Question

Apply the formula of Problem 2 to evaluate the degree for the following mappings of the circle \(u^{2}+v^{2}=1\) into the circle \(x^{2}+y^{2}=1\) : a) \(x=\frac{3 u+4 v}{5}, y=\frac{4 u-3 v}{5}\) b) \(x=u^{2}-v^{2}, y=2 u v\) c) \(x=u^{3}-u v^{2}, y=3 u^{2} v-v^{3}\)

Short answer

Answer: The degrees of the given mappings are: a) Degree = 1 b) Degree = 2 c) Degree = 3

Step-by-step solution

Mapping a)

We are given the mapping \(x = \frac{3u + 4v}{5}\) and \(y = \frac{4u - 3v}{5}\). We need to find their degree. First, we will rewrite the expressions of \(x\) and \(y\) as: \(x(u,v) = \frac{3u}{5} + \frac{4v}{5}\) \(y(u,v) = \frac{4u}{5} - \frac{3v}{5}\) These expressions are linear functions of \(u\) and \(v\).

Computing \(x^2 + y^2\)

In order to evaluate the degree for this mapping, we have to check if \(x^2 + y^2 = 1\) for any values where \(u^2 + v^2 = 1\). This requires calculating \(x(u,v)^2 + y(u,v)^2\): \(x(u,v)^2 + y(u,v)^2 = (\frac{3u}{5} + \frac{4v}{5})^2 + (\frac{4u}{5} - \frac{3v}{5})^2\)

Simplifying and Factoring

Now we will simplify and factor this expression: \(= (\frac{9u^2}{25} + \frac{24uv}{25} + \frac{16v^2}{25}) + (\frac{16u^2}{25} - \frac{24uv}{25} + \frac{9v^2}{25})\) \(= \frac{25u^2}{25} + \frac{25v^2}{25}\) \(= u^2 + v^2\)

Degree of the Mapping

Since \(x^2 + y^2 = u^2 + v^2\), the given mapping in case a) maps the circle \(u^2 + v^2 = 1\) into the circle \(x^2 + y^2 = 1\), and its degree is 1. This is because the mapping is bijective and preserves the geometric structure of the circle.

Mapping b)

For case b, we are given the mapping \(x = u^2 - v^2\) and \(y = 2uv\). Substituting these expressions into the equation \(x^2 + y^2 = 1\), we get: \((u^2 - v^2)^2 + (2uv)^2 = 1\)

Computing and Simplifying

Expanding the expression, we have: \(u^4 - 2u^2v^2 + v^4 + 4u^2v^2 = 1\) Simplifying this expression, we get: \(u^4 + 2u^2v^2 + v^4 = 1\) Notice that this is equivalent to the equation \((u^2 + v^2)^2 = 1\). Since \((u^2 + v^2)^2\) is true if and only if \(u^2 + v^2 = 1\), the given mapping in case b) also maps the circle \(u^2 + v^2 = 1\) into the circle \(x^2 + y^2 = 1\).

Degree of the Mapping b)

The degree of the mapping is 2, as both \(x\) and \(y\) are quadratic functions of \(u\) and \(v\).

Mapping c)

For case c, we are given the mapping \(x = u^3 - uv^2\) and \(y = 3u^2v - v^3\). Substituting these expressions into the equation \(x^2 + y^2 = 1\), we get: \((u^3 - uv^2)^2 + (3u^2v - v^3)^2 = 1\) Notice that the mapping is much more complicated than the linear transformation in case a) and the quadratic transformation in case b). It involves determining the power of the functions of \(u\) and \(v\) that make \(x^2 + y^2 = 1\).

Degree of the Mapping c)

The degree of the mapping is 3, as both \(x\) and \(y\) are cubic functions of \(u\) and \(v\). In conclusion, the degrees of the given mappings are as follows: a) Degree = 1 b) Degree = 2 c) Degree = 3

Key concepts

Linear Transformation

A linear transformation is a function between two vector spaces that preserves the operations of vector addition and scalar multiplication. In simpler terms, it's a fancy way of saying that lines remain straight, and everything scales proportionally. In math terms, a linear transformation can be represented as:

• \(x(u,v) = au + bv\)
• \(y(u,v) = cu + dv\)

where \(a, b, c,\) and \(d\) are constants. These expressions are linear functions of \(u\) and \(v\), meaning they don't involve any powers higher than one. This concept becomes evident when looking at the mapping from the exercise:

• \(x = \frac{3u + 4v}{5}\)
• \(y = \frac{4u - 3v}{5}\)

Here, both \(x\) and \(y\) are linear functions of \(u\) and \(v\), confirming the transformation is linear. To verify the validity and degree of such mappings, you substitute the transformation into \(x^2 + y^2\) and check if it equals \(1\) when \(u^2 + v^2 = 1\). This concept is primarily used in geometry, physics, and computer graphics.

Quadratic Function

Quadratic functions are polynomial functions of degree 2, where the highest power of the variable is two. They form a parabola when graphed and have the general form:

• \(ax^2 + bx + c = 0\)

In the given exercise, the quadratic mapping involves the expressions:

• \(x = u^2 - v^2\)
• \(y = 2uv\)

These equations are derived from combining square terms and products of \(u\) and \(v\). When evaluated, these expressions map a circle onto another circle, maintaining the equation \(x^2 + y^2 = 1\) for \(u^2 + v^2 = 1\). The degree of such a mapping is 2 because the functions involve quadratic terms. Quadratics are fundamental in many areas, including statistics, where they describe relationships in quadratic regression, and physics, where they model acceleration.

Cubic Function

Cubic functions are polynomial functions with a degree of 3, which means the highest power of the variable is three. They produce more complex curves compared to linear or quadratic functions, often resembling an "S" shape in their simplest form. The general form of a cubic function is:

• \(ax^3 + bx^2 + cx + d = 0\)

In the given exercise, the cubic mapping is depicted in:

• \(x = u^3 - uv^2\)
• \(y = 3u^2v - v^3\)

These functions involve third-degree terms, which contribute to their complexity. Within the process of determining the degree of this mapping, it's crucial to verify that \(x^2 + y^2 = 1\) when the input \(u^2 + v^2 = 1\) holds true. The degree of this mapping is 3, linking back to the highest degree of the variable terms in the equations. Cubic functions are prevalent in economics to depict polynomial trend lines and in physics for analyzing moments and rotational dynamics.