MAT2615 ASSIGNMENT 2 2025

MAT2615
ASSIGNMENT 2
2025

, QUESTION 1


a).


f(x, y, z) = x yz


∂f ∂f ∂f
∇f(x, y, z) = ( , , )
∂x ∂y ∂z
∂ ∂ ∂
∇f(x, y, z) = ( [x yz ], [x yz ], [x yz ])
∂x ∂y ∂z
∇f(x, y, z) = (yzx yz−1 , zx yz ln(x) , yx yz ln(x))

∇f(e, 1,1) = ((1)(1)e(1)(1)−1 , (1)e(1)(1) ln(e) , (1)e(1)(1) ln(e))

∇f(e, 1,1) = (1, e, e)


Let ∶ u = (3, −2,1)



|u| = √(3)2 + (−2)2 + (1)2

|u| = √14


u
Unit vector , û =
|u|
(3, −2,1)
û =
√14
3 2 1
û = ( ,− , )
√14 √14 √14


3 2 1
Du ∇f(e, 1,1) = (1, e, e) ∙ ( ,− , )
√14 √14 √14
3 2 1
Du ∇f(e, 1,1) = − e+ e
√14 √14 √14
3 1
Du ∇f(e, 1,1) = − e
√14 √14

, 3−e
Du ∇f(e, 1,1) =
√14




b).


Maximum rate of increase = ‖∇f(e, 1,1)‖
= ‖(1, e, e)‖

= √(1)2 + (e)2 + (e)2

= √1 + 2e2




c).


Let ∶ d = (a, b, c)


Dd ∇f(e, 1,1) = 0
(1, e, e) ∙ (a, b, c) = 0
a + be + ce = 0


Let b = 1 and c = 0 , then


a + be + ce = 0
a + (1)(e) + (0)(e) = 0
a+e=0
a = −e


d = (a, b, c)
d = (−e, 1,0)
Direction = (−e, 1,0)