Review from class notes the notion of a surface. Now consider F: R³ → R, a differentiable (smooth) function. Thus, the partial derivatives of F, OF OF OF FT F = , บ მე and F₂ ду = Əz exist and are continuous on all of R³. We say a point p = (P1, P2, P3) Є R³ is a critical point of F if Fr(p) Fy(p) Fz(p) = 0. Otherwise p is said to be a regular point of F. For any cЄR, the level set F-1(c) is a regular level set if F-1(c) 0 and consists entirely of regular points. Regular Level Set Theorem. Let F: R³ → R be a differentiable function. Then every regular level set of F is a surface in R³. Note: The Regular Level Set Theorem does not say that non-regular level sets of F are never surfaces! It only says that regular level sets always are. In each case below decide if the set is (or is not) a surface in R³. Give a one-line explanation of your answer in each case. The Regular Level Set Theorem may be helpful in deciding that a set is a surface. It will not help you in arguing that a set is not a surface. (In the latter situation, just state where the problem points are. No further explanation is necessary.) (i.) C = {(x, y, z) Є R³ : x² — y² + z² = 0}; - (ii.) O = {(x, y, z) Є R³ : x² + y² + z² = 0}; (iii.) F-¹(c) where F(x, y, z) = x²+ y²+z² and c> 0; (iv.) X = Graph f where f(x, y) = x² - y²; (v.) Y the solution set to the equation zxy = 0; (vi.) PUP₁ where P₁ = {(x, y, z) = R³: x = 0} and Po = {(x, y, z) = R³ : y = 0}
Review from class notes the notion of a surface. Now consider F: R³ → R, a differentiable (smooth) function. Thus, the partial derivatives of F, OF OF OF FT F = , บ მე and F₂ ду = Əz exist and are continuous on all of R³. We say a point p = (P1, P2, P3) Є R³ is a critical point of F if Fr(p) Fy(p) Fz(p) = 0. Otherwise p is said to be a regular point of F. For any cЄR, the level set F-1(c) is a regular level set if F-1(c) 0 and consists entirely of regular points. Regular Level Set Theorem. Let F: R³ → R be a differentiable function. Then every regular level set of F is a surface in R³. Note: The Regular Level Set Theorem does not say that non-regular level sets of F are never surfaces! It only says that regular level sets always are. In each case below decide if the set is (or is not) a surface in R³. Give a one-line explanation of your answer in each case. The Regular Level Set Theorem may be helpful in deciding that a set is a surface. It will not help you in arguing that a set is not a surface. (In the latter situation, just state where the problem points are. No further explanation is necessary.) (i.) C = {(x, y, z) Є R³ : x² — y² + z² = 0}; - (ii.) O = {(x, y, z) Є R³ : x² + y² + z² = 0}; (iii.) F-¹(c) where F(x, y, z) = x²+ y²+z² and c> 0; (iv.) X = Graph f where f(x, y) = x² - y²; (v.) Y the solution set to the equation zxy = 0; (vi.) PUP₁ where P₁ = {(x, y, z) = R³: x = 0} and Po = {(x, y, z) = R³ : y = 0}
Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
6th Edition
ISBN:9781337111348
Author:Bruce Crauder, Benny Evans, Alan Noell
Publisher:Bruce Crauder, Benny Evans, Alan Noell
Chapter6: Rates Of Change
Section6.1: Velocity
Problem 12SBE
Related questions
Question
(ii)
Transcribed Image Text:Review from class notes the notion of a surface. Now consider F: R³ → R, a
differentiable (smooth) function. Thus, the partial derivatives of F,
OF
OF
OF
FT
F
=
, บ
მე
and F₂
ду
=
Əz
exist and are continuous on all of R³. We say a point p = (P1, P2, P3) Є R³ is a
critical point of F if
Fr(p)
Fy(p) Fz(p) = 0.
Otherwise p is said to be a regular point of F. For any cЄR, the level set F-1(c)
is a regular level set if F-1(c) 0 and consists entirely of regular points.
Regular Level Set Theorem. Let F: R³ → R be a differentiable function. Then
every regular level set of F is a surface in R³.
Note: The Regular Level Set Theorem does not say that non-regular level sets of
F are never surfaces! It only says that regular level sets always are.
In each case below decide if the set is (or is not) a surface in R³. Give a one-line
explanation of your answer in each case. The Regular Level Set Theorem may be
helpful in deciding that a set is a surface. It will not help you in arguing that a set
is not a surface. (In the latter situation, just state where the problem points are.
No further explanation is necessary.)
(i.) C = {(x, y, z) Є R³ : x² — y² + z² = 0};
-
(ii.) O = {(x, y, z) Є R³ : x² + y² + z² = 0};
(iii.) F-¹(c) where F(x, y, z) = x²+ y²+z² and c> 0;
(iv.) X = Graph f where f(x, y) = x² - y²;
(v.) Y the solution set to the equation zxy = 0;
(vi.) PUP₁ where P₁ = {(x, y, z) = R³: x = 0} and Po = {(x, y, z) = R³ : y = 0}
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