A 10-foot ladder leans against a wall. If the bottom slides away at 2 ft/s, how fast is the top sliding down when the bottom is 6 feet from the wall?

Question

Accepted Answer

1.5 ft/s. The correct answer is 1.5 ft/s. Using the Pythagorean relationship x² + y² = 100 and implicit differentiation: 2x(dx/dt) + 2y(dy/dt) = 0. When x = 6, y = √(100 - 36) = 8. Substituting: 2(6)(2) + 2(8)(dy/dt) = 0, so 24 + 16(dy/dt) = 0, giving dy/dt = -1.5 ft/s. The speed is 1.5 ft/s (the negative sign indicates downward motion).

The value 1 ft/s would result from an arithmetic error in the calculation, likely from incorrectly computing y or mishandling the equation. The value 2 ft/s incorrectly assumes the top slides at the same rate as the bottom, but the rates differ because of the geometric relationship. The value 3 ft/s may come from an error in the implicit differentiation step or from confusing which variable represents which rate.

Related rates problems are a signature application of derivatives on the CLEP Calculus exam. The strategy is always: (1) identify the relationship between variables, (2) differentiate implicitly with respect to time, (3) substitute known values. The ladder problem is the classic related rates scenario, and mastering it prepares students for similar problems involving cones, spheres, and other geometric relationships.

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