Calculus Notes: Fundamentals and Applications of Related Rates

Theoretical Foundation of Related Rates

• Definition: Related rates problems involve finding the rate of change of one quantity by relating it to other quantities whose rates of change are known.
• Mathematical Formulation:
  • Consider a system governed by an implicit relation among multiple time-dependent variables $x(t)$, $y(t)$, and $z(t)$, expressed generally as:     $f(x(t), y(t), z(t)) = C$     where $C$ is a constant.
  • Differentiating both sides of the relation with respect to time $t$ using the Chain Rule yields:     $\frac{d}{dt}\left(f(x(t), y(t), z(t))\right) = 0$
  • The resulting differentiated equation expresses a relationship involving the positions/states $(x, y, z)$ and their first-order time derivatives (rates of change) $\left(x' = \frac{dx}{dt}, y' = \frac{dy}{dt}, z' = \frac{dz}{dt}\right)$.
• Primary Objective: Infer an unknown rate of change (e.g., $x'$) at a specific instant by substituting known values of the variables ($x, y, z$) and the remaining known rates ($y', z'$) into the differentiated equation.

Problem-Solving Methodology

• Step 1: Translation of Language to Mathematics:
  • Word problems present scenarios in narrative or descriptive language rather than pre-packaged equations.
  • Identify all time-varying quantities and assign explicit variable names (e.g., height $h$, radius $r$, volume $V$, distances $x, y, z$).
  • Identify the given rates of change as time derivatives and express the target unknown as a differential quotient (e.g., $\frac{dV}{dt}$, $\frac{dr}{dt}$, or $\frac{dz}{dt}$).
• Step 2: Establishing Geometric or Physical Constraints:
  • Formulate a static equation $f = 0$ that relates the spatial or physical variables based on geometric formulas (e.g., volume of a cone, volume of a sphere, Pythagorean theorem).
  • Reduce variables if possible using proportional or geometric constraints defined in the problem description.
• Step 3: Implicit Differentiation:
  • Differentiate the simplified relational equation implicitly with respect to time $t$.
• Step 4: Evaluation at the Instant in Question:
  • Substitute the instantaneous numerical values of the variables and known rates into the differentiated equation.
  • Solve algebraically for the target rate.

Example 1: Conical Sand Pile

• Physical Scenario:
  • Sand leaks out of a container and falls onto a flat surface, accumulating gradually into a conical pile.
  • Let $h$ denote the height of the conical pile.
  • Let $r$ denote the radius of the circular base of the conical pile.
  • Let $V$ denote the volume of the sand pile.
• Physical Identity:
  • The rate at which sand leaves the container and adds to the conical pile is equal to the rate of change of the volume of the pile ($\frac{dV}{dt}$).
• Given Conditions:
  • The height $h$ of the pile is increasing at a rate of $\frac{dh}{dt} = 3\,\text{inches per minute}$.
  • The height is constrained to always equal twice the radius of the base: $h = 2r$.
  • Target instantaneous condition: Evaluate the rate at which sand is leaking out ($\frac{dV}{dt}$) when the height $h = 4\,\text{inches}$.
• Derivation & Solution:
  • General volume formula for a right circular cone:     $V = \frac{1}{3}\pi r^2 h$
  • Express radius $r$ in terms of height $h$ using the constraint $h = 2r \implies r = \frac{h}{2}$:     $V = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h$$V = \frac{1}{3}\pi \left(\frac{h^2}{4}\right) h$$V = \frac{1}{12}\pi h^3$
  • Differentiate $V$ implicitly with respect to time $t$:     $\frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \frac{dh}{dt}$$\frac{dV}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt}$
  • Substitute the instantaneous values $h = 4\,\text{inches}$ and $\frac{dh}{dt} = 3\,\text{inches per minute}$:     $\frac{dV}{dt} = \frac{\pi}{4} (4)^2 (3)$$\frac{dV}{dt} = \frac{\pi}{4} (16) (3)$$\frac{dV}{dt} = 12\pi\,\text{cubic inches per minute}$

Example 2: Inflatable Spherical Balloon

• Physical Scenario:
  • Gas is being pumped into a spherical balloon at a constant volumetric rate.
• Given Conditions:
  • Volumetric rate of increase: $\frac{dV}{dt} = 20\,\text{ft}^3/\text{min}$.
  • Radius variable denoted by $r$.
  • Target instantaneous condition: Find how fast the radius is increasing ($\frac{dr}{dt}$) when the radius $r = 1\,\text{ft}$.
• Derivation & Solution:
  • Volume formula for a sphere:     $V = \frac{4}{3}\pi r^3$
  • Differentiate implicitly with respect to time $t$:     $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$
  • Substitute $\frac{dV}{dt} = 20\,\text{ft}^3/\text{min}$ and $r = 1\,\text{ft}$ into the equation:     $20 = 4\pi (1)^2 \frac{dr}{dt}$$20 = 4\pi \frac{dr}{dt}$$\frac{dr}{dt} = \frac{20}{4\pi}$$\frac{dr}{dt} = \frac{5}{\pi}\,\text{ft/min}$

Example 3: Two Bicyclists Moving Perpendicularly

• Physical Scenario:
  • Two bicyclists start from the same origin point.
  • Rider 1 travels due East; let $x$ denote Rider 1's distance from the starting point.
  • Rider 2 travels due North; let $y$ denote Rider 2's distance from the starting point.
  • Let $z$ denote the straight-line distance between Rider 1 and Rider 2.
• Given Conditions:
  • Speed of Rider 1 (Eastward): $\frac{dx}{dt} = 16\,\text{mph}$.
  • Speed of Rider 2 (Northward): $\frac{dy}{dt} = 12\,\text{mph}$.
  • At the evaluation moment, Rider 1 has traveled a total distance of $x = 4\,\text{miles}$.
  • Target: Find the rate of change of the distance between the two riders ($\frac{dz}{dt}$) at this precise moment.
• Derivation & Solution:
  • Geometric constraint (Pythagorean Theorem for a right triangle):     $x^2 + y^2 = z^2$
  • Determine total elapsed time $t$ when $x = 4\,\text{miles}$:     $t = \frac{x}{\frac{dx}{dt}} = \frac{4\,\text{miles}}{16\,\text{mph}} = \frac{1}{4}\,\text{hour}$
  • Calculate Rider 2's position $y$ at $t = \frac{1}{4}\,\text{hour}$:     $y = \left(\frac{dy}{dt}\right) \cdot t = 12\,\text{mph} \times \frac{1}{4}\,\text{hour} = 3\,\text{miles}$
  • Calculate the hypotenuse distance $z$ at this moment:     $z = \sqrt{x^2 + y^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5\,\text{miles}$
  • Differentiate $x^2 + y^2 = z^2$ implicitly with respect to time $t$:     $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 2z \frac{dz}{dt}$$x \frac{dx}{dt} + y \frac{dy}{dt} = z \frac{dz}{dt}$
  • Substitute $x = 4$, $\frac{dx}{dt} = 16$, $y = 3$, $\frac{dy}{dt} = 12$, and $z = 5$ into the differentiated equation:     $(4)(16) + (3)(12) = 5 \frac{dz}{dt}$$64 + 36 = 5 \frac{dz}{dt}$$100 = 5 \frac{dz}{dt}$$\frac{dz}{dt} = 20\,\text{mph}$