calculus
posted by Luis .
A 6ft tall woman is walking away from a srteet lamp at 4 ft/sec that is 24ft tall. How fast is the length of her shadow changing?

It depends upon her distance from the swtreet lamp at the time. Did they not tell you that distance?

no time given

Make a sketch. Draw a line from the top of the street lamp to the end of her shadow
I see two similar rightangled triangles.
Let the length of her shadow be x ft
let her distance to the street lamp be y
then 24/(x+y) = 6/x
which simplifies to
3x = y
3dx/dt = dy/dt
dx/dt = 4/3 ft/s
In this type of question, the length of her shadow is growing at a uniform rate, independent of the time passed or her position.
Secondly, had the question been, "how fast is her shadow moving?" you would have to add her speed to the 4/3.