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January 31, 2015

January 31, 2015

Posted by **Luis** on Monday, December 14, 2009 at 4:08am.

- calculus -
**drwls**, Monday, December 14, 2009 at 4:58amIt depends upon her distance from the swtreet lamp at the time. Did they not tell you that distance?

- calculus -
**Luis**, Monday, December 14, 2009 at 5:12amno time given

- calculus -
**Reiny**, Monday, December 14, 2009 at 7:12amMake 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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