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......