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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

2007-09-07 20:46:11 · 4 answers · asked by Patricia 2 in Science & Mathematics Mathematics

4 answers

Let x be the horizontal distance between the man and the lightpole, and y be the horizontal distance between the his shadow tip and the lightpole.
By the similarity of two triangles,
y/15 = (y-x)/1.687
Differentiate with respect to time t,
y'/15 = (y'-x')/1.687
a.
Plug in x' = 6 m/min and solve for y',
y' = 6.76 m/min

b.
y'-x' = 0.76 m/min

2007-09-07 21:10:26 · answer #1 · answered by sahsjing 7 · 0 0

Sounds like kids in toilet training, but in order to do #2, you have to set up #1. So consider a figure with a street light on the left, a "stick"man some distance from it, and a ray from the light to the top of the man (of height 1.687m) which reaches the ground at the tip of the shadow. Let x be the horizontal distance from the lightpole base to the tip of the shadow and s the length of the shadow (on the ground between the man and the ray). At any time, x= 5*t+s.
Whatever x is, we know dx/dt=+6 m/min (a pretty slow-walking man, probably after a party).

Then, we can write dx/dt = 5+ds/dt. Then ds/dt=1 m/min

2007-09-07 21:27:46 · answer #2 · answered by cattbarf 7 · 0 0

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2016-10-19 23:08:43 · answer #3 · answered by ? 4 · 0 0

why dont you do your own homework instead of expecting us to do it for you

2007-09-07 20:54:53 · answer #4 · answered by IT'S ME AGAIN 6 · 0 3

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