Study Notes on Related Rates

Related Rates in Calculus

Introduction to Related Rates

  • Related rates are a real-world application of calculus, particularly in contexts requiring implicit differentiation.

  • They measure how one quantity changes in relation to another, often in scenarios where two variables are continuously changing.

Example 1: Inflating a Balloon

  • Scenario: A balloon is inflated, taking the shape of a perfect sphere.

    • The balloon's volume and radius are both increasing as it expands.

    • Rates of increase of these quantities are interconnected, justifying the term "related rates."

  • Measurement Context:

    • Volume Rate: Measured increase of volume, denoted as dV/dt = 100 cubic centimeters per second.

    • Radius Rate: Desired value to determine is dr/dt when the diameter of the balloon is 50 cm (thus, radius r = 25 cm).

  • Key Formula: Volume of a sphere is given by:

    • Formula: V = \frac{4}{3} \pi r^3.

  • Differentiation Steps:

    1. Differentiate the volume formula with respect to time t:

    • \frac{dV}{dt} = \frac{d}{dt}(\frac{4}{3} \pi r^3).

    1. Apply the chain rule:

    • \frac{dV}{dt} = 4 \pi r^2 \cdot \frac{dr}{dt}.

    1. Rearranging for dr/dt gives:

    • \frac{dr}{dt} = \frac{dV}{dt} \cdot \frac{1}{4 \pi r^2}.

  • Substituting Values:

    • Use known values:

    • dV/dt = 100,

    • r = 25.

    • Evaluation:

    • Plugging in:

      • \frac{dr}{dt} = \frac{100}{4 \pi (25^2)}.

    • Result:

      • \frac{dr}{dt} = \frac{1}{25\pi} centimeters per second.

Example 2: Ladder Against a Wall

  • Scenario: A ladder of length 10 feet slides away from the wall.

    • dx/dt = 1 foot per second (the rate at which the base of the ladder is moving away from the wall).

  • Objective: Find dy/dt (the rate at which the top of the ladder slides down the wall) when the bottom is 6 feet from the wall.

  • Diagram Representation:

    • Right triangle where:

    • Base (distance from the wall) is x,

    • Height (vertical distance from the ground to the top of the ladder) is y,

    • Hypotenuse is the ladder at a constant length of 10 feet.

  • Equation Formation:

    • Using the Pythagorean theorem:

    • x^2 + y^2 = 10^2.

    • Differentiate the equation with respect to time:

    • \frac{d}{dt}(x^2 + y^2) = \frac{d}{dt}(100).

  • Differentiation Steps:

    • Applying derivatives gives:

    • 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0.

  • Rearranging for dy/dt:

    • Rearranging yields:

    • \frac{dy}{dt} = -\frac{x}{y} \cdot \frac{dx}{dt}.

  • Substituting Known Values:

    • When x = 6, use the Pythagorean theorem to find y:

    • 6^2 + y^2 = 10^2,

    • Solving gives y = 8.

  • Final Calculation Steps:

    • Plug in the values:

    • \frac{dy}{dt} = -\frac{6}{8} \cdot 1,

      • Simplifying gives:

      • \frac{dy}{dt} = -\frac{3}{4} feet per second (indicating that y is decreasing as the ladder slides).

Conclusion on Related Rates Problems

  • Related rates problems hinge on understanding the relationship between changing quantities as functions of time.

  • Essential steps:

    • Draw a Diagram: Visualize the changes and relationships.

    • Write Equations: Relate the quantities mathematically.

    • Differentiate: Calculate derivatives concerning time, often requiring the chain rule.

  • The ultimate goal is to find d ext{something}/dt for the desired quantities, leading to practical solutions in scenarios like inflating balloons or moving ladders.

Comprehension Check

  • Engaging in related rates problems involves calculating derivatives of multi-variable functions and understanding temporal changes in physical quantities. The ability to visualize these relationships is crucial.