illumination inversely proportional.... Find pt where sum is

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littlejodo

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The illumination at a point is inversely proportional to the square of the distance of the point from the light source and directly proportional to the intensity of the light source. If two light sources are 20 feet apart and their intensities are 80 and 30 respectively, at what point between them will the sum of their illumination be a minimum?

Let x be the distance from the brighter source at which the sum of illumination is minimum.

x = ? feet

Drawing the diagram for this one was a lot easier, but figuring out the wording of this problem has my head spinning a bit. This is what I came up with, sadly, incorrectly:

If the illumination is inversely proportional to the distance, that says to me = 1/x^2
being directly proportional to the intensity says = y for either 80 or 30

I took a shot in the dark (no illumination here) and thought that you might multiply them together to get:
Illumination = y/x^2 and I could just add the two together like this:
80/x^2 + 30/(20-x^2) then take the derivative and get the x and y values. -- the 20 - x is the make up for only measuring distance from one side.

I ended up getting :
-160x - 60x^-3 = 0
-160 = 60x^-3
-160x^4 = 60 --- multiplied by x^3
x^4 = 60/-160
x = (60/-160)^1/4 ---- OOPS! can't take the 4th root of a negative....

Something tells me that I didn't have the right equation to start with... any help???
 
Re: illumination anyone??

"Let x be the distance from the brighter source at which the sum of illumination is minimum."

This is a good definition. Go with it.

Illumination from 80 is \(\displaystyle \frac{80k}{x^{2}}\)

Illumination from 30 is \(\displaystyle \frac{30j}{(20-x)^{2}}\)

Total Illumination \(\displaystyle \frac{80k}{x^{2}} + \frac{30j}{(20-x)^{2}}\)

j and k are constants. I'm expecting you to figure out if they can be ignored.
 
Re: illumination anyone??

I appreciate the help...

My teacher had said, "when finding the derivative, forget all constants." So I guess I would have to ignore 80k and 30j.. only then I am still in a tough spot.

I am not sure where k and j come from or what they really stand for... but when I eliminate them I guess I get:

1/x^2 + 1/(20-x)^2 = 0

Somehow I always get mixed up in the differentiation here, I get a different answer almost every time. I think it is because I don't really know what to do with the 1s in the quotient rule.. or I could make x^2 = x^-2 and then use the product and chain rules...?

x^-2 + (20-x)^-2 = 0
-2x^-3 - 2(20-x)^-3(-1) = 0
-2x^-3 + 2(20 -x)^-3 = 0
where do I go from here? what is the best way to get rid of the ^-3?

I had thought of moving the second term to the right and then multiplying both by x^4 but that wouldn't take care of the 20 in the denominator, right?

please help. thanks so much.
 
Re: illumination anyone??

littlejodo said:
... My teacher had said, "when finding the derivative, forget all constants." ...
Click to expand...


I certainly hope not! Otherwise, your teacher is sloppy. It seems probable to me that you either misunderstood, or have misapplied, what was actually said.
So I guess I would have to ignore 80k and 30j ...
Click to expand...


When TK writes the pronoun "they", he refers to the previous nouns, which are "j" and "k", not "80j" and "30k". Is that English not clear? TK did not write anything about ignoring the parameters in this exercise (80 and 30). If they could be ignored, then what exactly would this exercise be talking about? (This question is rhetorical.)

Is this a calculus-based physics course, and you're having trouble understanding the physics? Is this a calculus class, and you've been given this word problem without sufficient information to write a function definition for the total illumination?

If you tell me what class you're taking, then I may be able to help you.

~ Mark :|

 
Re: illumination anyone??

It is an online calculus course, so there are online lectures and the textbook. In the future I will never again take a math class online if I can help it.

I spend a lot ( i mean a LOT) of time trying to understand what is going on, but this word problem is an example of something I have difficulty with.

thank you for your help.
 
Re: illumination anyone??



Direct variation and inverse variation are pre-calculus topics.

DIRECT:

y = c * x, where c is a non-zero real constant

INVERSE:

y = c / x

Instead of y, c, and x, I now switch to:

L = illumination

i = intensity

d = distance from light source

You were on the right track with "inversely proportional to the square of the distance ... and directly proportional to the intensity"

L = i /d^2

In this exercise, there are two light sources.

L = L1 + L2

L1 = 80/x^2

L2 = 30/(20 - x)^2 because the distance from the dimmer light source is 20 - x. (You wrote that you drew a picture, so this should be clear.)

I calculated a point that is 8 feet 4.5 inches from the dimmer light source (rounded to the nearest half inch).

Thank you for continuing to show your work.

Cheers,

~ Mark :)

 
Re: illumination anyone??

So I have L = 80k/x^2 + 30j/(20-x)^2

When I differentiate:
-160k/x^3 - 60j/(20-x)^3

I can set them equal and cross multiply:
-160k/x^3 = 60j/(20-x)^3
(-160k)(20-x)^3 = (60j)(x^3) --- I know that somehow I need to do something with k and j and I checked out the linked example but I didn't follow at this point.

I tried moving things around to solve for j and k but things got deep:
(-160k)(20-x)^3 = (60j)(x^3) --- then I divided by all but the k
k = (60jx^3)/[(-160)(20-x)^3] --- then I went back up a step and did the same for j
j = (-160k)(20-x)^3 / 60x^3
now when I try to sub them back in i get miles-long equations...
(-160((60jx^3)/[(-160)(20-x)^3])(20-x)^3 = (60j)(x^3) --- is this right? Is there a better way to decrease the variables in play??

thank you so much!
 
Re: illumination anyone??

littlejodo said:
So I have L = 80k/x^2 + 30j/(20-x)^2

When I differentiate:
-160k/x^3 - 60j/(20-x)^3 <<<< That should be +60j/...

\(\displaystyle \frac {160k}{x^3} \, = \, \frac{60j}{(20-x)^3}\)

\(\displaystyle [\frac{20-x}{x}]^3 \, = \, \frac{60j}{160k} \,= \, C_1 \, ---- some \, constant\)

\(\displaystyle [\frac{20-x}{x}] \, = C_1^{1/3} \, = C\)

Now solve for 'x'....
Click to expand...
 
Re: illumination anyone??

When I differentiate:
-160k/x^3 - 60j/(20-x)^3 <<<< That should be +60j/...
Click to expand...

I didn't get this in the other example either, perhaps you can clue me in? I thought that the negative in the 160 came from the fact that x^2 in the denominator turns into a -2 when using the product rule, so why wouldn't it be the same for the (20-x)^2 in the denominator? What is the difference in these two terms that I am missing? (I'm sure this is something simple that I just missed along the way, better late than never, right?)

Thanks again!!
 
Re: illumination anyone??

littlejodo said:
When I differentiate:
-160k/x^3 - 60j/(20-x)^3 <<<< That should be +60j/...
Click to expand...

I didn't get this in the other example either, perhaps you can clue me in? I thought that the negative in the 160 came from the fact that x^2 in the denominator turns into a -2 when using the product rule, so why wouldn't it be the same for the (20-x)^2 in the denominator? What is the difference in these two terms that I am missing? (I'm sure this is something simple that I just missed along the way, better late than never, right?)

Thanks again!!
Click to expand...

Following chain rule, you'd have to differentiate (20 -x) and that would be (-1).
 
Re: illumination anyone??

solving for x, if I keep the constant I get that x = 20/(C+1)

If I turn the constant C into zero then I get x = 20

If I go with x = 20, and then look back at the origin of C as C^1/3 then would I take 20^3 to get the "real" C? and then put that equal to 60j/160k and try to get values for j and k?

If I take x = 0 and put it back into the original equation of 80k/x^2 + 30k/(20-x)^2 then I have a problem because 20 -20 = 0 and that would leave the second term undefined.

I'm running out of paper... :)
 
Re: illumination anyone??

are you saying that I should keep the answer of x = 20/c+1 ?

if j = k, wait, how does that help me? I am so sorry... by now I have no idea where I am in the process. I think I am still confused by the existence of j and k at all.. then add in the C ... I'm lost in alphabet soup.
 
Re: illumination anyone??

I've come back to this problem several times today. I have others like it and I think if I could just make sense of the process then I'd be on my way...

I appreciate all the help I've received so far, but if any more help could be offered it would be great. I do a lot better understanding plain english than mathematical symbols and such. Please?
 
Re: illumination anyone??

Slow, Careful, One Step at a Time

Find the derivative:

\(\displaystyle -\frac{160k}{x^{3}}-\frac{60j}{(x-20)^{3}}\)

Set this to zero and use some algebra to get this equivalent expression

\(\displaystyle -160k(x-20)^{3}-60jx^{3} = 0\)

This may seem a little magic, but it really comes only from the realization that the denominator does not contribute much to whether or not the entire expression is zero (0). That's why I picked only the numerator to ponder.

This is a GENERAL solution, so it may be more complicated than necessary. I have assumed that the two light sources have different parameters. It is reasonable to assume that j = k in the absence of additional information. This gives...(after a little algebra)

\(\displaystyle -20k(11x^{3}-480x^{2}+9600x-64000) = 0\)

NOW we get to pat ourselves on the back and say, "Hmmm...I think those constants out front don't help much."

Find x for which -20 = 0 -- None!

Find x for which k = 0 -- None!

Find x for which \(\displaystyle 11x^{3}-480x^{2}+9600x-64000 = 0\) -- Ah! Now THERE'S a problem we can solve.

There is only one Real Solutions. Find it! Of course, it is useful to note that it should be somewhere in this neighborhood [0,20].

When the day rolls around that we discover that j=k is FALSE, then we have to rework things a bit. For now we're done.
 
Re: illumination anyone??

The illumination at a point is inversely proportional to the square of the distance of the point from the light source and directly proportional to the intensity of the light source. If two light sources are 20 feet apart and their intensities are 80 and 30 respectively, at what point between them will the sum of their illumination be a minimum?

Let x be the distance from the brighter source at which the sum of illumination is minimum.
Click to expand...

Summary: From setting your derivative to zero, you had

(-160k)(20-x)^3 = (60j)(x^3)

which had a sign error that Subhotosh corrected. So then

(160k)(20-x)^3 = (60j)(x^3)

Subhotosh also pointed out that j=k, so they cancel:

(160)(20-x)^3 = (60)(x^3)

[Note: At this point, the tutors have given us everything we need to solve.]

***

New stuff: Solving for x,

x = 11.6203 feet

Luminosity here is about .99 luminosity units.
 
Re: illumination anyone??

I'll just complete the algebra - since good amount of time has passed...

wjm11 said:
The illumination at a point is inversely proportional to the square of the distance of the point from the light source and directly proportional to the intensity of the light source. If two light sources are 20 feet apart and their intensities are 80 and 30 respectively, at what point between them will the sum of their illumination be a minimum?

Let x be the distance from the brighter source at which the sum of illumination is minimum.
Click to expand...

Summary: From setting your derivative to zero, you had

(-160k)(20-x)^3 = (60j)(x^3)

which had a sign error that Subhotosh corrected. So then

(160k)(20-x)^3 = (60j)(x^3)

Subhotosh also pointed out that j=k, so they cancel:

(160)(20-x)^3 = (60)(x^3)

\(\displaystyle [\frac{20-x}{x}]^3 \, = \, \frac{60}{160} \, = \, 0.375\)

\(\displaystyle [\frac{20-x}{x}] \, = \, [0.375]^{\frac{1}{3}} \, = \, 0.721124785\)

\(\displaystyle \frac{20}{x} \, = \, 1.721124785\)

\(\displaystyle x \, = \, \frac{20}{1.721124785} \, = \, 11.62030794\)

Now prove that lumonisity is indeed minimum here (x = 11.62030794) ...

[Note: At this point, the tutors have given us everything we need to solve.]

***

New stuff: Solving for x,

x = 11.6203 feet

Luminosity here is about .99 luminosity units.
Click to expand...
 
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