Posted by **Anonymous** on Tuesday, October 23, 2012 at 4:47pm.

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 -
**bobpursley**, Tuesday, October 23, 2012 at 5:36pm
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 -
**Reiny**, Tuesday, October 23, 2012 at 5:51pm
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 -
**Calc**, Tuesday, October 23, 2012 at 9:36pm
Reiny is definitely correct.

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