Light between 2 towers reflecting off a lake

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In summary: You're supposed to use Fermat's principle of least time, which can be used to derive the law of reflection. For this exercise they want you to use calculus.In summary, the author is asking for help with a calculus problem, and suggests using Fermat's principle of least time to solve it. The author provides a sketch of the situation and explains that, in order to solve the problem, they need to find the equation of line BF. They also mention that they have forgotten something extremely simple, but they are not seeing where to go from here.
  • #1
ArmChairPhysicist
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Please forgive me if I'm in breach of some etiquette with asking for help here, but calculus isn't my strongest subject, and I learn best by example.

I am currently studying in the Life Of Fred Calculus book (do I need to state copyright or something?). I am attempting to work a problem that reads as such:

"Two towers are 120 feet apart. The one on the left is 7 feet tall and the one on the right is 21 feet tall. A beam of light from the top of the left tower bounces off of a lake (note: I assume the lake spans the distance between the two towers, perfectly in line with them and not off somewhere else.) and hits the top of the other tower. Assuming the light takes the shortest path, how far from the base of the left tower will the beam strike the lake?" Authors note: "The algebra- not the calculus- is tough. After you have taken the derivative and set it equal to zero, try x= 10, 20, 30... and one of those will work."
Things I know need to be done.
As the problem states, I'm looking for the length base of a triangle that has 7 feet (the tower) as its height and the beam of light as the hypotenuse. I know I'm going to need the almighty derivative of an equation relating to the triangle, but I'm not sure how to get that equation.
Is there anyone that has experience with this book, or calculus in general, that could help me?
Many thanks in advance for your time.
 
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  • #2
Forgot one thing, I know the answer, the distance from the base is fifty feet. But I'm stuck on figuring out how to solve this problem to get that answer in the first place.
 
  • #3
Can you make a sketch of the situation? From this you should be able to write an expression for the length of the light path.
 
  • #4
I can. Give me a minute to photograph it.
 
  • #5
Hint: Assume the light reflects off of the lake at a distance X from the base of the left tower. Think of the distance traveled by the light as having two parts: Top of left tower to lake (point X) and lake to top of right tower. Each part is the hypotenuse of a triangle: use a bit of trig to express the total distance traveled as a function of X.
 
  • #6
image.jpg
If I am understanding the problem, I need to find the equation of line BF, take the equation's derivative (praise the almighty derivative), set the derivative equal to zero, and then solve by algebra. Where I've gotten stuck is finding out how to get the equation of BF so that AF and DF connect to the two towers as described in the problem. I'm sure that I've forgotten something extremely simple, but I'm not seeing where to go from here.
 
  • #7
Ah there it is, trig. I'll try this
 
  • #8
If it's a reflection, doesn't that constrain the angles of incidence and reflection...?
 
  • #9
I thought about the laws of reflection, but given the confines of the book (calculus and not physics) I doubted the author would assume the reader would know or remember those rules, considering they aren't mentioned
 
  • #10
berkeman said:
If it's a reflection, doesn't that constrain the angles of incidence and reflection...?
You're not supposed to use that fact. But that will be the answer.
 
  • #11
ArmChairPhysicist said:
I thought about the laws of reflection, but given the confines of the book (calculus and not physics) I doubted the author would assume the reader would know or remember those rules, considering they aren't mentioned
Right. You are actually using Fermat's principle of least time, which can be used to derive the law of reflection. For this exercise they want you to use calculus.
 
  • #12
Doc Al said:
You're not supposed to use that fact
Oopsies! (Kobayashi-Maru...)
 
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  • #13
Your diagram is incomplete. Draw the point F on the surface of the lake (just guess where it is), then draw the lines BF and DF. You should then be able to write expressions for the length of the line BF in terms of x and the length of the line DF in terms of x.
 
  • #14
ArmChairPhysicist said:
If I am understanding the problem, I need to find the equation of line BF, take the equation's derivative (praise the almighty derivative), set the derivative equal to zero, and then solve by algebra. Where I've gotten stuck is finding out how to get the equation of BF so that AF and DF connect to the two towers as described in the problem. I'm sure that I've forgotten something extremely simple, but I'm not seeing where to go from here.
Draw the point F on your diagram. (The distance BF is what I called "X".) Then draw lines AF and DF; those are the two parts of the light's path that I referred to.

Use trig to find expressions for lengths AF and DF.
 
  • #15
I've done that, working on the trig now.
 
  • #16
In triangle abf , the only known values are angle abf = 90' and side AB = 7 feet. I don't know the hypotenuse, because to find that I need side BF, which is the whole reading I'm working this, to find BF. And I must be overlooking a trig function, because to use any of the six, I need at least another angle or side.
 
  • #17
And technically I already know BF= 50, but working the problem backwards isn't going to teach me much.
 
  • #18
You 'know' the side BF, it's distance is x. So write the hypotenuse AF in terms of AB (7 feet) and x.
 
  • #19
But can I even solve that with two out of three variables being unknown?
 
  • #20
7^2 + X^2 = C^2. C=AF
 
  • #21
ArmChairPhysicist said:
But can I even solve that with two out of three variables being unknown?
Just do as phyzguy says: Express the hypotenuse in terms of X. You're not solving for anything yet.
 
  • #22
ArmChairPhysicist said:
7^2 + X^2 = C^2. C=AF
Good. Express C in terms of X.

Now do the same for the other piece of the path, DF.
 
  • #23
If BF = X, what must CF equal?
 
  • #24
So we have
AF = sqrt(7^2 + X^2)
DF = sqrt(21^2 + (120-X)^2)
 
  • #25
CF is equal to 120-X
 
  • #26
ArmChairPhysicist said:
But can I even solve that with two out of three variables being unknown?

There is only one unknown, the distance BF, which I suggested you call X. Given that, the distance AF is given, as you said by AF^2 = 7^2 + X^2, or AF = √(7^2 + X^2).
 
  • #27
I have a feeling the chain rule will be needed soon.
 
  • #28
I'll try applying that and see if that gets me somewhere, unless there is another step that I should do first.
 
  • #29
ArmChairPhysicist said:
So we have
AF = sqrt(7^2 + X^2)
DF = sqrt(21^2 + (120-X)^2)
Perfect. Now you must minimize AF + DF.
 
  • #30
ArmChairPhysicist said:
I have a feeling the chain rule will be needed soon.
Oh yes.
 
  • #31
Stand by, mathematics in progress.
 
  • #32
If I recall correctly, because both expressions are of the same power: ^1/2, I can combine them without any issue, correct?
 
  • #33
ArmChairPhysicist said:
If I recall correctly, because both expressions are of the same power: ^1/2, I can combine them without any issue, correct?

Combine them how? If you mean [itex] \sqrt(a) + \sqrt(b) = \sqrt(a + b)[/itex], then no, you can't do that.
 
  • #34
So I have
(7^2 + X^2)^1/2 + (21^2 + (120-x)^2)^1/2 as the total length
 
  • #35
ArmChairPhysicist said:
So I have
(7^2 + X^2)^1/2 + (21^2 + (120-x)^2)^1/2 as the total length

Correct. Good job. Now use calculus to find the value of X which minimizes the length.
 
<h2>1. How does light reflect off a lake?</h2><p>When light hits the surface of a lake, it bounces off at an angle determined by the law of reflection. The angle of incidence (the angle between the incoming light ray and the normal, or perpendicular, to the surface) is equal to the angle of reflection (the angle between the reflected light ray and the normal).</p><h2>2. Why does the light appear to form a straight line between the two towers?</h2><p>When light reflects off the surface of a still body of water, it creates a mirror-like effect. This means that the reflected light appears to be coming from a virtual image of the towers located beneath the surface of the water. The virtual image is a straight line between the two towers, giving the appearance of a straight line of light.</p><h2>3. Does the color of the light change when it reflects off the lake?</h2><p>Yes, the color of the light can change when it reflects off a lake. This is because the water can absorb or scatter certain wavelengths of light, causing a change in the overall color of the reflected light. Additionally, the angle of the light and the properties of the water (such as clarity and depth) can also affect the color of the reflected light.</p><h2>4. Why does the light appear brighter on the lake's surface compared to the surrounding area?</h2><p>The light appears brighter on the lake's surface because the water acts as a reflective surface, directing more light towards the viewer's eyes. This creates a contrast between the bright reflection on the water and the darker surrounding area, making the light appear brighter on the lake's surface.</p><h2>5. Can the light between the two towers be seen from all angles?</h2><p>No, the light between the two towers can only be seen from certain angles. This is because the angle of reflection is equal to the angle of incidence, meaning that the light will only reflect towards an observer if they are in the correct position to receive the reflected light. If an observer is not in the correct position, they will not see the light between the two towers on the lake's surface.</p>

1. How does light reflect off a lake?

When light hits the surface of a lake, it bounces off at an angle determined by the law of reflection. The angle of incidence (the angle between the incoming light ray and the normal, or perpendicular, to the surface) is equal to the angle of reflection (the angle between the reflected light ray and the normal).

2. Why does the light appear to form a straight line between the two towers?

When light reflects off the surface of a still body of water, it creates a mirror-like effect. This means that the reflected light appears to be coming from a virtual image of the towers located beneath the surface of the water. The virtual image is a straight line between the two towers, giving the appearance of a straight line of light.

3. Does the color of the light change when it reflects off the lake?

Yes, the color of the light can change when it reflects off a lake. This is because the water can absorb or scatter certain wavelengths of light, causing a change in the overall color of the reflected light. Additionally, the angle of the light and the properties of the water (such as clarity and depth) can also affect the color of the reflected light.

4. Why does the light appear brighter on the lake's surface compared to the surrounding area?

The light appears brighter on the lake's surface because the water acts as a reflective surface, directing more light towards the viewer's eyes. This creates a contrast between the bright reflection on the water and the darker surrounding area, making the light appear brighter on the lake's surface.

5. Can the light between the two towers be seen from all angles?

No, the light between the two towers can only be seen from certain angles. This is because the angle of reflection is equal to the angle of incidence, meaning that the light will only reflect towards an observer if they are in the correct position to receive the reflected light. If an observer is not in the correct position, they will not see the light between the two towers on the lake's surface.

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