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Consider the mapping \(F: \mathbf{R}^{2} \rightarrow \mathbf{R}^{2}\) defined by \(F(x, y)=(3 y, 2 x) .\) Let \(S\) be the unit circle in \(\mathbf{R}^{2}\) that is, the solution set of \(x^{2}+y^{2}=1\) (a) Describe \(F(S)\). (b) Find \(F^{-1}(S)\)

Short Answer

Expert verified
(a) F(S) is an ellipse centered at the origin with semi-major axis a = (1/3) and semi-minor axis b = (1/2). (b) F鈦宦(S) is an ellipse centered at the origin with semi-major axis a = (1/2) and semi-minor axis b = (1/3).

Step by step solution

01

Apply the mapping F to the unit circle equation

We know the unit circle equation is x虏 + y虏 = 1. Since F maps (x, y) to (3y, 2x), let's substitute x = 2u and y = (1/3)v, where (u, v) are points on F(S). So the equation becomes: (2u)虏 + ((1/3)v)虏 = 1 Step 2: Simplify the equation
02

Simplifying the resulting equation

Now, we simplify the equation we just obtained to find the equation of F(S): (4u虏) + (1/9)v虏 = 1 This is the equation of an ellipse centered at the origin with semi-major axis a = (1/3) and semi-minor axis b = (1/2). (a) Hence, F(S) is an ellipse centered at the origin with semi-major axis a = (1/3) and semi-minor axis b = (1/2). Step 3: Finding F鈦宦(S)
03

Find the inverse of the mapping F

To find F鈦宦(S), we will first find the inverse mapping F鈦宦, and then apply it to the unit circle equation. In the mapping F, the roles of x and y are switched, and they are multiplied by different constants. To invert the mapping, we need to switch back and divide by the corresponding multipliers. Let F鈦宦(u,v) = (x, y): x = (1/2)v y = (1/3)u Step 4: Apply the inverse mapping to the unit circle equation
04

Applying F鈦宦 to the unit circle equation

Now that we have the F鈦宦 mapping, we will apply it to the unit circle equation (x虏 + y虏 = 1): ((1/2)v)虏 + ((1/3)u)虏 = 1 Step 5: Simplify the equation
05

Simplifying the resulting equation

Now, let's simplify the equation we got from the previous step to find the equation for F鈦宦(S): (1/4)v虏 + (1/9)u虏 = 1 This is the equation of an ellipse centered at the origin with semi-major axis a = (1/2) and semi-minor axis b = (1/3). (b) Hence, F鈦宦(S) is an ellipse centered at the origin with semi-major axis a = (1/2) and semi-minor axis b = (1/3).

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Key Concepts

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

Unit Circle
The unit circle is a fundamental concept in mathematics and especially useful in trigonometry and calculus. It is defined as the set of all points \( (x, y) \in \mathbb{R}^{2} \) such that \( x^{2} + y^{2} = 1 \). This specific equation represents a circle with a radius of one, centered at the origin of the coordinate plane, (0,0).

The beauty of the unit circle lies in its simplicity and symmetry. Every point on this circle can be associated with an angle measured from the positive x-axis, making it a powerful tool for solving trigonometric equations and for understanding the properties of sine, cosine, and other trigonometric functions.
  • Radius: 1
  • Center: (0, 0)
  • Equation: \( x^{2} + y^{2} = 1 \)
Understanding how the unit circle relates to transformations and mappings is crucial, as it provides a basic form to test and visualize these concepts.
Ellipse
An ellipse is a closed, symmetric shape that appears frequently in geometry. It resembles a stretched circle, defined formally as the set of all points where the sum of the distances to two fixed points, called foci, is constant. However, for our case involving a linear transformation, we have a simpler equation of the form:

\( \frac{u^{2}}{a^{2}} + \frac{v^{2}}{b^{2}} = 1 \)

Here, \( a \) and \( b \) represent the semi-major and semi-minor axes respectively. These axes are essentially the lengths from the center to the furthest and nearest edges of the ellipse.
  • When \( a > b \), it's wider than it is tall.
  • When \( b > a \), it's taller than it is wide.
    • An ellipse can be the outcome of applying specific transformations, like linear mappings to circles. In our example, transforming the unit circle using mapping \( F \) resulted in an ellipse with particular axes lengths.
Inverse Mapping
Inverse mapping is a way to reverse the effects of a specific transformation or mapping. Suppose a mapping \( F \) takes a point \( (x, y) \) in the plane and maps it to another point \( (u, v) \). The inverse mapping, denoted \( F^{-1} \), does the opposite. It maps \( (u, v) \) back to \( (x, y) \).

For this, it's important to find the correct expressions that reverse the operations done by \( F \). For example, if \( F(x, y) = (3y, 2x) \), then the inverse would be calculated by solving for \( x \) and \( y \) considering \( u = 3y \) and \( v = 2x \). Consequently:
  • \( x = \frac{1}{2}v \)
  • \( y = \frac{1}{3}u \)
Knowledge of inverse mapping is vital for solving problems where we need the original set back or to understand how initial conditions can be restored.
Mapping
In mathematics, mapping refers to the process of transforming each element in one set to another set through a specific rule or function. It's akin to a function where each input (an element from the first set) gives one output (an element in the second set).

Mappings can be linear or non-linear, and in our case, we focus on linear mappings that are easier to understand and manipulate mathematically. For example, the mapping
\( F: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2} \) given by \( F(x, y) = (3y, 2x) \), rearranges the coordinates of a point in a specific manner by transforming \( x \) into \( 2x \) and \( y \) into \( 3y \).
  • It's a way to systematically apply transformations across a whole geometric space.
  • Mappings can alter shapes, sizes, and orientations of geometric forms in space.
Understanding these transformations provides a significant advantage in fields like physics or engineering, where evaluating how systems change under transformations is crucial.

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Most popular questions from this chapter

Define \(F: \mathbf{R}^{3} \rightarrow \mathbf{R}^{2}\) and \(G: \mathbf{R}^{3} \rightarrow \mathbf{R}^{2}\) by \(F(x, y, z)=(2 x, y+z)\) and \(G(x, y, z)=(x-z, y) .\) Find formulas defining the maps: (a) \(F+G\) (b) \(3 F\) (c) \(2 F-5 G\)

Let \(v\) and \(w\) be elements of a real vector space \(V .\) The line segment \(L\) from \(v\) to \(v+w\) is defined to be the set of vectors \(v+t w\) for \(0 \leq t \leq 1 .\) (See Fig. 5.6.) (a) Show that the line segment \(L\) between vectors \(v\) and \(u\) consists of the points: (i) \((1-t) v+t u\) for \(0 \leq t \leq 1\), (ii) \(t_{1} v+t_{2} u\) for \(t_{1}+t_{2}=1, t_{1} \geq 0, t_{2} \geq 0\). (b) Let \(F: V \rightarrow U\) be linear. Show that the image \(F(L)\) of a line segment \(L\) in \(V\) is a line segment in \(U\).

For any mapping \(f: A \rightarrow B\), show that \(\mathbf{1}_{B} \circ f=f=f \circ \mathbf{1}_{A}\).

Let \(F\) and \(G\) be the linear operators on \(\mathbf{R}^{2}\) defined by \(F(x, y)=(y, x)\) and \(G(x, y)=(0, x) .\) Find formulas defining the following operators: (a) \(F+G\) (b) \(2 F-3 G\) (c) \(F G\) (d) \(G F\) (e) \(F^{2}\) (f) \(G^{2}\)

Consider the matrix mapping \(A: \mathbf{R}^{4} \rightarrow \mathbf{R}^{3}\), where \(A=\left[\begin{array}{rrrr}1 & 2 & 3 & 1 \\ 1 & 3 & 5 & -2 \\ 3 & 8 & 13 & -3\end{array}\right]\). Find a basis and the dimension of (a) the image of \(A,(b)\) the kernel of \(A .\) (a) The column space of \(A\) is equal to \(\operatorname{Im} A\). Now reduce \(A^{T}\) to echelon form: $$ A^{T}=\left[\begin{array}{rrr} 1 & 1 & 3 \\ 2 & 3 & 8 \\ 3 & 5 & 13 \\ 1 & -2 & -3 \end{array}\right] \sim\left[\begin{array}{rrr} 1 & 1 & 3 \\ 0 & 1 & 2 \\ 0 & 2 & 4 \\ 0 & -3 & -6 \end{array}\right] \sim\left[\begin{array}{lll} 1 & 1 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] $$ Thus, \(\\{(1,1,3),(0,1,2)\\}\) is a basis of \(\operatorname{Im} A\), and \(\operatorname{dim}(\operatorname{Im} A)=2\). (b) Here \(\operatorname{Ker} A\) is the solution space of the homogeneous system \(A X=0\), where \(X=\left\\{x, y, z_{1} t\right)^{T}\). Thus, reduce the matrix \(A\) of coefficients to echelon form: $$ \left[\begin{array}{rrrr} 1 & 2 & 3 & 1 \\ 0 & 1 & 2 & -3 \\ 0 & 2 & 4 & -6 \end{array}\right] \sim\left[\begin{array}{rrrr} 1 & 2 & 3 & 1 \\ 0 & 1 & 2 & -3 \\ 0 & 0 & 0 & 0 \end{array}\right] \quad \text { or } \quad \begin{array}{r} x+2 y+3 z+t=0 \\ y+2 z-3 t=0 \end{array} $$ The free variables are \(z\) and \(t\). Thus, \(\operatorname{dim}(\) Ker \(A)=2\). (i) Set \(z=1, t=0\) to get the solution \((1,-2,1,0)\). (ii) Set \(z=0, t=1\) to get the solution \((-7,3,0,1)\). Thus, \((1,-2,1,0)\) and \((-7,3,0,1)\) form a basis for Ker \(A\).

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