Light between 2 towers reflecting off a lake

[ArmChairPhysicist]

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.

 

[ArmChairPhysicist]

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.

 

[phyzguy]

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.

 

[ArmChairPhysicist]

I can. Give me a minute to photograph it.

 

[Doc Al]

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.

 

[ArmChairPhysicist]

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.

 

[ArmChairPhysicist]

Ah there it is, trig. I'll try this

 

[berkeman]

If it's a reflection, doesn't that constrain the angles of incidence and reflection...?

 

[ArmChairPhysicist]

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

 

[Doc Al]

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.

 

[Doc Al]

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.

 

[berkeman]

You're not supposed to use that fact

Oopsies! (Kobayashi-Maru...)

 

[phyzguy]

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.

 

[Doc Al]

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.

 

[ArmChairPhysicist]

I've done that, working on the trig now.

 

[ArmChairPhysicist]

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.

 

[ArmChairPhysicist]

And technically I already know BF= 50, but working the problem backwards isn't going to teach me much.

 

[phyzguy]

You 'know' the side BF, it's distance is x. So write the hypotenuse AF in terms of AB (7 feet) and x.

 

[ArmChairPhysicist]

But can I even solve that with two out of three variables being unknown?

 

[ArmChairPhysicist]

7^2 + X^2 = C^2. C=AF

 

[Doc Al]

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.

 

[Doc Al]

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.

 

[Doc Al]

If BF = X, what must CF equal?

 

[ArmChairPhysicist]

So we have
AF = sqrt(7^2 + X^2)
DF = sqrt(21^2 + (120-X)^2)

 

[ArmChairPhysicist]

CF is equal to 120-X

 

[phyzguy]

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).

 

[ArmChairPhysicist]

I have a feeling the chain rule will be needed soon.

 

[ArmChairPhysicist]

I'll try applying that and see if that gets me somewhere, unless there is another step that I should do first.

 

[Doc Al]

So we have
AF = sqrt(7^2 + X^2)
DF = sqrt(21^2 + (120-X)^2)

Perfect. Now you must minimize AF + DF.

 

[Doc Al]

I have a feeling the chain rule will be needed soon.

Oh yes.

 

[ArmChairPhysicist]

Stand by, mathematics in progress.

 

[ArmChairPhysicist]

If I recall correctly, because both expressions are of the same power: ^1/2, I can combine them without any issue, correct?

 

[phyzguy]

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.

 

[ArmChairPhysicist]

So I have
(7^2 + X^2)^1/2 + (21^2 + (120-x)^2)^1/2 as the total length

 

[phyzguy]

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.

 

[ArmChairPhysicist]

I can expand (120-x)^2 into its polynomial and combine within the expression,

 

[ArmChairPhysicist]

Ok, now the chain rule

 

[ArmChairPhysicist]

If I did it right
AF' is

2x/(2(7^2+X^2)^1/2)

 

[ArmChairPhysicist]

When taking the derivative of the DF expression, is it best to expand (120-X)^2 before taking the derivative?

 

[ArmChairPhysicist]

I have this now for DF'
282-2X / 2(21^2 +(120-X)^2)^1/2

 

[ArmChairPhysicist]

Then I simplify those, set them to zero and solve I believe?

 

[ArmChairPhysicist]

Could I square the fractions to eliminate the radicals?

 

[ArmChairPhysicist]

If not how should I go about simplifying what I have

 

[ArmChairPhysicist]

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

 

[ArmChairPhysicist]

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?

 

[Ray Vickson]

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.

 

[Dick]

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.

 

[ArmChairPhysicist]

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?

 

[ArmChairPhysicist]

Then when I reduced the fractions in my equation 282-2x became 141-x unless I messed up somewhere

 

[ArmChairPhysicist]

So from this