What is the equation of the tangent plane to the surface z = -x2 - 2y2 at the po

Question

What is the equation of the tangent plane to the surface z = -x2 - 2y2 at the point (1, 1, -3)? -2x - 4y + z = -9 2x + 4y + z = 6 x + 2y + z = 0 x + 2y - z = 6 2x + 4y + z = 3 None of these Let f(x, y) = ex2 - y2. Which of the following vectors describes the direction in which f increases most rapidly at the point (1,0)? d = (1,0) d = (2e,-2e) d = (0,-1) d = (2e,2e) d = (0,0) The rate of change is 0 in every direction Calculate , where f(x, y) = 3x2y2 - 6y2 9x2y2 18x2y - 12y 0 18x2y 6x3y - 12y None of these

Explanation / Answer

3) df/dx = 9 * x^2 *y^2 - 0

d^2 f /dx dy = 18 x^2 y

d is the answer

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