A man 5.50 ft tall approaches a street light 18.0 ft above the ground at a rate of 4.0 ft/s. at what rate is the end of the man's shadow moving when he is 9.0 ft from the base of the light?
Calculus - Reiny, Saturday, August 27, 2011 at 7:51am
Make a sketch.
Let the distance between the man and the post be x ft
let the length of his shadow be y ft
by similar triangles:
5.5/y = 18/(x+y)
18y = 5.5x + 5.5y
12.5y = 5.5x
12.5dy/dt = 5.5dx/dt
given: dx/dt = 4 ft/s
then dy/dt = 5.5/12.5 or .44 ft/s
So the shadow is lengthening at .44 ft/s
and the shadow is moving at 4+.44 or 4.44 ft/sec
(The end of the shadow is both lengthening and being moved along by the movement of the man.
Similar to somebody walking in a moving train.
Suppose a train is going at 60 mph, but somebody is walking inside the train at 4 mph in the same direction as the train. To an observer at a crossing, the person would be moving at 64 mph )
Calculus - wrongg, Wednesday, February 20, 2013 at 9:11pm