A 13 ft ladder is leaning against a house when its base begins to slide away. By

Question

A 13 ft ladder is leaning against a house when its base begins to slide away. By the time the base is 3 ft from the house, the base is moving at the rate of 2 ft/sec. Answer the following:

At what speed is the top of the ladder sliding down the wall at that time? Answer = (12)/(8sqrt10) ft/sec.

At what rate is the area of the triangle formed by the ladder, wall, and ground changing then? Answer = (1/2)((8sqrt10)+(-9sqrt10)/(20)) sq. ft/sec.

At what rate is the angle heta changing then? Answer =????

Explanation / Answer

x = distance of base of ladder from house
y = distance of top of ladder from ground

x^2 + y^2 = 13^2
2xdx/dt + 2ydy/dt = 0
dy/dt = -(x/y)dx/dt

when x = 3
y = sqrt(13^2 - 3^2)
y = sqrt( 160 ) ft ==>12.649 ft
dx/dt =2 ft/sec

dy/dt = -(x/y)dx/dt
dy/dt = -(3/12.649)* 2
dy/dt = -0.47434 ft/sec

b.

A = xy/2
dA/dt = (1/2)(xdy/dt + ydx/dt)
dA/dt = (1/2)(3*(-0.47434) + 12.649*)
dA/dt = -59.5 ft^2/sec

c.

B = angle between ladder and ground
cos(B) = x/13
-sin(B)dB/dt = (1/13)dx/dt
dB/dt = (-1/13)(dx/dt)/sin(B)
dB/dt = (-1/13)(2)/(12.649/13)
dB/dt = -0.15811 rad/sec

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