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DHS 15.2 - Option 23. Decisions Ryabushko AP

DHS 15.2 - Option 23. Decisions Ryabushko AP

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  • Calculate and circulation of the vector field (M) over the contour of a triangle obtained by intersection of the plane (p): Ax + By + Cz = D with the coordinate planes, with respect to the positive direction of the normal vector bypass n = (A, B, C) this plane in two ways: 1) using the definition of circulation; 2) using the Stokes formula.

1.23. a (M) = (2y + z) i + (x - y) j - 2zk, (p): x - y + z = 2

  • Find the magnitude and direction of the greatest changes in the function u (M) = u (x, y, z) at the point M0 (x0, y0, z0)

2.23. u (M) = x (y2 + z2), M0 (1, -2, 1)

  • Find the greatest density of the circulation of the vector field a (M) = (x, y, z) at the point M0 (x0, y0, z0)

3.23. a (M) = z2i - xzj + z2k, M0 (1, -2, 1)

  • Determine whether the vector field a (M) = (x, y, z) the potential

4.23. a (M) = 3 (x - z) i + (x2 - y2) j + 3zk

Additional information

Detailed solution. Designed in PDF format for easy viewing of IDZ solutions on smartphones and PCs. In MS Word (doc format) sent additionally.