Related Rates: Fresh Take

  • Show how quantities change using derivatives and explore how these changes are connected
  • Apply the chain rule to calculate how one changing quantity affects another

Related-Rates Problem-Solving

The Main Idea 

  • Related rates problems involve finding the rate of change of one quantity when it’s related to another quantity whose rate of change is known.
  • Use the chain rule to relate the rates of change of different quantities.
  • Common Applications:
    • Geometric problems (e.g., expanding circles, changing triangles)
    • Physical scenarios (e.g., water flowing, objects moving)
  • Important Considerations:
    • Draw a diagram if possible
    • Don’t substitute values too early (before differentiation)
    • Pay attention to units and sign conventions

Problem-Solving Strategy

  1. Assign variables to all relevant quantities
  2. Establish the given information and what needs to be found
  3. Develop an equation relating the variables
  4. Differentiate the equation with respect to time
  5. Substitute known values and solve for the unknown rate

A conical water tank is being filled at a rate of [latex]10 \text{ft}^3/\text{min}[/latex]. The tank has a height of [latex]12 \text{ft}[/latex] and a base radius of [latex]6 \text{ft}[/latex]. At what rate is the water level rising when the water is [latex]4 \text{ft}[/latex] deep?

A street lamp is mounted at the top of a [latex]15[/latex] foot tall pole. A man [latex]6[/latex] feet tall walks away from the pole with a speed of [latex]5 \text{ft}/\text{sec}[/latex] along a straight path. At what rate is the tip of his shadow moving when he is [latex]40 \text{feet}[/latex] from the pole?

A circular oil slick is expanding at a constant rate of [latex]2[/latex] m2 /sec. How fast is the radius of the slick increasing when the radius is [latex]10[/latex] meters?

Assign variables:

[latex]A[/latex] = area of oil slick
[latex]r[/latex] = radius of oil slick

Given information:

[latex]\frac{dA}{dt} = 2 , \text{m}^2/\text{sec}[/latex]
[latex]r = 10 , \text{m}[/latex]
Need to find [latex]\frac{dr}{dt}[/latex]

Develop equation:

Area of a circle: [latex]A = \pi r^2[/latex]

Differentiate with respect to time:

[latex]\frac{dA}{dt} = 2\pi r\frac{dr}{dt}[/latex]

Substitute and solve:

[latex]2 = 2\pi(10)\frac{dr}{dt}[/latex]
[latex]\frac{dr}{dt} = \frac{2}{20\pi} = \frac{1}{10\pi} \approx 0.0318 , \text{m}/\text{sec}[/latex]

Therefore, when the radius of the oil slick is [latex]10 \text{meters}[/latex], it’s expanding at a rate of approximately [latex]0.0318 \text{m}/\text{sec}[/latex] or [latex]3.18 \text{cm}/\text{sec}[/latex].