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DHS 15.2 - Option 28. Decisions Ryabushko AP

DHS 15.2 - Option 28. 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.

1.28. a (M) = xi + (x + z) j + (y + z) k, (p): 3x + 3y + z = 3

  • 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)

2.28. u (M) = xy2 - z, 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)

3.28. a (M) = (y - z) i + yj - z2k, M0 (-1, 2, 1)

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

4.28. a (M) = x / yi + y / zj + z / xk

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.