MAT2615 ASSIGNMENT 3 2022

MAT2615
ASSIGNMENT 3
2022

,Solution:


𝑦 1
𝑓(𝑥, 𝑦) = − ln(𝑥) + 𝑦 2
𝑥 2


𝑦 1
𝑓𝑥 (𝑥, 𝑦) = − −
𝑥2 𝑥
2𝑦 1
𝑓𝑥𝑥 (𝑥, 𝑦) = +
𝑥3 𝑥2


1
𝑓𝑥𝑦 (𝑥, 𝑦) = −
𝑥2


1
𝑓𝑦 (𝑥, 𝑦) = +𝑦
𝑥
𝑓𝑦𝑦 (𝑥, 𝑦) = 1



𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑝𝑜𝑖𝑛𝑡𝑠:
𝑓𝑥 (𝑥, 𝑦) = 0 𝑎𝑛𝑑 𝑓𝑦 (𝑥, 𝑦) = 0



∴ 𝑓𝑥 (𝑥, 𝑦) = 0
𝑦 1
− − =0
𝑥2 𝑥
−𝑦 − 𝑥 = 0

𝑥+𝑦 =0 1



∴ 𝑓𝑦 (𝑥, 𝑦) = 0
1
+𝑦 =0
𝑥

, 1 + 𝑥𝑦 = 0 2



𝐹𝑟𝑜𝑚 1 𝑥 = −𝑦 𝑎𝑛𝑑 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑖𝑛𝑡𝑜 2



1 + 𝑥𝑦 = 0 2

1 + (−𝑦)𝑦 = 0

1 − 𝑦2 = 0

𝑦2 = 1

𝑦 = ±√1

𝑦 = 1 𝑜𝑟 𝑦 = −1



𝐹𝑜𝑟 𝑦 = 1:
𝑥 = −𝑦
𝑥 = −1



𝐹𝑜𝑟 𝑦 = −1:
𝑥 = −𝑦
𝑥 = −(−1)
𝑥=1



𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑝𝑜𝑖𝑛𝑡𝑠: (−1,1) 𝑎𝑛𝑑 (1, −1)


2
𝐷(𝑎, 𝑏) = 𝑓𝑥𝑥 (𝑎, 𝑏) ∙ 𝑓𝑦𝑦 (𝑎, 𝑏) − [𝑓𝑥𝑦 (𝑎, 𝑏)]



2𝑦 1
𝑓𝑥𝑥 (𝑥, 𝑦) = +
𝑥3 𝑥2
1
𝑓𝑥𝑦 (𝑥, 𝑦) = −
𝑥2
𝑓𝑦𝑦 (𝑥, 𝑦) = 1