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So while reading the solution (which uses implicit differentiation) of the following related rates ladder problem I thought about the algebraic way of solving it, but it gives wrong result so I want to know what is wrong with it. The problem is this "A ladder 10 feet long rests against a vertical wall. If the bottom of the ladder slides away from the wall at the rate of 1 ft/s. How fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 ft from the wall ?". So my algebraic solution is this, till the end of the ladder's fall bottom must go 4 feet, so it will do it in 4 seconds, and this means that the top of the ladder goes down with the speed of 2 ft/s (because height is 8 ft by pythagorean theorem). But the right answer is 3/4 ft/s. So I think the fault in my solution is the assumption that velocity is constant, so is it? If not, what is the reason for this ?

Юрій Ярош
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  • That's right, just because the velocity of the bottom of the ladder is constant, it doesn't follow that the velocity of the top of the ladder is constant. That might be true if you were carrying the ladder from one place to another, but if you were (somehow) spinning the ladder around the top, the bottom could be moving very fast, while the top was stationary. – saulspatz Sep 09 '18 at 17:10
  • @saulspatz OK, is velocity of the bottom of the ladder constant? – Юрій Ярош Sep 09 '18 at 17:18
  • @saulspatz And another question which arises, why is the velocity of the top changing? – Юрій Ярош Sep 09 '18 at 17:21
  • Yes, that is what the problem means when it says the velocity is one ft/sec. – saulspatz Sep 09 '18 at 17:21

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