Difficult calculus problem. Related rates?

Question

How to answer this?

Math help! Related rates?
2. A Street light is 15 meters high. A man 1.687 meters tall walks away from the street light at a rate of 6 meters per minute.

a. At what rate is the tip of his shadow moving when he is 5 meters from the street light?
b. At what rate is the length of his shadow increasing?

Solution:
Please teach me how to do no. 2

Answer 1

Height of street lamp is H = 15
Height of man = h
If the man is distance x from the street light base, by similar triangles, the tip of the shadow is at y, where:
y/x = H/(H-h)
Therefore y = (H/(H-h)*x = 1.126718*x
dy/dt = 1.126718*dx/dt

Now let x = 5:

a) What is the speed of motion of the tip?
This is dy/dt = 1.126718*dx/dt = 1.126718*6 (meters/min)
= 6.76 (meters/minute)

b) The length of his shadow is:
L = y - x = 1.126718*x - x = 0.126718*x , so:
dL/dt = 0.126718*dx/dt = 0.126718*6 (meters/minute)
= 0.760308 (meters/minute)

Answer 2

Draw this figure: plot the points (0,15) and (M,0) with M > 0. The point (0,15) represents the street light and the point (M,0) is the man's position. The line segment from (M,0) to (M,1.687) represents the man. Now find the equation of the line through (0,15) and (M,1.687), and find (S,0), the point where this line hits the x-axis. The point (S,0) represents the tip of the shadow, and the segment from (M,0) to (S,0) represents the shadow.

In a, you are asked to find dS/dt when M = 5, given that dM/dt = 6. In b you are asked to find d(S-M)/dt.

Answer 3

You can use law of cosines to do it. Let x be the distance between the buddies. x^2 = 200^2 + 100^2 - 2*100*200cos(θ) Differentiate w.r.t. time, 2xx' = 2*100*200sin(θ)θ' where θ' = 7/100 rad/sec, x = 200 m, θ = cos^-1(50/200) Can you finish it now? --------- To find θ', you should know the relation v = θ' r, where v is the speed, and r is the radius.