5.6 Related Rates of Change
Key Takeaways
- Related rates problems calculate the unknown rate of change of one variable with respect to time (d/dt) by differentiating an underlying geometric relationship connecting multiple time-dependent variables.
- The Chain Rule with respect to time dictates that every time-varying variable x(t) generates a derivative term dx/dt upon differentiation (e.g. d/dt[x^2] = 2x dx/dt).
- The Golden Rule of Related Rates is: NEVER substitute numerical snapshot values for time-varying quantities before differentiating; variables must remain symbolic until d/dt is computed.
- Secondary geometric relations (such as similar triangles r/h = R/H in conical tanks or shadow problems) are used to eliminate extraneous variables prior to differentiation.
- Sign conventions must be rigorously maintained: quantities that grow with time have positive rates (dx/dt > 0), whereas quantities that shrink have negative rates (dy/dt < 0).
Last updated: August 2026
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 ($\frac{dx}{dt}, \frac{dy}{dt}, \frac{dr}{dt}, \frac{dh}{dt}, \frac{dV}{dt}, \frac{d\theta}{dt}$).
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:
\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}$$ --- ## 2. The 6-Step Master Related Rates Protocol ``` 6-Step Related Rates Protocol │ 1. Draw Diagram & Identify Variables ┼───> Distinguish CONSTANTS from VARIABLES 2. Record Rates & Snapshot Values ──┼───> Given: dx/dt = k; Target: dy/dt at t = t0 3. Formulate Primary Equation ──────┼───> Geometric link: Pythagorean, Volume, Trig 4. Eliminate Unwanted Variables ────┼───> Use secondary constraints (similar triangles) 5. Differentiate Implicitly (d/dt) ─┼───> Apply Chain Rule to every variable term 6. Substitute Snapshot & Solve ─────┴───> Plug in numbers AFTER differentiating & attach units ``` > [!IMPORTANT] > **The Cardinal Rule of Related Rates:** *NEVER substitute a numerical value for a changing quantity BEFORE differentiating!* If a ladder is sliding down a wall, its height $y(t)$ is changing. If you plug in $y = 12$ before differentiating, $\frac{d}{dt}[12] = 0$, which falsely destroys the rate $\frac{dy}{dt}$. Constants that NEVER change (such as the length of a rigid ladder $L = 13$) may be substituted before differentiating; all changing variables must remain symbolic until *after* differentiation. --- ## 3. Classic Problem Archetypes ### Archetype 1: The Sliding Ladder **Problem:** A $13\text{-foot}$ ladder leans against a vertical wall. The bottom of the ladder slides away from the wall at a constant rate of $0.5\text{ ft/s}$. How fast is the top of the ladder sliding down the wall at the instant when the bottom is $5\text{ feet}$ from the wall? ``` |\ | \ | \ Ladder of fixed length L = 13 ft | \ y | \ | \ | \ +-------+-----> x x ``` 1. **Variables & Constants:** Let $x$ be the distance from wall to ladder base, $y$ be height on wall. Fixed length $L = 13\text{ ft}$ (constant). 2. **Given & Target Rates:** $\frac{dx}{dt} = +0.5\text{ ft/s}$. Find $\frac{dy}{dt}$ when $x = 5\text{ ft}$. 3. **Primary Equation (Pythagorean Theorem):** $$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}$$ --- ### Archetype 2: The Draining / Filling Inverted Conical Tank Conical tank problems require using **similar triangles** to eliminate the radius $r$ before differentiating. **Problem:** An inverted conical water tank has a height of $H = 12\text{ meters}$ and a base radius at the top of $R = 4\text{ meters}$. Water is pumped into the tank at a constant rate of $\frac{dV}{dt} = 2\text{ m}^3\text{/min}$. How fast is the water level rising when the water is $6\text{ meters}$ deep? ``` |<--- 2R = 8m --->| +-----------------+ \ r / \ +-----+ / \ | | / h \ | | / H = 12m \| |/ \ * / \ / \ / V ``` 1. **Primary Volume Equation:** $V = \frac{1}{3}\pi r^2 h$. 2. **Secondary Relationship (Similar Triangles):** The cross-sectional triangle of the water cone is similar to the full tank cone: $$\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}$$ --- ### Archetype 3: The Moving Shadow (Streetlight Problem) **Problem:** A person $6\text{ feet}$ tall walks away from a $15\text{-foot}$ lamppost at a speed of $5\text{ ft/s}$. 1. At what rate is the length of the person's shadow increasing? 2. At what speed is the tip of the shadow moving? ``` | 15m |\ | \ | \ | |\ | |6\ +---+--+--------> x s |<-- x+s ------>| ``` 1. **Variables:** Let $x$ be distance from pole to person, $s$ be length of shadow. Tip of shadow distance from pole is $(x + s)$. 2. **Given Rate:** $\frac{dx}{dt} = 5\text{ ft/s}$. 3. **Similar Triangles Equation:** $$\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}$$ *(Note: Both shadow rates are completely constant and independent of the person's position $x$!)* --- ### Archetype 4: Expanding Spheres & Geometric Shapes **Problem:** A spherical balloon is being inflated with gas such that its volume increases at a constant rate of $100\pi\text{ cm}^3\text{/s}$. How fast is the surface area of the balloon increasing at the instant when the radius is $5\text{ cm}$? 1. **Equations:** $V = \frac{4}{3}\pi r^3$ and $S = 4\pi r^2$. 2. **Differentiate Volume:** $$\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}$$ --- ## 4. Master Reference Guide to Common Related Rates Relations | Geometric Shape | Primary Equation | Differentiated Rate Equation | | :--- | :--- | :--- | | **Right Triangle (Ladder)** | $x^2 + y^2 = L^2$ | $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$ | | **Circle Area** | $A = \pi r^2$ | $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$ | | **Sphere Volume** | $V = \frac{4}{3}\pi r^3$ | $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$ | | **Sphere Surface Area** | $S = 4\pi r^2$ | $\frac{dS}{dt} = 8\pi r \frac{dr}{dt}$ | | **Inverted Cone Tank** | $V = \frac{\pi R^2}{3H^2}h^3$ | $\frac{dV}{dt} = \frac{\pi R^2}{H^2}h^2 \frac{dh}{dt}$ | | **Cylinder Volume** | $V = \pi r^2 h$ | $\frac{dV}{dt} = \pi r^2 \frac{dh}{dt} + 2\pi r h \frac{dr}{dt}$ | | **Elevation Angle** | $\tan\theta = \frac{y}{x}$ | $\sec^2\theta \frac{d\theta}{dt} = \frac{x\frac{dy}{dt} - y\frac{dx}{dt}}{x^2}$ |Loading diagram...
Test Your Knowledge
A 10-foot ladder is leaning against a vertical wall. If the bottom of the ladder slides away from the wall at a speed of 2 ft/s, how fast is the top of the ladder sliding down the wall when the bottom is 6 feet from the wall?
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An inverted conical tank with height 10 feet and top radius 5 feet is leaking water. If the water depth h is 4 feet and the water level is dropping at a rate of 0.2 ft/min, at what rate is water leaking out of the tank in cubic feet per minute?
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A spotlight on the ground shines on a vertical wall 12 meters away. If a man 2 meters tall walks from the spotlight toward the building at a speed of 1.6 m/s, how fast is the length of his shadow on the wall decreasing when he is 4 meters from the spotlight?
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The radius of a sphere is increasing at a constant rate of 3 cm/s. At what rate is the volume of the sphere increasing at the instant when its surface area is 100 cm^2?
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