The function f(x,y) only has one critical point. Find the critical point of f(x,y) and use the Second Derivative Test to determine whether it corresponds to a local maximum, local minimum, saddle point or if the Second Derivative Test is inconclusive. f (x, y) = 2+4x4+6y4 C There is a local minimum at (0,0). There is a local maximum at (0,0). C There is a saddle point at (0,0). O The second derivative test is inconclusive at (0,0).The function f(x,y) has an absolute maximum and absolute minimum subject to the given constraint. Use Lagrange multipliers to find these values. f ( x, y) = 6x2 + y2 x2 + 5y + y2 = 24 O Maximum is 21. Minimum is 9. Maximum is 189. Minimum is 0. Maximum is 49. Minimum is -21. O Maximum is 21. Minimum is 0. O Maximum is 189. Minimum is 9.Compute the gradient of f(x,y) and evaluate it at the point P. f ( x, y) = x2 - 8x2y - 6xy2 P(-2,3) O(-10,-40) O (38,40) O(-38,40) O (38,-40) O(-10,40)Find the unit tangent vector T and the principal unit normal vector N of the parameterized curve r(t) = (10 sint , 10 cost , 24t) O T (t ) = (- 5 5 12 13 cost , sint , 13 13 N(t) = (sint, - cost, 0) O 12 12 T(t) = 5 13 - cost , - - sint 13 13 N(t) = (- sint, - cost , 0) O T (t ) = (- 5 12 `13 - cost, - 13 Sint 13 N(t) = (- sint, - cost, 0) OT (t ) = (cost, sint, 5 13 5 5 N(t) = (- sint , cost , 0) 13 13 O T(t ) = 12 12 5 cost , - sin t 13 13 13'\fConsider the following trajectory of a moving object. Find the tangential and normal components of the acceleration. r(t) = (+2 -6,2t - 2,9) Oa) 2t 2 a = 12 4 1 T + 12 4 1 N Ob) a= T 4 N V12 + 1 Od 2t 2 a T + N V12 4 1 V12 4 1 Od) a = T+7241 N 12 + 1