Posted by **Z32** on Thursday, September 24, 2009 at 3:40am.

A street light is at the top of a 17 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 6 ft/sec along a straight path. How fast is her shadow lengthening when she is 30 ft from the base of the pole?

- Calculus -
**Reiny**, Thursday, September 24, 2009 at 8:47am
I hope you made a diagram.

let her distance from the pole be x ft

let the length of her shadow be y ft

by similar triangles,

6/y = 17/(x+y)

simplifying,

11y = 6x

11 dy/dt = 6 dx/dt

but dx/dt = 6

so dy/dt = 6/11 ft/sec

notice the 30 ft is irrelevant.

Also be careful to notice the wording.

You asked, "how fast is her shadow lengthening" and that answer is 6/11

Had you asked, " how fast is her shadow moving", we would have had to add her own velocity to it, namely 6 + 6/11 ft/sec

- Calculus -
**Jarred**, Thursday, October 7, 2010 at 5:05pm
There is 1 error in the above answer.

11dy/dt = 6 dx/dt is true

dx/dt = 6 is also true

but dy/dt is not 6/11 ft/sec

the 6 must be multiplied by the 6 from dx/dt giving dy/dt = 36/11

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