A 6 foot tall man walks at a rate of 5 feet per second along one edge of a road that is 30 feet wide. On the other edge of the road is a light atop a pole 18 feet high. How fast is the length of the man's shadow increasing when he is 40 feet beyond the point directly across the road from the pole?

...I drew a diagram but I can't figure out the equations to find the length of the shadow, I distinguished as s, in terms of the distance away from the pole (as if on the same side of the road as the man), distinguished as x

Wow. This one gets tricky, since we're dealing in 3D, rather than the usual 2D.

If the bottom of the pole is at B, and the top is at T, and the point directly across the road is at Q, then if the man is x from Q, the diagonal distance across the road (BQ) is y, then

y^2 = x^2+30^2

Now, using similar triangles, if the man's shadow has length s, then

s/6 = (s+y)/18

Now, when x=40, y=50 and s=25
Taking derivatives, we have

y dy/dt = x dx/dt
50 dy/dt = 40*5
dy/dt = 8

1/6 ds/dt = 1/18 (ds/dt + dy/dt)
3 ds/dt = ds/dt + 8
2 ds/dt = 8
ds/dt = 4

To solve this problem, we can use similar triangles and the concept of rates. Let's break down the problem step by step.

1. First, let's label the given information on our diagram:
- The height of the man is 6 feet.
- The height of the pole is 18 feet.
- The width of the road is 30 feet.
- The distance of the man from the point directly across the road from the pole is x feet.

2. Now, we need to find the length of the man's shadow when he is x feet away from the pole. Let's label this length as s.

3. Looking at the diagram, we can see that we have two similar triangles: the triangle formed by the man, his shadow, and the road (let's call this triangle 1), and the triangle formed by the pole, its shadow, and the road (triangle 2).

4. Let's define some variables:
- Let h be the length of the man's shadow that lies on the road.
- Let s1 be the length of the man's entire shadow.
- Let s2 be the length of the pole's shadow.

5. We can see that the ratio of the respective sides of triangle 1 and triangle 2 will be the same:
- h / s1 = 6 / 18 (height of man / height of pole)
- h / s1 = 1 / 3

6. We know that the width of the road (30 feet) is equal to the sum of h and s2:
- h + s2 = 30

7. We can solve equations 5 and 6 simultaneously to find the values of h and s2:
- h / s1 = 1 / 3 (equation 5)
- h + s2 = 30 (equation 6)

Rearranging equation 5, we get:
- h = (1/3) * s1

Substituting the value of h in equation 6, we get:
- (1/3) * s1 + s2 = 30

8. We can now express s2 in terms of s1:
- s2 = 30 - (1/3) * s1 (equation 7)

9. Next, we need to find the rate at which the man's distance from the pole, x, is changing. We are given that the man is walking at a rate of 5 feet per second. Therefore, the rate of change of x with respect to time (dx/dt) is 5.

10. We are asked to find the rate at which the length of the man's shadow is changing with respect to time, ds1/dt, when the man is 40 feet beyond the point directly across the road from the pole.

11. Using similar triangles, we can relate the rates of change of the corresponding sides. Let's differentiate equation 7 implicitly with respect to time (t):
- d(s2)/dt = d(30 - (1/3) * s1)/dt
- d(s2)/dt = 0 - (1/3) * ds1/dt
- d(s2)/dt = -(1/3) * ds1/dt

12. We want to find ds1/dt when x = 40. From the diagram, we can observe that s1 = x + h. Substituting this into equation 11:
- d(s2)/dt = -(1/3) * d(x + h)/dt
- d(s2)/dt = -(1/3) * (dx/dt + dh/dt)

13. We are given that the man's rate of change in height (dh/dt) is 0 since he is not growing or shrinking. Therefore, equation 12 simplifies to:
- d(s2)/dt = -(1/3) * dx/dt

14. Substituting the given values into equation 13:
- d(s2)/dt = -(1/3) * 5
- d(s2)/dt = -5/3

15. Finally, we have the value of d(s2)/dt. Since s2 represents the length of the pole's shadow, d(s2)/dt is the rate at which the length of the pole's shadow is changing. However, we are asked to find the rate at which the length of the man's shadow (s1) is changing. We can now relate d(s1)/dt to d(s2)/dt:
- d(s1)/dt = -d(s2)/dt
- d(s1)/dt = -(-5/3)
- d(s1)/dt = 5/3

16. Therefore, the rate at which the length of the man's shadow is increasing when he is 40 feet beyond the point directly across the road from the pole is 5/3 feet per second.