/*! 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 20 Determine the umbilical points o... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Determine the umbilical points of the elipsoid $$ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1 $$

Short Answer

Expert verified
The umbilical points are at \((\pm a, 0, 0)\), \((0, \pm b, 0)\), and \((0, 0, \pm c)\).

Step by step solution

01

- Recall the definition of an umbilical point

An umbilical point on a surface in three-dimensional space is a point where the principal curvatures are equal.
02

- Identify the equation of the ellipsoid

The ellipsoid is given by the equation: \[ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1. \]
03

- Use the parameters to find key points

For the ellipsoid, the key points that are potential candidates for umbilical points are the coordinate axes intersections. These occur where two of the variables equal zero.
04

- Evaluate the principal curvatures

For the ellipsoid, the principal curvatures at a point on the surface can be found by evaluating the second fundamental form. Upon evaluation, it is found that the intersection points on the coordinate axes (i.e., \((\frac{x}{a}, \frac{y}{b}, \frac{z}{c}) = (\pm 1,0,0)\), \((0,\pm 1,0)\), and \((0,0,\pm 1))\)) are umbilical.
05

- List the umbilical points

The umbilical points of the ellipsoid are: \[ (\pm a, 0, 0), (0, \pm b, 0), (0, 0, \pm c). \]

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.

Principal Curvatures
To understand umbilical points on an ellipsoid, we need to first grasp the idea of principal curvatures. Principal curvatures are the maximum and minimum curvatures at a given point on a surface. Imagine standing on a hill - in one direction, the slope might be steep (high curvature), while in another, it might be gentle (low curvature).

These two extremes are the principal curvatures. At an umbilical point, these curvatures are equal. This means that the surface bends equally in all directions at that point. For an ellipsoid, finding these points means identifying where the surface curvature is uniform.
Ellipsoid Equation
An ellipsoid is a three-dimensional shape defined by a specific equation. The general form of this equation is:

\[ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1 \]

Here, \( a \), \( b \), and \( c \) are the semi-principal axes of the ellipsoid. This equation tells us how the shape stretches in the x, y, and z directions. If all three values are equal, the ellipsoid becomes a sphere. That’s important because umbilical points are easier to find in spheres, where every point is an umbilical point. However, in an ellipsoid, they are located at specific points where the curvature properties align.
Coordinate Axes Intersections
For an ellipsoid, one useful property is its intersections with the coordinate axes. These points occur where two of the variables \(x, y, z \) are zero, simplifying our calculations. Specifically, the points of intersection are:

- (\[ \pm 1, 0, 0 \]),
- (0, \[ \pm 1, 0 \]),
- (0, 0, \[ \pm 1 \]).

When transferred to the general form of an ellipsoid, these points translate to:

- (\[ \pm a, 0, 0 \]),
- (0, \[ \pm b, 0 \]),
- (0, 0, \[ \pm c \]). These points are significant because they are potential candidates for being umbilical points.
Second Fundamental Form
The second fundamental form is a mathematical tool used to study the curvature of surfaces. By evaluating it at different points on an ellipsoid, we can determine the principal curvatures. This involves calculating second-order partial derivatives of the ellipsoid's equation.

When these calculations are done at the points where the ellipsoid intersects with the coordinate axes, we find that the principal curvatures are equal. Thus, these intersection points are umbilical points.

So, for our ellipsoid equation:

- (\[ \pm a, 0, 0 \]),
- (0, \[ \pm b, 0 \]),
- (0, 0, \[ \pm c \]),

All are identified as umbilical points where the surface bends uniformly in all directions.

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 the surface \(S_{1}\) intersects the surface \(S_{2}\) along the regular curve \(C\), then the curvature \(k\) of \(C\) at \(p \in C\) is given by $$ k^{2} \sin ^{2} \theta=\lambda_{1}^{2}+\lambda_{2}^{2}-2 \lambda_{1} \lambda_{2} \cos \theta $$ where \(\lambda_{1}\) and \(\lambda_{2}\) are the normal curvatures at \(p\), along the tangent line to \(C\), of \(S_{1}\) and \(S_{2}\), respectively, and \(\theta\) is the angle made up by the normal vectors of \(S_{1}\) and \(S_{2}\) at \(p\).

(Local Convexity and Curvature). A surface \(S \subset R^{3}\) is locally convex at a point \(p \in S\) if there exists a neighborhood \(V \subset S\) of \(p\) such that \(V\) is contained in one of the closed half-spaces determined by \(T_{p}(S)\) in \(R^{3}\). If, in addition, \(V\) has only one common point with \(T_{p}(S)\), then \(S\) is called striclly locally convex at \(p\). a. Prove that \(S\) is strictly locally convex at \(p\) if the principal curvatures of \(S\) at \(p\) are nonzero with the same sign (that is, the Gaussian curvature \(K(p)\) satisfies \(K(p)>0\) ). b. Prove that if \(S\) is locally convex at \(p\), then the principal curvatures at \(p\) do not have different signs (thus, \(K(p) \geq 0\) ). c. To show that \(K \geq 0\) does not imply local convexity, consider the surface \(f(x, y)=x^{3}\left(1+y^{2}\right)\), defined in the open set \(U=\left\\{(x, y) \in R^{2} ; y^{2}<\frac{1}{2}\right\\} .\) Show that the Gaussian curvature of this surface is nonnegative on \(U\) and yet the surface is not locally convex at \((0,0) \in U\) (a deep theorem, due to \(\mathrm{R}\). Sacksteder, implies that such an example cannot be extended to the entire \(R^{2}\) if we insist on keeping the curvature nonnegative; cf. Remark 3 of Sec. \(5-6\) ). *d. The example of part \(\mathrm{c}\) is also very special in the following local sense. Let \(p\) be a point in a surface \(S\), and assume that there exists a neighborhood \(V \subset S\) of \(p\) such that the principal curvatures on \(V\) do not have different signs (this does not happen in the example of part c). Prove that \(S\) is locally convex at \(p\).

(Theorem of Joachimstahl.) Suppose that \(S_{1}\) and \(S_{2}\) intersect along a regular curve \(C\) and make an angle \(\theta(p), p \in C\). Assume that \(C\) is a line of curvature of \(S_{1}\). Prove that \(\theta(p)\) is constant if and only if \(C\) is a line of curvature of \(S_{2}\).

Show that a surface which is compact (i.e., it is bounded and closed in \(R^{3}\) ) has an elliptic point.

5\. Let \(S\) be a surface and \(\mathbf{x}: U \rightarrow S\) be a parametrization of \(S\). If \(a c-b^{2}<0\), show that $$ a(u, v)\left(u^{\prime}\right)^{2}+2 b(u, v) u^{\prime} v^{\prime}+c(u, v)\left(v^{\prime}\right)^{2}=0 $$ can be factored into two distinct equations, each of which determines a field of directions on \(\mathbf{x}(U) \subset S\). Prove that these two fields of directions are orthogonal if and only if $$ E c-2 F b+G a=0 . $$

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.