Gmail YouTube Maps Translate G= News my myucdavis Bookshelf: Verify... The first two questions are about the tangent plane to the surface z = 5x + y + In(x2y) in(x0, yo ) = (1, 1). The equation of the tangent plane to the curve z = f (x, y) in (20, yo ) is z - fx (xo, yo) (x - x0) + fy (x0, yo) (y - yo) + f(x0, yo). D Question 1 1 pts Find the equation of the tangent plane to the surface z = 5x + y + In(x2y) in(zo, yo ) - (1, 1). O z = 5(2 - 1) + 2(y - 1) +6 O z= 7(2 - 1) - 2(y - 1) + 6 O z = 5(x - 1) + 2(y - 1) + 4 Oz =7(x - 1) + 2(y - 1) + 6 Oz = 7(2 - 1) + (y - 1) +6Gmail YouTube Maps Translate G= News my myucdavis Bookshelf: Verify. D Question 2 1 pts Use your answer to the previous question to estimate In(0.92 x 1.05). In general, we assume that near point (x0, yo ) the surface almost coincides with the tangent plane: For any (x, y) near (x0 , yo ) f ( x, y) ~ fx (x0, yo) (x x0) + fy (x0, 40) (y - 30) + f(20, 90). So for our problem, for any (x, y) near (1, 1), 5x + y + In(x2y) ~ fx (1, 1) (x - 1) + fy (1, 1) (y - 1) + f(1, 1). O In(0.92 x 1.05) ~ 7(0.9 - 1) + 2(1.05 - 1) + 6 In(0.92 x 1.05) ~ 7(0.9 1) + 2(1.05 - 1) +6 - (5 . 0.9 + 1.05) In(0.92 x 1.05) ~ 7(0.9 1) 2(1.05 - 1) + 6 - (5 . 0.9 + 1.05) In(0.92 x 1.05) ~ 7(0.9 1) + (1.05 - 1) +6 - (5 . 0.9 + 1.05) In(0.92 x 1.05) ~ 5(0.9 - 1) + 2(1.05 - 1) + 6 - (5 . 0.9 + 1.05) D Question 3 1 pts