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Consider the function \(f\) defined by \(f(x, y)=\frac{x y}{x^{2}+y^{2}+1}\) a. What is the domain of \(f ?\) b. Evaluate limit of \(f\) at (0,0) along the following paths: \(x=0, y=0\), \(y=x,\) and \(y=x^{2}\) c. What do you conjecture is the value of \(\lim _{(x, y) \rightarrow(0,0)} f(x, y) ?\) d. Is \(f\) continuous at (0,0)\(?\) Why or why not? e. Use appropriate technology to sketch both surface and contour plots of \(f\) near (0,0) . Write several sentences to say how your plots affirm your findings in (a) - (d).

Short Answer

Expert verified
a. The domain of \(f(x,y)\) is 鈩澛, or all real number pairs (x, y). b. Along x=0 and y=0, the limit of \(f(x,y)\) is 0. Along the path y=x, the limit is \(\frac{1}{2}\), and along y=\(x^2\), the limit is 0. c. Since we obtained different values of the limit along different paths, there is no limit to the function as (x, y) approaches (0, 0). d. The function \(f(x,y)\) is not continuous at (0, 0) because the limit does not exist at that point. e. Plotting the surface plots and contour plots of \(f(x,y)\) confirms that the function does not have a limit as (x, y) approaches (0, 0), and thus, is not continuous at (0, 0).

Step by step solution

01

Find the domain of f(x, y)

The domain of the function is the set of all allowed values of x and y for which the function has a defined meaning. To find the domain of f(x, y), we need to find the values of x and y for which the denominator is not zero, i.e., x^2+y^2+1 鈮 0. Since the sum of squares of x and y cannot be negative, the only possible solution to x^2+y^2+1 = 0 is if the sum of squares is equal to -1. As this is not possible, we can safely say that the domain of f(x, y) is 鈩漗2, or all real number pairs (x, y).
02

Evaluate the limit along x=0 and y=0

To evaluate the limit of the function along the paths x=0 and y=0, we set x = 0 and then y=0. Along x=0: \( \lim_{(x, y) \rightarrow (0, 0)} f(0, y) = \lim_{y \rightarrow 0} \frac{0 * y}{0 + y^2 + 1} = \lim_{y \rightarrow 0} 0 = 0 \) Along y=0: \( \lim_{(x, y) \rightarrow (0, 0)} f(x, 0) = \lim_{x \rightarrow 0} \frac{x * 0}{x^2 + 0 + 1} = \lim_{x \rightarrow 0} 0 = 0 \)
03

Evaluate the limit along y=x and y=x^2

To evaluate the limit of the function along the paths y=x and y=x^2, we substitute y with x and y with x^2 respectively. Along y=x: \( \lim_{(x, y) \rightarrow (0, 0)} f(x, x) = \lim_{x \rightarrow 0} \frac{x * x}{x^2 + x^2 + 1} = \lim_{x \rightarrow 0} \frac{x^2}{2x^2+1} \) Since the numerator and denominator both have x^2, we divide both the numerator and denominator by x^2: \( \lim_{x \rightarrow 0} \frac{1}{2 + \frac{1}{x^2}} = \frac{1}{2+0} = \frac{1}{2} \) Along y=x^2: \( \lim_{(x, y) \rightarrow (0, 0)} f(x, x^2) = \lim_{x \rightarrow 0} \frac{x * x^2}{x^2 + x^4 + 1} = \lim_{x \rightarrow 0} \frac{x^3}{x^2 + x^4 + 1} \) Since the numerator has x^3, we divide both the numerator and the denominator by x^2: \( \lim_{x \rightarrow 0} \frac{x}{1 + x^2 + \frac{1}{x^2}} = \frac{0}{1 + 0 + \infty} = 0 \)
04

Conjecture the limit as (x, y) approaches (0, 0)

Since we obtained different values of the limit along the different paths (y=x and y=x^2), we can say that there is no limit to the function as (x, y) approaches (0, 0).
05

Determine the continuity of f at (0, 0)

A function is continuous at a point if the limit of the function as (x, y) approaches the point exists and is equal to the value of the function at that point. Since we determined that the limit does not exist as (x, y) approaches (0, 0), the function is not continuous at (0, 0).
06

Sketch the surface and contour plots

Use appropriate software, like Mathematica, Matlab or any other suitable graphing calculator, to create both surface plots and contour plots of the given function f(x, y). Analyze the generated plots, and you can observe that the function does not have a limit as (x, y) approaches (0, 0), which confirms our findings that f is not continuous at (0, 0). Furthermore, the surface plot and the contour plot will provide a visual understanding of the function and allows you to see how the function behaves near (0, 0).

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Key Concepts

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

Limits
In multivariable calculus, limits are fundamental for analyzing the behavior of functions as inputs approach a certain point. For a function of two variables, say \( f(x,y) \), we are often interested in the value that \( f(x,y) \) approaches as \((x,y)\) moves closer to a specific point, such as \((0,0)\). To fully understand a multivariable limit, it's important to consider the limit along various paths.

- When calculating the limit of a function along different paths, the path essentially means the way or route in which \((x,y)\) approaches the target point.- If the limit of \( f(x,y) \) as \((x,y)\) approaches a point is different along different paths, the limit does not exist at that point.- Consistent limit values along all paths indicate the existence of the limit.

In the given exercise, the limits of \( f(x, y) = \frac{xy}{x^2 + y^2 + 1} \) were calculated along multiple paths such as \( y = x \) and \( y = x^2 \), revealing different results. This disparity confirms that there is no single well-defined limit as \((x,y)\) approaches \((0,0)\).
Continuity
Continuity in multivariable calculus extends the concept from single-variable calculus to functions of several variables. A function \( f(x,y) \) is continuous at a point \((a, b)\) if the following conditions are met:

  • The function is defined at \((a, b)\).
  • The limit of the function as \((x,y)\) approaches \((a,b)\) exists.
  • The limit equals the value of the function at that point.


For the function \( f(x,y) = \frac{xy}{x^2 + y^2 + 1} \), the limit as \((x,y)\) approaches \((0,0)\) does not exist due to differing limits along different paths. Therefore, the function is not continuous at \((0,0)\), because one of the fundamental conditions of continuity is violated. This illustrates how continuity can be impacted in multivariable functions by the directionality of approaching the point of interest.
Surface and Contour Plots
Surface plots and contour plots are excellent tools for visualizing functions of two variables, offering crucial insights into the functions' behavior.

- **Surface Plots**: These are 3D plots that display the function's surface over a certain range. They help you see how the function changes with various combinations of \( x \) and \( y \). In this exercise, the surface plot of \( f(x, y) \) would illustrate its behavior as you move closer to the origin, confirming that the surfaces do not approach a single height at \((0,0)\).

- **Contour Plots**: These are 2D plots featuring lines that connect points of equal function value. They help visualize the elevation or depth of function values in a plane. It's useful in understanding how function values change over a region.

By using these graphs, you can visually affirm mathematical findings. For instance, they reveal that near the origin, the function values aren't consistent, thus reinforcing the conclusion that the limit does not exist at \((0,0)\).
Function Domain
A function's domain refers to all possible input values for which the function is defined. For a multivariable function \( f(x, y) \), the domain consists of all \((x, y)\) pairs that do not make the function undefined.

- For the function \( f(x, y) = \frac{xy}{x^2 + y^2 + 1} \), the denominator is always nonzero because \( x^2 + y^2 + 1 \) is always positive. Thus, there are no restrictions on the values of \( x \) and \( y \), ensuring that the domain of \( f \) is all real numbers, \( \mathbb{R}^2 \).

- Understanding the domain is crucial for finding limits, assessing continuity, and accurately visualizing surface and contour plots. Knowing that the entire plane is the domain confirms there are no holes or gaps where the function is undefined.

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Most popular questions from this chapter

One mole of ammonia gas is contained in a vessel which is capable of changing its volume (a compartment sealed by a piston, for example). The total energy \(U\) (in Joules) of the ammonia is a function of the volume \(V\) (in cubic meters) of the container, and the temperature \(T\) (in degrees Kelvin) of the gas. The differential \(d U\) is given by \(d U=840 d V+27.32 d T\). (a) How does the energy change if the volume is held constant and the temperature is decreased slightly? \(\odot\) it increases slightly \(\odot\) it does not change \(\odot\) it decreases slightly (b) How does the energy change if the temperature is held constant and the volume is increased slightly? \(\odot\) it does not change \(\odot\) it increases slightly \(\odot\) it decreases slightly (c) Find the approximate change in energy if the gas is compressed by 150 cubic centimeters and heated by 3 degrees Kelvin. Change in energy = ___________. Please include units in your answer.

In this exercise we consider how to apply the Method of Lagrange Multipliers to optimize functions of three variable subject to two constraints. Suppose we want to optimize \(f=f(x, y, z)\) subject to the constraints \(g(x, y, z)=c\) and \(h(x, y, z)=k\). Also suppose that the two level surfaces \(g(x, y, z)=c\) and \(h(x, y, z)=k\) intersect at a curve \(C\). The optimum point \(P=\left(x_{0}, y_{0}, z_{0}\right)\) will then lie on \(C\). a. Assume that \(C\) can be represented parametrically by a vector-valued function \(\mathbf{r}=\mathbf{r}(t) .\) Let \(\overrightarrow{O P}=\mathbf{r}\left(t_{0}\right) .\) Use the Chain Rule applied to \(f(\mathbf{r}(t)), g(\mathbf{r}(t)),\) and \(h(\mathbf{r}(t)),\) to explain why $$\begin{array}{l} \nabla f\left(x_{0}, y_{0}, z_{0}\right) \cdot \mathbf{r}^{\prime}\left(t_{0}\right)=0 \\ \nabla g\left(x_{0}, y_{0}, z_{0}\right) \cdot \mathbf{r}^{\prime}\left(t_{0}\right)=0, \text { and } \\ \nabla h\left(x_{0}, y_{0}, z_{0}\right) \cdot \mathbf{r}^{\prime}\left(t_{0}\right)=0\end{array}$$ Explain how this shows that \(\nabla f\left(x_{0}, y_{0}, z_{0}\right), \nabla g\left(x_{0}, y_{0}, z_{0}\right),\) and \(\nabla h\left(x_{0}, y_{0}, z_{0}\right)\) are all orthogonal to \(C\) at \(P\). This shows that \(\nabla f\left(x_{0}, y_{0}, z_{0}\right), \nabla g\left(x_{0}, y_{0}, z_{0}\right),\) and \(\nabla h\left(x_{0}, y_{0}, z_{0}\right)\) all lie in the same plane. b. Assuming that \(\nabla g\left(x_{0}, y_{0}, z_{0}\right)\) and \(\nabla h\left(x_{0}, y_{0}, z_{0}\right)\) are nonzero and not parallel, explain why every point in the plane determined by \(\nabla g\left(x_{0}, y_{0}, z_{0}\right)\) and \(\nabla h\left(x_{0}, y_{0}, z_{0}\right)\) has the form \(s \nabla g\left(x_{0}, y_{0}, z_{0}\right)+t \nabla h\left(x_{0}, y_{0}, z_{0}\right)\) for some scalars \(s\) and \(t\). c. Parts (a.) and (b.) show that there must exist scalars \(\lambda\) and \(\mu\) such that $$\nabla f\left(x_{0}, y_{0}, z_{0}\right)=\lambda \nabla g\left(x_{0}, y_{0}, z_{0}\right)+\mu \nabla h\left(x_{0}, y_{0}, z_{0}\right)$$ So to optimize \(f=f(x, y, z)\) subject to the constraints \(g(x, y, z)=c\) and \(h(x, y, z)=k\) we must solve the system of equations $$\nabla f(x, y, z)=\lambda \nabla g(x, y, z)+\mu \nabla h(x, y, z)$$ $$\begin{array}{l}g(x, y, z)=c, \text { and } \\\h(x, y, z)=k .\end{array}$$ for \(x, y, z, \lambda,\) and \(\mu .\) Use this idea to find the maximum and minimum values of \(f(x, y, z)=\) \(x+2 y\) subject to the constraints \(y^{2}+z^{2}=8\) and \(x+y+z=10\).

The temperature at any point in the plane is given by \(T(x, y)=\frac{100}{x^{2}+y^{2}+3}\). (a) What shape are the level curves of \(T ?\) \(\odot\) hyperbolas \(\odot\) circles \(\odot\) lines \(\odot\) ellipses \(\odot\) parabolas \(\odot\) none of the above (b) At what point on the plane is it hottest? __________ What is the maximum temperature? ___________. (c) Find the direction of the greatest increase in temperature at the point (-3,-3) ____________. What is the value of this maximum rate of change, that is, the maximum value of the directional derivative at (-3,-3)\(?\) ___________. (d) Find the direction of the greatest decrease in temperature at the point (-3,-3) ___________. What is the value of this most negative rate of change, that is, the minimum value of the directional derivative at (-3,-3)\(?\) __________.

Find the absolute maximum and minimum of the function \(f(x, y)=x^{2}+\) \(y^{2}\) subject to the constraint \(x^{4}+y^{4}=6561\). As usual, ignore unneeded answer blanks, and list points in lexicographic order. Absolute minimum value: ____________________. attained at ( ____________, ______________) ( ____________, ) Absolute minimum value: ____________________. attained at ( ____________, ______________) ( ____________, )

Find all directions in which the directional derivative of \(f(x, y)=y e^{-x y}\) is 1 at the point (0,2) .

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