5.6 Related Rates of Change

5.6 Related Rates of Change

Core CLEP Concept: In related rates problems, two or more geometric quantities change simultaneously over time $t$. By establishing an algebraic equation that connects these variables at all times and differentiating implicitly with respect to time using the Chain Rule, we create a dynamic rate equation relating their derivatives ($\frac{dx}{dt}, \frac{dy}{dt}, \frac{dr}{dt}, \frac{dh}{dt}, \frac{dV}{dt}, \frac{d\theta}{dt}$).

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1. Foundational Theory & Implicit Time Differentiation

When quantities vary with time, each variable is an implicit function of $t$: $x = x(t), y = y(t), r = r(t), V = V(t)$. Applying the operator $\frac{d}{dt}$ requires using the Chain Rule on every variable term:

$$\begin{aligned} \frac{d}{dt}[x^2] &= 2x\frac{dx}{dt} \\[4pt] \frac{d}{dt}[r^3] &= 3r^2\frac{dr}{dt} \\[4pt] \frac{d}{dt}[x\cdot y] &= x\frac{dy}{dt} + y\frac{dx}{dt} \quad \text{(Product Rule)} \\[4pt] \frac{d}{dt}\left[\frac{x}{y}\right] &= \frac{y\frac{dx}{dt} - x\frac{dy}{dt}}{y^2} \quad \text{(Quotient Rule)} \\[4pt] \frac{d}{dt}[\sin\theta] &= \cos\theta \frac{d\theta}{dt} \\[4pt] \frac{d}{dt}\left[\frac{1}{h}\right] &= -\frac{1}{h^2}\frac{dh}{dt} \end{aligned}$$

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2. The 6-Step Master Related Rates Protocol

                           6-Step Related Rates Protocol
                                         │
     1. Draw Diagram & Identify Variables ┼───> Distinguish CONSTANTS from VARIABLES
     2. Record Rates & Snapshot Values ──┼───> Given: dx/dt = k; Target: dy/dt at t = t0
     3. Formulate Primary Equation ──────┼───> Geometric link: Pythagorean, Volume, Trig
     4. Eliminate Unwanted Variables ────┼───> Use secondary constraints (similar triangles)
     5. Differentiate Implicitly (d/dt) ─┼───> Apply Chain Rule to every variable term
     6. Substitute Snapshot & Solve ─────┴───> Plug in numbers AFTER differentiating & attach units

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3. Classic Problem Archetypes

Archetype 1: The Sliding Ladder

Problem: A $13\text{-foot}$ ladder leans against a vertical wall. The bottom of the ladder slides away from the wall at a constant rate of $0.5\text{ ft/s}$. How fast is the top of the ladder sliding down the wall at the instant when the bottom is $5\text{ feet}$ from the wall?

      |\ 
      | \ 
      |  \  Ladder of fixed length L = 13 ft
      |   \ 
    y |    \ 
      |     \ 
      |      \ 
      +-------+-----> x
          x

1. Variables & Constants: Let $x$ be the distance from wall to ladder base, $y$ be height on wall. Fixed length $L = 13\text{ ft}$ (constant).
2. Given & Target Rates: $\frac{dx}{dt} = +0.5\text{ ft/s}$. Find $\frac{dy}{dt}$ when $x = 5\text{ ft}$.
3. Primary Equation (Pythagorean Theorem): $$x^2 + y^2 = 13^2 = 169$$
4. Snapshot Height: When $x = 5$, $y = \sqrt{169 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\text{ ft}$.
5. Differentiate implicitly with respect to $t$: $$\frac{d}{dt}[x^2 + y^2] = \frac{d}{dt}[169] \implies 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \implies x\frac{dx}{dt} + y\frac{dy}{dt} = 0$$
6. Substitute snapshot values and solve: $$(5)(0.5) + (12)\frac{dy}{dt} = 0 \implies 2.5 + 12\frac{dy}{dt} = 0 \implies \frac{dy}{dt} = -\frac{2.5}{12} = -\frac{5}{24}\text{ ft/s}$$ $$\text{Conclusion: The top of the ladder is sliding down at } \frac{5}{24}\text{ ft/s} \approx 0.208\text{ ft/s}.$$

Extended Ladder Questions:

• Rate of change of enclosed triangle area ($A = \frac{1}{2}xy$): $$\frac{dA}{dt} = \frac{1}{2}\left(x\frac{dy}{dt} + y\frac{dx}{dt}\right) = \frac{1}{2}\left(5\left(-\frac{5}{24}\right) + 12(0.5)\right) = \frac{1}{2}\left(-\frac{25}{24} + 6\right) = \frac{119}{48}\text{ ft}^2\text{/s}$$
• Rate of change of angle with ground ($\cos\theta = x/13$): $$-\sin\theta \frac{d\theta}{dt} = \frac{1}{13}\frac{dx}{dt} \implies -\left(\frac{12}{13}\right)\frac{d\theta}{dt} = \frac{1}{13}(0.5) \implies \frac{d\theta}{dt} = -\frac{0.5}{12} = -\frac{1}{24}\text{ rad/s}$$

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Archetype 2: The Draining / Filling Inverted Conical Tank

Conical tank problems require using similar triangles to eliminate the radius $r$ before differentiating.

Problem: An inverted conical water tank has a height of $H = 12\text{ meters}$ and a base radius at the top of $R = 4\text{ meters}$. Water is pumped into the tank at a constant rate of $\frac{dV}{dt} = 2\text{ m}^3\text{/min}$. How fast is the water level rising when the water is $6\text{ meters}$ deep?

          |<--- 2R = 8m --->|
          +-----------------+
           \       r       /
            \   +-----+   /
             \  |     |  /
            h \ |     | / H = 12m
               \|     |/
                \  *  /
                 \   /
                  \ /
                   V

1. Primary Volume Equation: $V = \frac{1}{3}\pi r^2 h$.
2. Secondary Relationship (Similar Triangles): The cross-sectional triangle of the water cone is similar to the full tank cone: $$\frac{r}{h} = \frac{R}{H} = \frac{4}{12} = \frac{1}{3} \implies r = \frac{h}{3}$$
3. Substitute $r$ into volume formula to eliminate $r$: $$V = \frac{1}{3}\pi \left(\frac{h}{3}\right)^2 h = \frac{1}{3}\pi \left(\frac{h^2}{9}\right) h = \frac{\pi}{27} h^3$$
4. Differentiate implicitly with respect to $t$: $$\frac{dV}{dt} = \frac{\pi}{27} \left(3h^2\frac{dh}{dt}\right) = \frac{\pi}{9} h^2 \frac{dh}{dt}$$
5. Substitute snapshot values ($h = 6, \frac{dV}{dt} = 2$) and solve: $$2 = \frac{\pi}{9} (6)^2 \frac{dh}{dt} \implies 2 = \frac{\pi}{9}(36) \frac{dh}{dt} \implies 2 = 4\pi \frac{dh}{dt} \implies \frac{dh}{dt} = \frac{2}{4\pi} = \frac{1}{2\pi}\text{ m/min}$$

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Archetype 3: The Moving Shadow (Streetlight Problem)

Problem: A person $6\text{ feet}$ tall walks away from a $15\text{-foot}$ lamppost at a speed of $5\text{ ft/s}$.

1. At what rate is the length of the person's shadow increasing?
2. At what speed is the tip of the shadow moving?

      |
  15m |\ 
      | \ 
      |  \ 
      |   |\ 
      |   |6\ 
      +---+--+--------> 
        x    s
      |<-- x+s ------>|

1. Variables: Let $x$ be distance from pole to person, $s$ be length of shadow. Tip of shadow distance from pole is $(x + s)$.
2. Given Rate: $\frac{dx}{dt} = 5\text{ ft/s}$.
3. Similar Triangles Equation: $$\frac{15}{x + s} = \frac{6}{s} \implies 15s = 6(x + s) = 6x + 6s \implies 9s = 6x \implies s = \frac{2}{3}x$$
4. Rate shadow is lengthening: $$\frac{ds}{dt} = \frac{2}{3}\frac{dx}{dt} = \frac{2}{3}(5) = \frac{10}{3}\text{ ft/s}$$
5. Speed of the tip of the shadow: $$\frac{d}{dt}[x + s] = \frac{dx}{dt} + \frac{ds}{dt} = 5 + \frac{10}{3} = \frac{25}{3}\text{ ft/s}$$

(Note: Both shadow rates are completely constant and independent of the person's position $x$!)

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Archetype 4: Expanding Spheres & Geometric Shapes

Problem: A spherical balloon is being inflated with gas such that its volume increases at a constant rate of $100\pi\text{ cm}^3\text{/s}$. How fast is the surface area of the balloon increasing at the instant when the radius is $5\text{ cm}$?

1. Equations: $V = \frac{4}{3}\pi r^3$ and $S = 4\pi r^2$.
2. Differentiate Volume: $$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \implies 100\pi = 4\pi (5)^2 \frac{dr}{dt} \implies 100\pi = 100\pi \frac{dr}{dt} \implies \frac{dr}{dt} = 1\text{ cm/s}$$
3. Differentiate Surface Area: $$\frac{dS}{dt} = 8\pi r \frac{dr}{dt} = 8\pi (5) (1) = 40\pi\text{ cm}^2\text{/s}$$

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4. Master Reference Guide to Common Related Rates Relations