How to find a gradient

How to find a gradient

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By consideration of the questions including a concept of a gradient, most often functions perceive as scalar fields. Therefore it is necessary to enter the corresponding designations.

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Instruction

1. Let function be set by three arguments of u=f (x, y, z). A private derivative of function, on an example on x, define as the derivative on this argument received at fixation of other arguments. For other arguments similarly. Designations of a private derivative registers in a look: df/dh = u’x …

2. The full differential will be equal to du=(дf/дх) dx + (df/dy) to dy+(дf/дz) dz. Private derivatives can be understood how derivatives in the directions of coordinate axes. Therefore there is a question of finding of a derivative in the direction of the set s vector in a point of M (x, y, z) (do not forget that direction s sets a single vector-ort of s^o). At the same time vector differential of arguments {dx, dy, dz } = {dscos (alpha), dssos (beta), dssos (scale) }.

3. Considering a type of full differential of du, it is possible to draw a conclusion that the derivative in direction s is equal in a point of M: (du/ds) |M=((дf/дх)| Mcos (alpha) + ((df/dy) | M) cos (beta) + (df/dz) | M) cos (scale). If s = s (sx, sy, sz) then directing cosines {cos (alpha), cos (beta), cos (scale) } are calculated (see fig. 1a).

4. Definition of a derivative in the direction, including a variable M point, can be rewritten in the form of a scalar product: (du/ds)= ({df/dh, df/dy, df/dz }, {cos (alpha), cos (beta), cos (scale) })= (grad u, s^o). This expression will be fair for the scalar field. If function is considered just, then gradf is the vector having the coordinates coinciding with private derivative f (x, y, z). (x, y, z) {{df/dh, df/dy, df/dz } =) = (df/dh) i+(дf/дy)j + (df/dz) k. Here (i, j, k) – Horta of coordinate axes in the rectangular Cartesian system of coordinates.

5. If to use a differential vector operator of Hamilton a nabla, then gradf can be written down as multiplication of this vector operator by f scalar (see fig. 1b). In terms of communication of gradf with a derivative in the direction, equality (gradf, s^o) =0 is possible if these vectors of an ortogonalna. Therefore gradf often define how the direction of the fastest change of the scalar field. And in terms of differential operations (gradf - one of them), gradf properties in accuracy repeat properties of differentiation of functions. In particular, if f=uv, then gradf=(vgradu+u gradv).

Author: «MirrorInfo» Dream Team

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