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DHS 15.2 - Option 19. Decisions Ryabushko AP

DHS 15.2 - Option 19. 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.

1.19. a (M) = xi + (y - 2z) j + (2x - y + 2z) k, (p): x + 2y + 2z = 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)

2.19. u (M) = x2 (y2 + z), M0 (4, 1, -3)

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

3.19. a (M) = (x + y) i + xyzj - xk, M0 (4, 1, -3)

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

4.19. a (M) = z2i + (xz + y) j + x2yk

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.