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The Speed of Change of a Shadow


Date: 12/12/95 at 21:39:16
From: Anonymous
Subject: Calculus, Rates

A street lamp whose light is 5 meters from the ground is 6 meters 
from a vertical brick wall.  A person 2 meters tall walks at right 
angles away from the wall directly towards the street lamp at a speed 
of 1.2 meters per second.  At what speed is the shadow of the top of 
the person's head descending down the wall when the person is 1.5 
meters from the wall.

As with most of these problems the only difficulty I have occurs 
when I try to create an equation that will allow me for solve for 
dx/dt.  I'm not sure how to start it but I tried using triangle laws to 
get some substitutions but nothing seemed to work.

Thanks for your help.


Date: 5/29/96 at 1:54:45
From: Doctor Anthony
Subject: Re: Calculus, Rates

Let x be the distance of the man from the wall, and y be the height of 
his shadow up the wall.  

Draw a horizontal line through the top of the man's head to give two 
similar triangles, which then allow us to write the following ratios as 
equal.

3/(6-x) = (2-y)/x

Multiply this out to give 5x + 6y - xy = 12

Differentiate with respect to time

5(dx/dt)+6(dy/dt)-x(dy/dt)-y(dx/dt)=0

Now when x=1.5, the above ratio equation gives y=1. 
Also dx/dt = 1.2
Substitute these values the into the differential equation

5(1.2)+6(dy/dt)-1.5(dy/dt)-1.2 = 0
6 + 4.5(dy/dt) -1.2 = 0
    4.5(dy/dt) = - 4.8
        dy/dt = -4.8/4.5
              = -1.06667

The negative value is correct since the shadow is moving down the 
wall and y is decreasing.  

-Doctor Anthony,  The Math Forum

    
Associated Topics:
High School Calculus

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