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(2) Let f(x1, x2, x3) = (x13+x2, x12+x3). Find df(1,2,3)(-3,-2,-1).

(2) Let f(x1, x2, x3) = (x13+x2, x12+x3). Find df(1,2,3)(-3,-2,-1).

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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Definitions are given in the images, hope it helps to clarify the vector calculus notation.

(2) Let f(x1, x2, x3) = (x13+x2, x12+x3). Find df(1,2,3)(-3,-2,-1).

(3) Let f(x, y, z) = (xy, yz, zx). Find df(0,1,2).

(4) Let f(t) = (1, t, t2, t3). Find df1(t).

(5) Let f(t) = (1, t, t2, t3). Find dft(1).

Theorem 1.8. (Inverse Function Theorem) Let U be an open sub- set of R", and let p e U. Let g U R" be a smooth map. If dg, e Hom(R", R") is a linear isomorphism, then there exist open neighbourhoods Uo, Vo of p, g(p) such that glu, : U → Vo is a dif- feomorphism. For example, consider the map f : R3 → R3 given by f(x1, x2, 13) = (x1F2C3, xỉ + x3 + x, 21 + x2 + x3). We have X2X3 X1X3 x1x2 2x2 df (12,73) = 2.x1 2.x3 1 1 1 If the determinant of this matrix is nonzero, then f must a diffeomor- phism when restricted to an open neighbourhood of the point (1, 2, 23) (although we cannot immediately predict how large this neighbourhood can be).
Transcribed Image Text:Theorem 1.8. (Inverse Function Theorem) Let U be an open sub- set of R", and let p e U. Let g U R" be a smooth map. If dg, e Hom(R", R") is a linear isomorphism, then there exist open neighbourhoods Uo, Vo of p, g(p) such that glu, : U → Vo is a dif- feomorphism. For example, consider the map f : R3 → R3 given by f(x1, x2, 13) = (x1F2C3, xỉ + x3 + x, 21 + x2 + x3). We have X2X3 X1X3 x1x2 2x2 df (12,73) = 2.x1 2.x3 1 1 1 If the determinant of this matrix is nonzero, then f must a diffeomor- phism when restricted to an open neighbourhood of the point (1, 2, 23) (although we cannot immediately predict how large this neighbourhood can be).
The map df : U → R", PH dfp af (d). af is called the derivative of f. If f is a C1 function, then df is a Co function, i.e. a continuous map. Another notation for the vector function (,..., f) is Vf (the gradient of f). Here we are regarding ( .. an as a vector in R". If we regard ( mation from R" to R (i.e. an element of the dual vector space (R")*), then we can apply it to a fixed vector v in R": ,..., ) as the 1 x n matrix of a linear transfor- V1 dfp(v) of (p). fe se i=1 It can also be written Vfp v or (Vfp, v). We shall usually regard df, as a linear transformation from R" to R (rather than the vector Vf, in R"). Note that we have written the vectorv = (v1,..., Un) E R" as a column vector, in order to use the usual matrix multiplication rule. Although there is no conceptual difference between row vectors and column vectors, it will be convenient to use column vectors in matrix calculations. If v is a unit vector (i.e. v| = 1), dfp(v) has an important meaning: it is the directional derivative of f at p in the direction of the vector v. As a concrete example, consider the function f : U → R given by f(x, y, z) = (x+3, zy–x, sin x, 3, y-7), where U = {(x, y, z) E R³ | y # 0}. For any (a, b, c) E U, the linear transformation df(a.b.c) has matrix 1 -1 b. COs a -76-8 0 (a 5 x 3 matrix). For any vector (v1, v2, v3) E R3 we have 1 -V1 + cv2 + bv3 Vị cos a -1 df (a,b.e) (V1, V2, V3) = cos a V2 V3 -76-8 0 -7vzb-8 We could write df (a,b.c) (V1, V2, V3) = (v1, -vi+cv2+bv3, vị cos a, 0, –7vzb-8). We could write (v1, v2, v3) as a column vector in df(a.b.e) (v1, v2, V3). (However, when matrices are multiplied together, we should use rows or columns consistently with the usual rule of matrix multiplication!)
Transcribed Image Text:The map df : U → R", PH dfp af (d). af is called the derivative of f. If f is a C1 function, then df is a Co function, i.e. a continuous map. Another notation for the vector function (,..., f) is Vf (the gradient of f). Here we are regarding ( .. an as a vector in R". If we regard ( mation from R" to R (i.e. an element of the dual vector space (R")*), then we can apply it to a fixed vector v in R": ,..., ) as the 1 x n matrix of a linear transfor- V1 dfp(v) of (p). fe se i=1 It can also be written Vfp v or (Vfp, v). We shall usually regard df, as a linear transformation from R" to R (rather than the vector Vf, in R"). Note that we have written the vectorv = (v1,..., Un) E R" as a column vector, in order to use the usual matrix multiplication rule. Although there is no conceptual difference between row vectors and column vectors, it will be convenient to use column vectors in matrix calculations. If v is a unit vector (i.e. v| = 1), dfp(v) has an important meaning: it is the directional derivative of f at p in the direction of the vector v. As a concrete example, consider the function f : U → R given by f(x, y, z) = (x+3, zy–x, sin x, 3, y-7), where U = {(x, y, z) E R³ | y # 0}. For any (a, b, c) E U, the linear transformation df(a.b.c) has matrix 1 -1 b. COs a -76-8 0 (a 5 x 3 matrix). For any vector (v1, v2, v3) E R3 we have 1 -V1 + cv2 + bv3 Vị cos a -1 df (a,b.e) (V1, V2, V3) = cos a V2 V3 -76-8 0 -7vzb-8 We could write df (a,b.c) (V1, V2, V3) = (v1, -vi+cv2+bv3, vị cos a, 0, –7vzb-8). We could write (v1, v2, v3) as a column vector in df(a.b.e) (v1, v2, V3). (However, when matrices are multiplied together, we should use rows or columns consistently with the usual rule of matrix multiplication!)
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