Posted by Johnny on Thursday, November 8, 2012 at 6:37am.
A 40-foot ladder is leaning against a house when its base starts to slide away. By the time the base is 32 feet from the house the base is moving at a rate of 4ft/sec.
a. How fast is the top of the ladder moving down the wall at that moment?
b. At what rate is the area of the triangle formed by the ladder, wall, and ground changing at the moment?
c. At what rate is the angle between the ladder and the ground changing at that moment?
I wasn't here the day the teacher taught us this...plz help me i have no idea where to start
- Calculus - Steve, Thursday, November 8, 2012 at 10:54am
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.
x^2 + y^2 = 40^2
2x dx/dt + 2y dy/dt = 0
2(32)(4) + 2(24) dy/dt = 0
dy/dt = -16/3
so, the top is sliding down at 16/3 ft/s
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
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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