The equation of the tangent plane to the surface z 3x2y 2x

The equation of the tangent plane to the surface z = 3x^2y - 2xy^4 at the point (x, y, z) = (-1, 1, 5) is z = x+ y+. The unit normal line at this same point oriented to make an acute angle with the positive z axis is i+ j+ k

Solution

The equation of the tangent plane to the surface z = f(x,y) at the point (x₀ , y₀ , z₀ ) is z - z₀ = f \'_x ( x₀ , y₀ )( x - x₀) + f \' _y ( x₀ , y₀ )( y - y₀). Now, At the point ( -1 , 1 , 5), we have f \' _x = 6xy - 2y⁴ = - 6 - 2 = - 8 and f \' _y = 3x² - 8xy³ = 3 + 8 = 11. Thus, the equation of the required tangent plane is z - 5 = -8(x +1) + 11(y -1) = - 8x+11y -19 or, 8x - 11y + z +14 = 0.

Any non-zero vector that is orthogonal to the direction vectors of a plane is normal vector to the plane. Therefore, the coefficient vector ( 8, -11 ,1) is normal to the plane 8x - 11y + z +14 = 0 at the point ( -1, 1, 5). The magnitude of tis vector is { ( 8 )² + (-11)² + (1)²} = 186. Thus a unit normal vector to the above plane at the point (-1,1,5) is 8/ 186 i -11/186j +1/186k. Further, if is the angle between this vector and the unit positive z-axis vector, then cos = 1/ 186 = 1/13.6381817 = 0.073323557 so that = cos^-1 0.073323557 = 85.79509605 or, 85.8^o which is an acute angle. Thus, the required vector is   8/ 186 i -11/186j +1/186k.