IDZ 10.2 - Option 6. Decisions Ryabushko AP
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- Find the equation of the tangent plane and the normal to the given surface S at the point M0 (x0, y0, z0)
1.6 S: x2 + y2 + z2 - 6y + 4z + 4 = 0, M0 (2, 1, - 1)
- Find the second partial derivatives of the functions. Ensure that z "xy = z" yx
2.6 z = arctg (x + y)
- To verify whether the above equation the function u.
- Examine the following function extremum.
4.6 z = 2x3 + 2y3 - 6xy + 5
- Find the maximum and minimum values of the function z = z (x, y) in D, given the limited lines.
5.6 z = x2 + y2 - 2x - 2y +8, D: x = 0, y = 0, x + y - 1 = 0
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