Posted by Johnny on Thursday, November 8, 2012 at 6:37am.
This is just an exercise in implict differentiation.
At the moment in question, we have a 24-32-40 right triangle, with height=24.
Let θ be the angle between the ladder and the ground.
Let the base be x, and the height be y.
a)
x^2 + y^2 = 40^2
2x dx/dt + 2y dy/dt = 0
when x=32,
2(32)(4) + 2(24) dy/dt = 0
dy/dt = -16/3
so, the top is sliding down at 16/3 ft/s
b)
a = 1/2 xy
da/dt = 1/2 (y dx/dt + x dy/dt)
da/dt = 1/2 (24(4) + 32(-16/3))
= -112/3 ft^2/s
c)
y/x = tanθ, so
y = x tanθ
dy/dt = tanθ dx/dt + x sec^2θ dθ/dt
-112/3 = 3/4 (4) + 32 (5/4)^2 dθ/dt
dθ/dt = -121/150 rad/s
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