Light between 2 towers reflecting off a lake

[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

 

[ArmChairPhysicist]

To this?

 

[ArmChairPhysicist]

Cancel out the denominator, then square both sides?

 

[ArmChairPhysicist]

Which gets me this.

 

[Ray Vickson]

View attachment 196047 To this?

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.

 

[ArmChairPhysicist]

Now I simply solve for x?

 

[ArmChairPhysicist]

Ok thank you

 

[ArmChairPhysicist]

Thank you so much for your help and time.

 

[Dick]

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.