11.3 Surface Area, Volume & Rates in Context (Distance, Speed, Time)

Surface Area, Volume & Rates in Context (Distance, Speed, Time)

Quick Answer: WEST-B Mathematics Objective 0014 evaluates candidates on three-dimensional geometry (surface area and volume), geometric scaling relationships, and real-world rate word problems. Essential 3D formulas include rectangular prism volume ($V = lwh$) and surface area ($SA = 2lw + 2lh + 2wh$), cylinder volume ($V = \pi r^2 h$) and surface area ($SA = 2\pi r^2 + 2\pi rh$), cone volume ($V = \frac{1}{3}\pi r^2 h$), pyramid volume ($V = \frac{1}{3}Bh$), and sphere volume ($V = \frac{4}{3}\pi r^3$) and surface area ($SA = 4\pi r^2$). When linear dimensions scale by $k$, surface area scales by $k^2$ and volume scales cubically by $k^3$. In rate problems, remember: $d = rt$, average speed is always $\frac{\text{Total Distance}}{\text{Total Time}}$ (never average rates directly), and joint work problems follow $\frac{1}{t_1} + \frac{1}{t_2} = \frac{1}{T_{\text{together}}}$.

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1. Surface Area of Three-Dimensional Solids

Surface Area ($SA$) is the total two-dimensional area of all exterior faces and curved surfaces enclosing a three-dimensional solid. It is measured in square units ($\text{in}^2$, $\text{cm}^2$, $\text{m}^2$).

+---------------------------------------------------------------------------------------------------+
|                                SURFACE AREA MASTER FORMULA TABLE                                  |
|                                                                                                   |
|   +-----------------------+-------------------------------+-----------------------------------+   |
|   | 3D SOLID              | TOTAL SURFACE AREA (SA)       | COMPONENT DECOMPOSITION           |   |
|   +-----------------------+-------------------------------+-----------------------------------+   |
|   | Rectangular Prism     | SA = 2lw + 2lh + 2wh          | Top/Bottom (2lw) + Front/Back     |   |
|   |                       |    = 2(lw + lh + wh)          | (2lh) + Left/Right (2wh)          |   |
|   | Cube                  | SA = 6s^2                     | 6 congruent square faces          |   |
|   | Right Circular        | SA = 2πr^2 + 2πrh             | 2 circular bases (2πr^2) +        |   |
|   | Cylinder              |                               | lateral rectangular jacket (2πrh) |   |
|   | Right Circular        | SA = πr^2 + πrl               | Circular base (πr^2) + lateral    |   |
|   | Cone                  | (l = √(r^2 + h^2) slant ht)   | cone surface (πrl)                |   |
|   | Sphere                | SA = 4πr^2                    | Exactly 4 times the area of its   |   |
|   |                       |                               | great circle                      |   |
|   +-----------------------+-------------------------------+-----------------------------------+   |
+---------------------------------------------------------------------------------------------------+

Unfolding 3D Nets

A net is a two-dimensional planar pattern that can be folded along its edges to construct a three-dimensional polyhedron or solid. Analyzing nets is a primary conceptual strategy tested on the WEST-B to calculate surface area without memorizing isolated formulas.

The Cylinder Net Breakdown:

When a right circular cylinder is unrolled:

1. The top and bottom bases form two congruent flat circles, each with area $A = \pi r^2$ (Total base area $= 2\pi r^2$).
2. The lateral curved body unrolls into a flat rectangle whose height equals the cylinder height $h$ and whose width equals the circumference of the circular base ($C = 2\pi r$).
3. Lateral Area: $A_{\text{lateral}} = \text{width} \times \text{height} = (2\pi r) \times h = 2\pi rh$.
4. Total Surface Area: $SA = 2\pi r^2 + 2\pi rh$.

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2. Volume of Three-Dimensional Solids

Volume ($V$) is the amount of three-dimensional space enclosed within a solid boundary, measured in cubic units ($\text{in}^3$, $\text{cm}^3$, $\text{m}^3$, liters, gallons).

+---------------------------------------------------------------------------------------------------+
|                                    VOLUME MASTER FORMULA TABLE                                    |
|                                                                                                   |
|   +-----------------------+-----------------------+-------------------------------------------+   |
|   | 3D SOLID              | VOLUME FORMULA        | RELATIONSHIP TO BASE AREA (B)             |   |
|   +-----------------------+-----------------------+-------------------------------------------+   |
|   | Rectangular Prism     | V = l * w * h = B * h | B = lw (area of rectangular base)         |   |
|   | Cube                  | V = s^3               | Special prism where l = w = h = s         |   |
|   | Right Circular Cyl.   | V = πr^2 * h = B * h  | B = πr^2 (area of circular base)          |   |
|   | Right Circular Cone   | V = (1/3)πr^2 * h     | Exactly 1/3 the volume of matching cylinder|
|   | Pyramid               | V = (1/3) * B * h     | Exactly 1/3 the volume of matching prism  |   |
|   | Sphere                | V = (4/3)πr^3         | Enclosed spherical volume                 |   |
|   +-----------------------+-----------------------+-------------------------------------------+   |
+---------------------------------------------------------------------------------------------------+

The "One-Third Rule" for Cones and Pyramids

A fundamental geometric relationship frequently highlighted on teacher certification exams is that any pointed solid (cone or pyramid) occupies exactly one-third ($\frac{1}{3}$) of the volume of the uniform prism or cylinder sharing the identical base area $B$ and perpendicular height $h$: $V_{\text{pyramid}} = \frac{1}{3} V_{\text{prism}} = \frac{1}{3}Bh$ $V_{\text{cone}} = \frac{1}{3} V_{\text{cylinder}} = \frac{1}{3}\pi r^2 h$

Worked Example: Solid Geometry Calculations

Problem: A science teacher has a cylindrical container with radius $r = 4\text{ cm}$ and height $h = 10\text{ cm}$, and a solid metal sphere with radius $r = 3\text{ cm}$.

1. What is the total volume of the cylindrical container?
2. What is the volume of the metal sphere?
3. If the sphere is submerged in water inside the cylinder, how much water volume does it displace?

Solution:

• Cylinder Volume: $V_{\text{cyl}} = \pi r^2 h = \pi (4^2)(10) = 160\pi\text{ cm}^3 \approx 160 \times 3.1416 = 502.65\text{ cm}^3$
• Sphere Volume (Displaced Volume): $V_{\text{sphere}} = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (3^3) = \frac{4}{3}\pi (27) = 36\pi\text{ cm}^3 \approx 36 \times 3.1416 = 113.10\text{ cm}^3$
• Displaced Volume: By Archimedes' principle, the submerged sphere displaces exactly its own volume: $36\pi\text{ cm}^3 \approx 113.10\text{ mL}$ ($1\text{ cm}^3 = 1\text{ mL}$).
• Remaining Air/Liquid Space: $160\pi - 36\pi = 124\pi\text{ cm}^3 \approx 389.56\text{ cm}^3$.

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3. The 3D Dimensional Scaling Law (k, k², k³)

When all linear dimensions of a three-dimensional object are scaled by a factor of $k$:

+---------------------------------------------------------------------------------------------------+
|                                 3D SCALING RELATIONSHIPS MATRIX                                   |
|                                                                                                   |
|   Dimension Level       Geometric Measure              Scaling Multiplier     Example (k = 3)     |
|   +-------------------+------------------------------+----------------------+-----------------+   |
|   | 1D (Linear)       | Length, Radius, Height, Peri | * k^1                | 3x longer       |   |
|   | 2D (Area)         | Surface Area, Base Area      | * k^2                | 9x larger area  |   |
|   | 3D (Volume/Mass)  | Volume, Capacity, Weight     | * k^3                | 27x larger vol  |   |
|   +-------------------+------------------------------+----------------------+-----------------+   |
+---------------------------------------------------------------------------------------------------+

Practical Classroom Example:

If a small storage crate measuring $2\text{ ft} \times 2\text{ ft} \times 2\text{ ft}$ has a volume of $8\text{ cu ft}$ and a surface area of $24\text{ sq ft}$, what happens when a larger crate is built with all dimensions doubled ($k = 2$)?

• New Linear Dimensions: $4\text{ ft} \times 4\text{ ft} \times 4\text{ ft}$ ($2 \times 2 = 4\text{ ft}$).
• New Surface Area: $SA_{\text{new}} = 24 \times (2^2) = 24 \times 4 = 96\text{ sq ft}$ (Surface area quadruples).
• New Volume: $V_{\text{new}} = 8 \times (2^3) = 8 \times 8 = 64\text{ cu ft}$ (Volume increases by a factor of $8$).

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4. Rate Word Problems: Distance, Speed & Time (d = rt)

Rate represents a ratio comparing two quantities with different units (e.g., miles per hour, dollars per pound, gallons per minute).

The Fundamental Motion Formulas

$\text{Distance} = \text{Rate} \times \text{Time} \quad (d = rt)$ $\text{Rate} = \frac{\text{Distance}}{\text{Time}} \quad \left(r = \frac{d}{t}\right)$ $\text{Time} = \frac{\text{Distance}}{\text{Rate}} \quad \left(t = \frac{d}{r}\right)$

The "Average Speed" Trap on Multi-Leg Trips

Critical Rule: You can NEVER find the average speed of a multi-leg trip by taking the simple arithmetic mean of the speeds (i.e., $\frac{r_1 + r_2}{2}$ is FALSE whenever the travel times over the legs are unequal).

$\text{Average Speed } (r_{\text{avg}}) = \frac{\text{Total Distance Travelled}}{\text{Total Time Elapsed}} = \frac{d_1 + d_2 + \dots}{t_1 + t_2 + \dots}$

Worked Derivation: The Classic Round-Trip Problem

Problem: A school district bus travels $60\text{ miles}$ to an athletic competition at $60\text{ mph}$, and returns along the exact same $60\text{ miles}$ route during rush hour at $30\text{ mph}$. What is the average speed for the entire round trip?

• Step 1: Calculate outbound travel time ($t_1$): $t_1 = \frac{d_1}{r_1} = \frac{60\text{ miles}}{60\text{ mph}} = 1.0\text{ hour}$
• Step 2: Calculate return travel time ($t_2$): $t_2 = \frac{d_2}{r_2} = \frac{60\text{ miles}}{30\text{ mph}} = 2.0\text{ hours}$
• Step 3: Calculate total distance and total time: $d_{\text{total}} = 60 + 60 = 120\text{ miles}$ $t_{\text{total}} = 1.0 + 2.0 = 3.0\text{ hours}$
• Step 4: Compute true average speed: $r_{\text{avg}} = \frac{120\text{ miles}}{3.0\text{ hours}} = 40\text{ mph}$ (Notice that $40\text{ mph}$ is significantly lower than the incorrect arithmetic average $\frac{60 + 30}{2} = 45\text{ mph}$, because the vehicle spent twice as much time travelling at the slower speed!)

Pursuing and Opposing Direction Problems

1. Objects Moving in Opposite Directions (Separating or Closing In):
   • Their combined relative speed is the sum of their individual speeds: $r_{\text{combined}} = r_1 + r_2$.
   • Total separation distance after time $t$: $d = (r_1 + r_2)t$.
2. One Object Pursuing Another (Catch-Up Motion):
   • The closing speed is the difference between their speeds: $r_{\text{closing}} = r_{\text{fast}} - r_{\text{slow}}$.
   • Time required to close a head-start gap ($d_{\text{gap}}$): $t = \frac{d_{\text{gap}}}{r_{\text{fast}} - r_{\text{slow}}}$

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5. Work-Rate Problems & Simultaneous Productivity

Work-rate problems apply rate reasoning to tasks completed over time. If a person or machine completes an entire job in $t$ hours, their work rate (portion of job completed per hour) is $\frac{1}{t}$.

The Collaborative Work Equation

When two entities work together simultaneously without interfering with each other's productivity, their individual rates add together to form the combined rate:

$\text{Rate}_A + \text{Rate}_B = \text{Combined Rate}$ $\frac{1}{t_A} + \frac{1}{t_B} = \frac{1}{T_{\text{together}}}$

$\text{Solving for } T_{\text{together}}: \quad T_{\text{together}} = \frac{t_A \times t_B}{t_A + t_B} = \frac{\text{Product}}{\text{Sum}}$

Worked Example: Collaborative Grading

Problem: Teacher A can grade a set of standardized classroom assessments in $3\text{ hours}$. Teacher B can grade the same set of assessments in $6\text{ hours}$. If both teachers grade together, how many hours will it take to finish the grading?

• Step 1: Express individual hourly rates:
  • $\text{Rate of Teacher A} = \frac{1}{3}\text{ of the set per hour}$
  • $\text{Rate of Teacher B} = \frac{1}{6}\text{ of the set per hour}$
• Step 2: Add rates using a common denominator: $\text{Combined Rate} = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}\text{ of the set per hour}$
• Step 3: Solve for combined time: $T_{\text{together}} = \frac{1}{\text{Combined Rate}} = \frac{1}{1/2} = 2\text{ hours}$ (Alternatively using the Product-over-Sum formula: $T = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2\text{ hours}$.)