Homework SourseLadder against the wall A 13ft ladder is leaning against a v
Ladder against the wall. A 13-ft ladder is leaning against a vertical wall when jack begins pulling the foot of the ladder away from the wall at a rate of 0.5 ft/s. How fast is the top of the ladder sliding down the wall when the foot of the ladder is 5 ft from the wall?.
Solution
the equation for a right triangle:
A2 + B2 = C2
differentiating:
A*(dA/dt) + B*(dB/dt) = C*(dC/dt)
C = 13 ft
dC/dt = 0 ft (the ladder isn\'t changing length)
dB/dt = 0.5 ft/s (how fast its being pulled away from the wall)
B = 5 ft (how far away it is from the wall at a particular instance)
A = (C2 - B2) = (169 - 25) = 144 = 12 ft (how high up on the wall it is at a particular instant
(12 ft)*(dA/dt) + (5 ft)*(0.5 ft/s) = 0
dA/dt = -2.5/12 ft/s = - 5/24 ft/s = - 0.20833 ft/s
Note: its negative because its moving down the wall (i.e. the height on the wall is decreasing)
