Calculus
posted by Lost One
a 5m ladder rests against a vertical wall. The top of the ladder begins sliding down the wall at 2m/sec, while the foot of the ladder moves away from the base of the wall. How fast is the foot of the ladder moving away from the wall at the instant the foot of the ladder is 4m from the base of the wall?

Steve
when x=4, y=3
x^2+y^2 = 25
2x dx/dt + 2y dy/dt = 0
2(4) dx/dt + 2(3)(2) = 0
dx/dt = 12/8 = 1.5 m/s
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