math word problem... 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.)

uh... sorry, the wall is 22 feet away from the foot of the lightpost.

Answer 1

I think this can be done using similar triangles. Assume the distance between the base of the lamp and the base of the wall is d. For starters, place the man so that his shadow just reaches the base of the wall. So we're on the same page, suppose the lamp post is on the left and the wall is on the right. His distance from the base of the lamp is x. (Draw that diagram. You'll have similar triangles.)

Then, 9/d = 6/(d-x) ==> 9d - 9x = 6d ==> 3d = 9x ==> x = d/3. His shadow first appears on the wall when he's walked a third of the way toward it.

Now let's try to get an expression for the height of the shadow on the wall. Draw another diagram, this time placing the man further toward the wall -- maybe halfway or two-thirds between the lamp post and the wall. You need to extend the diagram to the right to see where the shadow would end .

Your diagram will have three similar right triangles with these distances: from the post to the man is again x; from the post to the wall is again d; and this time, from the post to the end of the man's shadow is a. The height of the shadow on the wall is h.

Using smilar triangles, we have these relations:

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

Eliminate the a to get h = f(x) in this manner:

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

9(a-d) = ha ==> (9-h)a = 9d ==> a = 9d/(9-h)

Set those two equal to each other, eliminating a:

3x = 9d/(9-h) ==> 9-h = 3d/x ==> h = 9 - 3d/x

This is what we want. We have h as a function of x. Now take the derivative with respect to time:

dh/dt = 3d/x^2 dx/dt

You know that dx/dt is 5 feet per second, so

dh/dt = (3/5) d/x^2

That might be your answer; the rate of change of the shadow varies inversely with the square of the man's distance from the light post.

But it occurs to me that you can do more with this problem ... you have two boundary conditions. You know that the shadow h equals zero when x = d/3, and you know that h equals 6 when x = d.

Although I'm tired of working on this, I might redefine it so x = d/3 when t=0 (when the shadow first appears on the wall), and h=0 when t=0. (I said that awkwardly. To restate: x(0) = d/3, and h(0) = 0, where x and h are functions of t.)

The guy walks a distance d - d/3 = 2d/3 in s seconds at a rate of 5 feet per second. Since distance equals rate times time, we have 2d/3 (distance) = 5s, so s = 2d/15 seconds. Then we have x(s) = d and h(s) = 6.

That's your second boundary condition. If I were to continue working on this interesting problem, I'd reframe it in terms of time rather than the distance x. That way, I'd get an expression for h(t) -- the height of the shadow as a function of time, and then all you have to do is take the derivative to get your answer dh/dt directly.

P.S. I just might go ahead and work on this some more later, but right now I need a break. In any event, I hope this helps.

Answer 2

How far away is the light from the wall? Give me that info and I might be able to answer your question.