have powers. A2A^2 is AAAA, not the entries squared.

    e.g. (1101)2=(1201)\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^2 = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}

    An=AA⋯A⏟n factors\displaystyle A^n = \underbrace{AA \cdots A}_{n \text{ factors}}
  6. 6.22.6

    Multiplying a Matrix by the Identity Matrix

    Multiplying by an identity matrix of the right size leaves a matrix unchanged. For an m×nm \times n matrix AA, use ImI_m on the left and InI_n on the right.

    e.g. (3125)(1001)=(3125)\begin{pmatrix} 3 & 1 \\ 2 & 5 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 1 \\ 2 & 5 \end{pmatrix}

    AI=IA=A\displaystyle AI = IA = A
    ImA=AIn=A\displaystyle I_m A = A I_n = A
  7. 6.22.7

    Properties of Matrix Multiplication

    Matrix multiplication is associative and distributive, and scalars can move anywhere. It is not commutative: usually AB≠BAAB \ne BA, so keep the order.

    e.g. A=(1101)A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, B=(1011)B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}: AB=(2111)AB = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} but BA=(1112)BA = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}

    (AB)C=A(BC)\displaystyle (AB)C = A(BC)
    A(B+C)=AB+AC\displaystyle A(B + C) = AB + AC
    k(AB)=(kA)B=A(kB)\displaystyle k(AB) = (kA)B = A(kB)
    AB≠BA  in general\displaystyle AB \ne BA \;\text{in general}
  8. 6.22.8

    Representing 2x2 Systems of Equations Using a Matrix Product

    Put the coefficients in a matrix AA, the variables in a column vector x\mathbf x, and the constants in a column vector b\mathbf b. The system becomes Ax=bA\mathbf x = \mathbf b.

    e.g. 2x+3y=72x + 3y = 7, x−y=1x - y = 1 becomes (231−1)(xy)=(71)\begin{pmatrix} 2 & 3 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 7 \\ 1 \end{pmatrix}

    {ax+by=ecx+dy=f\displaystyle \begin{cases} ax + by = e \\ cx + dy = f \end{cases}
    ⇔  (abcd)(xy)=(ef)\displaystyle \Leftrightarrow\; \begin{pmatrix} a & b \\ c & d \end{pmatrix} \binom{x}{y} = \binom{e}{f}
  9. 6.22.9

    Representing 3x3 Systems of Equations Using a Matrix Product

    Same idea with a 3×33 \times 3 coefficient matrix and x=(x,y,z)\mathbf x = (x, y, z) as a column. Line the variables up in the same order in every equation, and write 00 for a missing variable.

    e.g. x+2y−z=3x + 2y - z = 3, 2x+y=12x + y = 1, y+3z=4y + 3z = 4 becomes (12−1210013)(xyz)=(314)\begin{pmatrix} 1 & 2 & -1 \\ 2 & 1 & 0 \\ 0 & 1 & 3 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ 4 \end{pmatrix}

    Ax=b\displaystyle A\mathbf x = \mathbf b

6.23 Determinants

  1. 6.23.1

    The Determinant of a 2x2 Matrix

    Main diagonal product minus the other diagonal product. It is written det⁡A\det A or with vertical bars. Only square matrices have determinants.

    e.g. det⁡(3214)=12−2=10\det\begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix} = 12 - 2 = 10

    ∣abcd∣=ad−bc\displaystyle \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc
  2. 6.23.2

    The Geometric Interpretation of the 2x2 Determinant

    ∣det⁡A∣\lvert \det A \rvert is the area of the parallelogram formed by the two column vectors of AA. A negative determinant means the vectors are in clockwise order. A determinant of 00 means the columns are parallel, so the area is 00.

    e.g. Columns (3,0)(3, 0) and (1,2)(1, 2): det⁡(3102)=6−0=6\det\begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix} = 6 - 0 = 6, so area 66

    Area=∣ad−bc∣\displaystyle \text{Area} = \lvert ad - bc \rvert
  3. 6.23.3

    The Minors of a 3x3 Matrix

    The minor MijM_{ij} is the determinant of the 2×22 \times 2 matrix left after deleting row ii and column jj.

    e.g. For (123456789)\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}: M11=∣5689∣=45−48=−3M_{11} = \begin{vmatrix} 5 & 6 \\ 8 & 9 \end{vmatrix} = 45 - 48 = -3

    Mij=det⁡(delete row i, column j)\displaystyle M_{ij} = \det(\text{delete row } i,\ \text{column } j)
  4. 6.23.4

    The Determinant of a 3x3 Matrix

    Expand along the top row: each entry times its minor, with signs +, −, ++,\,-,\,+. For rows (a,b,c)(a, b, c), (d,e,f)(d, e, f), (g,h,i)(g, h, i):

    e.g. det⁡(1234567810)=1(50−48)−2(40−42)+3(32−35)=2+4−9=−3\det\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 10 \end{pmatrix} = 1(50 - 48) - 2(40 - 42) + 3(32 - 35) = 2 + 4 - 9 = -3

    det⁡A=aM11−bM12+cM13\displaystyle \det A = aM_{11} - bM_{12} + cM_{13}
    =a(ei−fh)−b(di−fg)\displaystyle = a(ei - fh) - b(di - fg)
    +c(dh−eg)\displaystyle \quad + c(dh - eg)

6.24 The Inverse of a Matrix

  1. 6.24.1

    Introduction to the Inverse of a Matrix

    The inverse A−1A^{-1} undoes AA: their product either way is the identity. Only square matrices can have inverses, and A−1A^{-1} exists exactly when det⁡A≠0\det A \ne 0. A matrix with det⁡A=0\det A = 0 is singular.

    e.g. (2004)−1=(120014)\begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}^{-1} = \begin{pmatrix} \tfrac12 & 0 \\ 0 & \tfrac14 \end{pmatrix}

    AA−1=A−1A=I\displaystyle AA^{-1} = A^{-1}A = I
  2. 6.24.2

    Inverses of 2x2 Matrices

    Swap aa and dd, negate bb and cc, then divide by the determinant. If ad−bc=0ad - bc = 0 there is no inverse.

    e.g. (4726)−1=110(6−7−24)\begin{pmatrix} 4 & 7 \\ 2 & 6 \end{pmatrix}^{-1} = \tfrac{1}{10}\begin{pmatrix} 6 & -7 \\ -2 & 4 \end{pmatrix}, since det⁡=24−14=10\det = 24 - 14 = 10

    A−1=1det⁡A(d−b−ca)\displaystyle A^{-1} = \frac{1}{\det A}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}
    det⁡A=ad−bc≠0\displaystyle \det A = ad - bc \ne 0
  3. 6.24.3

    Calculating the Inverse of a 3x3 Matrix Using the Cofactor Method

    Find all nine minors. Apply the checkerboard signs (++ in the top-left corner, alternating) to get the cofactors. Transpose the cofactor matrix to get the adjugate, then divide by det⁡A\det A.

    e.g. A=(120010001)A = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}: det⁡A=1\det A = 1, diagonal cofactors are 11, C21=−2C_{21} = -2, the rest 00; transposing gives A−1=(1−20010001)A^{-1} = \begin{pmatrix} 1 & -2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}

    Cij=(−1)i+jMij\displaystyle C_{ij} = (-1)^{i+j} M_{ij}
    adj⁡A=CT\displaystyle \operatorname{adj} A = C^T
    A−1=1det⁡Aadj⁡A\displaystyle A^{-1} = \frac{1}{\det A}\operatorname{adj} A
  4. 6.24.4

    Solving 2x2 Systems of Equations Using Inverse Matrices

    Write the system as Ax=bA\mathbf x = \mathbf b, then multiply both sides on the left by A−1A^{-1}. This works only when det⁡A≠0\det A \ne 0.

    e.g. 2x+y=52x + y = 5, x+y=3x + y = 3: A−1=(1−1−12)A^{-1} = \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}, so (xy)=(5−3−5+6)=(21)\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 - 3 \\ -5 + 6 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}

    Ax=b\displaystyle A\mathbf x = \mathbf b
    ⇒  x=A−1b\displaystyle \Rightarrow\; \mathbf x = A^{-1}\mathbf b
  5. 6.24.5

    Solving Systems of Equations Using Inverse Matrices

    Any square system Ax=bA\mathbf x = \mathbf b with det⁡A≠0\det A \ne 0 has the unique solution x=A−1b\mathbf x = A^{-1}\mathbf b. Order matters: A−1A^{-1} goes on the left of b\mathbf b. If det⁡A=0\det A = 0 there is no unique solution.

    e.g. x+2y=5x + 2y = 5, y=2y = 2, z=3z = 3: with A−1A^{-1} from the cofactor example, x=(5−423)=(123)\mathbf x = \begin{pmatrix} 5 - 4 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}

    x=A−1b\displaystyle \mathbf x = A^{-1}\mathbf b

6.25 Linear Transformations

xyîĵ(2, 0)(1, 1)area 2A =2101AîAĵThe columns arewhere î and ĵ land.det = 2·1 − 1·0 = 2det = 2 = area scale factor
A matrix moves î and ĵ to its columns, and every other point follows along. This one shears and stretches the unit square into a parallelogram with twice the area, which is its determinant.
  1. 6.25.1

    Introduction to Linear Transformations

    A linear transformation TT maps vectors to vectors and preserves addition and scalar multiples. It always sends the origin to the origin and lines to lines (or to a single point). Every linear transformation of the plane is multiplication by a 2×22 \times 2 matrix.

    e.g. T(x,y)=(2x,x+y)T(x, y) = (2x, x + y) is linear. T(x,y)=(x+1,y)T(x, y) = (x + 1, y) is not, since T(0,0)=(1,0)T(0, 0) = (1, 0).

    T(u+v)=T(u)+T(v)\displaystyle T(\mathbf u + \mathbf v) = T(\mathbf u) + T(\mathbf v)
    T(kv)=k T(v)\displaystyle T(k\mathbf v) = k\,T(\mathbf v)
  2. 6.25.2

    The Standard Matrix of a Linear Transformation

    The columns of the matrix are the images of the unit vectors i=(1,0)\mathbf i = (1, 0) and j=(0,1)\mathbf j = (0, 1), in that order.

    e.g. T(x,y)=(2x,x+y)T(x, y) = (2x, x + y): T(i)=(2,1)T(\mathbf i) = (2, 1), T(j)=(0,1)T(\mathbf j) = (0, 1), so A=(2011)A = \begin{pmatrix} 2 & 0 \\ 1 & 1 \end{pmatrix}

    A=(T(i)T(j))\displaystyle A = \begin{pmatrix} T(\mathbf i) & T(\mathbf j) \end{pmatrix}
  3. 6.25.3

    Linear Transformations of Points and Lines in the Plane

    To transform a point, multiply its position vector by the matrix. To transform a line, transform a general point on it (or two points) and find the line through the images.

    e.g. A=(2011)A = \begin{pmatrix} 2 & 0 \\ 1 & 1 \end{pmatrix}: (3,−1)↦(6,2)(3, -1) \mapsto (6, 2). The line y=2xy = 2x: (t,2t)↦(2t,3t)(t, 2t) \mapsto (2t, 3t), so its image is y=32xy = \tfrac32 x.

    (x′y′)=A(xy)\displaystyle \binom{x'}{y'} = A\binom{x}{y}
  4. 6.25.4

    Linear Transformations of Objects in the Plane

    Transform each vertex and join the images in the same order. Faster: put the vertices as the columns of one matrix and multiply once.

    e.g. Unit square under (2011)\begin{pmatrix} 2 & 0 \\ 1 & 1 \end{pmatrix}: (0,0),(1,0),(1,1),(0,1)↦(0,0),(2,1),(2,2),(0,1)(0,0), (1,0), (1,1), (0,1) \mapsto (0,0), (2,1), (2,2), (0,1), a parallelogram

    A(x1x2⋯y1y2⋯)\displaystyle A\begin{pmatrix} x_1 & x_2 & \cdots \\ y_1 & y_2 & \cdots \end{pmatrix}
  5. 6.25.5

    Dilations and Reflections as Linear Transformations

    A dilation about the origin with scale factor kk multiplies every coordinate by kk. The four standard reflections are in the xx-axis, the yy-axis, y=xy = x, and y=−xy = -x.

    e.g. Reflecting (2,5)(2, 5) in y=xy = x gives (5,2)(5, 2)

    scale k:(k00k)\displaystyle \text{scale } k: \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}
    x-axis:(100−1)\displaystyle x\text{-axis}: \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}
    y-axis:(−1001)\displaystyle y\text{-axis}: \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}
    y=x:(0110)\displaystyle y = x: \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}
    y=−x:(0−1−10)\displaystyle y = -x: \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}
  6. 6.25.6

    Shear and Stretch as Linear Transformations

    A stretch parallel to the xx-axis with factor kk multiplies xx by kk and keeps the yy-axis fixed (likewise for yy). An xx-shear keeps the xx-axis fixed and slides each point sideways by kk times its yy-coordinate (likewise for a yy-shear).

    e.g. The shear (1201)\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} sends (1,3)(1, 3) to (1+6,3)=(7,3)(1 + 6, 3) = (7, 3)

    x-stretch:(k001)\displaystyle x\text{-stretch}: \begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix}
    y-stretch:(100k)\displaystyle y\text{-stretch}: \begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}
    x-shear:(1k01)\displaystyle x\text{-shear}: \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}
    y-shear:(10k1)\displaystyle y\text{-shear}: \begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix}
  7. 6.25.7

    Right-Angle Rotations as Linear Transformations

    Rotations about the origin are counterclockwise unless stated. 90∘90^\circ sends (x,y)(x, y) to (−y,x)(-y, x); 180∘180^\circ sends it to (−x,−y)(-x, -y); 270∘270^\circ (90∘90^\circ clockwise) sends it to (y,−x)(y, -x).

    e.g. Rotating (3,1)(3, 1) by 90∘90^\circ gives (−1,3)(-1, 3)

    90∘:(0−110)\displaystyle 90^\circ: \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}
    180∘:(−100−1)\displaystyle 180^\circ: \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}
    270∘:(01−10)\displaystyle 270^\circ: \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}
  8. 6.25.8

    Rotations as Linear Transformations

    Rotation about the origin by angle θ\theta counterclockwise. For a clockwise rotation use −θ-\theta.

    e.g. Rotating (2,0)(2, 0) by 60∘60^\circ: (2cos⁡60∘,2sin⁡60∘)=(1,3)(2\cos 60^\circ, 2\sin 60^\circ) = (1, \sqrt3)

    Rθ=(cos⁡θ−sin⁡θsin⁡θcos⁡θ)\displaystyle R_\theta = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}
  9. 6.25.9

    Combining Linear Transformations Using 2x2 Matrices

    To apply AA first and then BB, use the single matrix BABA. The first transformation goes on the right, next to the vector.

    e.g. Reflect in the xx-axis, then rotate 90∘90^\circ: (0−110)(100−1)=(0110)\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, a reflection in y=xy = x

    x↦B(Ax)=(BA)x\displaystyle \mathbf x \mapsto B(A\mathbf x) = (BA)\mathbf x
  10. 6.25.10

    Inverting Linear Transformations

    The inverse transformation has matrix A−1A^{-1} and maps each image back to its original point. It exists only when det⁡A≠0\det A \ne 0. To undo a combination, undo the steps in reverse order.

    e.g. Under A=(2111)A = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} the image is (5,3)(5, 3): A−1(53)=(1−1−12)(53)=(21)A^{-1}\begin{pmatrix} 5 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} 5 \\ 3 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}

    x=A−1x′\displaystyle \mathbf x = A^{-1}\mathbf x'
    (BA)−1=A−1B−1\displaystyle (BA)^{-1} = A^{-1}B^{-1}
  11. 6.25.11

    Area Scale Factors of Linear Transformations

    Every area is multiplied by ∣det⁡A∣\lvert \det A \rvert. A negative determinant also means the shape is flipped (orientation reversed).

    e.g. A=(3112)A = \begin{pmatrix} 3 & 1 \\ 1 & 2 \end{pmatrix} has det⁡A=6−1=5\det A = 6 - 1 = 5, so a triangle of area 44 maps to area 2020

    image area=∣det⁡A∣×area\displaystyle \text{image area} = \lvert \det A \rvert \times \text{area}
  12. 6.25.12

    Singular Linear Transformations in the Plane

    A transformation is singular when det⁡A=0\det A = 0. It squashes the whole plane onto a line through the origin (or onto the origin itself), areas become 00, many points share an image, and there is no inverse.

    e.g. (1224)\begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}: det⁡=4−4=0\det = 4 - 4 = 0, and (x,y)↦(x+2y, 2x+4y)(x, y) \mapsto (x + 2y,\, 2x + 4y) always lands on the line y=2xy = 2x

    det⁡A=ad−bc=0\displaystyle \det A = ad - bc = 0

Unit 7 20 topics

Conic Sections

Ellipses as Conic Sections, and Hyperbolas as Conic Sections.

7.26 Ellipses as Conic Sections

xycbad1d2F1F2(5, 0)(−5, 0)(0, 3)(0, −3)x2/25 + y2/9 = 1a = 5 (half the long axis)b = 3 (half the short axis)c2 = a2 − b2 = 25 − 9c = 4, foci (±4, 0)d1 + d2 = 2a = 10for every point on the ellipseCo-vertex to focus is a: theshaded right triangle.
For an ellipse, a is half the long axis, b is half the short axis, and the foci sit c from the center with c² = a² − b². From any point on the curve, the two distances to the foci add to 2a.
  1. 7.26.1

    Introduction to Ellipses

    An ellipse is the set of points whose distances to two fixed points, the foci, add to a constant 2a2a. The major axis is the longest diameter and the minor axis is the shortest. The center is their midpoint.

    e.g. Foci (±3,0)(\pm 3, 0) and distance sum 10 give a=5a = 5.

    d1+d2=2a\displaystyle d_1 + d_2 = 2a
  2. 7.26.2

    Equations of Ellipses Centered at the Origin

    Here a>b>0a > b > 0, where aa is half the major axis and bb is half the minor axis. The larger denominator tells you the direction of the major axis.

    e.g. x225+y29=1\tfrac{x^2}{25} + \tfrac{y^2}{9} = 1: horizontal, vertices (±5,0)(\pm 5, 0), co-vertices (0,±3)(0, \pm 3)

    x2a2+y2b2=1  horizontal\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \;\text{horizontal}
    x2b2+y2a2=1  vertical\displaystyle \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \;\text{vertical}
  3. 7.26.3

    Equations of Ellipses Centered at a General Point

    Replace xx with x−hx - h and yy with y−ky - k to move the center to (h,k)(h, k). The axes keep their lengths 2a2a and 2b2b.

    e.g. (x−2)216+(y+1)24=1\tfrac{(x-2)^2}{16} + \tfrac{(y+1)^2}{4} = 1: center (2,−1)(2, -1), vertices (−2,−1)(-2, -1) and (6,−1)(6, -1)

    (x−h)2a2+(y−k)2b2=1\displaystyle \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1
  4. 7.26.4

    Finding the Center and Axes of Ellipses by Completing the Square

    Group the xx terms and the yy terms, factor out each squared term's coefficient, and complete the square in each group. Add the same amounts to the other side, then divide so the right side is 1.

    e.g. 4x2+9y2−16x+18y=11⇒4(x−2)2+9(y+1)2=364x^2 + 9y^2 - 16x + 18y = 11 \Rightarrow 4(x-2)^2 + 9(y+1)^2 = 36: center (2,−1)(2, -1), a=3a = 3, b=2b = 2

    Ax2+Cy2+Dx+Ey+F=0\displaystyle Ax^2 + Cy^2 + Dx + Ey + F = 0
    ⇒  (x−h)2a2+(y−k)2b2=1\displaystyle \Rightarrow\; \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1
  5. 7.26.5

    Finding Intercepts of Ellipses

    For xx-intercepts set y=0y = 0; for yy-intercepts set x=0x = 0. Solve each, and note that an ellipse may miss an axis entirely. Centered at the origin, the intercepts are the vertices and co-vertices.

    e.g. (x−1)24+y29=1\tfrac{(x-1)^2}{4} + \tfrac{y^2}{9} = 1: y=0⇒(x−1)2=4⇒x=−1,3y = 0 \Rightarrow (x - 1)^2 = 4 \Rightarrow x = -1, 3

    x2a2+y2b2=1\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
    ⇒  (±a,0),  (0,±b)\displaystyle \Rightarrow\; (\pm a, 0),\; (0, \pm b)
  6. 7.26.6

    Finding Intersections of Ellipses and Lines

    Substitute the line into the ellipse and solve the quadratic. Two real solutions mean two intersection points, one means the line is tangent, and none means they do not meet.

    e.g. (5x−6)(x−2)=0(5x - 6)(x - 2) = 0, so the points are (2,0)(2, 0) and (65,−45)\left(\tfrac65, -\tfrac45\right).

    x2+4y2=4,  y=x−2\displaystyle x^2 + 4y^2 = 4,\; y = x - 2
    ⇒  5x2−16x+12=0\displaystyle \Rightarrow\; 5x^2 - 16x + 12 = 0
  7. 7.26.7

    Foci of Ellipses

    The foci lie on the major axis, cc units from the center on each side, where aa is the semi-major axis.

    e.g. x225+y216=1\tfrac{x^2}{25} + \tfrac{y^2}{16} = 1: c2=9c^2 = 9, foci (±3,0)(\pm 3, 0)

    c2=a2−b2\displaystyle c^2 = a^2 - b^2
  8. 7.26.8

    Vertices and Eccentricity of Ellipses

    The vertices are the ends of the major axis, aa units from the center. Eccentricity measures how stretched the ellipse is: ee is between 0 and 1, with 0 a circle and values near 1 long and thin.

    e.g. x29+y225=1\tfrac{x^2}{9} + \tfrac{y^2}{25} = 1: vertices (0,±5)(0, \pm 5), c=4c = 4, e=45e = \tfrac45

    e=ca\displaystyle e = \frac{c}{a}
  9. 7.26.9

    Directrices of Ellipses

    The two directrices are lines perpendicular to the major axis, ae=a2c\tfrac{a}{e} = \tfrac{a^2}{c} from the center. For any point on the ellipse, its distance to a focus divided by its distance to the matching directrix equals ee.

    e.g. x225+y216=1\tfrac{x^2}{25} + \tfrac{y^2}{16} = 1: c=3c = 3, directrices x=±253x = \pm\tfrac{25}{3}

    x=h±a2c  horizontal\displaystyle x = h \pm \frac{a^2}{c} \;\text{horizontal}
    y=k±a2c  vertical\displaystyle y = k \pm \frac{a^2}{c} \;\text{vertical}
  10. 7.26.10

    The Area of an Ellipse

    Multiply π\pi by the two semi-axes. A circle is the case a=b=ra = b = r.

    e.g. x225+y29=1\tfrac{x^2}{25} + \tfrac{y^2}{9} = 1: A=15π≈47.1A = 15\pi \approx 47.1

    A=πab\displaystyle A = \pi a b

7.27 Hyperbolas as Conic Sections

xyF1F2(−4, 0)(4, 0)y = ¾xy = −¾xd1d2x2/16 − y2/9 = 1a = 4: vertices (±4, 0)b = 3c2 = a2 + b2 = 16 + 9c = 5, foci (±5, 0)asymptotes y = ±(b/a)x = ±¾x|d1 − d2| = 2a = 8for every point on the hyperbola
A hyperbola opens toward its foci and hugs its dashed asymptotes far from the center. Here c² = a² + b², and from any point on the curve the two focal distances differ by 2a.
  1. 7.27.1

    Equations of Hyperbolas Centered at the Origin

    The positive squared term sets the direction: positive x2x^2 opens left and right, positive y2y^2 opens up and down. a2a^2 always sits under the positive term.

    e.g. x29−y216=1\frac{x^2}{9} - \frac{y^2}{16} = 1 opens left and right with vertices (±3,0)(\pm 3, 0)

    x2a2−y2b2=1\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1
    y2a2−x2b2=1\displaystyle \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1
  2. 7.27.2

    Equations of Hyperbolas Centered at a General Point

    Shift the origin form so the center is (h,k)(h, k). The signs inside the parentheses are opposite the center's coordinates.

    e.g. (x−1)24−(y+2)29=1\frac{(x-1)^2}{4} - \frac{(y+2)^2}{9} = 1: center (1,−2)(1, -2), vertices (−1,−2)(-1, -2) and (3,−2)(3, -2)

    (x−h)2a2−(y−k)2b2=1\displaystyle \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1
    (y−k)2a2−(x−h)2b2=1\displaystyle \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1
  3. 7.27.3

    Asymptotes of Hyperbolas Centered at the Origin

    Two lines through the center that the branches approach. Shortcut: replace the 1 with 0 and solve for yy.

    e.g. x29−y216=1\frac{x^2}{9} - \frac{y^2}{16} = 1: y=±43xy = \pm\tfrac43 x

    y=±bax    (opens left/right)\displaystyle y = \pm\frac{b}{a}x \;\;\text{(opens left/right)}
    y=±abx    (opens up/down)\displaystyle y = \pm\frac{a}{b}x \;\;\text{(opens up/down)}
  4. 7.27.4

    Asymptotes of Hyperbolas Centered at a General Point

    Same slopes as the origin case, but the lines pass through the center (h,k)(h, k).

    e.g. (x−1)24−(y+2)29=1\frac{(x-1)^2}{4} - \frac{(y+2)^2}{9} = 1: y+2=±32(x−1)y + 2 = \pm\tfrac32(x - 1)

    y−k=±ba(x−h)\displaystyle y - k = \pm\frac{b}{a}(x - h)
    (use ±ab if it opens up/down)\displaystyle \text{(use } \pm\tfrac{a}{b} \text{ if it opens up/down)}
  5. 7.27.5

    Finding Intercepts and Intersections of Hyperbolas

    Set y=0y = 0 for xx-intercepts and x=0x = 0 for yy-intercepts. A negative square means no intercept on that axis. For an intersection with another curve, substitute one equation into the other and solve.

    e.g. x2−y2=5x^2 - y^2 = 5 meets y=2y = 2 where x2=9x^2 = 9: points (±3,2)(\pm 3, 2)

  6. 7.27.6

    Transverse Axes of Hyperbolas

    The segment joining the two vertices. It passes through the center and the foci and points the way the hyperbola opens.

    e.g. y225−x24=1\frac{y^2}{25} - \frac{x^2}{4} = 1: vertical, from (0,−5)(0, -5) to (0,5)(0, 5), length 1010

    length=2a\displaystyle \text{length} = 2a
  7. 7.27.7

    Conjugate Axes of Hyperbolas

    The segment through the center perpendicular to the transverse axis, reaching bb units each way. It does not touch the hyperbola.

    e.g. x29−y216=1\frac{x^2}{9} - \frac{y^2}{16} = 1: from (0,−4)(0, -4) to (0,4)(0, 4), length 88

    length=2b\displaystyle \text{length} = 2b
  8. 7.27.8

    Foci of Hyperbolas

    The foci lie on the transverse axis, cc units from the center. For any point on the hyperbola, the difference of its distances to the foci is 2a2a. Note the plus sign, unlike an ellipse.

    e.g. x29−y216=1\frac{x^2}{9} - \frac{y^2}{16} = 1: c=25=5c = \sqrt{25} = 5, foci (±5,0)(\pm 5, 0)

    c2=a2+b2\displaystyle c^2 = a^2 + b^2
  9. 7.27.9

    Eccentricity and Vertices of Hyperbolas

    The vertices are aa units from the center along the transverse axis. Eccentricity measures how wide the branches open, and for every hyperbola e>1e > 1.

    e.g. x29−y216=1\frac{x^2}{9} - \frac{y^2}{16} = 1: vertices (±3,0)(\pm 3, 0), e=53e = \tfrac53

    e=ca>1\displaystyle e = \frac{c}{a} > 1
    c=a2+b2\displaystyle c = \sqrt{a^2 + b^2}
  10. 7.27.10

    Directrices of Hyperbolas

    Two lines perpendicular to the transverse axis, a2c\tfrac{a^2}{c} units from the center, between the center and each vertex. For any point on the hyperbola, distance to a focus divided by distance to the matching directrix equals ee.

    e.g. x29−y216=1\frac{x^2}{9} - \frac{y^2}{16} = 1: c=5c = 5, directrices x=±95x = \pm\tfrac95

    x=h±a2c    (opens left/right)\displaystyle x = h \pm \frac{a^2}{c} \;\;\text{(opens left/right)}
    y=k±a2c    (opens up/down)\displaystyle y = k \pm \frac{a^2}{c} \;\;\text{(opens up/down)}

Unit 8 10 topics

Parametric Equations

Parametric Equations.

8.28 Parametric Equations

xyt = −2t = −1t = 0t = 1t = 2x = t2y = t3 − 3t−2 ≤ t ≤ 2Arrows point the way thecurve is traced as t grows.The curve crosses itselfat (3, 0), when t = ±√3.Eliminate t to get oneequation: y2 = x(x − 3)2.
A parametric curve gives x and y separately as functions of t. Plotting points in order of t shows the path and its direction, which an ordinary y = f(x) graph cannot show.
  1. 8.28.1

    Graphing Curves Defined Parametrically

    Both xx and yy are given in terms of a parameter tt. Make a table of tt values, plot the points (x,y)(x, y), and connect them in order of increasing tt. Arrows show the direction (orientation).

    e.g. x=t+1x = t + 1, y=2ty = 2t: t=0,1,2t = 0, 1, 2 gives (1,0),(2,2),(3,4)(1, 0), (2, 2), (3, 4)

    x=f(t),y=g(t)\displaystyle x = f(t), \quad y = g(t)
  2. 8.28.2

    Cartesian Equations of Parametric Curves

    Eliminate the parameter. Solve one equation for tt and substitute into the other, or use an identity when tt sits inside sine and cosine. Keep any restriction the parameter places on xx or yy.

    e.g. x=t+1x = t + 1, y=t2⇒t=x−1⇒y=(x−1)2y = t^2 \Rightarrow t = x - 1 \Rightarrow y = (x - 1)^2

    cos⁡2t+sin⁡2t=1\displaystyle \cos^2 t + \sin^2 t = 1
  3. 8.28.3

    Finding Intercepts of Curves Defined Parametrically

    For the yy-intercept, set x(t)=0x(t) = 0, solve for tt, and find yy. For the xx-intercept, set y(t)=0y(t) = 0, solve for tt, and find xx.

    e.g. x=t−2x = t - 2, y=t2−1y = t^2 - 1: x=0x = 0 at t=2t = 2 gives (0,3)(0, 3); y=0y = 0 at t=±1t = \pm1 gives (−1,0),(−3,0)(-1, 0), (-3, 0)

    x(t)=0→y-intercept\displaystyle x(t) = 0 \to y\text{-intercept}
    y(t)=0→x-intercept\displaystyle y(t) = 0 \to x\text{-intercept}
  4. 8.28.4

    Finding Intersections of Parametric Curves and Lines

    Substitute x(t)x(t) and y(t)y(t) into the equation of the line, solve for tt, and plug each tt back in to get the points.

    e.g. x=tx = t, y=t2y = t^2, line y=x+2y = x + 2: t2−t−2=0⇒t=2,−1t^2 - t - 2 = 0 \Rightarrow t = 2, -1, points (2,4),(−1,1)(2, 4), (-1, 1)

  5. 8.28.5

    Parametric Equations of Circles

    A circle with center (h,k)(h, k) and radius rr, traced counterclockwise once for 0≤t<2π0 \le t < 2\pi, starting at (h+r,k)(h + r, k).

    e.g. x=1+2cos⁡tx = 1 + 2\cos t, y=−2+2sin⁡ty = -2 + 2\sin t: center (1,−2)(1, -2), radius 22

    x=h+rcos⁡t\displaystyle x = h + r\cos t
    y=k+rsin⁡t\displaystyle y = k + r\sin t
    (x−h)2+(y−k)2=r2\displaystyle (x - h)^2 + (y - k)^2 = r^2
  6. 8.28.6

    Parametric Equations of Ellipses

    An ellipse with center (h,k)(h, k), horizontal semi-axis aa, and vertical semi-axis bb.

    e.g. x=4cos⁡tx = 4\cos t, y=2sin⁡t⇒x216+y24=1y = 2\sin t \Rightarrow \tfrac{x^2}{16} + \tfrac{y^2}{4} = 1

    x=h+acos⁡t\displaystyle x = h + a\cos t
    y=k+bsin⁡t\displaystyle y = k + b\sin t
    (x−h)2a2+(y−k)2b2=1\displaystyle \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1
  7. 8.28.7

    Parametric Equations of Parabolas

    A parabola with vertex at the origin. For y2=4axy^2 = 4ax (opening sideways) use x=at2x = at^2, y=2aty = 2at. For x2=4ayx^2 = 4ay (opening up or down) use x=2atx = 2at, y=at2y = at^2.

    e.g. y2=12xy^2 = 12x: a=3a = 3, so x=3t2x = 3t^2, y=6ty = 6t

    y2=4ax:  x=at2,  y=2at\displaystyle y^2 = 4ax:\; x = at^2,\; y = 2at
    x2=4ay:  x=2at,  y=at2\displaystyle x^2 = 4ay:\; x = 2at,\; y = at^2
  8. 8.28.8

    Parametric Equations of Parabolas Centered at (h,k)

    Shift the vertex to (h,k)(h, k) by adding hh to xx and kk to yy.

    e.g. (y−2)2=8(x+1)(y - 2)^2 = 8(x + 1): a=2a = 2, so x=−1+2t2x = -1 + 2t^2, y=2+4ty = 2 + 4t

    (y−k)2=4a(x−h)\displaystyle (y - k)^2 = 4a(x - h)
    ⇒  x=h+at2,  y=k+2at\displaystyle \Rightarrow\; x = h + at^2,\; y = k + 2at
    (x−h)2=4a(y−k)\displaystyle (x - h)^2 = 4a(y - k)
    ⇒  x=h+2at,  y=k+at2\displaystyle \Rightarrow\; x = h + 2at,\; y = k + at^2
  9. 8.28.9

    Parametric Equations of Horizontal Hyperbolas

    A hyperbola opening left and right, centered at (h,k)(h, k). It works because sec⁡2t−tan⁡2t=1\sec^2 t - \tan^2 t = 1.

    e.g. x=3sec⁡tx = 3\sec t, y=2tan⁡t⇒x29−y24=1y = 2\tan t \Rightarrow \tfrac{x^2}{9} - \tfrac{y^2}{4} = 1

    x=h+asec⁡t\displaystyle x = h + a\sec t
    y=k+btan⁡t\displaystyle y = k + b\tan t
    (x−h)2a2−(y−k)2b2=1\displaystyle \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1
  10. 8.28.10

    Parametric Equations of Vertical Hyperbolas

    A hyperbola opening up and down, centered at (h,k)(h, k). Secant goes with yy and tangent goes with xx.

    e.g. x=4tan⁡tx = 4\tan t, y=5sec⁡t⇒y225−x216=1y = 5\sec t \Rightarrow \tfrac{y^2}{25} - \tfrac{x^2}{16} = 1

    x=h+btan⁡t\displaystyle x = h + b\tan t
    y=k+asec⁡t\displaystyle y = k + a\sec t
    (y−k)2a2−(x−h)2b2=1\displaystyle \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1

Unit 9 6 topics

Polar Equations

Polar Coordinates.

9.29 Polar Coordinates

0π/6π/3π/22π/35π/6π7π/64π/33π/25π/311π/6123θ(r, θ)r = 3 cos 2θa four-petal rosex = r cos θy = r sin θr2 = x2 + y2tan θ = y/xr is the distance from the pole;θ is the angle from the x-axis.
A polar point is a distance r from the pole at an angle θ. The conversion formulas come straight from the right triangle with legs x and y and hypotenuse r.
  1. 9.29.1

    Introduction to Polar Coordinates

    A point (r,θ)(r, \theta) is rr units from the origin (the pole) at angle θ\theta measured counterclockwise from the positive xx-axis. A negative rr points the opposite way. Every point has infinitely many polar names.

    e.g. (2,π3)=(2,7π3)=(−2,4π3)(2, \tfrac{\pi}{3}) = (2, \tfrac{7\pi}{3}) = (-2, \tfrac{4\pi}{3})

    (r,θ)=(r,θ+2πn)\displaystyle (r, \theta) = (r, \theta + 2\pi n)
    (r,θ)=(−r,θ+π)\displaystyle (r, \theta) = (-r, \theta + \pi)
  2. 9.29.2

    Converting from Polar Coordinates to Cartesian Coordinates

    Use cosine for xx and sine for yy. To convert an equation, substitute these and use r2=x2+y2r^2 = x^2 + y^2.

    e.g. (4,π6)→(4⋅32,4⋅12)=(23,2)(4, \tfrac{\pi}{6}) \to (4 \cdot \tfrac{\sqrt3}{2}, 4 \cdot \tfrac12) = (2\sqrt3, 2)

    x=rcos⁡θ\displaystyle x = r\cos\theta
    y=rsin⁡θ\displaystyle y = r\sin\theta
    r2=x2+y2\displaystyle r^2 = x^2 + y^2
  3. 9.29.3

    Polar Equations of Circles Centered at the Origin

    r=ar = a is a circle centered at the origin with radius ∣a∣\lvert a \rvert.

    e.g. r=3r = 3: circle x2+y2=9x^2 + y^2 = 9

    r=a  ⟺  x2+y2=a2\displaystyle r = a \iff x^2 + y^2 = a^2
  4. 9.29.4

    Polar Equations of Radial Lines

    θ=α\theta = \alpha is the whole line through the origin at angle α\alpha, because rr can be positive or negative.

    e.g. θ=π4\theta = \tfrac{\pi}{4} is y=xy = x. θ=π2\theta = \tfrac{\pi}{2} is x=0x = 0.

    θ=α  ⟺  y=(tan⁡α) x\displaystyle \theta = \alpha \iff y = (\tan\alpha)\, x
  5. 9.29.5

    Polar Equations of Circles Centered on the Coordinate Axes

    These circles pass through the origin and have radius ∣a∣\lvert a \rvert. Cosine puts the center on the xx-axis, sine puts it on the yy-axis.

    e.g. r=6sin⁡θ⇒x2+y2=6y⇒x2+(y−3)2=9r = 6\sin\theta \Rightarrow x^2 + y^2 = 6y \Rightarrow x^2 + (y - 3)^2 = 9, center (0,3)(0, 3), radius 33

    r=2acos⁡θ:  center (a,0)\displaystyle r = 2a\cos\theta:\; \text{center } (a, 0)
    r=2asin⁡θ:  center (0,a)\displaystyle r = 2a\sin\theta:\; \text{center } (0, a)
  6. 9.29.6

    Finding Intersections of Polar Curves

    Set the two expressions for rr equal and solve for θ\theta. Check the pole separately, since both curves can pass through it at different angles.

    e.g. r=sin⁡θr = \sin\theta and r=cos⁡θr = \cos\theta: tan⁡θ=1\tan\theta = 1 gives (22,π4)(\tfrac{\sqrt2}{2}, \tfrac{\pi}{4}), plus the pole

    r1(θ)=r2(θ)\displaystyle r_1(\theta) = r_2(\theta)

Unit 10 28 topics

Complex Numbers

The Complex Plane, Further Complex Numbers, Complex Numbers in Polar Form, and more.

10.30 The Complex Plane

realimaginary−11234−ii2i3i3 + 2iPowers of i repeat every 4i¹ = ii² = −1i³ = −ii⁴ = 1× i
The number 3 + 2i sits 3 units along the real axis and 2 units up the imaginary axis. Each multiplication by i turns a quarter turn, so the powers of i cycle i, −1, −i, 1.
  1. 10.30.1

    The Complex Plane

    The complex number a+bia + bi is plotted as the point (a,b)(a, b). The horizontal axis is the real axis and the vertical axis is the imaginary axis.

    e.g. 3−2i3 - 2i is the point (3,−2)(3, -2)

    a+bi  →  (a,b)\displaystyle a + bi \;\to\; (a, b)
  2. 10.30.2

    The Magnitude of a Complex Number

    The magnitude (modulus) ∣z∣\lvert z \rvert is the distance from zz to the origin. It is always a nonnegative real number.

    e.g. ∣3+4i∣=9+16=5\lvert 3 + 4i \rvert = \sqrt{9 + 16} = 5

    ∣a+bi∣=a2+b2\displaystyle \lvert a + bi \rvert = \sqrt{a^2 + b^2}
  3. 10.30.3

    The Argument of a Complex Number

    The argument arg⁡z\arg z is the angle from the positive real axis to zz. Use tan⁡θ=ba\tan\theta = \tfrac{b}{a}, then pick the angle in the correct quadrant. The principal argument satisfies −π<θ≤π-\pi < \theta \le \pi.

    e.g. arg⁡(−1+i)=3π4\arg(-1 + i) = \tfrac{3\pi}{4}, arg⁡(1−i)=−π4\arg(1 - i) = -\tfrac{\pi}{4}

    tan⁡θ=ba\displaystyle \tan\theta = \frac{b}{a}
    z=∣z∣(cos⁡θ+isin⁡θ)\displaystyle z = \lvert z \rvert(\cos\theta + i\sin\theta)
  4. 10.30.4

    Arithmetic in the Complex Plane

    Adding complex numbers adds them like vectors (parallelogram rule), and subtracting works the same way. Multiplying by a positive real kk scales the distance from the origin by kk. Multiplying by ii rotates 90∘90^\circ counterclockwise.

    e.g. (2+i)+(1+3i)=3+4i(2 + i) + (1 + 3i) = 3 + 4i; i(2+i)=−1+2ii(2 + i) = -1 + 2i, a 90∘90^\circ turn

    (a+bi)+(c+di)\displaystyle (a + bi) + (c + di)
    =(a+c)+(b+d)i\displaystyle = (a + c) + (b + d)i
  5. 10.30.5

    Geometry in the Complex Plane

    The distance between z1z_1 and z2z_2 is ∣z1−z2∣\lvert z_1 - z_2 \rvert. The midpoint is their average. The points with ∣z−z0∣=r\lvert z - z_0 \rvert = r form a circle with center z0z_0 and radius rr.

    e.g. Distance from 1+i1 + i to 4+5i4 + 5i: ∣3+4i∣=5\lvert 3 + 4i \rvert = 5

    d=∣z1−z2∣\displaystyle d = \lvert z_1 - z_2 \rvert
    M=z1+z22\displaystyle M = \frac{z_1 + z_2}{2}

10.31 Further Complex Numbers

  1. 10.31.1

    The Complex Conjugate

    The conjugate of z=a+biz = a + bi is zˉ=a−bi\bar z = a - bi: flip the sign of the imaginary part. In the complex plane it is the reflection of zz across the real axis.

    e.g. −2+5i‾=−2−5i\overline{-2 + 5i} = -2 - 5i

    a+bi‾=a−bi\displaystyle \overline{a + bi} = a - bi
  2. 10.31.2

    Special Properties of the Complex Conjugate

    A number times its conjugate is a nonnegative real number. Conjugation passes through sums and products.

    e.g. (3+2i)(3−2i)=9+4=13(3 + 2i)(3 - 2i) = 9 + 4 = 13

    zzˉ=∣z∣2=a2+b2\displaystyle z\bar z = \lvert z \rvert^2 = a^2 + b^2
    z+zˉ=2a,z−zˉ=2bi\displaystyle z + \bar z = 2a,\quad z - \bar z = 2bi
    z+w‾=zˉ+wˉ\displaystyle \overline{z + w} = \bar z + \bar w
    zw‾=zˉ wˉ\displaystyle \overline{zw} = \bar z\,\bar w
  3. 10.31.3

    The Complex Conjugate and the Roots of a Quadratic Equation

    If a quadratic with real coefficients has a negative discriminant, its roots are a conjugate pair p±qip \pm qi. Their sum is 2p2p and their product is p2+q2p^2 + q^2.

    e.g. x2−4x+13=0⇒x=4±−362=2±3ix^2 - 4x + 13 = 0 \Rightarrow x = \tfrac{4 \pm \sqrt{-36}}{2} = 2 \pm 3i

    roots p±qi:\displaystyle \text{roots } p \pm qi:
    x2−2px+(p2+q2)=0\displaystyle x^2 - 2px + (p^2 + q^2) = 0
  4. 10.31.4

    Dividing Complex Numbers

    Multiply the top and bottom by the conjugate of the denominator. The denominator becomes the real number c2+d2c^2 + d^2.

    e.g. 5+5i1+2i=(5+5i)(1−2i)5=15−5i5=3−i\tfrac{5 + 5i}{1 + 2i} = \tfrac{(5 + 5i)(1 - 2i)}{5} = \tfrac{15 - 5i}{5} = 3 - i

    a+bic+di=(a+bi)(c−di)c2+d2\displaystyle \frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{c^2 + d^2}
  5. 10.31.5

    Solving Equations by Equating Real and Imaginary Parts

    Two complex numbers are equal only when their real parts are equal and their imaginary parts are equal. This turns one complex equation into two real equations.

    e.g. (x+y)+(x−y)i=5+i⇒x+y=5(x + y) + (x - y)i = 5 + i \Rightarrow x + y = 5, x−y=1x - y = 1, so x=3x = 3, y=2y = 2

    a+bi=c+di  ⟺  a=c,  b=d\displaystyle a + bi = c + di \iff a = c,\; b = d
  6. 10.31.6

    Extending Polynomial Identities to the Complex Numbers

    The usual identities still hold with complex numbers. With ii, a sum of squares factors like a difference of squares.

    e.g. x2+9=(x+3i)(x−3i)x^2 + 9 = (x + 3i)(x - 3i), so the roots are ±3i\pm 3i

    a2+b2=(a+bi)(a−bi)\displaystyle a^2 + b^2 = (a + bi)(a - bi)
    (a+b)2=a2+2ab+b2\displaystyle (a + b)^2 = a^2 + 2ab + b^2

10.32 Complex Numbers in Polar Form

ReImθrabz = a + bir = |z| = √(a² + b²)θ = arg ztan θ = b/az = r(cos θ + i sin θ)= r eiθHere z = 4 + 3i: r = 5.Multiply: multiply r, add θ.160°z6 = 16 roots, 60° apart
A complex number a + bi is a point in the plane, so it also has a length r and an angle θ. In that polar form the n-th roots of 1 sit evenly around the unit circle, like the corners of a regular hexagon.
  1. 10.32.1

    The Polar Form of a Complex Number

    Write z=a+biz = a + bi using its distance from the origin and its angle. The modulus r=∣z∣r = \lvert z \rvert is the length, and the argument θ\theta is measured counterclockwise from the positive real axis. Use the signs of aa and bb to put θ\theta in the right quadrant.

    e.g. 1+i1 + i: r=2r = \sqrt2, θ=π4\theta = \tfrac{\pi}{4}, so z=2(cos⁡π4+isin⁡π4)z = \sqrt2\left(\cos\tfrac{\pi}{4} + i\sin\tfrac{\pi}{4}\right)

    z=r(cos⁡θ+isin⁡θ)\displaystyle z = r(\cos\theta + i\sin\theta)
    r=a2+b2\displaystyle r = \sqrt{a^2 + b^2}
    tan⁡θ=ba\displaystyle \tan\theta = \frac{b}{a}
  2. 10.32.2

    Products of Complex Numbers Expressed in Polar Form

    To multiply, multiply the moduli and add the arguments.

    e.g. 2(cos⁡30∘+isin⁡30∘)⋅3(cos⁡60∘+isin⁡60∘)=6(cos⁡90∘+isin⁡90∘)=6i2(\cos 30^\circ + i\sin 30^\circ) \cdot 3(\cos 60^\circ + i\sin 60^\circ) = 6(\cos 90^\circ + i\sin 90^\circ) = 6i

    ∣z1z2∣=r1r2\displaystyle \lvert z_1 z_2 \rvert = r_1 r_2
    arg⁡(z1z2)=θ1+θ2\displaystyle \arg(z_1 z_2) = \theta_1 + \theta_2
  3. 10.32.3

    Quotients of Complex Numbers Expressed in Polar Form

    To divide, divide the moduli and subtract the arguments. The divisor cannot be 00.

    e.g. 8(cos⁡100∘+isin⁡100∘)2(cos⁡10∘+isin⁡10∘)=4(cos⁡90∘+isin⁡90∘)=4i\dfrac{8(\cos 100^\circ + i\sin 100^\circ)}{2(\cos 10^\circ + i\sin 10^\circ)} = 4(\cos 90^\circ + i\sin 90^\circ) = 4i

    ∣z1z2∣=r1r2\displaystyle \left\lvert \frac{z_1}{z_2} \right\rvert = \frac{r_1}{r_2}
    arg⁡z1z2=θ1−θ2\displaystyle \arg\frac{z_1}{z_2} = \theta_1 - \theta_2
  4. 10.32.4

    The CIS Notation

    cis⁡θ\operatorname{cis}\theta is shorthand for cos⁡θ+isin⁡θ\cos\theta + i\sin\theta. Products and quotients become r1r2cis⁡(θ1+θ2)r_1 r_2 \operatorname{cis}(\theta_1 + \theta_2) and r1r2cis⁡(θ1−θ2)\tfrac{r_1}{r_2}\operatorname{cis}(\theta_1 - \theta_2).

    e.g. 3cis⁡π2=3(0+1i)=3i3\operatorname{cis}\tfrac{\pi}{2} = 3(0 + 1i) = 3i

    cis⁡θ=cos⁡θ+isin⁡θ\displaystyle \operatorname{cis}\theta = \cos\theta + i\sin\theta
    z=rcis⁡θ\displaystyle z = r\operatorname{cis}\theta

10.33 De Moivre's Theorem

  1. 10.33.1

    De Moivre's Theorem

    To raise a complex number in polar form to an integer power nn, raise the modulus to the nnth power and multiply the argument by nn.

    e.g. (cos⁡π6+isin⁡π6)3=cos⁡π2+isin⁡π2=i\left(\cos\tfrac{\pi}{6} + i\sin\tfrac{\pi}{6}\right)^3 = \cos\tfrac{\pi}{2} + i\sin\tfrac{\pi}{2} = i

    [r(cos⁡θ+isin⁡θ)]n\displaystyle [r(\cos\theta + i\sin\theta)]^n
    =rn(cos⁡nθ+isin⁡nθ)\displaystyle = r^n(\cos n\theta + i\sin n\theta)
  2. 10.33.2

    Finding Powers of Complex Numbers Using De Moivre's Theorem

    Convert a+bia + bi to polar form, apply De Moivre's theorem, then convert back to a+bia + bi if asked. Reduce the final angle by multiples of 2π2\pi (or 360∘360^\circ).

    e.g. (1+i)8(1 + i)^8: r=2r = \sqrt2, θ=π4\theta = \tfrac{\pi}{4}, so (2)8cis⁡2π=16(\sqrt2)^8 \operatorname{cis} 2\pi = 16

    zn=rncis⁡(nθ)\displaystyle z^n = r^n \operatorname{cis}(n\theta)
  3. 10.33.3

    The Power-Reducing Formulas for Sine and Cosine

    Rewrite a squared sine or cosine using the cosine of the double angle.

    e.g. cos⁡215∘=1+cos⁡30∘2=2+34\cos^2 15^\circ = \tfrac{1 + \cos 30^\circ}{2} = \tfrac{2 + \sqrt3}{4}

    sin⁡2θ=1−cos⁡2θ2\displaystyle \sin^2\theta = \frac{1 - \cos 2\theta}{2}
    cos⁡2θ=1+cos⁡2θ2\displaystyle \cos^2\theta = \frac{1 + \cos 2\theta}{2}
  4. 10.33.4

    Euler's Formula

    Links the exponential function to cosine and sine, with θ\theta in radians. It gives the exponential form z=reiθz = re^{i\theta}.

    e.g. 2eiπ/3=2(12+32i)=1+3 i2e^{i\pi/3} = 2\left(\tfrac12 + \tfrac{\sqrt3}{2}i\right) = 1 + \sqrt3\,i

    eiθ=cos⁡θ+isin⁡θ\displaystyle e^{i\theta} = \cos\theta + i\sin\theta
    z=reiθ\displaystyle z = re^{i\theta}
    eiπ=−1\displaystyle e^{i\pi} = -1
  5. 10.33.5

    The Roots of Unity

    The nnth roots of unity are the nn solutions of zn=1z^n = 1. They all have modulus 11 and are spaced evenly around the unit circle, 2πn\tfrac{2\pi}{n} apart, starting at 11.

    e.g. Cube roots of unity: 11, −12+32i-\tfrac12 + \tfrac{\sqrt3}{2}i, −12−32i-\tfrac12 - \tfrac{\sqrt3}{2}i

    zn=1\displaystyle z^n = 1
    zk=cis⁡2πkn\displaystyle z_k = \operatorname{cis}\frac{2\pi k}{n}
    k=0,1,…,n−1\displaystyle k = 0, 1, \dots, n - 1
  6. 10.33.6

    Properties of Roots of Unity

    With ω=cis⁡2πn\omega = \operatorname{cis}\tfrac{2\pi}{n}, the nnth roots are 1,ω,ω2,…,ωn−11, \omega, \omega^2, \dots, \omega^{n-1}. For n≥2n \ge 2 they add to 00. Powers of ω\omega repeat every nn steps, and the conjugate of a root of unity is also a root.

    e.g. Cube roots: ω3=1\omega^3 = 1 and 1+ω+ω2=01 + \omega + \omega^2 = 0, so ω10=ω\omega^{10} = \omega

    1+ω+⋯+ωn−1=0\displaystyle 1 + \omega + \dots + \omega^{n-1} = 0
    ωn=1\displaystyle \omega^n = 1
    ωk+n=ωk\displaystyle \omega^{k + n} = \omega^k
  7. 10.33.7

    Square Roots of Complex Numbers

    Every nonzero complex number has two square roots, which are opposites. In polar form, take r\sqrt r and halve the argument, then add π\pi for the second root. Or set (a+bi)2=z(a + bi)^2 = z and match real and imaginary parts.

    e.g. Square roots of 2i2i: r=2r = 2, θ=π2\theta = \tfrac{\pi}{2}, so the roots are ±2cis⁡π4=±(1+i)\pm\sqrt2\operatorname{cis}\tfrac{\pi}{4} = \pm(1 + i)

    rcis⁡θ2\displaystyle \sqrt r \operatorname{cis}\frac{\theta}{2}
    rcis⁡(θ2+π)\displaystyle \sqrt r \operatorname{cis}\left(\frac{\theta}{2} + \pi\right)
  8. 10.33.8

    Higher Roots of Complex Numbers

    A nonzero complex number has exactly nn distinct nnth roots. They share modulus rn\sqrt[n]{r} and are spaced 2πn\tfrac{2\pi}{n} apart around a circle.

    e.g. Cube roots of 8i8i (r=8r = 8, θ=π2\theta = \tfrac{\pi}{2}): 2cis⁡π62\operatorname{cis}\tfrac{\pi}{6}, 2cis⁡5π62\operatorname{cis}\tfrac{5\pi}{6}, 2cis⁡3π22\operatorname{cis}\tfrac{3\pi}{2}, which are 3+i\sqrt3 + i, −3+i-\sqrt3 + i, −2i-2i

    zk=rn cis⁡θ+2πkn\displaystyle z_k = \sqrt[n]{r}\,\operatorname{cis}\frac{\theta + 2\pi k}{n}
    k=0,1,…,n−1\displaystyle k = 0, 1, \dots, n - 1

10.34 The Fundamental Theorem of Algebra

  1. 10.34.1

    The Fundamental Theorem of Algebra for Quadratic Equations

    Every quadratic has exactly 2 complex roots, counting a repeated root twice. With real coefficients, a negative discriminant gives a pair of complex conjugate roots.

    e.g. x2+2x+5=0x^2 + 2x + 5 = 0: x=−2±−162=−1±2ix = \tfrac{-2 \pm \sqrt{-16}}{2} = -1 \pm 2i

    ax2+bx+c=a(x−r1)(x−r2)\displaystyle ax^2 + bx + c = a(x - r_1)(x - r_2)
    b2−4ac<0⇒p±qi\displaystyle b^2 - 4ac < 0 \Rightarrow p \pm qi
  2. 10.34.2

    The Fundamental Theorem of Algebra with Cubic Equations

    Every cubic has exactly 3 complex roots, counting multiplicity. With real coefficients, nonreal roots come in conjugate pairs, so a real cubic has either 3 real roots or 1 real root and 1 conjugate pair.

    e.g. x3−x2+4x−4=(x−1)(x2+4)x^3 - x^2 + 4x - 4 = (x - 1)(x^2 + 4): roots 11, 2i2i, −2i-2i

    a(x−r1)(x−r2)(x−r3)\displaystyle a(x - r_1)(x - r_2)(x - r_3)
  3. 10.34.3

    Solving Cubic Equations With Complex Roots

    Find one root (test rational candidates, or use a given root). Divide it out with synthetic division, then solve the leftover quadratic. With real coefficients, if a+bia + bi is a root, so is a−bia - bi.

    e.g. x3−3x2+4x−2=0x^3 - 3x^2 + 4x - 2 = 0: x=1x = 1 works, the quotient is x2−2x+2x^2 - 2x + 2, so x=1±ix = 1 \pm i

    p(r)=0  ⟺  (x−r) is a factor\displaystyle p(r) = 0 \iff (x - r) \text{ is a factor}
  4. 10.34.4

    The Fundamental Theorem of Algebra with Quartic Equations

    Every quartic has exactly 4 complex roots, counting multiplicity. With real coefficients, nonreal roots come in conjugate pairs, so there are 4, 2, or 0 real roots.

    e.g. x4−1=(x−1)(x+1)(x2+1)x^4 - 1 = (x - 1)(x + 1)(x^2 + 1): roots ±1\pm 1, ±i\pm i

    a(x−r1)(x−r2)\displaystyle a(x - r_1)(x - r_2)
    ⋅(x−r3)(x−r4)\displaystyle \cdot (x - r_3)(x - r_4)
  5. 10.34.5

    Solving Quartic Equations With Complex Roots

    If a+bia + bi is a root, multiply its factor by its conjugate's factor to get a real quadratic factor, then divide it out. Quartics with only x4x^4, x2x^2, and constant terms factor like quadratics in x2x^2.

    e.g. x4−5x2−36=(x2−9)(x2+4)=0x^4 - 5x^2 - 36 = (x^2 - 9)(x^2 + 4) = 0, so x=±3x = \pm 3, ±2i\pm 2i

    (x−(a+bi))(x−(a−bi))\displaystyle (x - (a + bi))(x - (a - bi))
    =x2−2ax+(a2+b2)\displaystyle = x^2 - 2ax + (a^2 + b^2)

Unit 11 48 topics

Probability & Combinatorics

Introduction to Probability, Compound Events in Probability, Conditional Probability, and more.

11.35 Introduction to Probability

ABA onlyB onlyA ∩ Bneithersample spaceAddition lawP(A ∪ B) =P(A) + P(B) − P(A ∩ B)Subtract the overlap, orit gets counted twice.If A and B cannot bothhappen, P(A ∩ B) = 0.
A ∪ B is everything in either circle, and A ∩ B is the shaded overlap. Adding P(A) and P(B) counts the overlap twice, so subtract it once.
  1. 11.35.1

    Sets

    A set is a collection of distinct objects called elements. x∈Ax \in A means xx is an element of AA. The empty set ∅\varnothing has no elements, and A⊆BA \subseteq B means every element of AA is also in BB.

    e.g. Even numbers from 1 to 10: {2,4,6,8,10}\{2, 4, 6, 8, 10\}, a set with 5 elements

    A={1,2,3}\displaystyle A = \{1, 2, 3\}
    2∈A,  5∉A\displaystyle 2 \in A,\; 5 \notin A
  2. 11.35.2

    Experimental Probability

    The fraction of trials in which the event actually happened. More trials usually bring it closer to the theoretical probability.

    e.g. 27 heads in 50 flips: P(heads)≈2750=0.54P(\text{heads}) \approx \tfrac{27}{50} = 0.54

    P(A)≈successestrials\displaystyle P(A) \approx \frac{\text{successes}}{\text{trials}}
  3. 11.35.3

    Using Experimental Probability to Make Predictions

    Multiply the experimental probability by the number of future trials to predict how often the event will happen.

    e.g. Made 18 of 30 free throws. In 50 more shots, expect 1830×50=30\tfrac{18}{30} \times 50 = 30 makes

    expected count=P(A)×n\displaystyle \text{expected count} = P(A) \times n
  4. 11.35.4

    Sample Spaces and Events in Probability

    The sample space SS is the set of all possible outcomes. An event is any subset of SS.

    e.g. Flip two coins: S={HH,HT,TH,TT}S = \{HH, HT, TH, TT\}, and "at least one head" ={HH,HT,TH}= \{HH, HT, TH\}

    S={1,2,3,4,5,6}\displaystyle S = \{1, 2, 3, 4, 5, 6\}
    even={2,4,6}\displaystyle \text{even} = \{2, 4, 6\}
  5. 11.35.5

    The Likelihood of an Event

    Probability runs from 0 (impossible) to 1 (certain). Near 0 is unlikely, 12\tfrac12 is as likely as not, and near 1 is likely.

    e.g. One die: P(7)=0P(7) = 0 is impossible, and P(less than 7)=1P(\text{less than } 7) = 1 is certain

    0≤P(A)≤1\displaystyle 0 \le P(A) \le 1
  6. 11.35.6

    Single Events in Probability

    When all outcomes are equally likely, divide the number of favorable outcomes by the total number of outcomes.

    e.g. One die: P(prime)=36=12P(\text{prime}) = \tfrac{3}{6} = \tfrac12 (2, 3, 5)

    P(A)=n(A)n(S)\displaystyle P(A) = \frac{n(A)}{n(S)}
  7. 11.35.7

    Using Probability to Make Predictions

    Expected number of occurrences = probability × number of trials. It is a long-run average, not a guarantee.

    e.g. Roll a die 120 times: expect 120⋅16=20120 \cdot \tfrac16 = 20 sixes

    expected count=n⋅P(A)\displaystyle \text{expected count} = n \cdot P(A)
  8. 11.35.8

    The Complement of an Event

    The complement A′A' (also written AcA^c) is every outcome in SS that is not in AA. Its probability is 1 minus P(A)P(A).

    e.g. P(rain)=0.3⇒P(no rain)=0.7P(\text{rain}) = 0.3 \Rightarrow P(\text{no rain}) = 0.7

    P(A′)=1−P(A)\displaystyle P(A') = 1 - P(A)
  9. 11.35.9

    Venn Diagrams in Probability

    The rectangle is the sample space and each circle is an event. Fill each region with a count or probability; all regions together total n(S)n(S) or 1.

    e.g. 30 students: 12 soccer only, 8 both, 6 basketball only, 4 neither. P(soccer)=12+830=23P(\text{soccer}) = \tfrac{12 + 8}{30} = \tfrac23

  10. 11.35.10

    Geometric Probability

    When a point is chosen at random in a region, the probability is the favorable length, area, or volume divided by the total.

    e.g. A 4 by 4 square target on a 10 by 10 board: P=16100=0.16P = \tfrac{16}{100} = 0.16

    P=favorable areatotal area\displaystyle P = \frac{\text{favorable area}}{\text{total area}}

11.36 Compound Events in Probability

  1. 11.36.1

    The Union of Sets

    A∪BA \cup B ("A or B") is everything in AA, in BB, or in both. Each element is listed once.

    e.g. {1,2,3}∪{3,4}={1,2,3,4}\{1, 2, 3\} \cup \{3, 4\} = \{1, 2, 3, 4\}

    A∪B={x:x∈A or x∈B}\displaystyle A \cup B = \{x : x \in A \text{ or } x \in B\}
  2. 11.36.2

    The Intersection of Sets

    A∩BA \cap B ("A and B") is the elements in both AA and BB. If they share nothing, A∩B=∅A \cap B = \varnothing and the sets are disjoint.

    e.g. {1,2,3}∩{2,3,4}={2,3}\{1, 2, 3\} \cap \{2, 3, 4\} = \{2, 3\}

    A∩B={x:x∈A and x∈B}\displaystyle A \cap B = \{x : x \in A \text{ and } x \in B\}
  3. 11.36.3

    Compound Events in Probability From Experimental Data

    For "A and B," count the trials where both happened. For "A or B," count the trials where either happened, counting the overlap once. Divide by the total number of trials.

    e.g. 50 people: 20 own a dog, 15 a cat, 5 both. P(dog or cat)=20+15−550=0.6P(\text{dog or cat}) = \tfrac{20 + 15 - 5}{50} = 0.6

    P(A∩B)≈n(A∩B)total\displaystyle P(A \cap B) \approx \frac{n(A \cap B)}{\text{total}}
  4. 11.36.4

    Computing Probabilities for Compound Events Using Venn Diagrams

    Add the regions that make up the event. A∪BA \cup B is every region inside either circle; A∩BA \cap B is only the overlap.

    e.g. A only 0.3, both 0.1, B only 0.2, neither 0.4: P(A∪B)=0.3+0.1+0.2=0.6P(A \cup B) = 0.3 + 0.1 + 0.2 = 0.6, P(A∩B)=0.1P(A \cap B) = 0.1

  5. 11.36.5

    Computing Probabilities of Events Containing Complements Using Venn Diagrams

    A′A' is every region outside circle AA, including the region outside both circles. Work region by region.

    e.g. A only 0.3, both 0.1, B only 0.2, neither 0.4: P(A∩B′)=0.3P(A \cap B') = 0.3, P(A′)=0.2+0.4=0.6P(A') = 0.2 + 0.4 = 0.6

    A∩B′=A only\displaystyle A \cap B' = A \text{ only}
    (A∪B)′=neither\displaystyle (A \cup B)' = \text{neither}
  6. 11.36.6

    Computing Probabilities for Three Events Using Venn Diagrams

    Three circles make 8 regions. Fill from the center out: the triple overlap first, then each pair overlap minus the center, then each "only" region.

    e.g. 40 students: 20 Spanish, 15 French, 10 art; 5 Spanish and French, 4 Spanish and art, 3 French and art, 2 all three. P(at least one)=45−12+240=78P(\text{at least one}) = \tfrac{45 - 12 + 2}{40} = \tfrac78

    P(A∪B∪C)=P(A)+P(B)+P(C)\displaystyle P(A \cup B \cup C) = P(A) + P(B) + P(C)
    −P(A∩B)−P(A∩C)−P(B∩C)\displaystyle - P(A \cap B) - P(A \cap C) - P(B \cap C)
    +P(A∩B∩C)\displaystyle + P(A \cap B \cap C)
  7. 11.36.7

    The Addition Law of Probability

    For "A or B," add the two probabilities and subtract the overlap so it is not counted twice.

    e.g. One card: P(heart or king)=1352+452−152=413P(\text{heart or king}) = \tfrac{13}{52} + \tfrac{4}{52} - \tfrac{1}{52} = \tfrac{4}{13}

    P(A∪B)=P(A)+P(B)\displaystyle P(A \cup B) = P(A) + P(B)
    −P(A∩B)\displaystyle \quad - P(A \cap B)
  8. 11.36.8

    Applying the Addition Law With Event Complements

    The addition law works with complements too. Use P(A′)=1−P(A)P(A') = 1 - P(A), and "at least one" is 1 minus "neither."

    e.g. P(A)=0.6P(A) = 0.6, P(B)=0.5P(B) = 0.5, P(A∩B)=0.3P(A \cap B) = 0.3: P(A′∩B)=0.2P(A' \cap B) = 0.2, so P(A′∪B)=0.4+0.5−0.2=0.7P(A' \cup B) = 0.4 + 0.5 - 0.2 = 0.7

    P(A′∪B)=P(A′)+P(B)\displaystyle P(A' \cup B) = P(A') + P(B)
    −P(A′∩B)\displaystyle \quad - P(A' \cap B)
    P(A∪B)=1−P(A′∩B′)\displaystyle P(A \cup B) = 1 - P(A' \cap B')
  9. 11.36.9

    Mutually Exclusive Events

    Events that cannot happen together: A∩B=∅A \cap B = \varnothing. The addition law then has no overlap term. Mutually exclusive events with nonzero probabilities are never independent.

    e.g. One die: P(1 or 6)=16+16=13P(1 \text{ or } 6) = \tfrac16 + \tfrac16 = \tfrac13

    P(A∩B)=0\displaystyle P(A \cap B) = 0
    P(A∪B)=P(A)+P(B)\displaystyle P(A \cup B) = P(A) + P(B)

11.37 Conditional Probability

3/52/52/42/43/41/4RBRRR 3/5 × 2/4 = 6/20BRB 3/5 × 2/4 = 6/20RBR 2/5 × 3/4 = 6/20BBB 2/5 × 1/4 = 2/20Bag: 3 red,2 blue.Draw 2, noreplacing.Total:20/20 = 1Multiply along a branch, add across branches.P(one of each color) = P(RB) + P(BR) = 6/20 + 6/20 = 12/20 = 3/5
The second-draw branches change because the first marble is not put back. Multiply along a path to get one outcome; add outcomes to combine them.
  1. 11.37.1

    Conditional Probabilities From Venn Diagrams

    P(A∣B)P(A \mid B) is the probability of AA given that BB happened. Look only inside circle BB: divide the overlap by all of BB.

    e.g. A only 0.3, both 0.1, B only 0.2: P(A∣B)=0.10.1+0.2=13P(A \mid B) = \tfrac{0.1}{0.1 + 0.2} = \tfrac13

    P(A∣B)=P(A∩B)P(B)\displaystyle P(A \mid B) = \frac{P(A \cap B)}{P(B)}
  2. 11.37.2

    Conditional Probabilities From Tables

    Use only the row or column of the given condition. Divide the cell by that row or column total, not the grand total.

    e.g. 25 students play a sport, and 10 of them are in band: P(band∣sport)=1025=0.4P(\text{band} \mid \text{sport}) = \tfrac{10}{25} = 0.4

    P(A∣B)=n(A∩B)n(B)\displaystyle P(A \mid B) = \frac{n(A \cap B)}{n(B)}
  3. 11.37.3

    The Multiplication Law for Conditional Probability

    The probability that both happen is the probability of the first times the probability of the second given the first.

    e.g. 5 red and 3 blue marbles, draw 2 without replacement: P(both red)=58⋅47=514P(\text{both red}) = \tfrac58 \cdot \tfrac47 = \tfrac{5}{14}

    P(A∩B)=P(A)⋅P(B∣A)\displaystyle P(A \cap B) = P(A) \cdot P(B \mid A)
  4. 11.37.4

    The Law of Total Probability

    Split into cases that do not overlap and cover everything, such as BB and B′B'. Find P(A)P(A) in each case and add.

    e.g. 60 percent of days are sunny. You bike on 90 percent of sunny days and 30 percent of others: P(bike)=0.9(0.6)+0.3(0.4)=0.66P(\text{bike}) = 0.9(0.6) + 0.3(0.4) = 0.66

    P(A)=P(A∣B)P(B)\displaystyle P(A) = P(A \mid B)P(B)
    +P(A∣B′)P(B′)\displaystyle \quad + P(A \mid B')P(B')
  5. 11.37.5

    Tree Diagrams for Dependent Events

    Each branch is a conditional probability given the path so far, so later branches change. Multiply along a path for that outcome, and add paths for an event. Branches from one point sum to 1.

    e.g. 4 red and 2 blue, draw 2 without replacement: P(one of each)=46⋅25+26⋅45=1630=815P(\text{one of each}) = \tfrac46 \cdot \tfrac25 + \tfrac26 \cdot \tfrac45 = \tfrac{16}{30} = \tfrac{8}{15}

  6. 11.37.6

    Tree Diagrams for Dependent Events: Applications

    Use a tree for multi-step situations such as medical tests. To reverse a condition, divide the path you want by the sum of all paths that give the observed result.

    e.g. 2 percent have a condition; the test is positive for 90 percent of them and 5 percent of others. P(has it∣+)=0.0180.018+0.049≈0.27P(\text{has it} \mid +) = \tfrac{0.018}{0.018 + 0.049} \approx 0.27

    P(B∣A)=P(B) P(A∣B)P(A)\displaystyle P(B \mid A) = \frac{P(B)\,P(A \mid B)}{P(A)}
  7. 11.37.7

    Independent Events

    AA and BB are independent if one happening does not change the probability of the other. Check with either equation.

    e.g. P(A)=0.5P(A) = 0.5, P(B)=0.4P(B) = 0.4, P(A∩B)=0.2=0.5⋅0.4P(A \cap B) = 0.2 = 0.5 \cdot 0.4, so independent

    P(A∣B)=P(A)\displaystyle P(A \mid B) = P(A)
    P(A∩B)=P(A)⋅P(B)\displaystyle P(A \cap B) = P(A) \cdot P(B)
  8. 11.37.8

    Tree Diagrams for Independent Events

    The branch probabilities repeat at every stage because earlier results do not affect later ones (for example, drawing with replacement). Multiply along each path.

    e.g. Makes 70 percent of free throws, shoots 2: P(exactly one)=0.7(0.3)+0.3(0.7)=0.42P(\text{exactly one}) = 0.7(0.3) + 0.3(0.7) = 0.42

11.38 Discrete Random Variables

Number of heads in 6 fair coin flips1/6406/64115/64220/64315/6446/6451/646kmean = np = 3n = 6 trials, p = 0.5P(X = k) =C(n, k) pᵏ(1 − p)ⁿ⁻ᵏC(6, 3) = 20, soP(X = 3) = 20/64 ≈ 0.31mean μ = npσ = √(np(1 − p)) ≈ 1.22
A binomial distribution counts successes in n independent trials. With p = 0.5 the bars are symmetric and peak at the mean, np = 3.
  1. 11.38.1

    Probability Mass Functions of Discrete Random Variables

    A probability mass function (PMF) gives P(X=x)P(X = x) for each value a discrete random variable can take. Every probability is between 0 and 1, and all of them add to 1.

    e.g. P(X=1)=0.2P(X = 1) = 0.2, P(X=2)=0.5P(X = 2) = 0.5, P(X=3)=kP(X = 3) = k: k=1−0.7=0.3k = 1 - 0.7 = 0.3

    0≤P(X=x)≤1\displaystyle 0 \le P(X = x) \le 1
    ∑P(X=x)=1\displaystyle \sum P(X = x) = 1
  2. 11.38.2

    Cumulative Distribution Functions for Discrete Random Variables

    The CDF F(x)F(x) is the probability that XX is at most xx. Add the PMF values up to and including xx. It never decreases and ends at 1.

    e.g. PMF 0.2,0.5,0.30.2, 0.5, 0.3 on 1,2,31, 2, 3: F(2)=0.2+0.5=0.7F(2) = 0.2 + 0.5 = 0.7, so P(X>2)=0.3P(X > 2) = 0.3

    F(x)=P(X≤x)\displaystyle F(x) = P(X \le x)
    P(a<X≤b)=F(b)−F(a)\displaystyle P(a < X \le b) = F(b) - F(a)
  3. 11.38.3

    Expected Values of Discrete Random Variables

    Multiply each value by its probability and add. The expected value is the long-run average of XX, and it need not be a possible value.

    e.g. PMF 0.2,0.5,0.30.2, 0.5, 0.3 on 1,2,31, 2, 3: E(X)=0.2+1+0.9=2.1E(X) = 0.2 + 1 + 0.9 = 2.1

    E(X)=∑x P(X=x)\displaystyle E(X) = \sum x\,P(X = x)
  4. 11.38.4

    The Binomial Distribution

    X∼B(n,p)X \sim B(n, p) counts the successes in nn independent trials, each with the same success probability pp. kk can be 0,1,…,n0, 1, \ldots, n.

    e.g. n=4n = 4, p=12p = \tfrac12: P(X=2)=6⋅116=38P(X = 2) = 6 \cdot \tfrac{1}{16} = \tfrac38

    P(X=k)=(nk)pk(1−p)n−k\displaystyle P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
    E(X)=np\displaystyle E(X) = np
    Var(X)=np(1−p)\displaystyle \text{Var}(X) = np(1-p)
  5. 11.38.5

    Modeling With the Binomial Distribution

    Use it when there is a fixed number of trials, each trial is a success or failure, trials are independent, and pp stays the same. For "at least one," use the complement.

    e.g. 3 free throws, p=0.8p = 0.8: P(all 3)=0.83=0.512P(\text{all 3}) = 0.8^3 = 0.512; P(at least 1)=1−0.23=0.992P(\text{at least 1}) = 1 - 0.2^3 = 0.992

    P(X≥1)=1−P(X=0)\displaystyle P(X \ge 1) = 1 - P(X = 0)
    P(X≤k)=∑i=0kP(X=i)\displaystyle P(X \le k) = \sum_{i=0}^{k} P(X = i)
  6. 11.38.6

    The Geometric Distribution

    X∼Geo(p)X \sim \text{Geo}(p) is the number of the trial on which the first success happens, with independent trials and constant success probability pp. kk can be 1,2,3,…1, 2, 3, \ldots

    e.g. First 6 on the 3rd roll of a die: (56)2⋅16=25216\left(\tfrac56\right)^2 \cdot \tfrac16 = \tfrac{25}{216}

    P(X=k)=(1−p)k−1p\displaystyle P(X = k) = (1-p)^{k-1} p
    P(X>k)=(1−p)k\displaystyle P(X > k) = (1-p)^k
    E(X)=1p\displaystyle E(X) = \frac1p
  7. 11.38.7

    Modeling With the Geometric Distribution

    Use it when you repeat independent trials with the same pp until the first success. "Within kk tries" means X≤kX \le k, which is easiest by the complement.

    e.g. p=0.3p = 0.3: success within 2 tries is 1−0.72=0.511 - 0.7^2 = 0.51; expected tries =10.3≈3.33= \tfrac{1}{0.3} \approx 3.33

    P(X≤k)=1−(1−p)k\displaystyle P(X \le k) = 1 - (1-p)^k

11.39 The Normal Distribution

μ − 3σμ − 2σμ − σμμ + σμ + 2σμ + 3σ34%34%13.5%13.5%68%95%99.7%Empirical rule68% within 1σ of μ95% within 2σ99.7% within 3σz = (x − μ) / σz counts standarddeviations from the mean.
In a normal distribution, about 68% of values fall within one standard deviation of the mean, 95% within two, and 99.7% within three. A z-score says how many standard deviations a value sits from the mean.
  1. 11.39.1

    The Standard Normal Distribution

    Z∼N(0,1)Z \sim N(0, 1) is the bell curve with mean 0 and standard deviation 1. The total area under it is 1. Φ(z)=P(Z<z)\Phi(z) = P(Z < z) is the area to the left of zz, from a table or calculator.

    e.g. P(0<Z<1)=0.8413−0.5=0.3413P(0 < Z < 1) = 0.8413 - 0.5 = 0.3413

    Z∼N(0,1)\displaystyle Z \sim N(0, 1)
    P(Z<z)=Φ(z)\displaystyle P(Z < z) = \Phi(z)
    P(a<Z<b)=Φ(b)−Φ(a)\displaystyle P(a < Z < b) = \Phi(b) - \Phi(a)
  2. 11.39.2

    Symmetry Properties of the Standard Normal Distribution

    The curve is symmetric about 0. The area to the left of −z-z equals the area to the right of zz, and each half has area 0.5.

    e.g. Φ(1)=0.8413\Phi(1) = 0.8413: Φ(−1)=0.1587\Phi(-1) = 0.1587 and P(−1<Z<1)=0.6826P(-1 < Z < 1) = 0.6826

    Φ(−z)=1−Φ(z)\displaystyle \Phi(-z) = 1 - \Phi(z)
    P(Z>z)=1−Φ(z)\displaystyle P(Z > z) = 1 - \Phi(z)
    P(−z<Z<z)=2Φ(z)−1\displaystyle P(-z < Z < z) = 2\Phi(z) - 1
  3. 11.39.3

    The Normal Distribution

    X∼N(μ,σ2)X \sim N(\mu, \sigma^2) is a bell curve centered at μ\mu with standard deviation σ\sigma. Find probabilities by converting to zz-scores. P(X=a)=0P(X = a) = 0, so << and ≤\le give the same answer.

    e.g. X∼N(50,42)X \sim N(50, 4^2): P(X<54)=Φ(1)≈0.8413P(X < 54) = \Phi(1) \approx 0.8413

    X∼N(μ,σ2)\displaystyle X \sim N(\mu, \sigma^2)
    P(X<x)=Φ(x−μσ)\displaystyle P(X < x) = \Phi\left(\frac{x - \mu}{\sigma}\right)
  4. 11.39.4

    Mean and Variance of the Normal Distribution

    For X∼N(μ,σ2)X \sim N(\mu, \sigma^2), the mean is μ\mu, the variance is σ2\sigma^2, and the standard deviation is σ\sigma. The mean, median, and mode all sit at the peak. To find an unknown μ\mu or σ\sigma, get zz from the given probability and solve x=μ+zσx = \mu + z\sigma.

    e.g. σ=3\sigma = 3, P(X<26)=0.9772P(X < 26) = 0.9772: z=2z = 2, so 26=μ+626 = \mu + 6 and μ=20\mu = 20

    E(X)=μ\displaystyle E(X) = \mu
    Var(X)=σ2\displaystyle \text{Var}(X) = \sigma^2
    x=μ+zσ\displaystyle x = \mu + z\sigma
  5. 11.39.5

    Percentage Points of the Standard Normal Distribution

    Work backward from an area to a zz-value: find zz with P(Z<z)=pP(Z < z) = p using an inverse normal table or invNorm. For an area to the right, use 1−p1 - p on the left.

    e.g. P(Z<z)=0.975⇒z≈1.96P(Z < z) = 0.975 \Rightarrow z \approx 1.96; P(Z>z)=0.05⇒z≈1.645P(Z > z) = 0.05 \Rightarrow z \approx 1.645

    P(Z<z)=p  ⇒  z=Φ−1(p)\displaystyle P(Z < z) = p \;\Rightarrow\; z = \Phi^{-1}(p)
  6. 11.39.6

    Modeling With the Normal Distribution

    Standardize with z=x−μσz = \tfrac{x - \mu}{\sigma}, then use the standard normal. To turn a zz-value back into a data value, use x=μ+zσx = \mu + z\sigma.

    e.g. Heights N(64,32)N(64, 3^2) in: P(X>67)=P(Z>1)≈0.1587P(X > 67) = P(Z > 1) \approx 0.1587

    z=x−μσ\displaystyle z = \frac{x - \mu}{\sigma}
    x=μ+zσ\displaystyle x = \mu + z\sigma
  7. 11.39.7

    The Empirical Rule for the Normal Distribution

    For any normal distribution, about 68%, 95%, and 99.7% of values lie within 1, 2, and 3 standard deviations of the mean.

    e.g. μ=100\mu = 100, σ=15\sigma = 15: about 95% lie between 70 and 130; P(X>115)≈1−0.682=0.16P(X > 115) \approx \tfrac{1 - 0.68}{2} = 0.16

    μ±σ:  68%\displaystyle \mu \pm \sigma: \; 68\%
    μ±2σ:  95%\displaystyle \mu \pm 2\sigma: \; 95\%
    μ±3σ:  99.7%\displaystyle \mu \pm 3\sigma: \; 99.7\%

11.40 Combinatorics

  1. 11.40.1

    The Rules of Sum and Product

    Product rule: a task done in steps with mm ways then nn ways can be done mnmn ways. Sum rule: choosing one option from mm choices or nn other choices (no overlap) gives m+nm + n ways.

    e.g. 4 shirts and 3 pants: 4×3=124 \times 3 = 12 outfits. One of 5 movies or 3 games: 5+3=85 + 3 = 8 choices

    steps (and): m×n\displaystyle \text{steps (and): } m \times n
    no overlap (or): m+n\displaystyle \text{no overlap (or): } m + n
  2. 11.40.2

    Factorials

    n!n! is the product of the whole numbers from nn down to 1. By definition 0!=10! = 1.

    e.g. 5!=5⋅4⋅3⋅2⋅1=1205! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120, and 8!6!=8⋅7=56\tfrac{8!}{6!} = 8 \cdot 7 = 56

    n!=n(n−1)(n−2)⋯1\displaystyle n! = n(n-1)(n-2)\cdots 1
    0!=1\displaystyle 0! = 1
  3. 11.40.3

    Factorials in Variable Expressions

    Expand the larger factorial down to the smaller one so they cancel.

    e.g. (n+2)!n!=(n+2)(n+1)=n2+3n+2\tfrac{(n+2)!}{n!} = (n+2)(n+1) = n^2 + 3n + 2

    n!=n(n−1)!\displaystyle n! = n(n-1)!
    (n+1)!(n−1)!=(n+1)n\displaystyle \frac{(n+1)!}{(n-1)!} = (n+1)n
  4. 11.40.4

    Ordering Objects

    nn different objects can be arranged in a row in n!n! ways. If some objects are identical, divide by the factorial of each repeat count.

    e.g. 4 books on a shelf: 4!=244! = 24 orders. Letters of BOOK: 4!2!=12\tfrac{4!}{2!} = 12

    n!\displaystyle n!
    n!a! b!⋯  (repeats)\displaystyle \frac{n!}{a!\,b!\cdots} \;\text{(repeats)}
  5. 11.40.5

    Permutations

    Arrange rr of nn different objects when order matters.

    e.g. President, VP, and treasurer from 10 people: 10P3=10⋅9⋅8=720{}_{10}P_3 = 10 \cdot 9 \cdot 8 = 720

    nPr=n!(n−r)!\displaystyle {}_nP_r = \frac{n!}{(n-r)!}
  6. 11.40.6

    Combinations

    Choose rr of nn different objects when order does not matter. It is the permutation count divided by r!r!.

    e.g. A 3-person committee from 10: (103)=10⋅9⋅83⋅2⋅1=120\binom{10}{3} = \tfrac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1} = 120

    (nr)=nCr=n!r! (n−r)!\displaystyle \binom{n}{r} = {}_nC_r = \frac{n!}{r!\,(n-r)!}
  7. 11.40.7

    Computing Probabilities Using Combinatorics

    Count the favorable outcomes and the total outcomes with combinations or permutations, then divide.

    e.g. 5 boys and 4 girls, pick 3 at random: P(all girls)=(43)(93)=484=121P(\text{all girls}) = \tfrac{\binom43}{\binom93} = \tfrac{4}{84} = \tfrac{1}{21}

    P=favorable counttotal count\displaystyle P = \frac{\text{favorable count}}{\text{total count}}

Unit 12 20 topics

Statistics

Analyzing Data, Correlation, and Regression.

12.41 Analyzing Data

2468102597410102030405060minutescountmedian ≈ 27.8mean ≈ 28.2 (balance point)skewed right:mean > medianThe long tail on the rightpulls the mean toward it.Median: splits the area in halfMean: where the bars balance
In a right-skewed histogram the few large values in the tail drag the mean to the right of the median. Here the mean is estimated from bin midpoints; the median is less affected by the tail.
  1. 12.41.1

    Sampling

    A sample is part of a population used to learn about the whole. A random sample (simple random, systematic, stratified, or cluster) avoids bias; convenience and voluntary samples often do not represent the population.

    e.g. Survey a random 100 of 1,200 students; 35 ride the bus, so estimate 0.35×1200=4200.35 \times 1200 = 420 riders

  2. 12.41.2

    The Mean of a Data Set

    Add all the values and divide by how many there are. Write μ\mu for a population mean and xˉ\bar x for a sample mean.

    e.g. 3,5,7,93, 5, 7, 9: xˉ=244=6\bar x = \tfrac{24}{4} = 6

    xˉ=∑xin\displaystyle \bar x = \frac{\sum x_i}{n}
  3. 12.41.3

    Variance and Standard Deviation

    Variance is the mean of the squared distances from the mean. Standard deviation is its square root, in the data's units. For a sample, divide by n−1n - 1 instead of nn.

    e.g. 2,4,6,82, 4, 6, 8: μ=5\mu = 5, squares 9,1,1,99, 1, 1, 9, σ2=204=5\sigma^2 = \tfrac{20}{4} = 5, σ=5≈2.24\sigma = \sqrt5 \approx 2.24 (sample: s2=203s^2 = \tfrac{20}{3})

    σ2=∑(xi−μ)2n\displaystyle \sigma^2 = \frac{\sum (x_i - \mu)^2}{n}
    σ=σ2\displaystyle \sigma = \sqrt{\sigma^2}
    s2=∑(xi−xˉ)2n−1\displaystyle s^2 = \frac{\sum (x_i - \bar x)^2}{n - 1}
  4. 12.41.4

    Estimating Means and Variances For Grouped Data

    Treat every value in a class as the class midpoint mm, weighted by its frequency ff. The results are estimates.

    e.g. Classes 0 to 10, 10 to 20, 20 to 30 with f=2,3,5f = 2, 3, 5: midpoints 5,15,255, 15, 25, xˉ≈10+45+12510=18\bar x \approx \tfrac{10 + 45 + 125}{10} = 18, σ2≈338+27+24510=61\sigma^2 \approx \tfrac{338 + 27 + 245}{10} = 61

    xˉ≈∑fm∑f\displaystyle \bar x \approx \frac{\sum f m}{\sum f}
    σ2≈∑f(m−xˉ)2∑f\displaystyle \sigma^2 \approx \frac{\sum f (m - \bar x)^2}{\sum f}
  5. 12.41.5

    Covariance

    Measures how two variables move together. Positive means they tend to rise together, negative means one rises as the other falls, and near 0 means no linear pattern. For a sample, divide by n−1n - 1.

    e.g. (1,1),(2,3),(3,5)(1,1), (2,3), (3,5): xˉ=2\bar x = 2, yˉ=3\bar y = 3, products 2,0,22, 0, 2, σxy=43\sigma_{xy} = \tfrac43 (sample: 22)

    σxy=∑(xi−xˉ)(yi−yˉ)n\displaystyle \sigma_{xy} = \frac{\sum (x_i - \bar x)(y_i - \bar y)}{n}
  6. 12.41.6

    Sums of Squares

    SxxS_{xx} and SyyS_{yy} add the squared deviations from the mean; SxyS_{xy} adds the products of paired deviations. The shortcut forms skip computing each deviation. Variance is Sxx/nS_{xx}/n and covariance is Sxy/nS_{xy}/n.

    e.g. x=1,2,3x = 1, 2, 3: Sxx=14−363=2S_{xx} = 14 - \tfrac{36}{3} = 2

    Sxx=∑(x−xˉ)2\displaystyle S_{xx} = \sum (x - \bar x)^2
    =∑x2−(∑x)2n\displaystyle = \sum x^2 - \frac{(\sum x)^2}{n}
    Sxy=∑xy−∑x∑yn\displaystyle S_{xy} = \sum xy - \frac{\sum x \sum y}{n}
  7. 12.41.7

    The Z-Score

    A zz-score says how many standard deviations a value is from the mean. Positive means above the mean and negative means below.

    e.g. μ=70\mu = 70, σ=5\sigma = 5, x=80x = 80: z=105=2z = \tfrac{10}{5} = 2

    z=x−μσ\displaystyle z = \frac{x - \mu}{\sigma}

12.42 Correlation

r ≈ 0.9strong positiver ≈ 0no linear patternr ≈ −0.8negative
r runs from −1 to 1. The sign gives the direction; the closer to ±1, the tighter the points hug the trend line.
  1. 12.42.1

    Scatter Plots

    Plot paired data (x,y)(x, y) as points to look for a relationship. The explanatory variable goes on the xx-axis.

  2. 12.42.2

    Trend Lines

    A straight line through the middle of the cloud, with about as many points above it as below.

    y^=mx+b\displaystyle \hat{y} = mx + b
  3. 12.42.3

    Making Predictions Using Trend Lines

    Substitute xx into the trend line. Predicting inside the data’s range (interpolation) is safer than beyond it (extrapolation).

  4. 12.42.4

    Interpreting Trend Line Coefficients

    The slope is the predicted change in yy for each one-unit increase in xx. The intercept is the predicted yy at x=0x = 0, which may be meaningless.

    e.g. y^=2.5x+60\hat{y} = 2.5x + 60, hours studied vs. score: each hour adds about 2.5 points

  5. 12.42.5

    Linear Correlation

    rr measures how tightly points hug a line, from −1-1 to 11. The sign gives direction; near 0 means no linear pattern. Correlation does not show cause.

    −1≤r≤1\displaystyle -1 \le r \le 1
  6. 12.42.6

    Residuals and Residual Plots

    A residual is actual minus predicted. Positive means the point is above the line. A residual plot with random scatter around 0 means a linear model fits; a curve or other pattern means it does not.

    e.g. Line y^=2x+1\hat y = 2x + 1, point (3,8)(3, 8): y^=7\hat y = 7, residual =8−7=1= 8 - 7 = 1

    e=y−y^\displaystyle e = y - \hat y
  7. 12.42.7

    The Linear Correlation Coefficient

    rr measures the strength and direction of a linear relationship. Near 11 or −1-1 is strong, near 0 is weak or none. Correlation does not prove causation.

    e.g. (1,1),(2,3),(3,2)(1,1), (2,3), (3,2): Sxx=2S_{xx} = 2, Syy=2S_{yy} = 2, Sxy=1S_{xy} = 1, so r=14=0.5r = \tfrac{1}{\sqrt4} = 0.5

    r=SxySxx Syy\displaystyle r = \frac{S_{xy}}{\sqrt{S_{xx}\,S_{yy}}}
    −1≤r≤1\displaystyle -1 \le r \le 1
  8. 12.42.8

    Correlation vs. Causation

    A strong correlation shows two variables move together, not that one causes the other. A third variable may drive both, the cause may run the other way, or it may be chance. A controlled experiment is what isolates a cause.

    e.g. Ice cream sales and drownings rise together because hot weather drives both

12.43 Regression

2468123456789xyresidual =actual − predictedŷ = 1.49 + 0.71xleast-squares lineresidual = y − ŷPoints above the line havepositive residuals.The line makes the sum ofsquared residuals smallest.
A residual is the vertical gap between a data point and the fitted line. The least-squares line is the one that makes the squared gaps add up to as little as possible.
  1. 12.43.1

    Selecting a Regression Model

    Match the shape: linear for steady change, exponential for steady percent change, quadratic for a curve that turns. Leftover patterns in the residuals mean a poor fit.

  2. 12.43.2

    Linear Regression

    The least-squares regression line is the line that best fits the data by making the sum of the squared residuals as small as possible. It always passes through (xˉ,yˉ)(\bar x, \bar y). Find it with a calculator or Desmos.

    e.g. r=0.8r = 0.8, sx=2s_x = 2, sy=5s_y = 5, xˉ=4\bar x = 4, yˉ=10\bar y = 10: b=0.8⋅52=2b = 0.8 \cdot \tfrac52 = 2, a=10−8=2a = 10 - 8 = 2, so y^=2x+2\hat y = 2x + 2

    y^=a+bx\displaystyle \hat y = a + bx
    b=r sysx\displaystyle b = r\,\frac{s_y}{s_x}
    a=yˉ−bxˉ\displaystyle a = \bar y - b\bar x
  3. 12.43.3

    Quadratic Regression

    Fits a parabola to data that rise then fall, or fall then rise. Use a calculator or Desmos (type y1∼ax12+bx1+cy_1 \sim ax_1^2 + bx_1 + c). An R2R^2 close to 1 means a good fit.

    e.g. Points (0,1)(0, 1), (1,2)(1, 2), (2,5)(2, 5): c=1c = 1, a+b=1a + b = 1, 4a+2b=44a + 2b = 4, so a=1a = 1, b=0b = 0 and y=x2+1y = x^2 + 1

    y^=ax2+bx+c\displaystyle \hat y = ax^2 + bx + c
  4. 12.43.4

    Semi-Log Scatter Plots

    Plot log⁡y\log y against xx. If the points lie close to a straight line, the original data are exponential. The intercept gives log⁡a\log a and the slope gives log⁡b\log b.

    e.g. Line log⁡y=0.3x+1\log y = 0.3x + 1: a=101=10a = 10^1 = 10, b=100.3≈2b = 10^{0.3} \approx 2, so y≈10⋅2xy \approx 10 \cdot 2^x

    y=abx\displaystyle y = ab^x
    ⇒  log⁡y=log⁡a+xlog⁡b\displaystyle \Rightarrow\; \log y = \log a + x \log b
  5. 12.43.5

    Exponential Regression

    Fits y=abxy = ab^x to data that grow or shrink by a constant factor. Use a calculator or Desmos, or fit a line to the points (x,ln⁡y)(x, \ln y) and convert back. b>1b > 1 means growth and 0<b<10 < b < 1 means decay.

    e.g. Points (0,3)(0, 3), (1,6)(1, 6), (2,12)(2, 12) double each step: y=3⋅2xy = 3 \cdot 2^x

    y^=abx\displaystyle \hat y = ab^x
    ln⁡y=ln⁡a+xln⁡b\displaystyle \ln y = \ln a + x \ln b

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