math word problem involving derivatives.?

Question

a six-foot tall man is walking away from a street lamp 9 feet high and towards a wall that is 7 feet high at a constant rate of 5 feet per second. What is the rate of his shadow's lengthening on the wall? (this question needs a knowledge of derivatives. It also uses similar triangles. This is a question in HIGH SCHOOL. LOL...)

Answer 1

I answered this once over in the math category, but I have a better way to do it. We'll get everything in terms of time.

Draw this similar-triangle diagram: Light post on the left at x=0, wall in center-right at x=22. On the ground, mark off the point x = 22/3. That's where the man's shadow first reaches the base of the wall. Put the man somewhere in the middle at x > 22/3, so that his shadow extends beyond the wall. The end of the shadow is at x = d.

The similar-triangle relationships are:

9/d = 6/(d-x) = h/(d-22)

Eliminate d this way:

9(d-x) = 6d ==> 3d = 9x ==> d = 3x

9(d-22) = hd ==> (9-h)d = 22 x 9 = 198 ==> d = 198/(9-h)

3x = 198/(9-h) ==> 9-h = 198/3x = 66/x ==> h = 9 - 66/x

Now, x (the man's location) is a function of time: x = 5t.

Therefore, h(t) = 9 - 66/5t, subject to the condition that h is between 0 and 6 feet.

To get the rate of change of the shadow, take the derivative:

dh/dt = 66/5t^2

The limits on t are:

(a) 0 = h = 9 - 66/5t ==> t = 66/45 seconds

(b) 6 = h = 9 - 66/5t ==> t = 22/5 seconds

So when the man starts walking away from the lamp post, his shadow does not appear on the wall for 66/45 seconds (about a second and a half). Then the shadow on the wall lengthens at a rate dh/dt = 66/5t^2 until t = 22/5 seconds (4.4 seconds away from the post), at which time the man reaches the wall, and the length of his shadow is h = 9 - 66/[5(22/5)] = 9 - 3 = 6 feet.

In the approximately 3 seconds the shadow is growing, its rate of growth is dh/dt = 66/5t^2, where t=0 is at the base of the light post.

That's your answer. You could graph it.

Answer 2

Initially at least, his shadow won't be on the wall, so the answer must be discontinuous at 0 for some range. I can't be bothered working out the rest.