A point source of light is hung 30 feet directly above a straight horizontal path on which a man of 6 feet in height is walking. How fast is the man's shadow lengthening and how fast the tip of shadow is moving when he is walking away from the light at the rate of 100 ft/min.

Explanation

Let S be the position of the source of light.
Let AB be the position of the man at a time t.
Let OB = x ft be the distance of the man from the lamp post at time t and BC = y ft be the length of his shadow at that time.
Then `("dx")/("dt")=100` ft/min is the rate at which the man is walking away from the light.
`("dy")/("dt")` is the rate at which his shadow is lengthening.
From the figure, the triangles SOC and ABC are similar.
`:. ("OS")/("AB")=("OC")/("BC")`
`:.(30)/(6)=(x-y)/(y)`
`:.5y=x+y" ":.4y=x" ":.y=(x)/(4)`
`:.("dy")/("dt")=(1)/(4)("dx")/("dt")" "=(1)/(4)xx100=25` ft/min

Hence, the shadow of the man is lengthening at the rate of 25 ft/min.
Now, C is the tip of shadow and it is a distance of `x+y` from the light.
`:.("d")/("dt")(x+y)=("dx")/("dt")+("dy")/("dt")` is the rate at which the tip of the shadow is moving and `("dx")/("dt")+("dy")/("dt")=100+25=125`
Hence, the tip of shadow is moving at the rate of 125 ft/min.