/*! 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 1 On the plane \(\mathbb{R}^{2}\) ... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 1

On the plane \(\mathbb{R}^{2}\) with coordinates \(x\) and \(y\) a curve is defined by the relation \(F(x, y)=0\), where \(F \in C^{(2)}\left(\mathbb{R}^{2}, \mathbb{R}\right)\). Let \(\left(x_{0}, y_{0}\right)\) be a noncritical point of the function \(F(x, y)\) lying on the curve. a) Write the equation of the tangent to this curve at this point \(\left(x_{0}, y_{0}\right)\). b) Show that if \(\left(x_{0}, y_{0}\right)\) is a point of inflection of the curve, then the following equality holds: $$ \left(F_{x x}^{\prime \prime} F_{y}^{\prime 2}-2 F_{x y}^{\prime \prime} F_{x}^{\prime} F_{y}^{\prime}+F_{y y}^{\prime \prime} F_{x}^{\prime 2}\right)\left(x_{0}, y_{0}\right)=0 $$ c) Find a formula for the curvature of the curve at the point \(\left(x_{0}, y_{0}\right)\).

Short Answer

Expert verified
Answer: The equation of the tangent to the curve at point (x0, y0) is given by: F_x'(x_0, y_0)(x - x_0) + F_y'(x_0, y_0)(y - y_0) = 0. The condition for a point of inflection is given by: (F_{xx}'' F_y'^2-2F_{xy}'' F_{x}' F_{y}'+F_{yy}'' F_x'^2)(x_0, y_0) = 0. The formula for the curvature of the curve at point (x0, y0) is: k(x_0, y_0) = |F_x'(x_0, y_0)F_{yy}''(x_0, y_0) - 2F_{xy}''(x_0, y_0)F_x'(x_0, y_0)F_y'(x_0, y_0) + F_y'(x_0, y_0)F_{xx}''(x_0, y_0)| / (F_x'(x_0, y_0)^2 + F_y'(x_0, y_0)^2)^(3/2).

Step by step solution

01

a) Equation of the tangent

The gradient of the curve is given by the gradient of the function \(F(x, y)\), and we have $$ \nabla F(x, y) = \left(F_x'(x, y), F_y'(x, y)\right), $$ where \(F_x'(x, y)\) and \(F_y'(x, y)\) are the first partial derivatives of F with respect to x and y, respectively. Now, the gradient at the point \((x_0, y_0)\) is $$ \nabla F(x_0, y_0) = \left(F_x'(x_0, y_0), F_y'(x_0, y_0)\right). $$ The equation of the tangent line at this point can be written in terms of the dot product: $$ \left(F_x'(x_0, y_0), F_y'(x_0, y_0)\right) \cdot (x - x_0, y - y_0) = 0. $$ Expanding the dot product, we get $$ F_x'(x_0, y_0)(x - x_0) + F_y'(x_0, y_0)(y - y_0) = 0. $$
02

b) Inflection point condition

To show the condition for a point of inflection, we will evaluate the second derivative test. The second derivative test involves finding the mixed and pure second derivatives, which are $$ F_{xx}''(x, y),\ F_{yy}''(x, y),\ \text{and} \ F_{xy}''(x, y). $$ Now, let's consider that the given point \((x_0, y_0)\) is a point of inflection of the curve. In this case, the curvature of the curve changes its sign at this point, which means that the determinant of the Hessian matrix should be zero. The Hessian matrix is given by $$ H(F) = \begin{bmatrix} F_{xx}'' & F_{xy}''\\ F_{yx}'' & F_{yy}'' \end{bmatrix}. $$ Now, we must show that $$ \text{det}(H(F))=\left(F_{xx}'' F_y'^2-2F_{xy}''F_{x}'F_{y}'+F_{yy}'' F_x'^2\right)(x_0,y_0)=0. $$ Since \((x_0, y_0)\) is a point of inflection, we have $$ \text{det}(H(F(x_0, y_0))) = F_{xx}''(x_0, y_0)F_{yy}''(x_0, y_0) - (F_{xy}''(x_0, y_0))^2 = 0. $$ From the above equation, we obtain $$ F_{xx}''(x_0, y_0)F_{yy}''(x_0, y_0) = (F_{xy}''(x_0, y_0))^2. $$ Now, multiply both sides by \(F_x'^2(x_0, y_0)F_y'^2(x_0, y_0)\): $$ F_{xx}''(x_0, y_0)F_y'^2(x_0, y_0)F_x'^2(x_0, y_0)F_{yy}''(x_0, y_0) = F_x'^2(x_0, y_0)F_y'^2(x_0, y_0)(F_{xy}''(x_0, y_0))^2. $$ Rearranging, we have $$ \left(F_{xx}'' F_y'^2-2F_{xy}'' F_{x}' F_{y}'+F_{yy}'' F_x'^2\right)(x_0, y_0) = 0. $$
03

c) Curvature of the curve

The formula for the curvature of the curve is given by $$ k(x, y) = \frac{|F_x'(x, y)F_{yy}''(x, y) - 2F_{xy}''(x, y)F_x'(x, y)F_y'(x, y) + F_y'(x, y)F_{xx}''(x, y)|}{(F_x'(x, y)^2 + F_y'(x, y)^2)^{3/2}}, $$ So, the curvature of the curve at \((x_0, y_0)\) is given by $$ k(x_0, y_0) = \frac{|F_x'(x_0, y_0)F_{yy}''(x_0, y_0) - 2F_{xy}''(x_0, y_0)F_x'(x_0, y_0)F_y'(x_0, y_0) + F_y'(x_0, y_0)F_{xx}''(x_0, y_0)|}{(F_x'(x_0, y_0)^2 + F_y'(x_0, y_0)^2)^{3/2}}. $$

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Key Concepts

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

Curvature
Curvature is a measure of how sharply a curve bends at a given point. It offers insights into the geometry and nature of a curve in space.
The curvature at a point indicates the degree to which the curve deviates from being a straight line, or a flat part.

In mathematical terms, for a plane curve defined by the relation \[ F(x, y) = 0 \], the curvature \( k(x, y) \) at a point \( (x_0, y_0) \) can be expressed with the formula:
  • \[ k(x_0, y_0) = \frac{|F'_x(x_0, y_0) F''_{yy}(x_0, y_0) - 2 F''_{xy}(x_0, y_0) F'_x(x_0, y_0) F'_y(x_0, y_0) + F'_y(x_0, y_0) F''_{xx}(x_0, y_0)|}{(F'_x(x_0, y_0)^2 + F'_y(x_0, y_0)^2)^{3/2}} \]
This formula considers:
  • The first derivatives \( F'_x(x, y) \) and \( F'_y(x, y) \), which give the slope of the tangent plane.
  • The second derivatives \( F''_{xx} \), \( F''_{yy} \), and \( F''_{xy} \), describing how the slope itself changes with position.
Curvature helps in understanding properties like how fast the direction of a curve is changing .
Hessian Matrix
The Hessian matrix is a pivotal tool in calculus, mainly in optimization and understanding concavity/convexity.
For a function \( F(x, y) \), the Hessian matrix, denoted as \( H(F) \), is a square matrix of second-order partial derivatives:
  • \[ H(F) = \begin{bmatrix}F''_{xx} & F''_{xy} \F''_{yx} & F''_{yy} \end{bmatrix} \]
Each component represents how the curvature in one direction affects or is affected by the curvature in another.
The determinant of the Hessian matrix helps evaluate critical points:
- When the determinant is positive, it indicates either a local minimum or maximum.
- When negative, it points to a saddle point, where the direction changes.
  • A zero determinant suggests a possible point of inflection, where curvature changes sign.
This matrix provides a comprehensive view of the behavior of \( F(x, y) \) around a point. Understanding the Hessian matrix is crucial for exploring the nature and geometry of curves in multivariable functions.
Inflection Point
An inflection point on a curve is where the curve changes its concavity from concave up to concave down or vice versa.
This is the point where the curvature of the curve transitions, indicated by a change in sign.

For the curve defined as \( F(x, y) = 0 \), the presence of an inflection point at a given location can often be examined using the determinant of the Hessian matrix. This is because a zero determinant denotes that the curvature is switching its sign.

Inflection points are critical in the study of graphs and functions as they indicate a transition in behavior:
  • If you move along the curve, either side of an inflection point will show different curvatures.
  • Recognizing these points can provide significant insight into the oscillatory nature of functions.
Analyzing inflection points involves a deep dive into the second derivative test, where we study second-order changes in slope behavior.
Second Derivative Test
The second derivative test is a mathematical method used to determine the nature of critical points for differentiable functions. After identifying a stationary point, where the first derivative is zero, this test helps decide if it's a minimum, maximum, or an inflection point.

Here's how it works for a function \( F(x, y) \):
  • Calculate the Hessian matrix at the point, using second partial derivatives: \[ F''_{xx}, F''_{yy}, F''_{xy} \]
  • Find the determinant of the Hessian matrix. Its sign gives crucial insights:
    • Positive determinant implies a local extremum (min or max, depending on the diagonal elements).
    • Negative determinant indicates a saddle point.
    • Zero determinant means the test is inconclusive; often tying to inflection points.
The second derivative test is fundamental in calculus for examining stability and behavior at critical points, offering a clear framework for classifying the point's nature.

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Most popular questions from this chapter

a) Verify that for the function $$ f(x, y)= \begin{cases}x y \frac{x^{2}-y^{2}}{x^{2}+y^{2}}, & \text { if } x^{2}+y^{2} \neq 0 \\ 0, & \text { if } x^{2}+y^{2}=0\end{cases} $$ the following relations hold: $$ \frac{\partial f}{\partial x \partial y}(0,0)=1 \neq-1=\frac{\partial^{2} f}{\partial y \partial x}(0,0) $$ b) Prove that if the function \(f(x, y)\) has partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) in some neighborhood \(U\) of the point \(\left(x_{0}, y_{0}\right)\), and if the mixed derivative \(\frac{\partial^{2} f}{\partial x \partial y}\) (or \(\left.\frac{\partial^{2} f}{\partial y \partial x}\right)\) exists in \(U\) and is continuous at \(\left(x_{0}, y_{0}\right)\), then the mixed derivative \(\frac{\partial^{2} f}{\partial y \partial x}\) (resp. \(\left.\frac{\partial^{2} f}{\partial x \theta y}\right)\) also exists at that point and the following equality holds: $$ \frac{\partial^{2} f}{\partial x \partial y}\left(x_{0}, y_{0}\right)=\frac{\partial^{2} f}{\partial y \partial x}\left(x_{0}, y_{0}\right) $$

a) Show that the functions \(\pi^{i}(x)=x^{i}(i=1, \ldots, m)\), regarded as functions of the point \(x=\left(x^{1}, \ldots, x^{m}\right) \in \mathbb{R}^{m}\), form an independent system of functions in a neighborhood of any point of \(\mathbb{R}^{m}\). b) Show that, for any function \(f \in C\left(\mathbb{R}^{m} ; \mathbb{R}\right)\) the system \(\pi^{1}, \ldots, \pi^{m}, f\) is functionally dependent. c) If the system of smooth functions \(f^{1}, \ldots, f^{k}, k

Taylor's formula in multi-index notation. The symbol \(\alpha:=\left(\alpha_{1}, \ldots, \alpha_{m}\right)\) consisting of nonnegative integers \(\alpha_{i}, i=1, \ldots, m\), is called the multi-index \(\alpha\). The following notation is conventional: $$ \begin{aligned} |\alpha|: &=\alpha_{1}+\cdots+\alpha_{m} \\ \alpha ! &:=\alpha_{1} ! \cdots \alpha_{m} ! \end{aligned} $$ finally, if \(a=\left(a_{1}, \ldots, a_{m}\right)\), then $$ a^{\alpha}:=a_{1}^{\alpha_{1}} \cdots a_{m}^{\alpha_{m}} $$ a) Verify that if \(k \in \mathbb{N}\), then $$ \left(a_{1}+\cdots+a_{m}\right)^{k}=\sum_{|\alpha|=k} \frac{k !}{\alpha_{1} ! \cdots \alpha_{m} !} a_{1}^{\alpha_{1}} \cdots a_{m}^{\alpha_{m}} $$ or $$ \left(a_{1}+\cdots+a_{m}\right)^{k}=\sum_{|\alpha|=k} \frac{k !}{\alpha !} a^{\alpha} $$ where the summation extends over all sets \(\alpha=\left(\alpha_{1}, \ldots, \alpha_{m}\right)\) of nonnegative integers such that \(\sum_{i=1}^{m} \alpha_{i}=k\). b) Let $$ D^{\alpha} f(x):=\frac{\partial^{|\alpha|} f}{\left(\partial x^{1}\right)^{\alpha 1} \cdots\left(\partial x^{m}\right)^{\alpha_{m}}}(x) $$ Show that if \(f \in C^{(k)}(G ; \mathbb{R})\), then the equality $$ \sum_{i_{1}+\cdots+i_{m}=k} \partial_{i_{1} \cdots i_{k}} f(x) h^{i_{1}} \cdots h^{i_{k}}=\sum_{|\alpha|=k} \frac{k !}{\alpha !} D^{\alpha} f(x) h^{\alpha} $$ where \(h=\left(h^{1}, \ldots, h^{m}\right)\), holds at any point \(x \in G\). c) Verify that in multi-index notation Taylor's theorem with the Lagrange form of the remainder, for example, can be written as $$ f(x+h)=\sum_{|\alpha|=0}^{n-1} \frac{1}{\alpha !} D^{\alpha} f(x) h^{\alpha}+\sum_{|\alpha|=n} \frac{1}{\alpha !} D^{\alpha} f(x+\theta h) h^{\alpha} $$ d) Write Taylor's formula in multi-index notation with the integral form of the remainder (Theorem 4\()\).

Homogeneous functions and Euler's identity. A function \(f: G \rightarrow \mathbb{R}\) defined in some domain \(G \subset \mathbb{R}^{m}\) is called homogeneous (resp. positive-homogeneous) of degree \(n\) if the equality $$ f(\lambda x)=\lambda^{n} f(x) \quad\left(\text { resp. } f(\lambda x)=|\lambda|^{n} f(x)\right) $$ holds for any \(x \in \mathbb{R}^{m}\) and \(\lambda \in \mathbb{R}\) such that \(x \in G\) and \(\lambda x \in G\) A function is locally homogeneous of degree \(n\) in the domain \(G\) if it is a homogeneous function of degree \(n\) in some neighborhood of each point of \(G\). a) Prove that in a convex domain every locally homogeneous function is homogeneous. b) Let \(G\) be the plane \(\mathbb{R}^{2}\) with the ray \(L=\left\\{(x, y) \in \mathbb{R}^{2} \mid x=2 \wedge y \geq 0\right\\}\) removed. Verify that the function $$ f(x, y)= \begin{cases}y^{4} / x, & \text { if } x>2 \wedge y>0 \\ y^{3}, & \text { at other points of the domain, }\end{cases} $$ is locally homogeneous in \(G\), but is not a homogeneous function in that domain. c) Determine the degree of homogeneity or positive homogeneity of the following functions with their natural domains of definition: $$ \begin{aligned} f_{1}\left(x^{1}, \ldots, x^{m}\right) &=x^{1} x^{2}+x^{2} x^{3}+\cdots+x^{m-1} x^{m} \\ f_{2}\left(x^{1}, x^{2}, x^{3}, x^{4}\right) &=\frac{x^{1} x^{2}+x^{3} x^{4}}{x^{1} x^{2} x^{3}+x^{2} x^{3} x^{4}} \\ f_{3}\left(x^{1}, \ldots, x^{m}\right) &=\left|x^{1} \cdots x^{m}\right|^{l} \end{aligned} $$ d) By differentiating the equality \(f(t x)=t^{n} f(x)\) with respect to \(t\), show that if a differentiable function \(f: G \rightarrow \mathbb{R}\) is locally homogeneous of degree \(n\) in a domain \(G \subset \mathbb{R}^{m}\), it satisfies the following Euler identity for homogeneous functions: $$ x^{1} \frac{\partial f}{\partial x^{1}}\left(x^{1}, \ldots, x^{m t}\right)+\cdots+x^{m} \frac{\partial f}{\partial x^{m}}\left(x^{1}, \ldots, x^{m}\right) \equiv n f\left(x^{1}, \ldots, x^{m}\right) $$ e) Show that if Euler's identity holds for a differentiable function \(f: G \rightarrow \mathbb{R}\) in a domain \(G\), then that function is locally homogeneous of degree \(n\) in \(G\). Hint: Verify that the function \(\varphi(t)=t^{-n} f(t x)\) is defined for every \(x \in G\) and is constant in some neighborhood of \(1 .\)

The Legendre transform in \(m\) variables. The Legendre transform of \(x^{1}, \ldots, x^{m}\) and the function \(f\left(x^{1}, \ldots, x^{m}\right)\) is the transformation to the new variables \(\xi_{1}, \ldots, \xi_{m}\) and function \(f^{*}\left(\xi_{1}, \ldots, \xi_{m}\right)\) defined by the relations $$ \left\\{\begin{array}{l} \xi_{i}=\frac{\partial f}{\partial x^{i}}\left(x^{1}, \ldots, x^{m}\right) \quad(i=1, \ldots, m) \\ f^{*}\left(\xi_{1}, \ldots, \xi_{m}\right)=\sum_{i=1}^{m} \xi_{i} x^{i}-f\left(x^{1}, \ldots, x^{m}\right) \end{array}\right. $$ a) Give a geometric interpretation of the Legendre transform (8.106) as the transition from the coordinates \(\left(x^{1}, \ldots, x^{m}, f\left(x^{1}, \ldots, x^{m}\right)\right)\) of a point on the graph of the function \(f(x)\) to the parameters \(\left(\xi_{1}, \ldots, \xi_{m}, f^{*}\left(\xi_{1}, \ldots, \xi_{m}\right)\right.\) ) defining the equation of the plane tangent to the graph at that point. b) Show that the Legendre transform is guaranteed to be possible locally if \(f \in\) \(C^{(2)}\) and \(\operatorname{det}\left(\frac{\partial^{2} f}{\partial x^{i} \partial x^{j}}\right) \neq 0\) c) Using the same definition of convexity for a function \(f(x)=f\left(x^{1}, \ldots, x^{m}\right)\) as in the one-dimensional case (taking \(x\) to be the vector \(\left.\left(x^{1}, \ldots, x^{m}\right) \in \mathbb{R}^{m}\right)\), show that the Legendre transform of a convex function is a convex function. d) Show that $$ \mathrm{d} f^{*}=\sum_{i=1}^{m} x^{i} \mathrm{~d} \xi_{i}+\sum_{i=1}^{m} \xi_{i} \mathrm{~d} x^{i}-\mathrm{d} f=\sum_{i=1}^{m} x^{i} \mathrm{~d} \xi_{i} $$ and deduce from this relation that the Legendre transform is involutive, that is, verify the equality $$ \left(f^{*}\right)^{*}(x)=f(x) $$ e) Taking account of d), write the transform (8.106) in the following form, which is symmetric in the variables: $$ \left\\{\begin{array}{l} f^{*}\left(\xi_{1}, \ldots, \xi_{m}\right)+f\left(x^{1}, \ldots, x^{m}\right)=\sum_{i=1}^{m} \xi_{i} x^{i} \\ \xi_{i}=\frac{\partial f}{\partial x^{i}}\left(x^{1}, \ldots, x^{m}\right), \quad x^{i}=\frac{\partial f^{*}}{\partial \xi_{i}}\left(\xi_{1}, \ldots, \xi_{m}\right) \end{array}\right. $$ or, more briefly, in the form $$ f^{*}(\xi)+f(x)=\xi x, \quad \xi=\nabla f(x), \quad x=\nabla f^{*}(\xi) $$ where $$ \begin{aligned} \nabla f(x) &=\left(\frac{\partial f}{\partial x^{1}}, \ldots, \frac{\partial f}{\partial x^{m}}\right)(x), \quad \nabla f^{*}(\xi)=\left(\frac{\partial r^{*}}{\partial \xi_{1}}, \ldots, \frac{\partial f^{*}}{\partial \xi_{m}}\right)(\xi) \\ \xi x &=\xi_{i} x^{i}=\sum_{i=1}^{m} \xi_{i} x^{i} \end{aligned} $$ f) The matrix formed from the second-order partial derivatives of a function (and sometimes the determinant of this matrix) is called the Hessian of the function at a given point. Let \(d_{i j}\) and \(d_{i j}^{*}\) be the co-factors of the elements \(\frac{\partial^{2} f}{\partial x^{i} \partial x^{j}}\) and \(\frac{\partial^{2} f^{*}}{\partial \xi_{i} \partial \xi_{j}}\) of the Hessians $$ \left(\begin{array}{ccc} \frac{\partial^{2} f}{\partial x^{1} \partial x^{1}} & \cdots & \frac{\partial^{2} f}{\partial x^{1} \partial x^{m}} \\ \vdots & \ddots & \vdots \\ \frac{\partial^{2} f}{\partial x^{m} \partial x^{1}} & \cdots & \frac{\partial^{2} f}{\partial x^{m} \partial x^{m}} \end{array}\right)(x), \quad\left(\begin{array}{ccc} \frac{\partial^{2} f^{*}}{\partial \xi_{1} \partial \xi_{1}} & \cdots & \frac{\partial^{2} f^{*}}{\partial \xi_{1} \partial \xi_{m}} \\ \vdots & \ddots & \vdots \\ \frac{\partial^{2} f^{*}}{\partial \xi_{m} \partial \xi_{1}} & \cdots & \frac{\partial^{2} f^{*}}{\partial \xi_{m} \partial \xi_{m}} \end{array}\right)(\xi) $$ of the functions \(f(x)\) and \(f^{*}(\xi)\), and let \(d\) and \(d^{*}\) be the determinants of these matrices. Assuming that \(d \neq 0\), show that \(d \cdot d^{*}=1\) and that $$ \frac{\partial^{2} f}{\partial x^{i} \partial x^{j}}(x)=\frac{d_{i j}^{*}}{d^{*}}(\xi), \quad \frac{\partial^{2} f^{*}}{\partial \xi_{i} \partial \xi_{j}}(\xi)=\frac{d_{i j}}{d}(x) $$ g) A soap film spanning a wire frame forms a so-called minimal surface, having minimal area among all the surfaces spanning the contour. If that surface is locally defined as the graph of a function \(z=f(x, y)\), it turns out that the function \(f\) must satisfy the following equation for minimal surfaces: $$ \left(1+f_{y}^{\prime 2}\right) f_{x x}^{\prime \prime}-2 f_{x}^{\prime} f_{y}^{\prime} f_{x y}^{\prime \prime}+\left(1+f_{x}^{\prime 2}\right) f_{y y}^{\prime \prime}=0 $$ Show that after a Legendre transform is performed this equation is brought into the form $$ \left(1+\eta^{2}\right) f_{\eta \eta}^{* \prime \prime}+2 \xi \eta f_{\xi \eta}^{* \prime \prime}+\left(1+\xi^{2}\right) f_{\xi \xi}^{* \prime \prime}=0 $$
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