moving shadow: man walks at 5 ft/s toward light 16' high

[chargr10]

A 6' tall man walks at a rate of 5ft/se toward a light 16' above the ground. At what rate is the length of the shadow changing when he is 10ft from the base of the light.

I know to use similar triangles where 16/6=y/y-x and so dy/dt=8/5dx/dt therefore dy/dt=-8ft/sec. I was able to figure out dx/dt is -5ft/sec by process of elimination because I knew from the book the answer to the question is -3ft/sec, so d(y-x)/dt =-3ft/sec.

What I am having troubling figuring out is why dx/dt=-5ft/sec and not dy/dt=-5ft/sec. On the triangles, the base of the larger triangle is y and the base of the smaller triangle is y-x. What info in the problem tells me that dy/dt or is not equal to -5ft/sec. My trail an error method wont work on a test.

 

[soroban]

Re: moving shadow

chargr10!

It sounds like you don't understand the various parts of the diagram.

A 6' tall man walks at a rate of 5 ft/s toward a light 16' above the ground.
At what rate is the length of the shadow changing when he is 10 ft from the base of the light?

I would label the diagram like this:

    P ^
      |   *
      |       *   A
      |           *
   16 |           |   *
      |          6|       *
      |           |           *
      |           |               *
      * - - - - - * - - - - - - - - - *
      Q     x     B         s         C

\(\displaystyle \text{The light is at }P\!:\;\;PQ = 16\)

\(\displaystyle \text{The man is }AB = 6\)

\(\displaystyle \text{His distance from the base of the light is: }\:x = QB\)

\(\displaystyle \text{Since he is walking }toward\text{ the light, }x\text{ is decreasing at 5 ft/sec.}\)
. . \(\displaystyle \text{That is: }\;\frac{dx}{dt} = -5\text{ ft/sec}\)

\(\displaystyle \text{Let }s = BC\text{, the length of his shadow.}\)

. . \(\displaystyle \text{We want: }\:\frac{ds}{dt}\:\text{ when }x = 10.\)


\(\displaystyle \text{From similar triangles, we have: }\:\frac{s}{6} \:=\:\frac{x+s}{16} \quad\Rightarrow\quad s \:=\:\tfrac{3}{5}x\)

\(\displaystyle \text{Differentiate with respect to time: }\;\frac{ds}{dt} \:=\:\frac{3}{5}\cdot\frac{dx}{dt}\)

\(\displaystyle \text{Since }\frac{dx}{dt} = \text{-}5\text{, we have: }\;\frac{ds}{dt} \:=\:\frac{3}{5}(\text{-}5) \;=\; - 3\text{ ft/sec}\)