differential calculus
posted by jordan
a man 6 ft tall is walking toward a building at the rate of 4 ft/sec. if there is a light on the ground 40 ft from the building, how fast is the man's shadow on the building growing shorter when he is 30 ft from the building?

Steve
Using similar triangles, if the man is x from the light, and his shadow is h tall,
h/40 = 6/x
1/40 dh/dt = 6/x^2 dx/dt
So, when the man is 10 ft from the light,
1/40 dh/dt = 6/100 * 4
dh/dt = 9.6 ft/s
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