# calculus

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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?

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

• calculus -

no time given

• calculus -

Make a sketch. Draw a line from the top of the street lamp to the end of her shadow
I see two similar right-angled 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.

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