# Differentiation Story Problem

1. Nov 10, 2008

### =Lawrence=

1. The problem statement, all variables and given/known data

A man 6 feet tall walks at a rate of 5 feet per second toward a streetlight that is 30 ft high. The man's 3 ft tall child follows at the same speed, but 10 feet behind the man. At times, the shadow behind the child is caused by the man, and at other times, by the child.

a) Suppose the man is 90 feet from the streetlight. Show that the man's shadow extends beyond the child's shadow.

I solved this by drawing it out and using similar triangles and fairly simple algebra. Man's shadow is 112.5 ft from light. Kid's is 111+(1/9) ft from light.

b) The same as a, but with different answers.

I solved this the same way. Man's shadow is 75 ft from light. Kid's is 77+(7/9) ft from light.

c) Determine the distance from the man to the streetlight at which the tips of the two shadows are the same distance from the streetlight.

Same thing. d=80 ft

d) Determine how fast the tip of the shadow is moving as a function of x, the distance between the man and the street light. Discuss the continuity of this shadow speed function.

This is the part I don't know how to do. I have the answer, since it's in the back of the book, but that doesn't help me. I need to know how to do it.

2. Relevant equations

Rules of differentiation.

3. The attempt at a solution

I don't know where to start.

2. Nov 10, 2008

### Dick

If you've got the answers, I'll presume the previous ones are correct. If x is the distance from the man to the streetlight, then make d the distance from the man to the tip of his shadow. Use your similar triangles to find a relation between x and d. You must have already done this, right? Then what the relation between d/dt(x), the speed of the man, and d/dt(x+d) the speed of the shadow?

3. Nov 10, 2008

### =Lawrence=

I don't know. We've only done this kind of problem once before in class and I didn't understand it then. I could get the other stuff because it was geometry/algebra, but this part of the problem actually requires calculus, and at that, calculus I haven't done before.

dx/dt= -5 ft/s

24d=6x

24(dd/dt)=6(-5)

dd/dt=-30/24

Hmm...I just checked the back. That's not right. Apparently they came up with this:

Let x be the distance of the man from the light and let s be the distance from the light to the tip of the shadow. If 0 < x< 80, ds/dt= -50/9. If x>80, ds/dt= -25/4. There is a discontinuity at x = 80.

I have no idea where they're getting these numbers. They have basically the same format though, which makes me think I'm doing something right. I can't help but think it should be a lot more complicated though.

Last edited: Nov 10, 2008
4. Nov 11, 2008

### HallsofIvy

Staff Emeritus
What equation did you get connecting the distance the man is from the light, x, and the length of the shadow, say s? Let's call that s= f(x). Then ds/dt= f'(x) dx/dt and you are told that dx/dt= -5 ft/s. ds/dt= -5 f'(x)

5. Nov 11, 2008

### =Lawrence=

That's where I falter. I had 24x=6s. That's how I did the other problems.

6. Nov 11, 2008

### Dick

Ok, they want the speed of the tip of the shadow. Not the rate of change of the length of the shadow. So let s be the distance of the end of the shadow from the light. Then I get 6/(s-x)=30/s. Do you agree?