15 feet ladder is leaning against a wall when the base starts to slide away. By

Question

15 feet ladder is leaning against a wall when the base starts to slide away. By the time the base is 12 feet from the house, the base is moving at the rate of 6 ft / sec. How fast is the top of the ladder sliding down the wall? At what rate is the angle theta between the ladder and the ground changing then? Water is flowing at the rate of 50 m3 / min from a shallow conical reservoir (vertex down) of base radius 45 m and height 6 m. How fast is the water level falling when the water is 3 m deep? (write with proper unit) How fast is the radius of the water's surface changing then? (Volume of a cone is V = 1 / 3 pi r2 h)

Explanation / Answer

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

x^2 + y^2 = 15^2
2xdx/dt + 2ydy/dt = 0
dy/dt = -(x/y)dx/dt

when x = 12
y = sqrt(15^2 - 12^2)
y = 9 ft
dx/dt = 9 ft/sec

dy/dt = -(x/y)dx/dt
dy/dt = -(12/6)6
dy/dt = -12 ft/sec

b.

A = xy/2
dA/dt = (1/2)(xdy/dt + ydx/dt)
dA/dt = (1/2)(12*(-12) + 6*6)
dA/dt = -54ft^2/sec

c.

B = angle between ladder and ground
cos(B) = x/15
-sin(B)dB/dt = (1/15)dx/dt
dB/dt = (-1/15)(dx/dt)/sin(B)
dB/dt = (-1/15)(6)/(6/15)
dB/dt = -1 rad/sec

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