You are hanging a picture on the wall. The picture is 4 feet high and you want to place it at just the right height so that
if you are standing in front of it (let's say 3 feet away from it), at what height should we place it relative to eye level?
I am going to approagh this problem almost the same as I did for problem 1 except for that
I am going to change how I measure the distance of the picture from eye-level.
Before, eye-level was a constant at 2 feet below the base of the picture. Now, I am still going to measure the distance from the base of the picture, but now I am going to say that if eye-level is above the base of the picture, that distance is positive and if eye-level is below the base of the picture, that distance is negative.
Here are some examples of what that will look like:
1
In this example, eye level is 1 foot above the base of the painting.
2
In this example eye level is marked as -1 feet because it is below the base of the painting.
The formula to find the angle (where the location of eye level is our x-variable), is the following:
y = arctan ((4-x)/3) + arctan (x/3)
Below are illustrations of how this equation works. In both cases, the angles shaded in are being added together in order to get our desired angle:
When x is a positive number
When x is a negative number:
The difference here is that the orange triangle is a negative angle (because x<0), so it is actually being subtracted!
This is because arctan(-x) = -arctan(x).
It turns out that the viewer will get the biggest angular view when eye level is exactly at the center of the painting, or when x = 2.
I found this answer using derivatives, feel free to see how I did this.
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