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The Moving Shadow

Date: 03/24/99 at 20:27:32
From: Isaac
Subject: Related Rates of a Shadow's Movement

I am not sure I did this problem right, because here the man walks 
towards the light source, whereas in other problems of this nature he 
usually walks away from the light. Please check to see whether I did 
it right.

A man 6 feet tall walks at a rate of 5 feet per second toward a light 
that is 20 feet above the ground. 

   a) At what rate is the tip of his shadow moving?
   b) At what rate is the length of his shadow changing?

    ^ |`
   20 |  `
    v |    0
      |   /|\`
      |   / \   `
 ---- = 5feet/sec

 20      y
 --- =  -----  
  6     y - x 

20y - 20x = 6y

14y = 20x
 y = ------

y = ----- 
dy   10  dx
--- = ----------
dt    7  dt

 dy   10        50 
 -- =  --- (5) = ---- ft/sec, since he is walking towards the light
 dt    7           7
then it is negative, - ----  ft/sec  

a) -(50/7) ft/sec
b) d(y-x)   dy   dx    -(50)
     ------ = ---- - ---- =  ----- - (5) = -(85/7) ft/sec
       dt     dt    dt      7

Date: 03/26/99 at 13:59:44
From: Doctor Luis
Subject: Re: Related Rates of a Shadow's Movement

Your reasoning for the first part was correct; however, you could 
have made dx/dt negative from the start. That is why the second part 
(b) is not correct.

Here is a diagram:

     | \
     |  \
     |   \
     |    \
     |     \N
     |     |\
     |     | \
     |     |  \
     O     M    S

According to the problem, OP is 20 ft, MN is 6 ft. The man, at M, is 
walking TOWARD the light source at O. If x is the distance OM, then 
dx/dt HAS to be negative because the x-values are decreasing.

Now, if y is the distance OS, (where the tip of the shadow is at S),
then, as you explained above, by similar triangles you have:

       OP     OS
    ----- =  -----
       MN     MS

     20      y
   ---- = ------
     6     y - x

and so you found

  y = (10/7)x

 Now if you just differentiate with respect to time, you get

  dy/dt = (10/7)dx/dt
        = (10/7)(-5)
        = -50/7 ft/s

You can see that by making dx/dt negative from the start you avoid 
the complication of changing the sign of dy/dt in the middle of the
problem. (That is why you did not get the last part.)

Given that, we find how fast the length of the shadow is changing:

  --- (y - x) = dy/dt - dx/dt = (-50/7) - (-5) = -15/7 ft/sec

Notice that the term -dx/dt makes a positive contribution to the rate 
of change of the length of the shadow. This makes sense from a 
physical standpoint, because if the tip of the shadow were to be held 
still, the man, by walking away, would actually be increasing the 
length of the shadow (y-x would be increasing if y were constant). 
However, y is not constant, it is decreasing, and that's why it makes 
a negative contribution to the rate of change of the length of the 

My recommendation to you is that you choose a consistent set of 
directions from the start (for example in this problem we chose the 
direction of increasing x-values to the right) and then, based on 
that, give the time-derivatives the appropriate signs according to 
whether the values whose rate of change they represent are increasing 
or decreasing. If you do this from the beginning of the problem, you 
will not have to be changing the signs every time. 

What I am trying to say is you need to be consistent.

- Doctor Luis, The Math Forum   
Associated Topics:
High School Calculus

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