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Consider F: R³ → R² defined by F(x, y, z) = ((x + 1)² + y −1+z, xy+y+z) . Using the Implicit Function Theorem, we can conclude that set S = {(x, y, z) = R³: F(x, y, z) = (0,0)} can be locally described by the graph of C¹ functions in some neighbourhood of (0,0,0). By the theorem, we are guaranteed the existence of a suitable C¹ function of which form? [2 1 1 0 1 1, Hint: There are TWO correct options. Hint: DF (0, 0, 0) = x = y(y, z) □ y = y(x, z) □ z = y(x, y) □ (x, y) = 4(z) (x, z) = 4(y) (y, z)= y(x)

Consider F: R³ → R² defined by F(x, y, z) = ((x + 1)² + y −1+z, xy+y+z) . Using the Implicit Function Theorem, we can conclude that set S = {(x, y, z) = R³: F(x, y, z) = (0,0)} can be locally described by the graph of C¹ functions in some neighbourhood of (0,0,0). By the theorem, we are guaranteed the existence of a suitable C¹ function of which form? [2 1 1 0 1 1, Hint: There are TWO correct options. Hint: DF (0, 0, 0) = x = y(y, z) □ y = y(x, z) □ z = y(x, y) □ (x, y) = 4(z) (x, z) = 4(y) (y, z)= y(x)

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Consider F: R³ → R² defined by F(x, y, z) = ((x + 1)² + y −1+z, xy+y+z). Using the Implicit Function Theorem, we can conclude that set S = {(x, y, z) = R³ : F(x, y, z) = (0,0)} can be locally described by the graph of C¹ functions in some neighbourhood of (0, 0, 0). By the theorem, we are guaranteed the existence of a suitable C1 function of which form? 2 1 0 1 1 Hint: There are TWO correct options. Hint: DF (0,0,0) = x = y(y, z) Oy = y(x,x) z = y(x, y) □ (x, y) = 4(2) □ (x, z) = y(y) □ (y, z) = y(x)
Transcribed Image Text:Consider F: R³ → R² defined by F(x, y, z) = ((x + 1)² + y −1+z, xy+y+z). Using the Implicit Function Theorem, we can conclude that set S = {(x, y, z) = R³ : F(x, y, z) = (0,0)} can be locally described by the graph of C¹ functions in some neighbourhood of (0, 0, 0). By the theorem, we are guaranteed the existence of a suitable C1 function of which form? 2 1 0 1 1 Hint: There are TWO correct options. Hint: DF (0,0,0) = x = y(y, z) Oy = y(x,x) z = y(x, y) □ (x, y) = 4(2) □ (x, z) = y(y) □ (y, z) = y(x)
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