A 13 foot ladder is leaning against a wall. If the top slips down the wall at a rate of 4 ft/s, how fast will the foot be moving away from the wall when the top is 10 feet above the ground?


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The ladder against the wall is a right triangle: x^2 + y^2 = 13^2

Top slips down => dy/dt = -4 ft/s. When y = 10 ft. dx/dt = ?

1) Take the derivative with respect to time:

2x (dx/dt) + 2y (dy/dt) = 0

2) Substitute: (Note: when y = 10, x = sqrt(69), z = 13)

2(sqrt(69)) (dx/dt) + 2(10) (-4) = 0

3) Solve for dx/dt

dx/dt = 80/(2sqrt(69)) or 4.8154 ft/s


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