Calculus
posted by Nick .
A ladder 41 feet long that was leaning against a vertical wall begins to slip. Its top slides down the wall while its bottom moves along the level ground at a constant speed of 4 feet per second. How fast is the top of the ladder moving when its 9 ft above the ground?

x^2+y^2 = 41^2
2 x dx/dt + 2 y dy/dt = 0
dy/dt = (x/y) dx/dt
when y = 9
x^2 + 81 = 1681
x^2 = 1600
x = 40
so
dy/dy =  (40/9)(4)
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