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December 18, 2014

December 18, 2014

Posted by **Johnny** on Thursday, November 8, 2012 at 6:37am.

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 10:54amThis 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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