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A ladder leans against a vertical wall, with the bottom of the ladder 6 ft from the wall on a horizontal floor. At that time the bottom end of the ladder is being pulled away at the rate of 2 ft/s and the top of the ladder slips down the wall at the rate of 3 ft/s. How long is the ladder?

Guest Nov 27, 2018

#1**+1 **

Let x be the horizontal distance that the ladder is from the wall = 6 ft

At that time....the top of the ladder is sqrt (r^2 - [6ft]^2) = sqrt (r^2 - 36ft^2) ft from the floor

We have this

r^2 = x^2 + y^2

Take the derivative of each term with respect to time

2r dr/dt = 2x dx/ dt + 2y dy/dt

y = sqrt (r^2 - 6^2) = sqrt (r^2 - 36)

dr/dt = 0 [ the ladder length is constant ]

So when x = 6 ft , the distance the bottom of the ladder is from the wall is increasing and the distance the top of the ladder is from the flooor is decreasing

So dx / dt = 2 ft / s and dy / dt = -3ft / s

So we have

0 = 2(6ft) (2 ft/s) + 2sqrt (r^2 - 36ft^2)ft *(-3 ft/s)

Divide out ft/s

0 = 24ft - 6sqrt (r^2 - 36ft^2) ft

6 sqrt (r^2 - 36ft^2) ft = 24 ft divide both sides by 6

sqrt (r^2 - 36ft^2 ) = 4 ft square both sides

r^2 - 36ft^2 = 16ft^2

r^2 = (36 + 16) ft^2 take the square root of both sides

r = sqrt (52) ft ≈ 7.2 ft = the ladder length

CPhill Nov 27, 2018