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Find the largest possible domain and the corresponding range of each function. Determine the equation of the level curves \(f(x, y)=c\), together with the possible values of \(c .\) \(f(x, y)=\exp \left[-\left(x^{2}+y^{2}\right)\right]\)

Short Answer

Expert verified
Domain: all \( (x, y) \in \mathbb{R}^2 \), Range: \((0, 1]\). Level curves: \(x^2 + y^2 = -\ln(c)\), \(0 < c \leq 1\).

Step by step solution

01

Identify the Function Domain

The function given is \( f(x, y) = \exp \left[-\left(x^2 + y^2\right)\right] \). This is an exponential function, which is defined for all real numbers \(x\) and \(y\) because squares of real numbers (\(x^2\) and \(y^2\)) and their sum \(x^2 + y^2\) are all defined for all real numbers. Therefore, the largest domain for this function is all real pairs \((x, y) \in \mathbb{R}^2\).
02

Determine the Range

The exponential function \(\exp(u)\) where \(u = -\left(x^2 + y^2\right)\) always returns positive values because the exponential function is always positive irrespective of its input. The minimum value occurs as \(x^2 + y^2\to \infty\) leading to \( \exp(-\infty) = 0\), while the maximum occurs at \(x = 0, y = 0\), giving \( \exp(0) = 1\). Thus, the range of \(f(x, y)\) is \((0, 1]\).
03

Determine Level Curves

The level curves of a function are determined by setting \(f(x, y) = c\), where \(c\) is a constant. We have \(\exp\left[-(x^2 + y^2)\right] = c\). Taking the natural logarithm on both sides, we get \(- (x^2 + y^2) = \ln(c)\), hence \(x^2 + y^2 = - \ln(c)\). This represents a circle centered at the origin with radius \(\sqrt{-\ln(c)}\).
04

Determine Possible Values of \(c\)

For the equations of level curves to be valid, \(c\) must be within the range of the original function, that is, \(0 < c \leq 1\). This ensures \(- \ln(c)\) is a non-negative value since the logarithm of values less than or equal to 1 is non-positive. Thus, \(c\) is any positive number up to 1 exclusive of 0, ensuring real circles are formed.

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Key Concepts

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

Understanding Domain and Range in Multivariable Functions
When you're working with functions of multiple variables, understanding the domain and range is crucial. For our function, \( f(x, y) = \exp \left[-(x^2 + y^2)\right] \), the domain and range are relatively straightforward.

The "domain" of a function refers to all the possible input values. In this case, since the squares \( x^2 \) and \( y^2 \) are defined for all real numbers, the domain includes every pair of real numbers \((x, y)\). That's why we say the domain is \( \mathbb{R}^2 \), representing the entirety of the xy-plane.

Next, the "range" describes all possible outputs of the function. The exponential part always gives a positive result, since an exponential function never outputs a negative value. In this case, the range is \((0, 1]\). Why not exactly 0? Because no real value of \( x \) and \( y \) will make the sum \( x^2 + y^2 \to \infty \) in a finite manner. At the same time, the function can output its maximum value of 1 when \( x = 0 \) and \( y = 0 \), giving \( \exp(0) \).

So, think of the domain as all the spots where you can "plug in" values, and the range as all the results you could get.
Exponential Functions in Multivariable Calculus
Exponential functions are essential in both single variable and multivariable calculus. They're known for their rapid growth and decay characteristics. In our function \( f(x, y) = \exp \left[-(x^2 + y^2)\right] \), the exponential decay is particularly interesting.

Let's break down the negative sign before the square terms: \(-(x^2 + y^2)\). Here, the larger the sum of \( x^2 + y^2 \), the smaller your output since \( \exp(-u) \rightarrow 0 \) as \( u \rightarrow \infty \). Conversely, smaller values of \( x^2 + y^2 \) yield larger function values, maximizing at 1 when \( (x, y) = (0, 0) \).

Exponential functions in multivariable calculus can model real-world phenomena that involve areas like growth processes or decay models. Their derivatives are straightforward, often leading back to the original exponential function, but in our scenario, we're more interested in how the function values spread over 2D space.

In summary, exponential functions cover many everyday applications and interactions. When you see them as part of a multivariable function, envision how they grow outwards from the center, diminishing in intensity as you move further away.
Exploring Level Curves
Level curves are fascinating visual aids in multivariable calculus, offering insights into the shape and behavior of a function across a plane. For \( f(x, y) = \exp \left[-(x^2 + y^2)\right] \), these level curves are circles centered at the origin.

To find a level curve, you set the function equal to a constant \( c \):
\( \exp \left[-(x^2 + y^2)\right] = c \).
Taking the natural logarithm, \(-(x^2 + y^2) = \ln(c)\), leading to \( x^2 + y^2 = -\ln(c) \). This is a familiar equation of a circle with a radius of \( \sqrt{-\ln(c)} \).

What's intriguing is how these circles vary depending on \( c \):
  • When \( c \) is close to 0, the logarithm is very negative, resulting in a large circle.
  • As \( c \) approaches 1, the circles shrink, reflecting the declining function output as you approach the limit.
Only values \( 0 < c \leq 1 \) are valid for the range; \( c = 0 \) isn't defined since it implies an infinitely large radius. Visualize these curves as records of the function's behavior at different heights, akin to topographic lines on a map indicating elevation. They serve as a handy illustration of how function values distribute over space.

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Most popular questions from this chapter

Refer to the negative binomial host-parasitoid model. Problems \(7,8,11\), and 12 are best done with the help of a spreadsheet, but can also be done with a calculator. The negative binomial model is a discrete-generation host- parasitoid model of the form $$ \begin{array}{l} N_{t+1}=b N_{t}\left(1+\frac{a P_{t}}{k}\right)^{-k} \\ P_{t+1}=c N_{t}\left[1-\left(1+\frac{a P_{t}}{k}\right)^{ k}\right] \end{array} $$ for \(t=0,1,2, \ldots\) Evaluate the negative binomial model for the first 10 generations when \(a=0.02, c=3, k=0.75\), and \(b=1.5 .\) For the initial host density, choose \(N_{0}=5\), and for the initial parasitoid density, choose \(P_{0}=0\).

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Refer to the negative binomial host-parasitoid model. Problems \(7,8,11\), and 12 are best done with the help of a spreadsheet, but can also be done with a calculator. The negative binomial model is a discrete-generation host- parasitoid model of the form $$ \begin{array}{l} N_{t+1}=b N_{t}\left(1+\frac{a P_{t}}{k}\right)^{-k} \\ P_{t+1}=c N_{t}\left[1-\left(1+\frac{a P_{t}}{k}\right)^{ k}\right] \end{array} $$ for \(t=0,1,2, \ldots\) Show that when the initial parasitoid density is \(P_{0}=0\), the negative binomial model reduces to $$ N_{t+1}=b N_{t} $$ With \(N_{0}\) denoting the initial host density, find an expression for \(N_{t}\) in terms of \(N_{0}\) and the parameter \(b\).

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The functional responses of some predators are sigmoidal; that is, the number of prey attacked per predator as a function of prey density has a sigmoidal shape. If we denote the prey density by \(N\), the predator density by \(P\), the time available for searching for prey by \(T\), and the handling time of each prey item per predator by \(T_{h}\), then the number of prey encounters per predator as a function of \(N, T\), and \(T_{h}\) can be expressed as $$ f\left(N, T, T_{h}\right)=\frac{b^{2} N^{2} T}{1+c N+b T_{h} N^{2}} $$ where \(b\) and \(c\) are positive constants. (a) Investigate how an increase in the prey density \(N\) affects the function \(f\left(N, T, T_{h}\right)\) (b) Investigate how an increase in the time \(T\) available for search affects the function \(f\left(N, T, T_{h}\right)\). (c) Investigate how an increase in the handling time \(T_{h}\) affects the function \(f\left(N, T, T_{h}\right)\) (d) Graph \(f\left(N, T, T_{h}\right)\) as a function of \(N\) when \(T=2.4\) hours, \(T_{h}=0.2\) hours, \(b=0.8\), and \(c=0.5\)

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