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?
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