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Ladder Problem Explorer

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About This MicroSim

The sliding ladder problem is one of the most classic related rates problems in calculus. A ladder of fixed length \(L\) leans against a wall. The base slides away from the wall at a constant rate \(\frac{db}{dt}\). What is the rate at which the top of the ladder slides down the wall?

This simulation lets you see the answer visually and understand why the top speeds up as it approaches the ground.

The Mathematical Setup

The ladder, wall, and ground form a right triangle with:

  • \(b\) = distance from wall to base of ladder (horizontal leg)
  • \(h\) = height of top of ladder (vertical leg)
  • \(L\) = ladder length (hypotenuse, constant)

The Pythagorean theorem gives us the constraint:

\[b^2 + h^2 = L^2\]

To find \(\frac{dh}{dt}\), we differentiate both sides with respect to time \(t\):

\[\frac{d}{dt}(b^2 + h^2) = \frac{d}{dt}(L^2)\]
\[2b\frac{db}{dt} + 2h\frac{dh}{dt} = 0\]

Solving for \(\frac{dh}{dt}\):

\[\frac{dh}{dt} = -\frac{b}{h} \cdot \frac{db}{dt}\]

The Key Insight

Notice that \(\frac{dh}{dt}\) depends on the ratio \(\frac{b}{h}\). As the ladder slides:

  • When \(h\) is large (ladder nearly vertical), \(\frac{b}{h}\) is small, so the top moves slowly
  • When \(h\) is small (ladder nearly horizontal), \(\frac{b}{h}\) is large, so the top moves quickly
  • As \(h \to 0\), the speed \(|\frac{dh}{dt}| \to \infty\)

Delta Moment

"Watch that orange arrow grow as the ladder falls! The closer I get to the ground, the faster I'm moving. It's like the math is shouting 'SPEEEEED!' at me. That negative sign means I'm heading down, and the -b/h ratio tells me exactly how fast."

How to Use

  1. Ladder Length L: Adjust the total length of the ladder (6 to 15 feet)
  2. Base Speed db/dt: Set how fast the base slides away from the wall (0.5 to 4 ft/sec)
  3. Base Position b: Manually position the base to see instantaneous values
  4. Play/Pause: Watch the ladder slide in real-time
  5. Reset: Return to starting configuration

What to Observe

  • The green arrow shows the constant base velocity (db/dt)
  • The orange arrow shows the top velocity (dh/dt) - watch it grow!
  • The data panel updates with current positions and rates
  • The equation panel shows the calculation with substituted values
  • The Pythagorean check confirms \(b^2 + h^2 = L^2\) at all times

Lesson Plan

Learning Objectives

After using this MicroSim, students will be able to:

  1. Apply the related rates procedure to the ladder problem (Bloom Level 3)
  2. Calculate dh/dt given b, h, L, and db/dt using implicit differentiation
  3. Explain why the top of the ladder speeds up as it approaches the ground
  4. Verify solutions using the Pythagorean constraint

Prerequisite Knowledge

  • Implicit differentiation
  • Chain rule for derivatives
  • Pythagorean theorem
  • Understanding of rate of change

Suggested Activities

  1. Prediction First: Before pressing Play, ask students to predict: will the top move at constant speed, speed up, or slow down? Why?

  2. Critical Position: At what position is |dh/dt| = |db/dt|? (Answer: when b = h, i.e., when the ladder makes a 45-degree angle)

  3. Numerical Exploration: Set L = 10, db/dt = 2. Calculate dh/dt when b = 3, 6, and 9. Verify with the simulation.

  4. Limit Investigation: What happens mathematically as b approaches L? What does this mean physically?

  5. Parameter Sensitivity: How does changing L affect dh/dt at a fixed position? How does changing db/dt affect it?

Discussion Questions

  1. Why is dh/dt negative? What does the sign tell us about direction?

  2. The formula shows \(\frac{dh}{dt} = -\frac{b}{h} \cdot \frac{db}{dt}\). Why does the ratio b/h matter more than the individual values?

  3. In real life, what would prevent the infinite speed predicted as h approaches 0?

  4. How would this problem change if the ground were frictionless? (The ladder couldn't stay in contact with both surfaces.)

Assessment Questions

  1. A 13-foot ladder leans against a wall. The base is 5 feet from the wall and sliding away at 2 ft/sec. How fast is the top sliding down?

  2. Solution: h = 12, dh/dt = -(5/12)(2) = -5/6 ft/sec

  3. At what height is the top of the ladder when |dh/dt| = 3|db/dt|?

  4. Solution: When b/h = 3, so h = b/3. Combined with \(b^2 + h^2 = L^2\), solve for h.

  5. A 10-foot ladder has its base sliding at 1 ft/sec. At what position b is the top falling at 2 ft/sec?

  6. Solution: Need b/h = 2 with \(b^2 + h^2 = 100\). Substituting h = b/2: \(b^2 + b^2/4 = 100\), so \(b = 4\sqrt{5} \approx 8.94\) ft.

Embedding

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<iframe src="https://dmccreary.github.io/calculus/sims/ladder-problem-explorer/main.html"
        height="580px" width="100%" scrolling="no"></iframe>

References