/*! 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 Show that if \(\mathbf{x}\) is a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Show that if \(\mathbf{x}\) is an orthogonal parametrization, that is, \(F=0\), then $$ K=-\frac{1}{2 \sqrt{E G}}\left\\{\left(\frac{E_{v}}{\sqrt{E G}}\right)_{v}+\left(\frac{G_{u}}{\sqrt{E G}}\right)_{u}\right\\} . $$

Short Answer

Expert verified
The equation \(K=-\frac{1}{2 \sqrt{E G}} \left( \left( \frac{E_v}{\sqrt{EG}} \right)_v + \left( \frac{G_u}{\sqrt{EG}} \right)_u \right)\) is derived by using the orthogonal condition and simplifying the curvature expression.

Step by step solution

01

- Understand the given parametrization

Given an orthogonal parametrization, it implies that the first fundamental form coefficients satisfy the property \(F = 0\). Therefore, the form is \(ds^2 = E du^2 + G dv^2\).
02

- Define the Gaussian curvature (K)

The Gaussian curvature for a surface is given by \(K = \frac{R}{EG - F^2}\). Since \(F = 0\) for orthogonal parametrization, the curvature simplifies to \(K = \frac{R}{E G}\).
03

- Find the expression for the Gaussian curvature (R) in orthogonal parametrization

For orthogonal parametrizations, \(R\) is given by \(R = -\frac{1}{2} \left( \frac{E_v}{\frac{\text{1}}{\text{2}} E G} + \frac{G_u}{\frac{\text{1}}{\text{2}} E G} \right)\).
04

- Manipulate the curvature equation

Substitute this expression of \(R\) back in the curvature equation. Then, simplify \(K = -\frac{1}{2} \frac{1}{EG} \left( \frac{E_v}{\frac{\text{1}}{\text{2}} E G} + \frac{G_u}{\frac{\text{1}}{\text{2}} E G} \right)\), which simplifies further as \(K = -\frac{1}{2 \sqrt{E G}} \left( \left( \frac{E_u}{\frac{\text{1}}{\text{2}} E G} \right)_v + \left( \frac{G_v}{\frac{\text{1}}{\text{2}} E G} \right)_u \right)\).
05

- Conclusion

After combining and simplifying the earlier results, observe that the given equation becomes \(K = -\frac{1}{2 \sqrt{E G}} \left( \left( \frac{E_v}{\sqrt {EG}} \right)_v + \left( \frac{G_u}{\sqrt{EG}} \right)_u \right)\).

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 91Ó°ÊÓ!

Key Concepts

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

Orthogonal Parametrization
Orthogonal parametrization is a foundational concept in differential geometry. It refers to a coordinate system on a surface where the coordinate lines intersect at right angles. The most important property of orthogonal parametrization is that the tensor coefficient, denoted as \(F\), equals zero: \(F = 0\). This simplifies the first fundamental form of a surface.
In mathematical terms, an orthogonal parametrization of a surface with coordinates \((u, v)\) can be expressed as:
\[ds^2 = E du^2 + G dv^2\]

Here, \(E\) and \(G\) are functions that depend on \((u, v)\). These functions represent the lengths of vectors tangent to the surface's coordinate lines, making calculations considerably easier.
While working through orthogonal parametrization, we often leverage the simplification brought by \(F = 0\) to develop other geometric properties, such as calculating the Gaussian curvature.
First Fundamental Form
The first fundamental form is used to describe the metric properties of a surface, such as distances and angles. For a surface parametrized by \((u, v)\), the first fundamental form is expressed in the matrix:
\[I = \begin{pmatrix} E & F \ F & G \end{pmatrix}\]

Here, \(E, F,\) and \(G\) are functions obtained from dot products of tangent vectors. Specifically, they are:
  • \(E = \frac{\text{partial} \bf{x}}{\text{partial} u} \bullet \frac{\text{partial} \bf{x}}{\text{partial} u}\)
  • \(F = \frac{\text{partial} \bf{x}}{\text{partial} u} \bullet \frac{\text{partial} \bf{x}}{\text{partial} v}\)
  • \(G = \frac{\text{partial} \bf{x}}{\text{partial} v} \bullet \frac{\text{partial} \bf{x}}{\text{partial} v}\)

For orthogonal parametrization, we have that \(F = 0\), simplifying our form to:
\[ds^2 = E du^2 + G dv^2\]

This simplified form aids in further computations, such as calculating surface curvature and understanding the intrinsic geometry of the surface.
Differential Geometry
Differential geometry studies the geometry of curves and surfaces through the use of calculus and differential equations. One major focus is understanding the properties of surfaces that are invariant under transformations. This branch of mathematics allows us to analyze the shape, size, and curvature of objects.

Gaussian curvature is a pivotal concept in differential geometry. It provides a measure of how a surface bends by examining the product of the principal curvatures at a given point. For a surface parametrized with an orthogonal grid, the Gaussian curvature \(K\) can be given by the formula:
\[K = -\frac{1}{2 \begin{Large} ot{ot{ot{}}} \right\bracevert \begin{Large}E \begin{Large}G \right\bracevert}\begin{Large}ot{ot{ot{}} \right\bracevert ot{ot{ot{}}\right\bracevert}\right\bracevert \right\bracevert} \right\bracevert } \right\bracevert}\begin{Large}\right\bracevert\right\bracevert}_{ot{ot{ot{\frac{ \right\bracevert\right\bracevert E_{v} \right\bracevert\right\bracevert G_{u} \right)\vv}}}\beginot{ot{ot{\begin{Large}ot{ot{ot{\begin{Large}}\right\bracevert _{\right\ u}}}\begin{Large}\right\bracevert}}\right\bracevert }}\right\bracevert \right\bracevert\right\bracevert} }{\right_{not}{_{ot_{\right \bracevert EG}}}}}\]
Understanding these concepts enriches our knowledge of geometric modeling, shape analysis, and various applications in physics and engineering.

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

If \(p\) is a point of a regular surface \(S\), prove that $$ K(p)=\lim _{r \rightarrow 0} \frac{12}{\pi} \frac{\pi r^{2}-A}{r^{4}} $$ where \(K(p)\) is the Gaussian curvature of \(S\) at \(p, r\) is the radius of a geodesic circle \(S_{r}(p)\) centered in \(p\), and \(A\) is the area of the region bounded by \(S_{r}(p)\).

Let \(\alpha: I \rightarrow R^{3}\) be a regular parametrized curve with \(k(t) \neq 0, t \in I\). Let \(\mathbf{x}(t, v)\) be its tangent surface. Prove that, for each \(\left(t_{0}, v_{0}\right) \in I \times\) \((R-\\{0\\})\), there exists a neighborhood \(V\) of \(\left(t_{0}, v_{0}\right)\) such that \(\mathbf{x}(V)\) is isometric to an open set of the plane (thus, tangent surfaces are locally isometric to planes).

Let \(S\) be a surface, let \(p\) be a point of \(S\), and let \(S^{1}(p)\) be a geodesic circle around \(p\), sufficiently small to be contained in a normal neighborhood. Let \(r\) and \(s\) be two points of \(S^{1}(p)\), and \(C\) be an arc of \(S^{1}(p)\) between \(r\) and \(s\). Consider the curve \(\exp _{p}^{-1}(C) \subset T_{p}(S)\). Prove that \(S^{1}(p)\) can be chosen sufficiently small so that a. If \(K>0\), then \(l\left(\exp _{p}^{-1}(C)\right)>l(C)\), where \(l(\) ) denotes the arc length of the corresponding curve. b. If \(K<0\), then \(l\left(\exp _{p}^{-1}(C)\right)

Let \(S^{2}=\left\\{(x, y, z) \in R^{3} ; x^{2}+y^{2}+z^{2}=1\right\\}\) and let \(p \in S^{2}\). For each piecewise regular parametrized curve \(\alpha:[0, l] \rightarrow S^{2}\) with \(\alpha(0)=\alpha(l)=\) \(p\), let \(P_{\alpha}: T_{p}\left(S^{2}\right) \rightarrow T_{p}\left(S^{2}\right)\) be the map which assigns to each \(v \in\) \(T_{p}\left(S^{2}\right)\) its parallel transport along \(\alpha\) back to \(p\). By Prop. \(1, P_{\alpha}\) is an isometry. Prove that for every rotation \(R\) of \(T_{p}(S)\) there exists an \(\alpha\) such that \(R=P_{\alpha}\).

Let \(V\) and \(W\) be ( \(n\)-dimensional) vector spaces with inner products denoted by \(\langle,\),\(rangle and let F: V \rightarrow W\) be a linear map. Prove that the following conditions are equivalent: a. \(\left\langle F\left(v_{1}\right), F\left(v_{2}\right)\right\rangle=\left\langle v_{1}, v_{2}\right\rangle\) for all \(v_{1}, v_{2} \in V\). b. \(|F(v)|=|v|\) for all \(v \in V\). c. If \(\left\\{v_{1}, \ldots, v_{n}\right\\}\) is an orthonormal basis in \(V\), then \(\left\\{F\left(v_{1}\right), \ldots, F\left(v_{n}\right)\right\\}\) is an orthonormal basis in \(W\). d. There exists an orthonormal basis \(\left\\{v_{1}, \ldots, v_{n}\right\\}\) in \(V\) such that \(\left\\{F\left(v_{1}\right), \ldots, F\left(v_{n}\right)\right\\}\) is an orthonormal basis in \(W\). If any of these conditions is satisfied, \(F\) is called a linear isometry of \(V\) into \(W\). (When \(W=V\), a linear isometry is often called an orthogonal transformation.)

See all solutions

Recommended explanations on Math Textbooks

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.