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A 10-foot ladder. The base slides away from the wall at a steady 2 feet per second, and the top slides down the wall, but not steadily. It barely moves at first, then plunges. The base's rate never changes. The top's runs away.
The ladder never changes length, so x² + y² = 100 at every instant. Differentiate with respect to time, not x. That's episode 19's implicit differentiation with t in place of x. Both x and y depend on t, so each picks up a chain-rule factor: 2x·dx/dt + 2y·dy/dt = 0, which is x·dx/dt + y·dy/dt = 0.
At x = 6, y = 8: 6(2) + 8·dy/dt = 0, so dy/dt = −1.5 ft/s. Negative because the top is coming down. At x = 8 it's −8/3 ≈ −2.67. At x = 9.9 it's about −14. As the ladder flattens, y heads to zero and you're dividing by it. A real ladder's top leaves the wall before that. The model assumes it stays on.
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2026/10/04
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