# Calculus

posted by on .

A woman 5 ft tall walks at the rate of 5.5 ft/sec away from a streetlight that is 16 ft above the ground. At what rate is the tip of her shadow moving?

• Calculus - ,

Make the diagram. The streetlight height is 16. Let the distance from the streetlamp base to the tip of the shadow as x. Label the distance from the streetlamp base to the person as d.

Similar triangles:

16/x=5/(x-d)

5x=16x-16d
11x=16d
11 dx/dt=16 dd/dt

dd/dt =5.5ft/sec, solve for dx/dt

• Calculus - ,

I do not agree with bobpursley's solution

let the distance of the woman from the streetlight be x ft
let the length of her shadow be y ft

by ratios ...
5/y = 16/(x+y)
16y = 5x + 5y
11y = 5x
11 dy/dt = 5 dx/dt
but dx/dt = 5.5
dy/dt = (5/11)(5.5) = 2.5

So her shadow is lengthening at 2.5 ft/sec
but she is moving at 5.5 ft/sec, so her shadow is moving at 2.5+5.5 or 8 ft/sec

( I am walking along the inside of a moving train at 3 ft/sec, while the train is moving at 50 ft/sec
So I am moving at 53 ft/sec)

• Calculus - ,

Reiny is definitely correct.