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Assuming the lantern is in fact shining, we have
If the length of rope from the man to the pulley is z, and the man is x ft away from the wall, we have
x^2+30^2 = z^2
z dz/dt = x dx/dt
when x=40, z=50 and the lantern is at height 36-10 = 26
If the lantern is at height h, and the man's shadow is of length s, then using similar triangles, we have
s/6 = (s+x)/h
= (s+x)/(36 - (60-z)) = (s+x)/(z-24)
= (s+x)/(√(x^2+900)-24)
s(1/6 - 1/(√(x^2+900)-24)) = 6x/(√(x^2+900)-30)
ds/dt = -180/(x^2+900-30√(x^2+900)) dx/dt
= -180/(z^2-30z) dx/dt
So, plug in the numbers and we have
ds/dt = -180/(50^2-30*50 = -180/1000 = -0.18 ft/s
whew - better check my math
As a check, using z instead of all the radicals, we have
s/6 = (s+x)/(z-24)
I guess you can verify that ds/dt comes out the same. Just remember that both s and z are functions of x.
a building is 36 ft tall. A pulley is attached to the top of the building. A rope is looped through the pulley. One end of the rope is attached to the lantern that hangs vertically parallel to the side of the building. The rope passes up vertically from the lantern, through the pulley, and then diagonally downwards. The other end of the rope is being held by a man who is 6 ft tall. The man holds the rope on top of his head and walks away from the building, causing the lantern to rise. The rope is 60 ft. At a certain instant the man is walking away at a rate of 5 ft per second and the man is 40 ft away. Find the rate change of the length of the mans shadow.
1 answer
1 answer