In-Depth Notes on Related Rates

Related Rates

Learning Objectives

  • 4.1.1 Express changing quantities in terms of derivatives.
  • 4.1.2 Find relationships among the derivatives in a given problem.
  • 4.1.3 Use the chain rule to find the rate of change of one quantity that depends on the rate of change of other quantities.

Key Concepts

  • Associated rates of change can be derived using derivatives, especially in real-world applications where multiple quantities change over time.
  • Understanding the relationship between different rates of change is crucial in related rates problems.

Related Rates in Real-World Applications

Example: Inflating a Balloon

  • The volume of a spherical balloon can be expressed as:

    • V = (4/3)πr³ (Volume of a sphere given radius)

  • Rates of Change:

    • When inflating a balloon at a constant rate of dV/dt = 2 cm³/sec, find dr/dt when r = 3 cm.

  • Differentiation:

    • Differentiate both sides:

    (dV/dt = 4πr² * dr/dt)

    • Substitute known values:
      (2 = 4π(3²) * dr/dt)
    • Solving gives:
    • (dr/dt = 1/18π cm/sec)

Problem-Solving Strategy for Related Rates

  1. Assign Variables: Define all relevant variables.
  2. Draw a Figure: Visual representation helps clarify relationships.
  3. State Information: Write down given values and what rate needs to be found.
  4. Find Relationships: Establish equations connecting the variables.
  5. Differentiate: Apply the chain rule to relate rates of change.
  6. Solve: Substitute in known values and solve for the unknown.

Additional Examples

Example: An Airplane Flying at Constant Elevation

  • Known Values:
    • Plane's altitude: 4000 ft
    • Distance of observer from tower: 3000 ft
    • Plane's speed: 600 ft/sec
  • Use Pythagorean theorem:
    • x² + 4000² = s² (where x = horizontal distance, s = line-of-sight distance)
    • Differentiate to find the relationship between the rates:
    • x(dX/dt) + 0 = s(dS/dt)
    • Substitute known values and use the theorem to compute changes as the plane moves.
    • Result: When x = 3000 ft, dS/dt calculated to be 360 ft/sec.

Example: Rocket Launch

  • Identify quantities: height of rocket (h) and angle of camera (θ).
  • Using trigonometric relationships (tan θ = h/5000), derive related rates.
  • Solve for dθ/dt while substituting in known values:
    • Height of rocket at 1000 ft
    • Velocity known gives relationship to angle change.

Example: Water Draining from a Funnel

  • Rate of water draining = -0.03 ft³/sec, dimensions are set (height 2 ft, radius 1 ft).
  • Express volume in terms of height and radius relation using similar triangles: (r/h) = (1/2), thus simplify volume equation to:
    • V = (π/12)h³
  • Differentiate and solve for height change (dh/dt) based on volume change rate.