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Chapter 5: Problem 370

In the following exercises, the transformations \(T: S \rightarrow R\) are one- to-one. Find their related inverse transformations \(T^{-1}: R \rightarrow S\). $$ \begin{aligned} &\text x=e^{2 u+v}, y=e^{u-v}, \quad \text { where } \quad S=\mathrm{R}^{2} \text { and }\\\ &R=\\{(x, y) \mid x>0, y>0\\} \end{aligned} $$

Short Answer

Expert verified
The inverse transformation is \( u = \frac{1}{3} \ln(xy) \) and \( v = \frac{1}{3} \ln\left( \frac{x}{y^2} \right) \).

Step by step solution

01

Analyze the Given Transformation

The transformation given is \( x = e^{2u+v} \) and \( y = e^{u-v} \). These equations express \( x \) and \( y \) in terms of \( u \) and \( v \). Our goal is to find the inverse transformation, which means expressing \( u \) and \( v \) in terms of \( x \) and \( y \).
02

Solve for \( u \) in Terms of \( x \) and \( y \)

Start with the second equation: \( y = e^{u-v} \). Take the natural logarithm of both sides: \( \ln(y) = u - v \). Re-arrange this to express \( u \): \( u = \ln(y) + v \).
03

Substitute \( u \) in the First Equation

Substitute \( u = \ln(y) + v \) into the first equation: \( x = e^{2(\ln(y) + v) + v} \). This simplifies to \( x = e^{2\ln(y) + 3v} = y^2 e^{3v} \).
04

Solve for \( v \)

Rearrange the equation from Step 3 to solve for \( v \): \( x = y^2 e^{3v} \). Divide both sides by \( y^2 \): \( \frac{x}{y^2} = e^{3v} \). Taking the natural logarithm gives: \( 3v = \ln\left( \frac{x}{y^2} \right) \). Solve for \( v \) to obtain \( v = \frac{1}{3} \ln\left( \frac{x}{y^2} \right) \).
05

Substitute \( v \) Back to Find \( u \)

Now, substitute the expression for \( v \) into \( u = \ln(y) + v \). Thus, \( u = \ln(y) + \frac{1}{3} \ln\left( \frac{x}{y^2} \right) \). Simplify this to \( u = \frac{1}{3} \ln\left( y^3 \cdot \frac{x}{y^2} \right) = \frac{1}{3} \ln(\frac{xy}{y}) = \frac{1}{3} \ln(xy) \).
06

Write the Inverse Transformation

The inverse transformation \( T^{-1}: R \rightarrow S \) can be represented as: \( u = \frac{1}{3} \ln(xy) \) and \( v = \frac{1}{3} \ln\left( \frac{x}{y^2} \right) \).

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Key Concepts

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

One-to-One Transformation
A one-to-one transformation, also known as an injective function, is a type of mapping from one set to another where each element of the first set corresponds to a unique element of the second set. In other words, no two elements from the first set map to the same element in the second set.
This concept is crucial in transformations because it ensures that each input has a distinct output.
  • A one-to-one transformation can be reversed since the mapping is unique and exactly reversible.
  • It is the foundation for finding an inverse transformation.
  • For a transformation, such as the one given in the exercise where the equations are expressed as \(x = e^{2u+v}\) and \(y = e^{u-v}\), a one-to-one correspondence between \(u,v\) and \(x,y\) must be established to find the inverse.
In this case, showing it's one-to-one entails ensuring each distinct pair \(u, v\) produces a distinct \(x, y\). This allows for the reversal of the transformation, permitting us to roll back from \(x, y\) to \(u, v\). Understanding these properties paves the way for solving the inverse transformation.
Natural Logarithm
The natural logarithm, denoted as \(\ln\), is a logarithm with base \(e\), where \(e\) is an irrational and transcendental number approximately equal to 2.71828. It is a fundamental function in mathematics, particularly in calculus and complex analysis.
When dealing with transformations, the natural logarithm is essential for solving exponential equations.
  • In the given exercise, the natural logarithm helps in taking exponential expressions and transforming them into algebraic ones, which are easier to manage.
  • For instance, transforming a statement like \(y = e^{u-v}\) into \(\ln(y) = u - v\) allows isolation and manipulation of \(u\) and \(v\).
  • This property of turning multiplication into addition is one reason that logarithms are a valuable tool when solving for variables.
Understanding the natural logarithm's power and application is invaluable for reversing transformations and for solving the inverse of functions involving exponential expressions.
Solve for Variables
Solving for variables involves isolating a particular variable on one side of an equation. This process is fundamental in mathematics and often requires the consideration of algebraic manipulations and understanding the relationship between variables.
In the provided solution, solving for \(u\) and \(v\) involved several algebraic steps:
  • First, we used the equation \(y = e^{u-v}\) to express \(u\) in terms of \(v\) and \(y\) as \(u = \ln(y) + v\).
  • We then substituted this expression into the other equation \(x = e^{2u+v}\), allowing us to subsequently solve for \(v\) by using natural logarithms.
  • Finally, the expression for \(v\) was substituted back to determine \(u\).
These steps involved recognizing inverse operations (logarithm to exponential), performing substitutions, and algebraically simplifying expressions. Successfully solving for variables supports the broader aim of identifying the inverse transformation, echoing throughout various aspects of mathematics where expressions need to be rewritten in simpler or more convenient forms.

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Most popular questions from this chapter

[T] Use a C.AS to graph the solid whose volume is given by the iterated integral in spherical coordinates as \(\int_{0}^{2 \pi} \int_{3 \pi / 4}^{\pi / 4} \int_{0}^{1} \rho^{2} \sin \varphi d \rho d \varphi d \theta .\) Find the volume \(V\) of the solid. Round your answer to three decimal places.

In the following exercises, find the Jacobian \(J\) of the transformation. $$ x=e^{2 u-v}, y=e^{u+v} $$

In the following exercises, the function \(T: S \rightarrow R, T(u, v)=(x, y) \quad\) on the region \(S=\\{(u, v) \mid 0 \leq u \leq 1,0 \leq v \leq 1\\}\) bounded by the unit square is given, where \(R \subset \mathrm{R}^{2}\) is the image of \(S\) under \(T\). a. Justify that the function \(T\) is a \(C^{1}\) transformation. b. Find the images of the vertices of the unit square \(S\) through the function \(T\). c. Determine the image \(R\) of the unit square \(S\) and graph it. $$ x=2 u-v, y=u+2 v $$

The mass of a solid \(Q\) is given by \(\int_{0}^{2} \int_{0}^{\sqrt{4-x^{2}}} \sqrt{16-x^{2}-y^{2}}\left(x^{2}+y^{2}+z^{2}\right)^{n} d z d y d x\), where \(n\) is an integer. Determine \(n\) such the mass of the solid is \((2-\sqrt{2}) \pi\)

For the following problems, find the specified area or volume. The volume of the intersection between two spheres of radius 1 , the top whose center is (0,0,0.25) and the bottom, which is centered at (0,0,0) .
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