Comprehensive Study Notes on Related Rates

Fundamentals of Related Rates

• Core Definition: Related rates is a calculus technique used to analyze how variables in a formula change with respect to time ($t$). It relates the known rates of change of certain quantities to derive an unknown rate of change of another quantity at a specific instant.

• Primary Objective: To find the instantaneous rate of change of a geometric, physical, or economic quantity at a precise moment (e.g., determining the exact expansion velocity of a volume at $t = 3_\text{seconds}$).

• Implicit Functions of Time: Variables such as volume ($V$), radius ($r$), height ($h$), area ($A$), distance ($x$, $y$), or angle ($\theta$) are treated as implicit functions of time ($t$).

• Implicit Differentiation Requirement: Because these quantities are functions of $t$, taking the derivative of an equation containing these variables with respect to $t$ requires applying the chain rule. Differentiating any variable $v$ with respect to $t$ produces an accompanying differential term $\frac{dv}{dt}$.

General Four-Step Procedure for Related Rates Problems

• Step 1: Define Variables and Constants: Assign distinct symbols to all time-varying quantities and identify all constant values present in the system (e.g., assign $t$ for time, $A$ for area, $r$ for radius, $h$ for height, $\theta$ for angle of elevation).

• Step 2: Formulate the Primary Relational Equation: Identify or derive a geometric, physical, or trigonometric formula that links all non-time variables together.

• Step 3: Implicitly Differentiate with Respect to Time ($t$): Apply the differential operator $\frac{d}{dt}$ to both sides of the relational formula. Use differentiation rules (product rule, quotient rule, chain rule, power rule) as required, ensuring that every differentiated variable term generates a corresponding $\frac{d[\text{variable}]}{dt}$ differential factor.

• Step 4: Substitute Known Instantaneous Values and Solve: Plug in all given static values (distances, heights, radii) and known rates of change corresponding to the specific moment in time into the differentiated equation. Solve algebraically for the targeted unknown rate of change.

Mathematical Derivation: Volume Change in a Conical Reservoir

• Geometric Background and Formulas:

  • Area of a circle:         $A = \pi r^2$

  • Volume of a cylinder (a stack of circular cross-sections across height $h$):         $V = \pi r^2 h$

  • Volume of a cone (one-third the volume of a cylinder with equivalent base and height):         $V = \frac{1}{3}\pi r^2 h$         $V = \frac{\pi}{3} r^2 h$

• Dynamic System Model:

  • Consider a conical reservoir positioned under a water faucet. Water enters to a certain height $h$ and radius $r$. If a hole is opened at the apex/bottom of the cone, water drains out.

  • As water leaves the cone, the water level drops, causing both the liquid's height $h$ and upper radius $r$ to decrease continuously over time $t$.

  • Because $V$, $r$, and $h$ vary dynamically with time, all three are explicit functions of time: $V(t)$, $r(t)$, and $h(t)$.

• Implicit Differentiation wrt Time ($t$):

  • Apply $\frac{d}{dt}$ to both sides of the conical volume formula:         $\frac{d}{dt}(V) = \frac{d}{dt}\left(\frac{\pi}{3} r^2 h\right)$

  • The left side yields $\frac{dV}{dt}$, representing the instantaneous rate of change of volume with respect to time.

  • The right side requires the product rule between $r^2$ and $h$, treating $\frac{\pi}{3}$ as a constant factor:         $\frac{dV}{dt} = \frac{\pi}{3} \left( \left(\frac{d}{dt}(r^2)\right) h + r^2 \left(\frac{d}{dt}(h)\right) \right)$

  • Applying the chain rule yields:         $\frac{d}{dt}(r^2) = 2r \frac{dr}{dt}$         $\frac{d}{dt}(h) = \frac{dh}{dt}$

  • Combining these results yields the complete expanded equation:         $\frac{dV}{dt} = \frac{\pi}{3} \left( 2r h \frac{dr}{dt} + r^2 \frac{dh}{dt} \right)$         $\frac{dV}{dt} = \frac{2\pi}{3} r h \frac{dr}{dt} + \frac{\pi}{3} r^2 \frac{dh}{dt}$

• Necessary Data for Instantaneous Evaluation:     To evaluate $\frac{dV}{dt}$ at a specific instant, four independent pieces of information must be known:

  1. Instantaneous radius ($r$)

  2. Instantaneous height ($h$)

  3. Instantaneous rate of change of radius ($\frac{dr}{dt}$)

  4. Instantaneous rate of change of height ($\frac{dh}{dt}$)

Basic Algebraic Related Rates Examples

• Implicit Differentiation of $y = x^3$:

  • Given $y = x^3$, where both $x$ and $y$ are dynamic functions of time $t$.

  • Differentiate both sides with respect to $t$:         $\frac{d}{dt}(y) = \frac{d}{dt}(x^3)$         $\frac{dy}{dt} = 3x^2 \frac{dx}{dt}$

• Evaluation at Specific Instant ($t = 1$):

  • Given conditions at $t = 1$: $x = 2$ and $\frac{dx}{dt} = 4_\text{units/s}$.

  • Note: Quantitative parameters ($x$ and $\frac{dx}{dt}$) are context-dependent and change over time; values given for $t = 1$ apply strictly to that moment.

  • Substitute values into the differentiated formula:         $\frac{dy}{dt} = 3(2)^2 \times 4$         $\frac{dy}{dt} = 3(4) \times 4$         $\frac{dy}{dt} = 48_\text{units/s}$

  • Interpretation: At time $t = 1$, $y$ increases at an instantaneous rate of $48_\text{units}$ per unit of time.

Geometric Application: Circular Oil Spill Expansion

• Problem Statement: An oil tanker spill (modeled on events such as the Exxon Valdez incident) creates a thin circular slick on the surface of the water. The radius of the circular slick expands at a constant rate of $3_\text{ft/s}$. Determine the rate at which the surface area of the spill is increasing at the exact instant when the radius reaches $30_\text{ft}$.

• Identified Variables and Given Rates:

  • Time: $t$

  • Radius: $r = 30_\text{ft}$

  • Area: $A$

  • Rate of radius expansion: $\frac{dr}{dt} = 3_\text{ft/s}$

• Relating Formula:

  • Area of a circle:         $A = \pi r^2$

• Implicit Differentiation:

  • Differentiate with respect to $t$:         $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$

• Computation:

  • Substitute $r = 30_\text{ft}$ and $\frac{dr}{dt} = 3_\text{ft/s}$:         $\frac{dA}{dt} = 2\pi (30_\text{ft}) \times 3_\text{ft/s}$         $\frac{dA}{dt} = 180\pi_\text{ft}^2/\text{s}$

• Dimensional Verification & Numerical Result:

  • Units match surface area growth: $\text{ft} \times \text{ft/s} = \text{ft}^2/\text{s}$.

  • Exact rate of area expansion: $180\pi_\text{ft}^2/\text{s}$.

  • Decimal approximation: $180 \times 3.14159265... \approx 565.49_\text{ft}^2/\text{s}$.

Trigonometric Application: Tracking Camera Angle for a Rocket Launch

• Problem Setup: A rocket ascends vertically from a launchpad. An automated television tracking camera is situated at ground level at a fixed distance of $3000_\text{ft}$ from the launch point.

• Target Condition: When the rocket reaches an altitude of $h = 4000_\text{ft}$, its climbing speed is $\frac{dh}{dt} = 600_\text{ft/s}$. Calculate the required angular velocity $\frac{d\theta}{dt}$ of the tracking camera at that moment to keep the rocket centered in the frame.

• Assigned Variables:

  • Time: $t$

  • Rocket Height / Altitude: $h$

  • Camera Angle of Elevation: $\theta$

  • Horizontal Camera Distance (constant): $3000_\text{ft}$

• Formula Selection:

  • The Pythagorean theorem ($a^2 + b^2 = c^2$) relates side lengths but omits the camera angle $\theta$.

  • The tangent trigonometric ratio connects angle $\theta$, opposite side $h$, and adjacent side $3000_\text{ft}$:         $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{3000}$         $\tan(\theta) = \frac{1}{3000}h$

• Implicit Differentiation:

  • Differentiate both sides with respect to $t$:         $\frac{d}{dt}(\tan(\theta)) = \frac{d}{dt}\left(\frac{1}{3000}h\right)$         $\sec^2(\theta) \frac{d\theta}{dt} = \frac{1}{3000} \frac{dh}{dt}$

• Geometric Evaluation at Instant $h = 4000_\text{ft}$:

  • The camera, launch site, and rocket form a right triangle with adjacent side $3000_\text{ft}$ and opposite side $4000_\text{ft}$.

  • Calculate hypotenuse $c$ using the Pythagorean theorem:         $c = \sqrt{3000^2 + 4000^2} = \sqrt{9000000 + 16000000} = \sqrt{25000000} = 5000_\text{ft}$

  • Determine $\cos(\theta)$:         $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{3000}{5000} = \frac{3}{5}$

  • Determine $\sec(\theta)$ (reciprocal of cosine):         $\sec(\theta) = \frac{1}{\cos(\theta)} = \frac{5}{3}$

  • Determine $\sec^2(\theta)$:         $\sec^2(\theta) = \left(\frac{5}{3}\right)^2 = \frac{25}{9}$

• Substitution and Algebraic Solution:

  • Substitute $\sec^2(\theta) = \frac{25}{9}$ and $\frac{dh}{dt} = 600_\text{ft/s}$ into the differentiated formula:         $\frac{25}{9} \frac{d\theta}{dt} = \frac{1}{3000} (600)$         $\frac{25}{9} \frac{d\theta}{dt} = \frac{1}{5}$

  • Isolate $\frac{d\theta}{dt}$ by multiplying both sides by $\frac{9}{25}$:         $\frac{d\theta}{dt} = \frac{1}{5} \times \frac{9}{25}$         $\frac{d\theta}{dt} = \frac{9}{125}_\text{rad/s} = 0.072_\text{rad/s}$

• Conversion to Degrees per Second:

  • Convert radians per second to degrees per second using the conversion factor $\frac{180^\circ}{\pi}$:         $\frac{d\theta}{dt} = 0.072 \times \frac{180^\circ}{\pi}$         $\frac{d\theta}{dt} = \frac{12.96}{\pi} \approx 4.1253^\circ/\text{s} \approx 4.13^\circ/\text{s}$

• Final Interpretation: At the moment the rocket reaches an altitude of $4000_\text{ft}$, the tracking camera must tilt upwards at an instantaneous angular velocity of $0.072_\text{rad/s}$ (approximately $4.13^\circ/\text{s}$) to keep the rocket centered in view.