Section 2.3 Notes (Models and Applications)

Section 2.3: Models and Applications

• Focus: translate English into math, solve models (formulas) and apply to word problems.

• Models = formulas/methods to solve real-world quantities (distance, cost, area, volume, etc.).

• Language cues: map words to operations (see below).

• Quiz/Exam logistics (brief):

  • Quiz A: definitions/verbal; 15 points; grade may appear tentative until manual grading;
  • Quiz 1: problem-set style; timed (≈1 hour, ~10 questions); can use notes/calculator; retakes allowed with penalties (5% after 1st retry, 10% after 2nd, 20% after 3rd);
  • Quizzes due midnight before class; plan time to retake; open after class;
  • Exams: 1st exam = Chapter 2, 2nd exam = Chapter 3, etc.; final covers Chapter 6; you don’t need to retread all material each time.
  • After a quiz, wait for final grading before checking your score.

• Key vocabulary (translate to math):

  • Product = multiplication
  • Difference = subtraction
  • Quotient = division
  • Sum = addition

• Translating word problems: essential steps

  • Declare variables for unknowns (e.g., one number as x, another as y).
  • Create one equation per relationship described in the text.
  • When there are two unknowns, you typically get two equations.
  • Use a substitution/elimination approach to solve for one variable, then back-substitute to find the other.
  • Use organizational cues (circle the easy variable, arrow it into the other equation).
  • After solving, answer the question directly and check for reasonableness (does it make sense? do the numbers satisfy all conditions?).

• Quick problem-check habit: after solving, answer 2 questions: (1) Did we answer the question? (2) Does the answer make sense?

• Common setup technique: charting (distance = rate × time) helps organize multiple quantities.

• Calculator/notation notes: you may use a calculator; keep units consistent when converting between minutes/hours, etc.

Key concepts and core formulas

• Model concept: a formula that describes a relationship between quantities.

• Basic relation (distance problem): ext{Distance} = ext{Rate} imes ext{Time} \, (D = R \cdot T)

• Two-number problem (sum and a difference)

  • Example setup:
  • Let the numbers be x and y with one relation: x = y + 17 (one number exceeds the other by 17).
  • And their sum: x + y = 31.
  • Solve by substitution:
  • Substitute x = y + 17 into the second equation: y + 17 + y = 31.
  • Solve for y, then compute x from x = y + 17.
  • Solution pattern: one equation in one variable after substitution; then back-substitute.
  • Check: ensure both conditions are satisfied (difference and sum).

• Break-even cost problem (two plans)

  • Setup: 34 + 0.05m = 40 + 0.04m where m = minutes of talk time.
  • Solve: m = \frac{40 - 34}{0.05 - 0.04} = 600.
  • Decision rule: choose Plan A if minutes ≤ 600; Plan B if minutes ≥ 600.

• Rectangle perimeter problems

  • Perimeter: p = 2l + 2w
  • Given relation: l = w + 3 and p = 54 (for example).
  • Substitution: replace l with w + 3 → 54 = 2(w+3) + 2w = 4w + 6 → w = 12; then l = 15.

• Square/rectangle area from perimeter and dimensions

  • If length is related to width (e.g., l = w + 6) and p = 48, solve for w similarly; then compute area A = l\times w.
  • Example result: w = 9, l = 15 \Rightarrow A = 9\times 15 = 135\,\text{square inches}.

• Volume problems (rectangular box)

  • Volume: V = l \cdot w \cdot h
  • If l = 2w and h = 8 and V = 1600, then 1600 = (2w) \cdot w \cdot 8 = 16w^2\Rightarrow w = 10
  • Then l = 20, h = 8; dimensions: 20 \times 10 \times 8.

• Cylinder radius from volume

  • Volume: V = \pi r^2 h
  • Solve for r: r^2 = \frac{V}{\pi h}, \quad r = \sqrt{\frac{V}{\pi h}}
  • Example pattern: substitute numeric values for V and h to obtain r.

• Triangle area-related height (area formula)

  • Area: A = \frac{1}{2} b h
  • Solve for height: h = \frac{2A}{b}
  • Example: If A = 215 and base b = 15, then h = \frac{2 \cdot 215}{15} = \frac{430}{15} \approx 28.7\,\text{units}.

Representative worked examples (brief steps)

• Example: two numbers from sum and difference

  • Equations: x = y + 17,\, x + y = 31
  • Substitution: substitute x = y+17 into the second equation: y+17 + y = 31
  • Solve: y = 7,\quad x = 24
  • Check: 24 - 7 = 17 and 24 + 7 = 31.

• Example: cell-phone plan break-even (as above)

  • Equations: 34 + 0.05m = 40 + 0.04m → m = 600
  • Decision: Plan A if minutes ≤ 600; Plan B if minutes ≥ 600.

• Example: perimeter to complete dimensions

  • Given p = 54, l = w + 3 → substitute into p = 2l + 2w to get a single-variable equation; solve for w and then l.

• Example: area after perimeter (square/rectangle)

  • Perimeter p = 48, l = b with relation like l = b + 6 or similar; solve for one dimension; compute area A = l\times w.

• Example: shipping box (volume)

  • Given V = 1600, l = 2w,\, h = 8; substitute into V = l w h: 1600 = (2w) w 8 = 16 w^2; solve w = 10, then l = 20, h = 8.

• Example: radius from volume (cylinder)

  • Given values for V and h, compute r = \sqrt{\dfrac{V}{\pi h}}; plug numbers and evaluate.

• Example: triangle height from area

  • Given A = 215,\, b = 15; h = \dfrac{2A}{b} = \dfrac{430}{15} \approx 28.7\"" inches.