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A 10m ladder is leaning against a vertical wall with its foot on the same horizontal plane
as the base of the wall. If the lower end of the ladder is moving away from the wall
horizontally at 1m/sec, how fast is the top of the ladder descending when the lower end
is 4 meters from the wall?
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Let x be horizontal distance from the wall and y be vertical coordinate.
Then from Pythagoras
x^2 + y^2 = 10^2. (1)
Here x = x(t) and y = y(t) are functions of time, t.
Differentiate equation (1) over t. You will get
2x*x'(t) + 2y*y'(t) = 0,
hence
y'(t) = - (x*x'(t))/y.
Evaluate it at the given values x = 4 m, x'(t) = 1 m/s, y = = = .
You will get y'(t) = - = - 0.436 m/s (rounded). ANSWER
Solved.