Unformatted text preview: , 1) = ∇ f (0 , 1) · u = − 2 / √ 5. 3. The gradient is a 6 x, 2 y, 4 z A , which equals a 6 , − 8 , 4 A at (1 , − 4 , 1). Thus the equation of the plane is 0 = a 6 , − 8 , 4 A · a x − 1 , y + 4 , z − 1 A = 6 x − 8 y + 4 z − 42 , which can be rewritten as 3 x − 4 y + 2 z = 21. Any of these forms, including the dot product form, is acceptable. 4. Since we are given the critical points, we only need to compute f xx , f xy , f yy and D = f xx f yy − ( f xy ) 2 there. quantity at ( x, y ) at (0 , 0) at (1 , √ 2) at (1 , − √ 2) f xx − 2 − 2 − 2 − 2 f xy − 2 y − 2 √ 2 2 √ 2 f yy 2 x − 2 − 2 D 4(1 − x − y 2 ) 4 − 8 − 8 By the second derivative test, (0 , 0) is a local maximum and the other two are saddle points....
View
Full Document
 Fall '06
 staff
 Math, Calculus, Harshad number, B SOLUTIONS, dot product form, compute fx ﬁrst, 19 17 ws

Click to edit the document details