Calculus
posted by Julia .
A 10 foot ladder leans against a 20 foot wall. Someone begins pushing the base of the ladder toward the wall at the rate of one foot per second. How quickly is the top of the ladder moving up the wall after 2 seconds?

Let y be the wall height
let x be the horizontal distance
x^2+y^2=100
2x dx/dt+2y dy/dt=0
dy/dt=x dx/dt
so calculate the x,y positions after 2 seconds, I have no idea the starting position. 
This problem is the most popular leadin 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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