Question

The intensity of a light source is
S/x^2
, where S is the strength of the light source, and x is the distance from the source. Suppose we place a lamp of strength 9 units on a number line at the origin, and another of strength 1 unit at a position D > 0.

Answer 1

If the lamps shine in all directions, only the distance from the lamps matters.


equation for Lamp 1:
I = 9 / (x-0)^2 lamp 1 is at x=0, so the distance to that is the same as the
value of x. since it's squared, this is also for x on the left.

equation for Lamp 2:
I = 1/ (D-x)^2 distance formula. If D is at 13, and x is at 12, 13-12=1.
If D is at 13, and x is at 14, 13-14=-1.
the result is again squared, so we need no absolute.

The total intensity the object receives, is the light of lamp1 + light of lamp2.

Answer 2

So \[I_{total} = \frac{ 9 }{ x^2 } +\frac{ 1 }{ (D-x)^2 }\]

part (b) how can I get the critical number ?

Answer 3

differentiate the equation and equal it to zero ?

Answer 4

Yes - the function models the intensity, which will go down when you move to the right from the light with intensity "9", it will reach a lowest strength and will only go up when you get very close to D with the light intensity "1".

Since the curve always goes down then always goes up, the lowest point (where intensity stops to decrease and starts to increase, "turning point") has a derivative of 0.
At that position there is no longer a decrease but not yet an increase, and you find that point exactly like you described by taking derivative and equal it to zero.

Answer 5

sorry but i found some difficulties while differentiating :( @phanta_seea

Answer 6

@amistre64 , @experimentX any idea for part (b)

Answer 7

i think I am having trouble understanding question ... what does the question on b) say?

Answer 8

I think they want the critical points for the function
It is in the attachment

Answer 9

do you mean to say that you need to find the interval on which the function is increasing and on which the function is decreasing??

Answer 10

yes

Answer 11

well well ... do this

Answer 12

http://www.wolframalpha.com/input/?i=Solve%5BD%5B9%2Fx%5E2+%2B+1%2F%28x-D%29%5E2%2C+x%5D+%3D%3D+0%2C+x%5D

Answer 13

see that first value,
http://www4a.wolframalpha.com/Calculate/MSP/MSP3811e363a773hf2fg34000047e87bb11a3730ic?MSPStoreType=image/gif&s=41&w=501.&h=36.

you get something like this,