8. Find the tangent plane and the quadratic approximation to f(x, y) = (x + 2y)^5 at the point x = 1 and y = 0. Solution Given f(x, y) = (x + 2y)5 then f(1, 0) = 1 fx(x, y) = 5(x + 2y)4 then fx(1, 0) = 5 fy(x, y) = 5(x + 2y)4 * 2 = 10(x + 2y)^4 then fy(1, 0) = 10 fxx(x, y) = 20(x + 2y)3 then fxx(1, 0) = 20 fxy(x, y) = 20(x + 2y)3 * 2 then fxy(1, 0) = 40 fyy(x, y) = 10 * 4(x + 2y)3 * 2 then fyy(1, 0) = 80. So, the quadratic approximation at (1, 0) is 1 + [5(x - 1) + 10(y - 0)] + (1/2!) [20(x - 1)2 + 2 * 40(x - 1)(y - 0) + 80(y - 0)2] = 1 + 5(x - 1) + 10(y - 0) + 10(x - 1)2 + 40(x - 1)(y - 0) + 40(y - 0)2 We obtain the tangent plane (linear) approximation by ignoring the second degree terms: 1 + 5(x - 1) + 10(y - 0)..