Calculus II: Related Rates and Variable Differentiation

Introduction to Related Variables

• Conceptual Overview: In calculus, related rates problems involve variables that change with respect to time ($t$).

• Interdependence of Rates: If two or more variables are related to each other by an equation, their rates of change with respect to time are also mathematically related.

• Fundamental Proportion (Linear Case):

  • Suppose two variables are related by the equation $y = 2x$.

  • If both variables are changing with respect to time, the value of $y$ remains twice the value of $x$ at any given instance.

  • Consequently, the rate of change of $y$ with respect to time ($\frac{dy}{dt}$) is consistently twice the rate of change of $x$ with respect to time ($\frac{dx}{dt}$).

Procedural Demonstration: Example 1

• Problem Statement: The variables $x$ and $y$ are differentiable functions of $t$ and are related by the equation:

  • $y = x^2 + 3$

• Given Conditions:

  • When $x = 1$, the rate of change of $x$ is $\frac{dx}{dt} = 2$.

• Objective: Calculate $\frac{dy}{dt}$ when $x = 1$.

• Step-by-Step Solution:

  1. Differentiate with Respect to Time: Use the Chain Rule to differentiate both sides of the equation with respect to $t$.

     • $\frac{d}{dt}[y] = \frac{d}{dt}[x^2 + 3]$

     • $\frac{dy}{dt} = 2x \frac{dx}{dt}$

  2. Substitution: Insert the known values (variables and their rates) into the derived equation.

     • When $x = 1$ and $\frac{dx}{dt} = 2$:

     • $\frac{dy}{dt} = 2(1)(2)$

  3. Final Result:

     • $\frac{dy}{dt} = 4$

Case Study: Changing Area in Circular Ripples (Example 2)

• The Scenario: A pebble is dropped into a calm pool of water. This action causes ripples in the form of concentric circles.

• Physical Properties: As the outer radius of the ripple increases, the total area of the disturbed water increases.

• Quantitative Data:

  • The radius ($r$) of the outer ripple increases at a constant rate of $1\,\text{foot per second}$.

  • This is expressed as $\frac{dr}{dt} = 1$.

• Problem Objective: Determine the rate of change of the total area ($\frac{dA}{dt}$) specifically at the moment the radius $r$ is $4\,\text{feet}$.

• Mathematical Modeling:

  • Original Equation: The area of a circle is defined as $A = \pi r^2$.

  • Differentiation: Differentiate both sides with respect to $t$ using the Chain Rule:

    • $\frac{d}{dt}[A] = \frac{d}{dt}[\pi r^2]$

    • $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$

• Calculation:

  • Plug in $r = 4$ and $\frac{dr}{dt} = 1$:

  • $\frac{dA}{dt} = 2\pi(4)(1)$

  • $\frac{dA}{dt} = 8\pi$

• Conclusion: When the radius is $4\,\text{feet}$, the area is changing at a rate of $8\pi\,\text{square feet per second}$.

• Observation on Constancy: In this example, the radius changes at a constant rate ($\frac{dr}{dt} = 1$ for all $t$), however, the area changes at a nonconstant rate because it depends on the value of $r$ at a specific moment.

Methodology for Solving Related-Rate Problems

To solve tasks involving related variables, the following guidelines should be followed:

1. Identification: Identify all given quantities and all quantities that need to be determined. If possible, create a visual sketch and label the known and unknown quantities.

2. Equation Building: Write an equation that mathematically relates all variables whose rates of change are either provided or need to be found. (e.g., using physical formulas or geometric relationships).

3. Differentiation: Utilize the Chain Rule to differentiate both sides of the equation with respect to time ($t$).

4. Substitution and Resolution: After differentiation, substitute all known values for variables and their rates of change into the resulting equation. Finally, solve for the required rate of change.

Reference Formulas and Mathematical Models

• Geometric Formulas: Related rates problems frequently rely on standard volume and area formulas.

  • Volume of a Sphere: $V = \frac{4}{3}\pi r^3$, where $r$ is the radius.

• Mathematical Models for Rates: Tables in reference materials summarize common rates of change models to assist in Step 1 of the solving process.

Business Application: Increasing Production and Revenue (Example 4)

• Scenario Context: A company is increasing production. The weekly production level is denoted by $x$.

• Market Data:

  • Production Growth Rate: $\frac{dx}{dt} = 200\,\text{units per week}$.

  • Demand Function: $p = 100 - 0.001x$, where $p$ is the price per unit in dollars.

• Objective: Find the rate of change of revenue ($R$) with respect to time ($t$) when the weekly production level reaches $x = 2000\,\text{units}$. Determine if this rate exceeds $20,000\,\text{dollars per week}$.

• Formulating the Equation:

  • Revenue ($R$) is the product of price ($p$) and demand ($x$).

  • $R = xp$

  • $R = x(100 - 0.001x) = 100x - 0.001x^2$

• Applying the Chain Rule:

  • Differentiate the revenue equation with respect to $t$:

  • $\frac{dR}{dt} = (100 - 0.002x) \frac{dx}{dt}$

• Calculations for Specific Production Level:

  • Substitute $x = 2000$ and $\frac{dx}{dt} = 200$:

  • $\frac{dR}{dt} = (100 - 0.002(2000)) \times 200$

  • $\frac{dR}{dt} = (100 - 4) \times 200$

  • $\frac{dR}{dt} = 96 \times 200 = 19,200$

• Verdict: The rate of change of the revenue is $19,200\,\text{dollars per week}$. This is not greater than $20,000\,\text{dollars per week}$.