MATH-181 (Calculus I) Curriculum

Session 11 - Related Rates

Handout Worksheet

Fall Semester 2026

Session Objectives & Overview

By the end of this session, you will be able to:

Formulate time-derivative equations using the Chain Rule operator $\frac{d}{dt}$ on geometric single-variable models (circles, spheres, and rigid cylinders), distinguishing between static constants and dynamic variables, and interpreting sign conventions and dimensional units.
Solve planar kinematic and tracking problems governed by Pythagorean relations ( $x^2 + y^2 = z^2$ or $x^2 + y^2 = L^2$ ) and trigonometric relationships ( $\tan\theta = y/x$ , $\cos\theta = x/L$ ), computing instantaneous linear separation velocities and angular tracking rates ( $\frac{d\theta}{dt}$ ).
Resolve multi-variable geometric systems (inverted cones, triangular troughs, and dynamic shadow projections) by establishing geometric similarity constraints to eliminate redundant dependent variables before executing implicit differentiation with respect to time.

Micro-Lecture

1. Time as the Master Parameter & The Dynamic Tail

Mathematical Foundation: In Session 09, we mastered implicit differentiation with respect to spatial coordinate $x$ . In dynamic physical systems, geometric configurations evolve over time. Time ( $t$ ) serves as the master independent variable. Every physical dimension—radius $r(t)$ , height $h(t)$ , volume $V(t)$ , angle $\theta(t)$ , or separation distance $z(t)$ —behaves as an implicit composite function of $t$ . Applying the time differential operator $\frac{d}{dt}$ to any dynamic variable invokes the Chain Rule, appending a "dynamic tail" representing its instantaneous time derivative: $\frac{d}{dt}[u^n] = n u^{n-1} \cdot \frac{du}{dt}, \qquad \frac{d}{dt}[\tan\theta] = \sec^2\theta \cdot \frac{d\theta}{dt}$ (see Related Rates).
Engineering Context: Sensor telemetry systems (radar receivers, flowmeters, strain gauges) measure localized rates of change. Related rates equations translate these direct instrument readings into unmeasurable physical quantities, such as true vehicle intercept velocities, reservoir fill trajectories, or thermal boundary expansions.

2. The Static vs. Dynamic Rule of Substitution

Mathematical Foundation: The single most common student error in related rates is substituting numeric values into equations before differentiating. If a variable is replaced by a constant before differentiation, its derivative vanishes ( $\frac{d}{dt}[\text{constant}] = 0$ ). Engineers enforce a strict binary rule:
- Static Constants: Parameters whose physical values never change during the problem (e.g., the length of a rigid ladder, the fixed radius of a cylindrical tank, or the fixed altitude of an aircraft in level flight) must be substituted before taking the time derivative.
- Dynamic Variables: Parameters that change over time must remain symbols throughout differentiation. Their numeric "snapshot" values are substituted only after the derivative equation is established.

3. Geometric Reduction & The Multi-Variable Challenge

Mathematical Foundation: When a governing equation contains multiple dynamic variables (such as $V = \frac{1}{3}\pi r^2 h$ in a conical tank), direct differentiation produces multiple unknown rates ( $\frac{dr}{dt}$ and $\frac{dh}{dt}$ ). When telemetry only provides one rate, engineers must utilize geometric constraints—such as similar triangles ( $r/h = R/H$ )—to eliminate the redundant variable prior to differentiation (see Geometric Similarity in Related Rates).

Skill Block 1

Worked Example A: Rate of Surface Expansion for a Circular Spill

An offshore pipeline rupture produces a circular oil slick spreading across open water. Oceanographic survey telemetry confirms that the radius of the spill is increasing at a steady rate of $30\text{ m/hr}$ . Determine the instantaneous rate at which the surface area of the oil slick is expanding at the exact moment the radius reaches $100\text{ meters}$ .

Worked Solution:

Governing Formula: The Area of a Circle Formula: $A = \pi r^2$ (This formula and its time derivative principles are defined in Related Rates.)
Identify Knowns and Target Derivative:
- Dynamic variables: $A(t) = \text{area in }\text{m}^2$ , $r(t) = \text{radius in }\text{m}$ .
- Given rate: $\frac{dr}{dt} = +30\text{ m/hr}$ (positive because the slick is expanding).
- Snapshot condition: $r = 100\text{ m}$ .
- Target rate: Find $\frac{dA}{dt}$ .
Differentiate implicitly with respect to time ( $t$ ): $\frac{d}{dt}[A] = \frac{d}{dt}[\pi r^2] = 2\pi r \frac{dr}{dt}$
Substitute snapshot values and calculate: $\frac{dA}{dt} = 2\pi (100)(30) = 6000\pi\text{ m}^2/\text{hr} \approx 18849.56\text{ m}^2/\text{hr}$
Engineering Interpretation: At the moment the radius reaches $100\text{ meters}$ , the surface area of the oil spill is increasing at an instantaneous rate of $6000\pi\text{ m}^2/\text{hr}$ .

Worked Example B: Liquid Drainage in a Rigid Cylindrical Tank

A vertical cylindrical industrial water tank has a constant internal radius of $10\text{ inches}$ . A rupture in the discharge manifold causes water to drain out at a constant volumetric rate of $50\text{ in}^3/\text{s}$ . Determine the rate of change of the water level when the water depth in the tank is $25\text{ inches}$ .

Worked Solution:

Governing Formula: The Volume of a Cylinder Formula: $V = \pi r^2 h$ (This formula and its time derivative principles are defined in Related Rates.)
Classify Parameters (Static vs. Dynamic):
- The tank is a rigid steel structure, so the radius is a static constant: $r = 10\text{ in}$ at all times ( $\frac{dr}{dt} = 0$ ).
- The water depth $h(t)$ and volume $V(t)$ change continuously over time (dynamic variables).
- Substitute the static constant $r = 10$ into the primary equation before differentiation: $V = \pi (10)^2 h = 100\pi h$
Identify Knowns and Target Derivative:
- Given rate: $\frac{dV}{dt} = -50\text{ in}^3/\text{s}$ (negative because liquid is draining).
- Snapshot condition: $h = 25\text{ in}$ .
- Target rate: Find $\frac{dh}{dt}$ .
Differentiate implicitly with respect to time ( $t$ ): $\frac{d}{dt}[V] = \frac{d}{dt}[100\pi h] \implies \frac{dV}{dt} = 100\pi \frac{dh}{dt}$
Substitute known rate and solve: $-50 = 100\pi \frac{dh}{dt} \implies \frac{dh}{dt} = \frac{-50}{100\pi} = -\frac{1}{2\pi}\text{ in/s} \approx -0.159\text{ in/s}$
Engineering Interpretation: The water height is decreasing at a rate of $\frac{1}{2\pi}\text{ in/s}$ (approximately $0.159\text{ in/s}$ ). Notice that because the cylinder has vertical walls, this rate is completely independent of the water depth $h$ .

Practice Problem 1: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) In the Conceptual Framework, explain why static constants (like the cylinder radius $r = 10\text{ in}$ ) must be substituted into the equation before differentiating, while dynamic variables (like depth $h = 25\text{ in}$ ) must not be substituted until after differentiating.
(b) Deconstruct Worked Example A: Why does differentiating $\pi r^2$ with respect to time produce $2\pi r \frac{dr}{dt}$ instead of simply $2\pi r$ ? What role does time $t$ play as the master independent variable?
(c) Review the linked Engineering Applications (Environmental Plume Mapping (Environmental Engineering)). Why must the remediation team monitor the radial expansion rate $\frac{dr}{dt}$ rather than relying solely on the surface area rate $\frac{dA}{dt}$ to deploy containment barriers?

Solution Key:

(a) In physical modeling, static parameters are invariant properties of the container or boundary geometry. Because their values never change over time, their time derivatives are identically zero. Substituting static constants before taking the derivative correctly eliminates extra terms and prevents algebraic clutter. In contrast, dynamic variables evolve continuously with time. If a dynamic snapshot value (such as $h = 25\text{ in}$ ) is plugged in prior to differentiation, the equation degenerates into a relation of static numbers ( $V = 100\pi(25) = 2500\pi$ ), and taking the time derivative would yield $0 = 0$ , completely erasing the rate of change $\frac{dh}{dt}$ from the calculation.
(b) Time $t$ is the fundamental independent variable. In single-variable calculus with respect to $r$ , the rate of change of $r$ with respect to itself is unity ( $\frac{dr}{dr} = 1$ ). However, in related rates, $r$ is driven by time: $r = r(t)$ . By the Chain Rule, the derivative of any outer expression $f(r)$ with respect to time is the derivative of the outer function with respect to $r$ , multiplied by the inner rate of change with respect to time: $\frac{d}{dt}[\pi r^2] = \pi \cdot 2r \cdot \frac{dr}{dt} = 2\pi r \frac{dr}{dt}$ .
(c) As established in Environmental Plume Mapping (Environmental Engineering), field crews cannot physically deploy surface booms or trench cutoffs based on an abstract areal metric like $\text{m}^2/\text{hr}$ . Crews need a physical linear setback distance (radius $r$ ) and an advance velocity ( $\frac{dr}{dt}$ ) to schedule how many minutes remain before the toxic front breaches emergency containment corridors.

Show your steps and calculations here.

Practice Problem 2: Diameter Growth Rate of an Inflating Sphere

A spherical basketball is being inflated with a high-pressure electric pump that introduces compressed air at a constant volumetric rate of $3\text{ in}^3/\text{s}$ . Determine the instantaneous rate at which the diameter of the basketball is increasing at the exact moment the radius reaches $2\text{ inches}$ .

Solution Key:

Governing Formula: The Volume of a Sphere Formula: $V = \frac{4}{3}\pi r^3$ (This formula is defined in Related Rates.)
Relate Radius to Diameter: The diameter $D$ is twice the radius: $D = 2r \implies r = \frac{D}{2}$ . Alternatively, differentiate $V$ with respect to $r$ first and use $\frac{dD}{dt} = 2\frac{dr}{dt}$ .
Differentiate implicitly with respect to time ( $t$ ): $\frac{dV}{dt} = \frac{4}{3}\pi \left(3r^2 \frac{dr}{dt}\right) = 4\pi r^2 \frac{dr}{dt}$
Substitute snapshot values ( $r = 2\text{ in}$ , $\frac{dV}{dt} = 3\text{ in}^3/\text{s}$ ): $3 = 4\pi (2)^2 \frac{dr}{dt} = 16\pi \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{3}{16\pi}\text{ in/s}$
Compute the rate of diameter change $\frac{dD}{dt}$ : $\frac{dD}{dt} = 2\frac{dr}{dt} = 2\left(\frac{3}{16\pi}\right) = \frac{3}{8\pi}\text{ in/s} \approx 0.1194\text{ in/s}$
Engineering Interpretation: At the instant the radius reaches $2\text{ inches}$ , the diameter of the basketball is expanding at an instantaneous rate of $\frac{3}{8\pi}\text{ in/s}$ .

Show your steps and calculations here.

Practice Problem 3: Leaking Spherical Balloon & Dimensional Sign Conventions

A spherical weather balloon filled with helium develops a seam puncture causing gas to escape at a constant volumetric rate of $10\text{ in}^3/\text{min}$ . At the exact moment the radius is $5\text{ inches}$ , determine the rate of change of the radius and provide a complete physical summary sentence adhering to proper sign conventions.

Solution Key:

Governing Formula: The Volume of a Sphere Formula: $V = \frac{4}{3}\pi r^3$ (This formula is defined in Related Rates.)
Identify Knowns and Target Derivative:
- Volume is decreasing due to leaking: $\frac{dV}{dt} = -10\text{ in}^3/\text{min}$ .
- Snapshot radius: $r = 5\text{ in}$ .
- Target rate: Find $\frac{dr}{dt}$ .
Differentiate implicitly with respect to time ( $t$ ): $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$
Substitute snapshot values and solve: $-10 = 4\pi (5)^2 \frac{dr}{dt} = 100\pi \frac{dr}{dt}$ $\frac{dr}{dt} = \frac{-10}{100\pi} = -\frac{1}{10\pi}\text{ in/min} \approx -0.0318\text{ in/min}$
Dimensional & Contextual Interpretation:
- Dimensional Analysis: The numerator represents radius (inches), and the denominator represents time (minutes), yielding $\text{in/min}$ .
- Sign Convention: The algebraic negative sign signifies that the physical dimension is contracting.
- Summary Sentence: When the radius is $5\text{ inches}$ , the radius of the balloon is decreasing at a rate of $\frac{1}{10\pi}\text{ in/min}$ (approximately $0.0318\text{ in/min}$ ).

Show your steps and calculations here.

Regroup 1

The Early Substitution Trap: Emphasize why plugging in snapshot values (like $h = 25$ or $r = 5$ ) before differentiating completely destroys the mathematical derivative by turning dynamic variables into constants with zero rates.
Double Negatives in Engineering Reports: Clarify that a summary sentence stating "the height is decreasing at $-0.5\text{ m/s}$ " is technically claiming that the height is increasing. The verbal word "decreasing" absorbs the negative algebraic sign.
Leibniz Dimensional Check: Show how the fraction structure of $\frac{d(\text{Variable})}{d(\text{Time})}$ guarantees the correct engineering units for any rate.

Skill Block 2

Worked Example A: Sliding Ladder on a Vertical Wall

A rigid $13\text{-foot}$ ladder rests against a vertical wall. The base of the ladder is pulled horizontally away from the wall along the ground at a steady speed of $3\text{ ft/s}$ . Determine how fast the top of the ladder is sliding down the vertical wall at the exact moment the base of the ladder is $5\text{ feet}$ away from the wall.

Worked Solution:

Governing Formula: The Pythagorean Theorem: $x^2 + y^2 = L^2$ (This formula is defined in Related Rates.)
Identify Knowns and Parameters:
- The ladder length $L = 13\text{ ft}$ is a static constant ( $\frac{dL}{dt} = 0$ ).
- Base position: $x(t)$ with $\frac{dx}{dt} = +3\text{ ft/s}$ .
- Vertical contact height: $y(t)$ .
- Snapshot condition: $x = 5\text{ ft}$ .
- Target rate: Find $\frac{dy}{dt}$ .
Substitute Static Constant into Governing Equation: $x^2 + y^2 = 13^2 = 169$
Differentiate implicitly with respect to time ( $t$ ): $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \implies x\frac{dx}{dt} + y\frac{dy}{dt} = 0$
Determine Snapshot Height $y$ via Pythagorean Theorem: $5^2 + y^2 = 169 \implies 25 + y^2 = 169 \implies y^2 = 144 \implies y = 12\text{ ft}$
Substitute Snapshot Values and Solve for $\frac{dy}{dt}$ : $(5)(3) + (12)\frac{dy}{dt} = 0 \implies 15 + 12\frac{dy}{dt} = 0$ $\frac{dy}{dt} = -\frac{15}{12} = -\frac{5}{4} = -1.25\text{ ft/s}$
Engineering Interpretation: At the instant the base is $5\text{ feet}$ from the wall, the top of the ladder is sliding down the wall at a rate of $1.25\text{ ft/s}$ (the negative sign confirms the downward vertical motion).

Worked Example B: Angular Elevation Tracking of an Ascending Rocket

An automated telemetry optical tracking camera is mounted on a flat desert test range at a fixed ground distance of $4000\text{ meters}$ from a vertical rocket launch stand. A rocket lifts off vertically and climbs at an instantaneous vertical velocity of $300\text{ m/s}$ . Determine the rate at which the camera's angle of elevation $\theta$ must rotate to track the rocket when the vehicle reaches an altitude of $3000\text{ meters}$ .

Worked Solution:

Governing Formula: The Tangent Ratio for a Right Triangle: $\tan\theta = \frac{y}{x}$ (This formula and its time derivative principles are defined in Related Rates.)
Identify Knowns and Parameters:
- Ground distance to launch pad $x = 4000\text{ m}$ is a static constant ( $\frac{dx}{dt} = 0$ ).
- Vertical altitude $y(t)$ is a dynamic variable with vertical speed $\frac{dy}{dt} = +300\text{ m/s}$ .
- Snapshot condition: $y = 3000\text{ m}$ .
- Target rate: Find $\frac{d\theta}{dt}$ in $\text{rad/s}$ .
Substitute Static Constant before Differentiation: $\tan\theta = \frac{y}{4000} = \frac{1}{4000} y$
Differentiate implicitly with respect to time ( $t$ ): $\sec^2\theta \frac{d\theta}{dt} = \frac{1}{4000} \frac{dy}{dt} \implies \frac{d\theta}{dt} = \frac{1}{4000 \sec^2\theta} \frac{dy}{dt} = \frac{\cos^2\theta}{4000} \frac{dy}{dt}$
Evaluate Snapshot Trigonometric Values:
- Opposite side: $y = 3000\text{ m}$ .
- Adjacent side: $x = 4000\text{ m}$ .
- Line-of-sight hypotenuse: $z = \sqrt{4000^2 + 3000^2} = \sqrt{16000000 + 9000000} = \sqrt{25000000} = 5000\text{ m}$
- From the right triangle: $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4000}{5000} = \frac{4}{5} = 0.8$ $\sec^2\theta = \frac{1}{\cos^2\theta} = \frac{1}{(4/5)^2} = \frac{25}{16}$
Substitute Values to Compute Angular Slew Velocity $\frac{d\theta}{dt}$ : $\frac{d\theta}{dt} = \frac{(4/5)^2}{4000} (300) = \frac{16/25}{4000} (300) = \frac{16}{25} \cdot \frac{3}{40} = \frac{48}{1000} = 0.048\text{ rad/s}$
Convert to Degrees per Second: $\frac{d\theta}{dt} = 0.048\text{ rad/s} \times \left(\frac{180^\circ}{\pi\text{ rad}}\right) = \frac{8.64^\circ}{\pi} \approx 2.75^\circ/\text{s}$
Engineering Interpretation: The telemetry tracking camera must rotate upward at an instantaneous angular velocity of $0.048\text{ rad/s}$ (approximately $2.75^\circ/\text{s}$ ) to maintain optical track on the rocket.

Practice Problem 4: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) In the Conceptual Framework, differentiate the general Pythagorean relation $x^2 + y^2 = z^2$ . Why does this relation simplify to $x\frac{dx}{dt} + y\frac{dy}{dt} = 0$ in the sliding ladder scenario (Worked Example A), but retain all three terms in the separating ships scenario (Practice Problem 5)?
(b) Deconstruct Worked Example B: How did we evaluate $\sec^2\theta$ at the snapshot moment without ever calculating the angle $\theta$ in degrees or radians? What makes this algebraic method superior for real-time engineering code?
(c) Review the linked Engineering Applications (Radar-Guided Missile Intercept Tracking (Aerospace Engineering)). Why can radar hardware only measure slant range rate $\frac{dz}{dt}$ , and why is a related rates conversion algorithm mandatory to calculate true horizontal target speed $\frac{dx}{dt}$ ?

Solution Key:

(a) Differentiating the general Pythagorean Theorem $x^2 + y^2 = z^2$ with respect to time yields $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt} \implies x\frac{dx}{dt} + y\frac{dy}{dt} = z\frac{dz}{dt}$ . In the sliding ladder scenario, the ladder is a rigid structural member of fixed length $L$ . Because $L$ cannot physically stretch or compress, its time rate of change is identically zero ( $\frac{dL}{dt} = 0$ ), reducing the right side to $L(0) = 0$ . In contrast, when two ships depart from a port along perpendicular courses, the distance between them $z(t)$ increases continuously as they sail, so $\frac{dz}{dt} > 0$ and all three terms must be retained.
(b) In Worked Example B, the right triangle formed by the camera and the rocket has base $4000\text{ m}$ and altitude $3000\text{ m}$ , giving a line-of-sight hypotenuse of $5000\text{ m}$ via the 3-4-5 Pythagorean triple. By trigonometric definition, $\sec\theta = \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{5000}{4000} = \frac{5}{4}$ , which immediately yields $\sec^2\theta = \frac{25}{16}$ or $\cos^2\theta = \frac{16}{25}$ . This algebraic ratio avoids computing $\theta = \arctan(0.75) \approx 36.87^\circ$ and then taking its cosine, avoiding floating-point rounding errors and drastically accelerating execution in embedded guidance microcontrollers.
(c) Radar transceivers rely on radio frequency echoes. The sensor only measures the Doppler carrier compression and the propagation delay of the return pulse, which strictly quantifies radial line-of-sight velocity $\frac{dz}{dt}$ . As detailed in Radar-Guided Missile Intercept Tracking (Aerospace Engineering), the true horizontal velocity is $\frac{dx}{dt} = \frac{z}{x}\frac{dz}{dt}$ . Because the line-of-sight vector is angled relative to the ground, the raw radial velocity is smaller than true ground speed. Without this related rates coordinate transformation, automated tracking algorithms would underestimate target velocity and compute incorrect lead angles.

Show your steps and calculations here.

Practice Problem 5: Separation Rate of Perpendicular Moving Vessels

Two cargo ships leave the Port of Long Beach at the same time. Vessel A steams due east at a constant speed of $24\text{ knots}$ (nautical miles per hour), while Vessel B steams due north at a constant speed of $10\text{ knots}$ . At what rate is the straight-line separation distance between the two vessels changing at the moment Vessel A has traveled $12\text{ nautical miles}$ and Vessel B has traveled $5\text{ nautical miles}$ ?

Solution Key:

Governing Formula: The Pythagorean Theorem for Two-Axis Motion: $a^2 + b^2 = c^2$ (This formula is defined in Related Rates.)
Identify Knowns and Target Derivative:
- Both ships travel away from the origin (port), so their position coordinates are increasing: $\frac{da}{dt} = +24\text{ knots}$ and $\frac{db}{dt} = +10\text{ knots}$ .
- Snapshot distances: $a = 12\text{ nm}$ and $b = 5\text{ nm}$ .
- Target rate: Find $\frac{dc}{dt}$ .
Determine Snapshot Separation Distance $c$ : $c = \sqrt{a^2 + b^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\text{ nm}$
Differentiate implicitly with respect to time ( $t$ ): $2a\frac{da}{dt} + 2b\frac{db}{dt} = 2c\frac{dc}{dt} \implies a\frac{da}{dt} + b\frac{db}{dt} = c\frac{dc}{dt}$
Substitute snapshot values and solve for $\frac{dc}{dt}$ : $(12)(24) + (5)(10) = (13)\frac{dc}{dt}$ $288 + 50 = 13\frac{dc}{dt} \implies 338 = 13\frac{dc}{dt}$ $\frac{dc}{dt} = \frac{338}{13} = 26\text{ knots}$
Engineering Interpretation: At the specified snapshot moment, the straight-line distance between the two cargo ships is expanding at an instantaneous rate of $26\text{ knots}$ ( $26\text{ nautical miles per hour}$ ).

Show your steps and calculations here.

Practice Problem 6: The Leaning Ladder & Angular Displacement Rate

A $25\text{-foot}$ heavy industrial ladder leans against a warehouse wall. The base of the ladder is accidentally struck by a forklift and begins sliding toward the wall. At the instant the base is $12\text{ feet}$ from the wall, the base is moving inward toward the wall at a speed of $1\text{ ft/s}$ .

(a) At what rate is the top of the ladder sliding up the vertical warehouse wall at this moment?
(b) At what rate is the angle $\theta$ between the ladder and the horizontal ground changing at this moment?

Solution Key:

Governing Formulas:
- Vertical motion: Pythagorean Theorem $x^2 + y^2 = 25^2 = 625$ .
- Angular motion: Cosine definition $\cos\theta = \frac{x}{25}$ . (These formulas are defined in Related Rates.)
Snapshot Geometry:
- Given: $x = 12\text{ ft}$ , ladder length $L = 25\text{ ft}$ .
- Because the base moves toward the wall, horizontal distance is shrinking: $\frac{dx}{dt} = -1\text{ ft/s}$ .
- Compute vertical height $y$ : $12^2 + y^2 = 25^2 \implies 144 + y^2 = 625 \implies y^2 = 481 \implies y = \sqrt{481}\text{ ft} \approx 21.93\text{ ft}$
- Compute trigonometric ratios: $\sin\theta = \frac{y}{25} = \frac{\sqrt{481}}{25}, \qquad \cos\theta = \frac{x}{25} = \frac{12}{25}$
Part (a): Vertical Ladder Sliding Velocity ( $\frac{dy}{dt}$ ):
- Differentiate $x^2 + y^2 = 625$ : $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \implies x\frac{dx}{dt} + y\frac{dy}{dt} = 0$
- Solve for $\frac{dy}{dt}$ : $\frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt} = -\frac{12}{\sqrt{481}}(-1) = \frac{12}{\sqrt{481}}\text{ ft/s} \approx 0.547\text{ ft/s}$
- Answer (a): The top of the ladder is sliding up the wall at a rate of $\frac{12}{\sqrt{481}}\text{ ft/s}$ (approx $0.55\text{ ft/s}$ ).
Part (b): Rate of Change of Base Angle ( $\frac{d\theta}{dt}$ ):
- Differentiate $\cos\theta = \frac{x}{25}$ : $-\sin\theta \frac{d\theta}{dt} = \frac{1}{25}\frac{dx}{dt}$
- Substitute $\sin\theta = \frac{\sqrt{481}}{25}$ and $\frac{dx}{dt} = -1$ : $-\left(\frac{\sqrt{481}}{25}\right)\frac{d\theta}{dt} = \frac{1}{25}(-1) \implies \sqrt{481}\frac{d\theta}{dt} = 1$ $\frac{d\theta}{dt} = \frac{1}{\sqrt{481}}\text{ rad/s} \approx 0.0456\text{ rad/s}$
- Answer (b): The angle between the ladder and the ground is increasing at a rate of $\frac{1}{\sqrt{481}}\text{ rad/s}$ (approx $0.046\text{ rad/s}$ or $2.61^\circ/\text{s}$ ).

Show your steps and calculations here.

Regroup 2

Pythagorean Derivative Decomposition: Review why differentiating $x^2 + y^2 = z^2$ produces $x\frac{dx}{dt} + y\frac{dy}{dt} = z\frac{dz}{dt}$ after dividing out the constant factor of 2.
Trig Ratios vs. Angles in Radians: Remind students to construct the snapshot right triangle to evaluate $\cos\theta$ , $\sin\theta$ , and $\sec\theta$ as exact fractions rather than computing intermediate inverse trigonometric angle values.
Unit Protocol for Angular Velocities: Reinforce that all derivatives of trigonometric functions in calculus presuppose radian measure. If an engineering problem requests degrees per second, convert only at the very final numerical step via $\times \frac{180^\circ}{\pi\text{ rad}}$ .

Skill Block 3

Worked Example A: Fluid Level Drop in an Inverted Conical Reservoir

An inverted conical water cistern at an agricultural research station has a top radius of $R = 4\text{ meters}$ and a total vertical depth of $H = 12\text{ meters}$ . A base drainage valve is opened, releasing water at a constant volumetric rate of $2\text{ m}^3/\text{min}$ . Determine the instantaneous rate at which the water depth $h$ is dropping when the water level is $6\text{ meters}$ deep.

Worked Solution:

Governing Formula: The Volume of an Inverted Cone with Similar Triangle Reduction: $V = \frac{1}{3}\pi r^2 h, \qquad \frac{r}{h} = \frac{R}{H}$ (These formulas and reduction principles are defined in Geometric Similarity in Related Rates and Related Rates.)
Establish Geometric Similarity Ratio: The tank dimensions are static constants ( $R = 4\text{ m}$ , $H = 12\text{ m}$ ). By cross-sectional similar triangles: $\frac{r}{h} = \frac{4}{12} = \frac{1}{3} \implies r = \frac{1}{3}h$
Substitute into Volume Formula Before Differentiating: Because the problem asks for $\frac{dh}{dt}$ , eliminate $r$ in favor of $h$ : $V = \frac{1}{3}\pi \left(\frac{1}{3}h\right)^2 h = \frac{1}{3}\pi \left(\frac{1}{9}h^2\right) h = \frac{\pi}{27} h^3$
Differentiate implicitly with respect to time ( $t$ ): $\frac{dV}{dt} = \frac{\pi}{27} \cdot 3h^2 \frac{dh}{dt} = \frac{\pi}{9} h^2 \frac{dh}{dt}$
Substitute Known Inflow/Outflow Rate and Snapshot Depth:
- Water is draining out: $\frac{dV}{dt} = -2\text{ m}^3/\text{min}$ .
- Snapshot depth: $h = 6\text{ m}$ . $-2 = \frac{\pi}{9} (6)^2 \frac{dh}{dt} = \frac{\pi}{9} (36) \frac{dh}{dt} = 4\pi \frac{dh}{dt}$
Solve for $\frac{dh}{dt}$ : $\frac{dh}{dt} = \frac{-2}{4\pi} = -\frac{1}{2\pi}\text{ m/min} \approx -0.159\text{ m/min}$
Engineering Interpretation: When the water is $6\text{ meters}$ deep, the water surface level is falling at a rate of $\frac{1}{2\pi}\text{ m/min}$ (approximately $15.9\text{ cm/min}$ ).

Worked Example B: Dynamic Shadow Length vs. Shadow Tip Motion

A street lamp is mounted at the top of an $18\text{-foot}$ tall light pole. A $6\text{-foot}$ tall engineer walks away from the light pole along a straight, horizontal sidewalk at a steady speed of $4\text{ ft/s}$ .

(a) Determine how fast the length of the engineer's shadow is increasing.
(b) Determine how fast the tip of the shadow is moving along the sidewalk relative to the base of the pole.

Worked Solution:

Governing Formula: Similar Triangles for Ground Shadow Casting: $\frac{h_{\text{person}}}{s} = \frac{H_{\text{pole}}}{x + s}$ (This formula is defined in Geometric Similarity in Related Rates.)
Define Variables & Extract Given Data:
- Let $x(t) =$ distance from light pole base to the person ( $\frac{dx}{dt} = +4\text{ ft/s}$ ).
- Let $s(t) =$ length of the cast shadow on the ground.
- Let $z(t) = x(t) + s(t) =$ total distance from the pole to the tip of the shadow.
- Static parameters: $H_{\text{pole}} = 18\text{ ft}$ , $h_{\text{person}} = 6\text{ ft}$ .
Cross-Multiply and Simplify the Similar Triangle Ratio: $\frac{6}{s} = \frac{18}{x + s} \implies 6(x + s) = 18s \implies 6x + 6s = 18s \implies 6x = 12s \implies s = \frac{1}{2}x$
Part (a): Rate of Change of Shadow Length ( $\frac{ds}{dt}$ ):
- Differentiate $s = \frac{1}{2}x$ implicitly with respect to $t$ : $\frac{ds}{dt} = \frac{1}{2}\frac{dx}{dt}$
- Substitute $\frac{dx}{dt} = 4\text{ ft/s}$ : $\frac{ds}{dt} = \frac{1}{2}(4) = 2\text{ ft/s}$
- Answer (a): The length of the shadow is increasing at a rate of $2\text{ ft/s}$ .
Part (b): Velocity of the Shadow Tip ( $\frac{d(x+s)}{dt}$ ):
- The tip of the shadow coordinate is $x + s$ : $\frac{d}{dt}[x + s] = \frac{dx}{dt} + \frac{ds}{dt} = 4 + 2 = 6\text{ ft/s}$
- Alternatively, express tip coordinate as $x + \frac{1}{2}x = \frac{3}{2}x$ : $\frac{d}{dt}\left[\frac{3}{2}x\right] = \frac{3}{2}\frac{dx}{dt} = \frac{3}{2}(4) = 6\text{ ft/s}$
- Answer (b): The tip of the shadow is moving away from the light pole at a rate of $6\text{ ft/s}$ .

Practice Problem 7: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) In the Conceptual Framework, explain why attempting to differentiate the general cone volume formula $V = \frac{1}{3}\pi r^2 h$ directly using the Product Rule leads to a dead end in a standard draining cone problem. How does cross-sectional geometric similarity resolve this?
(b) Deconstruct Worked Example B: Contrast the rate at which the length of the shadow is stretching ( $\frac{ds}{dt}$ ) with the rate at which the tip of the shadow is traveling ( $\frac{d(x+s)}{dt}$ ). Why is the tip rate always strictly larger than walking speed?
(c) Review the linked Engineering Applications (Hydraulic Reservoir Surge Control (Civil Engineering)). How does the cross-sectional width-to-depth constraint ( $w = h$ ) convert a volumetric surge rate into a vertical water rise rate $\frac{dh}{dt}$ ?

Solution Key:

(a) Applying the Product Rule directly to $V = \frac{1}{3}\pi r^2 h$ yields $\frac{dV}{dt} = \frac{1}{3}\pi(2r\frac{dr}{dt}h + r^2\frac{dh}{dt})$ . In standard fluid drainage scenarios, flowmeters measure volumetric discharge $\frac{dV}{dt}$ , but surface sensors only measure depth $h$ , leaving $\frac{dr}{dt}$ unknown. Because the tank walls are rigid, the cross-sectional geometry forms similar right triangles ( $\frac{r}{h} = \frac{R}{H} \implies r = \frac{R}{H}h$ ). Substituting this ratio into the volume equation before differentiating eliminates $r$ , producing $V(h) = \frac{\pi R^2}{3H^2}h^3$ . Differentiating this single-variable cubic equation yields a relationship containing only the known $\frac{dV}{dt}$ and the requested $\frac{dh}{dt}$ .
(b) As demonstrated in Worked Example B, the shadow length $s$ measures the distance from the person's feet to the head of the shadow. The shadow stretches because the light rays strike the person at an increasingly shallow angle as they walk away. The position of the shadow tip is measured from the fixed origin at the base of the light pole, given by $z = x + s$ . By the sum rule of derivatives, the velocity of the shadow tip is the sum of the person's physical walking velocity ( $\frac{dx}{dt}$ ) and the shadow elongation rate ( $\frac{ds}{dt}$ ). Because both rates are positive, the shadow tip must always move faster than the person.
(c) In Hydraulic Reservoir Surge Control (Civil Engineering), the flood basin is an inverted triangular prism where width equals depth ( $w = h$ ). The cross-sectional area of trapped water is $A = \frac{1}{2}wh = \frac{1}{2}h^2$ . Over a fixed longitudinal length $L = 100\text{ m}$ , the total volume is $V = A \cdot L = \frac{1}{2}h^2(100) = 50h^2$ . Differentiating with respect to time yields $\frac{dV}{dt} = 100h \frac{dh}{dt}$ . This geometric constraint establishes that as water depth increases, each incremental cubic meter of stormwater produces a smaller rise in water height because the top surface width is wider.

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Practice Problem 8: Conical Granular Heap Inflow Rate

A pneumatic industrial conveyor unloads dry grain into a conical stockpile. As grain pours onto the apex, the heap maintains a geometry where the vertical height is always three times the base radius ( $h = 3r$ ). If grain is deposited at a constant rate of $20\text{ cm}^3/\text{s}$ , determine how fast the radius of the base is expanding at the exact instant the radius is $4\text{ centimeters}$ .

Solution Key:

Governing Formula: The Volume of a Cone Formula: $V = \frac{1}{3}\pi r^2 h$ (This formula is defined in Geometric Similarity in Related Rates.)
Apply Geometric Constraint Before Differentiating: The pile geometry enforces $h = 3r$ . Because we want to solve for $\frac{dr}{dt}$ , substitute $h = 3r$ directly into $V$ : $V = \frac{1}{3}\pi r^2 (3r) = \pi r^3$
Differentiate implicitly with respect to time ( $t$ ): $\frac{dV}{dt} = \frac{d}{dt}[\pi r^3] = 3\pi r^2 \frac{dr}{dt}$
Substitute Known Inflow Rate and Snapshot Radius:
- Grain is being added: $\frac{dV}{dt} = +20\text{ cm}^3/\text{s}$ .
- Snapshot radius: $r = 4\text{ cm}$ . $20 = 3\pi (4)^2 \frac{dr}{dt} = 3\pi (16) \frac{dr}{dt} = 48\pi \frac{dr}{dt}$
Solve for $\frac{dr}{dt}$ : $\frac{dr}{dt} = \frac{20}{48\pi} = \frac{5}{12\pi}\text{ cm/s} \approx 0.1326\text{ cm/s}$
Engineering Interpretation: At the instant the radius reaches $4\text{ centimeters}$ , the base of the grain heap is expanding outward at an instantaneous rate of $\frac{5}{12\pi}\text{ cm/s}$ (approximately $1.33\text{ mm/s}$ ).

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Practice Problem 9: Spotlight Shadow Projection on a Wall

A powerful stage spotlight is placed on the floor $30\text{ feet}$ away from a vertical backdrop wall. A $6\text{-foot}$ tall performer walks directly away from the spotlight toward the wall at a steady speed of $3\text{ ft/s}$ . Determine the instantaneous rate of change of the height of the performer's shadow cast on the wall when the performer is $10\text{ feet}$ from the wall.

Solution Key:

Governing Formula: Similar Triangles for Projected Shadows: $\frac{h_{\text{performer}}}{x} = \frac{y_{\text{shadow}}}{D_{\text{wall}}}$ (This formula is defined in Geometric Similarity in Related Rates.)
Identify Knowns and Coordinates:
- Distance from spotlight to wall is fixed: $D_{\text{wall}} = 30\text{ ft}$ (static constant).
- Performer height is fixed: $h_{\text{performer}} = 6\text{ ft}$ (static constant).
- Let $x(t) =$ distance from spotlight to the performer.
- The performer moves away from the spotlight: $\frac{dx}{dt} = +3\text{ ft/s}$ .
- Snapshot condition: Performer is $10\text{ ft}$ from the wall $\implies x = 30 - 10 = 20\text{ ft}$ .
- Let $y(t) =$ height of the shadow on the wall. Find $\frac{dy}{dt}$ .
Formulate the Primary Equation: $\frac{6}{x} = \frac{y}{30} \implies y = \frac{180}{x} = 180x^{-1}$
Differentiate implicitly with respect to time ( $t$ ): $\frac{dy}{dt} = \frac{d}{dt}[180x^{-1}] = -180x^{-2} \frac{dx}{dt} = -\frac{180}{x^2}\frac{dx}{dt}$
Substitute Snapshot Values ( $x = 20\text{ ft}$ , $\frac{dx}{dt} = 3\text{ ft/s}$ ): $\frac{dy}{dt} = -\frac{180}{(20)^2} (3) = -\frac{180}{400}(3) = -\frac{9}{20}(3) = -\frac{27}{20}\text{ ft/s} = -1.35\text{ ft/s}$
Engineering Interpretation: When the performer is $10\text{ feet}$ from the wall, the height of the cast shadow on the wall is decreasing at a rate of $1.35\text{ ft/s}$ .

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Practice Problem 10: Multivariable Thermal Expansion of a Solar Array

A large industrial photovoltaic solar panel array installed at the UNLV solar research park undergoes midday thermal expansion during peak desert summer heat. The rectangular panel has length $L = 50.0\text{ meters}$ and width $W = 30.0\text{ meters}$ . Precision optical strain gauges indicate that the length is expanding at a rate of $\frac{dL}{dt} = 0.04\text{ m/hr}$ and the width is expanding at a rate of $\frac{dW}{dt} = 0.02\text{ m/hr}$ . Determine the instantaneous rate at which the surface area of the solar array is increasing.

Solution Key:

Governing Formula: The Multivariable Product Rule for Rectangular Area: $A = L \cdot W$ (This rule is defined in Product Rule and Related Rates.)
Identify Knowns and Snapshot Values:
- Both dimensions vary simultaneously over time: $L(t)$ and $W(t)$ .
- Given expansion rates: $\frac{dL}{dt} = +0.04\text{ m/hr}$ and $\frac{dW}{dt} = +0.02\text{ m/hr}$ .
- Snapshot dimensions: $L = 50.0\text{ m}$ and $W = 30.0\text{ m}$ .
- Target rate: Find $\frac{dA}{dt}$ .
Differentiate implicitly with respect to time ( $t$ ) via Product Rule: $\frac{dA}{dt} = \frac{d}{dt}[L \cdot W] = \frac{dL}{dt} W + L \frac{dW}{dt}$
Substitute Snapshot Values and Calculate: $\frac{dA}{dt} = (0.04)(30.0) + (50.0)(0.02) = 1.20 + 1.00 = 2.20\text{ m}^2/\text{hr}$
Engineering Interpretation: Under midday solar heating, the total collector area of the photovoltaic array is expanding at an instantaneous rate of $2.20\text{ m}^2/\text{hr}$ . Mechanical engineers use this rate to design flexible expansion joints between panel modules to prevent thermal buckling and cell fracturing.

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Regroup 3

Pre-Derivative Variable Elimination: Re-emphasize that geometric similarity (such as $r = \frac{R}{H}h$ ) must always be substituted into formulas before taking the time derivative. Doing so reduces multi-variable chaos to an elementary single-variable power rule problem.
Shadow Geometry Pitfalls: Review the distinction between shadow length $s$ and shadow tip distance $x + s$ . Emphasize drawing clear geometric diagrams with distinct variable labels for every segment.
Coupled Rates via Product Rule: Remind students that when two dimensions of a 2D or 3D object both change independently (like length and width in Practice Problem 10), the Product Rule is mandatory ( $\frac{d}{dt}[xy] = \frac{dx}{dt}y + x\frac{dy}{dt}$ ).

Synthesis Wrap-up

The Universal Related Rates Blueprint:
1. Sketch the physical system and assign clear variables to all moving parts ( $x, y, r, h, \theta$ ).
2. Substitute static constants immediately; leave dynamic variables strictly symbolic.
3. Formulate the primary geometric constraint; use similar triangles to eliminate extra dynamic variables if needed.
4. Apply $\frac{d}{dt}$ to both sides, chaining every dynamic variable with its rate tail ( $\frac{du}{dt}$ ).
5. Substitute snapshot numbers and known rates; solve for the target rate and attach verified dimensional units.
Homework Assignment: Complete the Knewton Alta assignment: Related Rates. Practice sketching clean right triangles and identifying whether problem hypotenuses are static constants or dynamic variables.