Calculus
posted by Z32 .
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
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
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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