Problem 4 - A Gentle Introduction to the Art of Mathematics

Question

Determine a formula for the bijection from (-1,1) to the line \(y=1\) determined by vertical projection onto the upper half of the unit circle, followed by projection from the point (0,0) .

Step-by-Step Solution

Answer

\(\left( \frac{x}{\sqrt{1 - x^2}}, 1 \right)\)

Step 1 - Understanding the Projection

The bijection involves two projections. The first is vertical projection onto the upper half of the unit circle, and the second is projection from the origin (0,0) onto the line y=1.

Step 2 - Vertical Projection onto the Upper Half Unit Circle

For any point x in the interval (-1,1), its vertical projection onto the upper half unit circle is given by \( (x, \sqrt{1 - x^2}) \).

Step 3 - Projection from the Origin onto the Line y=1

Next, this point \( (x, \sqrt{1 - x^2}) \) is projected onto the line y=1. We use similar triangles to find the formula. A point (x_1, y_1) projected onto (x_2, 1) lies on the same line passing through the origin. Hence, \(\frac{x - 0}{\sqrt{1 - x^2} - 0} = \frac{x'}{1}\), which simplifies to: \(x' = \frac{x}{\sqrt{1 - x^2}}\).

Step 4 - Writing the Final Formula

Combining these results, the final bijection formula from (-1,1) onto the line y=1 is \((\frac{x}{\sqrt{1 - x^2}}, 1)\).

Key Concepts

vertical projection

When we talk about vertical projection in mathematics, we often refer to projecting points from one space vertically down or up to another space. In this problem, we aim to project points from the interval (-1,1) onto the upper half of the unit circle. Every point in the interval (-1,1) is of the form (x, 0). By performing a vertical projection, we move this point directly above (or below) to a new point on the unit circle. In this case, for a point x in (-1,1), its vertical projection lands on a point \( (x, \sqrt{1 - x^2}) \) on the upper half of the unit circle. This is due to the Pythagorean theorem, where the radius of the unit circle is always 1.

unit circle projection

The unit circle is a circle with a radius of one, centered at the origin (0,0). It’s a fundamental concept in trigonometry and calculus. In this exercise, the points projected onto the upper half of the unit circle then need to be mapped onto the line y=1. The unit circle is defined by the equation \(x^2 + y^2 = 1\). Once a point has been vertically projected onto the unit circle, each point \( (x, \sqrt{1 - x^2}) \) will be mapped onto the line y=1 using another type of projection, which leads us to the next step of using similar triangles.

similar triangles

Similar triangles are triangles that have the same shape but not necessarily the same size. They have equal corresponding angles and proportional corresponding sides. In the problem, once we have the point \( (x, \sqrt{1 - x^2}) \) on the unit circle, we project this point onto the line y=1 using similar triangles. We imagine a line passing through the origin (0,0) and the point \( (x, \sqrt{1 - x^2}) \). This line intersects the line y=1 at some new point \( (x', 1) \). By setting up similar triangles, we have the proportion: $$\frac{x - 0}{\sqrt{1 - x^2} - 0} = \frac{x'}{1}$$ Solving for x', we find that: \(x' = \frac{x}{\sqrt{1 - x^2}}\). Therefore, the final bijection formula is \( (\frac{x}{\sqrt{1 - x^2}}, 1) \).