See the diagram below, ALWAYS DRAW A DIAGRAM WHEN DOING A RELATED RATES PROBLEM.

you were right about forming a smaller triangle within the larger one with a height of 6 ft. where do you go from there? well, you use the concept of similar triangles. remember similar triangles are triangles whose sides are propotional, so the ration of any two sides of one triangle is equal to the ratio of the corresponding sides of a triangle similar to it.

so the height of the big triangle is the 16ft pole with the light on top of it. the 6 ft line x meters away from the pole is the woman (she's on the Atkin's diet which is why she's so thin).

Obviously, we will use similar triangles here.

Now 16/(x + y) = 6/y ……the ratio of height/base of the large triangle = ratio of height/base of small triangle

=> 16y = 6x + 6y

=> 10y = 6x …………….now differentiate implicitly

=> 10 dy/dt = 6 dx/dt = 6(4) = 24

=> dy/dt = 24/10 = 2.4 ft/sec

But the rate at which the tip of her shadow is moving away from the pole is equal to the rate at which she is moving plus the rate at which her shadow is extending in front of her. So then,

ds/dt = dx/dt + dy/dt = 4 + 2.4 = 6.4 ft/sec

so this is a slightly different way from how your professor did the problem, well not really, it is essentially the same, except he solved for y where i just left 10y.

i'm kind of in a rush now, i'll do the second problem later