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[Guest]

"A street light is mounted at the top of a 15-ft-tall pole. A man 6ft tall walks away from the pole with a speed of 5ft/s along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole?"
This is how I did it:
\(\displaystyle \frac{15}{6} = \frac{x+y}{y}\)
\(\displaystyle 0 = \frac{y(\frac{dy}{dx} + \frac{dy}{dt}) - (x + y)\frac{dy}{dt}}{y^2}\)
\(\displaystyle y\frac{dx}{dt} - x\frac{dy}{dt} = 0\)
\(\displaystyle \frac{dy}{dt} = \frac{y}{x} \frac {dx}{dt}\)
\(\displaystyle \frac{dy}{dt} = \frac{6}{9}(5)\)
\(\displaystyle \frac{dy}{dt} = \frac{30}{9} = \frac{10}{3} ft/s\)
then, since the guy is also moving at 5ft/s, add 5 to dy/dt, and the answer is:
\(\displaystyle \frac{25}{3} ft/s\)
This is the right answer. However, my solution manual does it another way. They find y = (2/3)x, and then say that the tip of the shadow moves at a rate of:
\(\displaystyle \frac{d}{dt}(x + y) = \frac{d}{dt}(\frac {5}{3} x) = \frac {5}{3} (5) = \frac {25}{3} ft/s.\)
Are both ways correct? I was trying to explain my solution to someone, but now I'm confused as to whether my solution is correct.

According to my way, why don't you simply stop at
\(\displaystyle \frac{10}{3} ft/s\)
I'm thinking that this is the rate of the shadow moving relative to the man. But to find the rate of the shadow relative to the street light (or the earth), you must add the speed of the man. Is my reasoning correct??

 

[tkhunny]

unfortunately, you did not write down clear definitions for 'x' and 'y', so one is left to guess what you mean.

x = distance from pole to man
y = distance from man to shadow tip

The important part is that 'x' and 'y' are NOT independent, y = f(x). You did this when you substituted y = 6 and x = 9. How did you know that? You're solutions are the same.

What's that dy/dx in step #2? Just a typo, I think.

 

[soroban]

Hello, asdfmaster!

Your work and your reasoning is absolutely correct . . . nice going!

And I'm delighted that you noticed that you had to add the man's speed.
\(\displaystyle \;\;\)Most people don't notice that omission.

Your first answer is "how fast is his shadow lengthening?"


Both methods are correct.
Since they wanted the speed of the tip of the shadow relative to the Earth,
\(\displaystyle \;\;\)they probably set up the diagram like this ( I would):

        *
        |   *
        |       *
        |       |   *
      15|      6|       *
        |       |           *
        |   x   |      y-x      *
       -+-------+-------------------*-
        : - - - - - y - - - - - - - :

where \(\displaystyle y\) is the distance from the tip of his shadow to the pole.


Then: \(\displaystyle \L\,\frac{15}{6}\:=\:\frac{y}{y\,-\,x}\;\;\Rightarrow\;\;15y\,-\,15x\:=\:6y\;\;\Rightarrow\;\;y\,=\,\frac{5}{3}x\)

Differentiate with respect to time: \(\displaystyle \L\,\frac{dy}{dt}\:=\:\frac{5}{3}\left(\frac{dx}{dt}\right)\)

Since \(\displaystyle \frac{dx}{dt}\,=\,5\) ft/s, we have: \(\displaystyle \L\,\frac{dy}{dt}\;=\;\frac{5}{3}(5)\;=\;\frac{25}{3}\) ft/sec.

 

[Guest]

Sorry about not defining the variables; I was thinking too fast.
You are right tkhunny, x is the length from the pole to the man, and y is the length from the man to the tip of the shadow.

soroban, I didn't think of setting up the problem that way. That's much easier.

thanks.