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The intensity of light varies inversely as the square of the distance from its source. If two searchlights are 600 meters apart...Asked by John
The intensity of light varies inversely as the square of the distance from its source. If two searchlights are 600 meters apart and one light is 8 times as strong as the other, where should a man cross the line between them in order to be illuminated as little as possible
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Answered by
Steve
As Tom showed you,
I = A/s^2 + 8A/(600-s^2)
Actually, that is not quite right. It should be
1/s^2 + 8/(600-s)^2
You should have caught that. In any case,
The A is just a constant, so it does not affect the solution, and we can say
I = 1/s^2 + 8/(600-s)^2
dI/ds = -2/s^3 + 16/(600-s)^3
= 18(200-s)(s^2+120000) / s^3(600-s)^3
clearly, dI/ds=0 when s=200 meters from the weaker source.
I = A/s^2 + 8A/(600-s^2)
Actually, that is not quite right. It should be
1/s^2 + 8/(600-s)^2
You should have caught that. In any case,
The A is just a constant, so it does not affect the solution, and we can say
I = 1/s^2 + 8/(600-s)^2
dI/ds = -2/s^3 + 16/(600-s)^3
= 18(200-s)(s^2+120000) / s^3(600-s)^3
clearly, dI/ds=0 when s=200 meters from the weaker source.