IDZ 10.2 - Option 9. Decisions Ryabushko AP
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1. Find the equation of the tangent plane and the normal to the given surface S at the point M0 (x0, y0, z0)
1.9 S: x2 + y2 + 2yz - z2 + y - 2z = 2, M0 (1, 1, 1)
2. Find the second partial derivatives of the functions. Ensure that z "xy = z" yx
2.9 z = arcctg (x - 3y)
3. To verify whether the above equation the function u.
4. Examine the following function extremum.
4 4.9 z = (x - y) - x2 - y2
5. Find the maximum and minimum values of the function z = z (x, y) in D, given the limited lines.
5.9 z = x2 - 2y2 + 4xy - 6x - 1, D: x = 0, y = 0, x + y - 3 = 0
1.9 S: x2 + y2 + 2yz - z2 + y - 2z = 2, M0 (1, 1, 1)
2. Find the second partial derivatives of the functions. Ensure that z "xy = z" yx
2.9 z = arcctg (x - 3y)
3. To verify whether the above equation the function u.
4. Examine the following function extremum.
4 4.9 z = (x - y) - x2 - y2
5. Find the maximum and minimum values of the function z = z (x, y) in D, given the limited lines.
5.9 z = x2 - 2y2 + 4xy - 6x - 1, D: x = 0, y = 0, x + y - 3 = 0
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