Question

6. Find an equation of the tangent plane to the surface z = 4x2-y2 +2y at (-1,2,4) = V20-下一77 at (2,1) 7. Find the linear approximation of f(z. y) and use it to approximte (1.95, 1.08). 8. Find the differential of the function

Answer 1

Equation of tangent plane at (- 1, 2, 4) to the surface z = 4x^2 - y^2 + 2y is (2 - z_0) = f_x (x_0, y_0) (x - x_0) + f_y (x_0, y_0) (y - y_0) Here (x, y_0, z_0 = (- 1, 2, 4) f_x = 8x Therefore f_x (x_0 y_0 = f_x (- 1, 2) = -8 f_y = -2y + 2 Therefore f_y (x_0, y_0) = f_y 9 - 1, 2) = -4 + 2 = -2 Therefore Eq^n of the tangent plane is (z - 4) = -8 (x + 1) - 2 (y - 2) = > 2 - 4 = -8x - 8 - 2y + 4 = > z = -8x - 2y = > 8x + 2y + z = 0 Linear approximation of f (x, y) = squareroot 20 - x^2 - y^2 at (2, 1) is given by L (x, y) = f (2, 1) + f_x (2, 1) (x, 2) + f_y (2, 1) (x, 1) Here f_x = 1/2 (20 x^2 - 7y^2)^- n1/2 (- 2x) = -x (20 - x^2 - 7 y^2)^- 1/2 \r\n Therefore f_x (2, 1) = -2 (20 - 2^2 - 7)^- 1/2 = -2/squareroot 9 = -2/3 f_y = -14 y