Question

A man 6 feet tall walks at a rate of 5 ft/sec toward a light that is 20 ft. above the ground. When he is 10 feet from the base of the ligh, at what rate is the tip of his shadow moving?

Answer 1

similar triangles gives us:

20/d+s = 6/s

20s = 6(d+s)

this is the relation between the stuff i beleive

Answer 2

[20s = 6(d+s)]'

20 s' = 6 (d'+s')

10s' = 3d'+3s'

10s'-3s' = 3d'

s'(7) = 3d'

s' = 3d'/7 is what I get for the rate of the shadow itself

Answer 3

d' = 5 since the man is walking at a rate of 5 ....

Answer 4

so if i did it right :) the shadow with respect to the mans walking is moving at 15/7 units per sec; and if we add that to the rate at which the person is walking we get:

15/7 + 5 ; and it might need to be negative since its towards the light

Answer 5

id prolly feel better if we did this with trig instead of similar triangles; i always feel like im missing something.

Answer 6

w = walking distance, and s = shadows length

tan(a) = (w+s)/20

[20 tan(a) = w+s]'

20 sec^2(a) a' = w' + s'

20 sec^2(a) a' - w' = s'

ewww, maybe not, since we dont know the rate of change of the angle, we could prolly find it out form the info, but it looks messier

Answer 7

the concern i have is that this seems to work out the same no matter how far from the light he is since there is no place to plug in a value for the distance he is from the light ....

Answer 8

\[\frac{ds}{dt}=\frac{ds}{dw}*\frac{dw}{dt}\]
\[\frac{ds}{dt}=\frac{ds}{dw}*5\]
as long as the rate of change of the shadow with repect to walking is 3/7 regardless of distance from the light, its good ; but i doubt that

Answer 9

Dw [20s = 6(w+s)]

20s' = 6(1+s')

20s' = 6+6s'

20s'-6s' = 6

s'(14) = 6 = 3/7 ....

Answer 10

Thanks a lot!