Let y be the wall height
let x be the horizontal distance
2x dx/dt+2y dy/dt=0
so calculate the x,y positions after 2 seconds, I have no idea the starting position.
This problem is the most popular lead-in question to "rate of change" questions for almost any Calculus text I have seen
Let the ladder be x ft from the wall, and y m up the wall
we know x^2 + y^2 = 100
2x(dx/dt) + 2y(dy/dt) = 0
given: dx/dt = 1 ft/s
find dy/dt when t = 2 or
in other words, when x = 2 ft since we know in 2 s it moved 2 ft.
also when x = 2
2^2 + y^2 = 100
y = √96
2(2)(1) + 2√96(dy/dt) = 0
dy/dt = -4/(2√96
= -.204 ft/s
at that moment the top of the ladder is dropping (note the - sign) at .204 ft/s
Just realized I read the question as the foot of the ladder moving away from the wall, so just change
dx/dt to -1,
the result will change to dy/dt = +.204
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