/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 Verify that the function $$ ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 7

Verify that the function $$ f(x, y)=\left(y-x^{2}\right)\left(y-3 x^{2}\right) $$ does not have an extremum at the origin, even though its restriction to each line passing through the origin has a strict local minimum at that point.

Short Answer

Expert verified
Short Answer: The function \(f(x, y) = (y - x^2)(y - 3x^2)\) does not have an extremum at the origin \((0,0)\) because neither of its partial derivatives is equal to zero at this point. However, it does have a strict local minimum along any line passing through the origin.

Step by step solution

01

Find the partial derivatives

Given the function \(f(x, y) = (y - x^2)(y - 3x^2)\). To find the partial derivatives, differentiate the function with respect to x and y: $$ \frac{\partial f}{\partial x} = -2x(y-3x^2) - 6x(y-x^2) $$ $$ \frac{\partial f}{\partial y} = (1 - x^2) - (3x^2 -1) = 2 - 4x^2 $$
02

Find the critical points

The critical points are the points where both partial derivatives are equal to zero. So, we have: $$ -2x(y-3x^2) - 6x(y-x^2) = 0 $$ $$ 2 - 4x^2 = 0 $$ From the second equation, we have \(x = \pm\frac12\). Hence, the partial derivatives are continuous around the origin \((0,0)\), but neither of them equals zero at this point, indicating that \((0,0)\) is not a critical point of the function.
03

Analyze the strict local minimum along any line through the origin

To analyze lines through the origin, we write them as \(y = mx\). Then, the restriction of the function \(f\) to those lines is given by: $$ h(m) = f(x, mx) = (mx - x^2)((1 - 3m)x^2) $$ Taking the derivative of \(h(m)\) with respect to \(x\) and setting it equal to zero, we have: $$ \frac{dh}{dx} = m(1-3m)x^2 + 2x^3(1 - 4m) = 0 $$ This equation is satisfied by \(x = 0\) for any \(m\). In addition, for any given \(m\), \(h''(0) > 0\), which indicates a strict local minimum along any line through the origin.
04

Conclusion

The function \(f(x, y) = (y - x^2)(y - 3x^2)\) does not have an extremum at the origin \((0,0)\), as neither of its partial derivatives is equal to zero at this point indicating that it is not a critical point. However, the function does have a strict local minimum along any line that passes through the origin, in accordance with the given exercise.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Partial Derivatives
When dealing with functions of multiple variables, like in our example with the function f(x, y) = (y - x^2)(y - 3x^2), it's essential to understand how these functions change with respect to each variable independently. This is where partial derivatives come into play.

Imagine you are holding a sheet that represents a smooth landscape, and you want to know how steep the hill is if you walk north. The partial derivative with respect to y will tell you this. Now, if you want to know the steepness as you walk east, the partial derivative with respect to x will provide that information. Mathematically, these are expressed with the symbols 鈭俧/鈭倄 and 鈭俧/鈭倅.

To find them, you treat all other variables as constants and take the derivative concerning the variable of interest. In our function, when we find 鈭俧/鈭倄, we treat y as a constant, and vice versa. These derivatives tell us about the slope of the function in each variable's direction. If the ground is flat along the north path, the partial derivative with respect to y is zero, suggesting no incline or decline 鈥� a momentary level path. Likewise, if 鈭俧/鈭倄 turns out to be zero, it indicates no elevation change as we move eastward, at least at that particular spot.
Critical Points
Now, let's discuss the concept of critical points, which are tied closely to the notion of partial derivatives. A critical point occurs when the partial derivatives of a function with respect to all variables are zero or when the partial derivatives do not exist.

In the context of our landscape analogy, a critical point is like a flat spot where the ground doesn't slope neither north nor east 鈥� for instance, the peak of a hill or the bottom of a valley. In other words, it's where the function's rate of change is momentarily halted in all directions.

In the provided exercise, we search for critical points by setting the calculated partial derivatives equal to zero. Critical points are important because they are potential locations of extrema, which are the highest or lowest points on the landscape. However, simply finding a critical point doesn鈥檛 automatically mean you鈥檝e found an extremum. Further tests, like the second derivative test, are often needed to determine the exact nature of the critical point 鈥� whether it's a peak, valley, or neither.
Local Minimum
A local minimum is a sort of oasis in the desert of mathematics 鈥� it's a point where the function value is lower than at all other points in the immediate vicinity. To visualize this, picture yourself standing at the bottom of a small dip in our metaphorical landscape. No matter which direction you step, you'll go uphill; that's your local minimum.

To determine if a critical point is a local minimum, we often use the second derivative test, which involves examining the curvature of the function at that point. For lines passing through the origin, like in our exercise example, we can examine the behavior of the function restricted to those lines. If, along any of these lines, the function has a valley at the origin, it means there's a strict local minimum there.

However, a function could still lack a local minimum at a particular point when considering its behavior in all directions, not just along a specific path. This situation is exemplified in our exercise: although the function exhibits a strict local minimum along any line through the origin, the overall behavior of the function does not yield a local minimum at the origin itself. It's a crucial reminder that functions with multiple variables may have complex landscapes, with their high and low points depending on the paths we take.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Show that the rank of a smooth mapping \(f: \mathbb{R}^{m} \rightarrow \mathbb{R}^{n}\) is a lower semicontinuous function, that is rank \(f(x) \geq \operatorname{rank} f\left(x_{0}\right)\) in a neighborhood of a point \(x_{0} \in \mathbb{R}^{m}\).

"If \(f(x, y, z)=0\), then \(\frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial x} \cdot \frac{\partial x}{\partial z}=-1 .\) " a) Give a precise meaning to this statement. b) Verify that it holds in the example of Clapeyron's ideal gas equation $$ \frac{P \cdot V}{T}=\mathrm{const} $$ and in the general case of a function of three variables. c) Write the analogous statement for the relation \(f\left(x^{1}, \ldots, x^{m}\right)=0\) among \(m\) variables. Verify that it is correct.

a) Verify that the tangent to a curve \(\Gamma: I \rightarrow \mathbb{R}^{m}\) is defined invariantly relative to the choice of coordinate system in \(\mathbb{R}^{m}\). b) Verify that the tangent plane to the graph \(S\) of a function \(y=f\left(x^{1}, \ldots, x^{m}\right)\) is defined invariantly relative to the choice of coordinate system in \(\mathbb{R}^{m}\). c) Suppose the set \(S \subset \mathbb{R}^{m} \times \mathbb{R}^{1}\) is the graph of a function \(y=f\left(x^{1}, \ldots, x^{m}\right)\) in coordinates \(\left(x^{1}, \ldots, x^{m}, y\right)\) in \(\mathbb{R}^{m} \times \mathbb{R}^{1}\) and the graph of a function \(\tilde{y}=\) \(\tilde{f}\left(\tilde{x}^{1}, \ldots, \tilde{x}^{m}\right)\) in coordinates \(\left(\tilde{x}^{1}, \ldots, \tilde{x}^{m}, \tilde{y}\right)\) in \(\mathbb{R}^{m} \times \mathbb{R}^{1}\). Verify that the tangent plane to \(S\) is invariant relative to a linear change of coordinates in \(\mathbb{R}^{m} \times \mathbb{R}^{1}\). d) Verify that the Laplacian \(\Delta f=\sum_{i=1}^{m} \frac{\partial^{2} f}{\partial x^{i^{2}}}(x)\) is defined invariantly relative to orthogonal coordinate transformations in \(\mathbb{R}^{m}\).

a) Draw the graph of the function \(z=x^{2}+4 y^{2}\), where \((x, y, z)\) are Cartesian coordinates in \(\mathbb{R}^{3}\) b) Let \(f: G \rightarrow \mathbb{R}\) be a numerically valued function defined on a domain \(G \subset\) \(\mathbb{R}^{m} .\) A level set ( \(c\)-level) of the function is a set \(E \subset G\) on which the function assumes only one value \((f(E)=c) .\) More precisely, \(E=f^{-1}(c) .\) Draw the level sets in \(\mathbb{R}^{2}\) for the function given in part a). c) Find the gradient of the function \(f(x, y)=x^{2}+4 y^{2}\), and verify that at any point \((x, y)\) the vector \(\operatorname{grad} f\) is orthogonal to the level curve of the function \(f\) passing through the point. d) Using the results of a), b), and c), lay out what appears to be the shortest path on the surface \(z=x^{2}+4 y^{2}\) descending from the point \((2,1,8)\) to the lowest point on the surface \((0,0,0)\). e) What algorithm, suitable for implementation on a computer, would you propose for finding the minimum of the function \(f(x, y)=x^{2}+4 y^{2} ?\)

Let \(f: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) be a smooth mapping satisfying the Cauchy-Riemann equations $$ \frac{\partial f^{1}}{\partial x^{1}}=\frac{\partial f^{2}}{\partial x^{2}}, \quad \frac{\partial f^{1}}{\partial x^{2}}=-\frac{\partial f^{2}}{\partial x^{1}} $$ a) Show that the Jacobian of such a mapping is zero at a point if and only if \(f^{\prime}(x)\) is the zero matrix at that point. b) Show that if \(f^{\prime}(x) \neq 0\), then the inverse \(f^{-1}\) to the mapping \(f\) is defined in a neighborhood of \(f\) and also satisfies the Cauchy- Riemann equations.
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.