in 14\"C Partly cloudy Problem 3. In our lectures, we've learned another crucial property of the gradient vector: it is perpendicular to the level curves or level surfaces of a function. This means that if we have a smooth curve described by the equation f (:r, y) :0 in the xyplane, then the gradient vector V f at every point on this curve is perpendicular to the curve. Similarly, for a smooth surface given by f(:r, y, z):0, the gradient Vf at each point on the surface is perpendicular to the surface itself. a) Utilizing this property of the gradient vector, write the equation of the tangent line to the curve g(:r, y) 2: $2 + my + 2y2 4 : 0 at the point (1, 1). (Recall that the slope of the tangent line is the negative reciprocal of the normal line). b) Applying the same property of the gradient vector, determine the equation of the tangent plane to the surface described by an, y, 2) 2:2 (a: yz) 63329'2 :0 at the point (71,, e_3/4). Additionally, discuss the nature or type of this point for the function f. 10:16 PM 2023710706