Light between 2 towers reflecting off a lake

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Homework Help Overview

The discussion revolves around a calculus problem involving two towers of different heights and a beam of light reflecting off a lake. The original poster is trying to determine the distance from the base of the left tower to the point where the light beam strikes the lake, given the heights of the towers and the distance between them.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to express the light path as a function of distance from the left tower and consider using trigonometric relationships to derive the necessary equations. There are questions about the laws of reflection and how they relate to the calculus approach required by the problem.

Discussion Status

Participants are actively working through the problem, with some providing hints and suggestions for sketching the scenario and deriving expressions for the lengths of the light path segments. There is a focus on using calculus to minimize the total distance traveled by the light, and various interpretations of the problem setup are being explored.

Contextual Notes

There is an emphasis on using calculus rather than physics principles, which has led to some uncertainty about how to incorporate the laws of reflection into the calculus framework. The original poster has indicated a known answer but is seeking to understand the process of arriving at that answer through calculus.

  • #31
Stand by, mathematics in progress.
 
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  • #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 \sqrt(a) + \sqrt(b) = \sqrt(a + b), 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.
 
  • #36
I can expand (120-x)^2 into its polynomial and combine within the expression,
 
  • #37
Ok, now the chain rule
 
  • #38
If I did it right
AF' is

2x/(2(7^2+X^2)^1/2)
 
  • #39
When taking the derivative of the DF expression, is it best to expand (120-X)^2 before taking the derivative?
 
  • #40
I have this now for DF'
282-2X / 2(21^2 +(120-X)^2)^1/2
 
  • #41
Then I simplify those, set them to zero and solve I believe?
 
  • #42
Could I square the fractions to eliminate the radicals?
 
  • #43
If not how should I go about simplifying what I have
 
  • #44
Currently I have this by cross multiplying out the denominators.
X•sqrt(21^2 +(120-X)^2) +
(141-X)•sqrt(7^2+X^2)=0
 
  • #45
image.jpg

This is my current equation, and is what I'm attempting to simplify.

I know that my end goal is to isolate X so I can solve, but I need to eliminate those radicals, and I can't figure out how. Any ideas?
 
  • #46
ArmChairPhysicist said:
View attachment 196037
This is my current equation, and is what I'm attempting to simplify.

I know that my end goal is to isolate X so I can solve, but I need to eliminate those radicals, and I can't figure out how. Any ideas?

Rewrite the derivative by combining the terms over the common denominator ##D = \sqrt{7^2+x^2} \sqrt{21^2+(120-x)^2}##. That will produce a numerator having the square roots in two terms. Now equate the numerator to zero, using the standard approach, which is to re-write the equation so that the two square-roots are on opposite sides; then square both sides. That gets rid of all the square roots.
 
Last edited:
  • #47
ArmChairPhysicist said:
View attachment 196037
This is my current equation, and is what I'm attempting to simplify.

I know that my end goal is to isolate X so I can solve, but I need to eliminate those radicals, and I can't figure out how. Any ideas?

I would check where that '141' in the numerator came from.
 
  • #48
The 141 came from me taking the derivative of 21^2+(120-x)2

2 • 21 ^2-1 =
42

(120-x)^2 = 2(120-X)^1 correct?
 
  • #49
Then when I reduced the fractions in my equation 282-2x became 141-x unless I messed up somewhere
 
  • #50
So from this
 

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  • #51
image.jpg
To this?
 
  • #52
Cancel out the denominator, then square both sides?
 
  • #53
image.jpg
Which gets me this.
 
  • #54
ArmChairPhysicist said:
Well, the answer is yes, since you have done exactly what I suggested. However: please do not continually ask questions as to whether your next step is correct or not---just work it all out, from beginning to the end (or, at least, as FAR as you can). Try to develop some confidence in your own work.
 
  • #55
Now I simply solve for x?
 
  • #56
Ok thank you
 
  • #57
Thank you so much for your help and time.
 
  • #58
ArmChairPhysicist said:
The 141 came from me taking the derivative of 21^2+(120-x)2

2 • 21 ^2-1 =
42

(120-x)^2 = 2(120-X)^1 correct?

Not correct. The derivative of ##21^2## is zero. It's a constant.
 

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