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.