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Chapter 5: Problem 3

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

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

Step by step solution

01

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 \(v\).
02

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 \(x(u,v)^2 + y(u,v)^2\): \(x(u,v)^2 + y(u,v)^2 = (\frac{3u}{5} + \frac{4v}{5})^2 + (\frac{4u}{5} - \frac{3v}{5})^2\)
03

Simplifying and Factoring

Now we will simplify and factor this expression: \(= (\frac{9u^2}{25} + \frac{24uv}{25} + \frac{16v^2}{25}) + (\frac{16u^2}{25} - \frac{24uv}{25} + \frac{9v^2}{25})\) \(= \frac{25u^2}{25} + \frac{25v^2}{25}\) \(= u^2 + v^2\)
04

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.
05

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: \((u^2 - v^2)^2 + (2uv)^2 = 1\)
06

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 \(x^2 + y^2 = 1\).
07

Degree of the Mapping b)

The degree of the mapping is 2, as both \(x\) and \(y\) are quadratic functions of \(u\) and \(v\).
08

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 \(x^2 + y^2 = 1\).
09

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Under the (invertible) change of coordinates in \(E^{3}: x^{1}=\bar{x}^{1}, x^{2}=\bar{x}^{1}+\bar{x}^{2}, z^{3}=\bar{x}^{1}+\bar{x}^{3}\), a smooth path \(C: x^{i}=x^{i}(t), a \leq t \leq b\) and a smooth surface \(S: x^{i}=x^{i}(u, v), i=\) \(1,2,3,(u, v)\) in \(R_{u v}\), become \(\bar{C}\) and \(\bar{S}\) respectively. Let \(\bar{X}_{i}\) denote \(X_{i}\left(x^{1}, x^{2}, x^{3}\right)\) with \(x^{1}, x^{2}, x^{3}\) expressed in terms of \(\bar{x}^{1}, \bar{x}^{2}, \bar{x}^{3}\). a) Express the line integral \(\int_{C} X_{1} d x^{1}+X_{2} d x^{2}+X_{3} d x^{3}\) as an integral over \(\bar{C}\). b) Express the surface integral \(\iint_{S} X_{1} d x^{2} d x^{3}+X_{2} d x^{3} d x^{1}+X_{3} d x^{1} d x^{2}\) as a surface integral over \(\bar{S}\).

Prove the validity of the transformation formula 3 $$ \iint_{R} F(x, y) d x d y=\int_{0}^{2 \pi} \int_{0}^{1} F(r \cos \theta, r \sin \theta) r d r d \theta, $$

Test for independence of path and evaluate the following integrals: a) \(\int_{(1,-2)}^{(3,4)} \frac{y d x-x d y}{x^{2}}\) on the line \(y=3 x-5\); b) \(\int_{(0,2)}^{(1,3)} \frac{3 x^{2}}{y} d x-\frac{x^{3}}{y^{2}} d y\) on the parabola \(y=2+x^{2}\). c) \(\int_{(1,0)}^{(-1,0)}(2 x y-1) d x+\left(x^{2}+6 y\right) d y\) on the circular arc \(y=\sqrt{1-x^{2}},-1 \leq x \leq 1\); d) \(\int_{(0,0)}^{\left(\frac{\pi}{4}, \frac{\pi}{4}\right)} \sec ^{2} x \tan y d x+\sec ^{2} y \tan x d y\) on the curve \(y=16 x^{3} / \pi^{2}\).

a) Show that if \(\mathbf{v}\) is one solution of the equation curl \(\mathbf{v}=\mathbf{u}\) for given \(\mathbf{u}\) in a simply connected domain \(D\), then all solutions are given by \(\mathbf{v}+\operatorname{grad} f\), where \(f\) is an arbitrary differentiable scalar in \(D\). b) Find all vectors \(\mathbf{v}\) such that curl \(\mathbf{v}=\mathbf{u}\) if $$ \mathbf{u}=\left(2 x y z^{2}+x y^{3}\right) \mathbf{i}+\left(x^{2} y^{2}-y^{2} z^{2}\right) \mathbf{j}-\left(y^{3} z+2 x^{2} y z\right) \mathbf{k} $$

Let \(S\) be an oriented surface in space that is planar; that is, \(S\) lies in a plane. With \(S\) one can associate the vector \(\mathbf{S}\), which has the direction of the normal chosen on \(S\) and has a length equal to the area of \(S\). a) Show that if \(S_{1}, S_{2}, S_{3}, S_{4}\) are the faces of a tetrahedron, oriented so that the normal is the exterior normal, then $$ \mathbf{S}_{1}+\mathbf{S}_{2}+\mathbf{S}_{3}+\mathbf{S}_{4}=\mathbf{0} \text {. } $$ [Hint: Let \(\mathbf{S}_{i}=A_{i} \mathbf{n}_{i}\left(A_{i}>0\right)\) for \(i=1, \ldots, 4\) and let \(\mathbf{S}_{1}+\cdots+\mathbf{S}_{4}=\mathbf{b}\). Let \(p_{1}\) be the foot of the altitude on face \(S_{1}\) and join \(p_{1}\) to the vertices of \(S_{1}\) to form three triangles of areas \(A_{12}, \ldots, A_{14}\). Show that, for proper numbering, \(A_{1 j}=\pm A_{j} \mathbf{n}_{j} \cdot \mathbf{n}_{1}\), with \(+\) or - according as \(\mathbf{n}_{j} \cdot \mathbf{n}_{1}>0\) or \(<0\), and \(A_{1 j}=0\) if \(\mathbf{n}_{j} \cdot \mathbf{n}_{1}=0(j=2,3,4)\). Hence deduce that \(\mathbf{b} \cdot \mathbf{n}_{j}=0\) for \(j=2,3,4\) and thus \(\mathbf{b} \cdot \mathbf{b}=0\).] b) Show that the result of (a) extends to an arbitrary convex polyhedron with faces \(S_{1}, \ldots, S_{n}\), that is, that $$ \mathbf{S}_{1}+\mathbf{S}_{2}+\cdots+\mathbf{S}_{n}=\mathbf{0}, $$ when the orientation is that of the exterior normal. c) Using the result of (b), indicate a reasoning to justify the relation $$ \iint_{S} \mathbf{v} \cdot d \boldsymbol{\sigma}=0 $$ for any convex closed surface \(S\) (such as the surface of a sphere or ellipsoid), provided that \(\mathbf{v}\) is a constant vector. d) Apply the result of (b) to a triangular prism whose edges represent the vectors \(\mathbf{a}, \mathbf{b}\), \(\mathbf{a}+\mathbf{b}\), c to prove the distributive law (Equation (1.19) $$ \mathbf{c} \times(\mathbf{a}+\mathbf{b})=\mathbf{c} \times \mathbf{a}+\mathbf{c} \times \mathbf{b} $$ for the vector product. This is the method used by Gibbs (cf. the book by Gibbs listed at the end of this chapter).
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