Out in the wild west a man has been injured in a duel. He steadies himself against a light post that is
three times as tall as he is. He summons his strength and begins to walk in a straight line away from the
light post. He accelerates initially but soon slows. After making it 6 meters he stops to catch his breath.
His speed v (measured in meters per second) throughout his journey is described by

-s^2/4+3×s/2

where s is his distance from the light post in meters.
(a) Find the rate of change of the length of his shadow when he has traveled 4 meters from the light post.
(b) Show that his shadow was changing the fastest when he reached his top speed.

use similar right triangles like this



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the bottom of the triangle will be described by 2 variables , s and one of your choice (i'll use x)
s for the distance from lamp to person
x as the distance from person to shadow's end
s+x is the whole bottom length of the lamp triangle

you know the height of the lamp and the person

set up similar triangles equation to solve for x (your shadow length)

take derivative and plug in ds/dt

This post was edited by saber_x3 on Mar 27 2014 12:05pm