A ladder 25 feet long is leaning against the wall of a house. The base of the la

Question

A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second. (a) What is the velocity of the top of the ladder when the base is given below? 15 feet away from the wall ft/sec 20 feet away from the wall ft/sec 24 feet away from the wall ft/sec Consider the triangle formed by the side of the house, ladder, and the ground. Find the rate at which the area of the triangle is changing when the base of the ladder is 24 feet from the wall. ft2/sec Find the rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 24 feet from the wall. rad/sec

Explanation / Answer

Solution:

l² = x² + y² : rightside is positive for x and upside is positive for y

x² + y² = (25)^2

x² + y² = 625

2x dx +2y dy =0 ;

dy = -(x/y)dx

dy/dt = -(x/y)dx/dt

we have dx/dt = +2 ft/sec

a)
x= 15 ft then y = (l² - x²) = (25² - 15²) = +20 ft
dy/dt = -(x/y)dx/dt = -(15/20)(2) = -1.5 ft/sec : downside

x= 20 ft then y = (l² - x²) = (25² - 20²) = +15 ft
dy/dt = -(x/y)dx/dt = -(20/15)(2) = -8/3 -2.667 ft/sec : downside

x= 24 ft then y = (l² - x²) = (25² - 24²) = +7 ft
dy/dt = -(x/y)dx/dt = -(24/7)(2) = -48/7 -6.857 ft/sec : downside

b)
A = xy/2
dA = d(xy/2)
dA = (ydx + xdy)/2
dA/dt = (ydx + xdy)/(2dt)
dA/dt = (y dx/dt + x dy/dt)/2
we have:
x= +24 ft then y = +7 ft
dx/dt = +2 ft/sec
dy/dt = -6.857 ft/sec
hence:
dA/dt = [ (7)(+2) + (24)(-6.857) ]/2
dA/dt = - 301/4 ft²/sec

c)
x = +24 ft then y = +7 ft
cos = y/l = y/25
sin = x/l = x/25
dsin = d(x/25)
cos d = dx/25
y/25 d = dx/25
d = dx/y
d/dt = (dx/y)/dt
d/dt = (dx/dt)/y
d/dt = (+2)/7 = +2/7 rad/sec

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