A street light is on top of a 6m pole.

GEE, that's brilliant.

Thanks Melody and CPhill!!

The answers are the same but they look different.  It will take me a while to work it all out!

EDIT:  This answer is correct just rediculously long.  I have redone it much more efficiently on a later post.

The two triangles are similar because they each have a right angle and they have a common angle.

Let the third angle be  $$\theta$$

You know that      $$\frac{dx}{dt}=1m/s$$

a) you are being asked to find  $$\frac{dy}{dt}$$  when x=6m

b) you are being asked to find   $$\frac{d(y-x)}{dt}=\frac{dy}{dt}-\frac{dx}{dt}$$

a)  I think it will be easier for me to get y and x in terms of $$\theta$$

$$\begin{array}{rllrll}
tan\theta&=&\frac{y}{6}\qquad \qquad \qquad \qquad & tan\theta&=&\frac{y-x}{1.8}\\\\
y&=&6tan\theta}\qquad& y-x&=&1.8tan\theta\\\\
&&&x&=&y-1.8tan\theta\\\\
&&&x&=&6tan\theta -1.8tan\theta\\\\
&&&x&=&4.2tan\theta \\\\
\frac{dy}{d\theta}&=&6sec^2\theta}\qquad& \frac{dx}{d\theta}&=&4.2sec^2\theta\\\\
\end{array}$$

 ---------------------------------------------------------------------------------------------------------

              $$\begin{array}{rll}
\frac{dy}{dt}&=&\frac{dy}{d\theta}\times \frac{d\theta}{dx}\times \frac{dx}{dt}\\\\
\frac{dy}{dt}&=&6sec^2\theta}\times \dfrac{1}{4.2sec^2\theta}\times 1\\\\
\frac{dy}{dt}&=&\frac{6}{4.2}\\\\
\frac{dy}{dt}&=&\frac{10}{7}\\\\
\frac{dy}{dt}&=&1\frac{3}{7}\;\;m/s\\\\
\end{array}$$

b)  $$\frac{dy}{dt}-\frac{dx}{dt}= 1\frac{3}{7}-1=\frac{3}{7}\;\;m/s$$

So the end of the shadow is moving away at a constant rate of      $$1\frac{3}{7}m/s$$

and

The length of the shadow is changing at a constant rate of     $$\frac{3}{7}\;\;m/s$$

Since both of these are constants, it makes no difference how foar away from the pole the man is.

The answers will always be the same.     

This is an old problem from my calculus class days....let me see if I remember how to solve it.

Let x be the distance that the tip of his shadow is from the pole. Let xp be the distance the person is from the pole, and let xs be the length of his shadow.

So......x = xp + xs

We want to determine x' when the person is 6m from the pole.

Note that x'p  = 1m/s

Let's eliminate xs from the above equation.

Using similar triangles, we have

1.8/6 = xs/x = xs/(xp + xs)  so

1.8/6 = xs/(xp + xs)     (  note that 1.8/6   = .3   )

(.3) (xp + xs)  = xs

.3xp = .7xs

xs = (3/7)xp

Now.....substituting back into the original equation, we have

x = xp + (3/7)xp  = (10/7)xp    ....   So.....differentiating both sides, we have....

x' = (10/7)x'p  = (10/7)(1m/s)  = (10/7)m/s

And that's how fast the tip of the shadow is moving away from the pole.

The second part is easy.....we want to find x's when xp = 6

Again, we can use

x = xp + xs      and differentiating, we have

x' = x'p  + x's

But, we know x'p = 1m/s  and x'= (10/7)m/s.....so we have

(10/7)m/s  = 1m/s  + x's

So  x's  =  (3/7)m/s  ....   And that's how fast the length of the shadow is changing!!

Here's a diagram of the problem......

Ok. Now I am going to give a much better more concise  answer.

The triangle on the top is similar to the full big triangle because they have two equal angles.  A common right angle and a common angle theta at the top.  Therefore their sides are in the same ratio therefore,

$$\begin{array}{rll}
\frac{y}{x}&=&\frac{6}{4.2}\\\\
y&=&\frac{6}{4.2}x\\\\
y&=&\frac{10}{7}x\\\\
\frac{dy}{dx}&=&\frac{10}{7}\\\\\\
\end{array}
\frac{dy}{dt}=\frac{dy}{dx}\times \frac{dx}{dt}\\\\
\frac{dy}{dt}=\frac{10}{7}\times 1 \qquad \mbox{We were given that }\frac{dx}{dt}=1\\\\
\frac{dy}{dt}=\frac{10}{7}$$

The end of the shadow is moving away from the light pole at a constant speed to 10/7 m/sec

The shadow is increasing in length at  constant rate of 3/7 m/s