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Related Rates

Solving problems where multiple variables change with time.

In the real world, variables rarely change in isolation. If you inflate a balloon, the radius expands, the volume increases, and the surface area grows—all at the same time. These changes are "related" to each other.

Related Rates problems ask us to calculate how fast one quantity is changing based on how fast another quantity is changing. The secret ingredient? Time (t).

1. The Concept: d/dt

In standard algebra, we usually differentiate with respect to x. But in related rates, everything is changing with respect to time t.

This means we use the Chain Rule implicitly. If you see an x in an equation, its derivative isn't just 1; it is dx/dt (the velocity of x).

[Image of diagram illustrating implicit differentiation with respect to time]
Example:
Equation: x² + y² = z²
Derivative with respect to t:
2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt)

2. General Strategy

Solving these problems can be tricky, but following a strict protocol helps avoid errors:

  • 1. Draw a Picture: Label quantities that change with variables (x, y) and quantities that stay constant with numbers.
  • 2. List Knowns & Unknowns: Write down what you know (e.g., dx/dt = 5) and what you want to find (e.g., dy/dt = ?).
  • 3. Write an Equation: Relate the variables using geometry (Pythagorean theorem, Volume formulas, Trig ratios).
  • 4. Differentiate: Take the derivative of both sides with respect to time t. Don't forget the Chain Rule!
  • 5. Substitute & Solve: Plug in your specific values at the "instant" in question and solve.

3. Classic Example: The Sliding Ladder

This is the most famous related rates problem.

Problem: A 10-meter ladder leans against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1 m/s, how fast is the top of the ladder sliding down the wall when the bottom is 6 meters from the wall?

[Image of ladder sliding down a wall diagram]

Step A: The Equation

This is a right triangle. We use the Pythagorean Theorem.

x² + y² = 10²
(Note: 10 is constant because the ladder's length doesn't change.)

Step B: Differentiate

2x(dx/dt) + 2y(dy/dt) = 0
Divide by 2:
x(dx/dt) + y(dy/dt) = 0

Step C: Find Missing Values

We are given x = 6. We need y. Use the original equation:

6² + y² = 100
36 + y² = 100
y² = 64 → y = 8

Step D: Solve

Plug in x=6, y=8, and dx/dt=1.

(6)(1) + (8)(dy/dt) = 0
6 + 8(dy/dt) = 0
8(dy/dt) = -6
dy/dt = -6/8 = -0.75 m/s

The negative sign confirms the top of the ladder is moving down.

4. Other Common Scenarios

  • Expanding Sphere: Relates volume (dV/dt) to radius (dr/dt). (V = 4/3 πr³)
  • Filling a Cone Tank: Relates water height to volume. (Use similar triangles to eliminate variables).
  • Shadow Problems: Use similar triangles to relate the speed of a person walking to the speed of their shadow tip.
[Image of expanding sphere diagram]