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Q2-946

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𝑧(π‘₯, 𝑦) =
𝑍(π‘₯, 1) =
π‘₯𝑠𝑖𝑛(π‘₯ − 1)
2 + π‘₯2
=≫ 𝐹 ′ π‘₯ (1,1,0) =
𝑍(1, 𝑦) =
π‘₯𝑦𝑠𝑖𝑛(π‘₯ − 𝑦)
, (1,1,0)
1 + π‘₯ 2 +𝑦 2
πœ•π‘§ ((sin⁑(π‘₯ − 1) + π‘₯π‘π‘œπ‘ (π‘₯ − 1)(2 + π‘₯ 2 ) − 2π‘₯ 2 (sin⁑(π‘₯ − 1) 1
=
=
(2 + π‘₯ 2 )2
πœ•π‘₯
3
𝑦𝑠𝑖𝑛(1 − 𝑦)
2 + 𝑦2
=≫ 𝐹 ′ 𝑦 (1,1,0) =
𝐹 ′ 𝑧 (1,1,0) =
πœ•π‘§ ((sin⁑(1 − 𝑦) − π‘¦π‘π‘œπ‘ (1 − 𝑦)(2 + 𝑦 2 ) − 2𝑦 2 (sin⁑(1 − 𝑦)
1
=
=−
2
2
(2 + 𝑦 )
πœ•π‘¦
3
πœ•π‘§
=1
πœ•π‘₯
We have the tangent plane:
𝐹 ′ 𝑧 (π‘₯, 𝑦, 𝑧)(𝑧 − 𝑧0 ) = 𝐹 ′ π‘₯ (π‘₯, 𝑦, 𝑧)(π‘₯ − π‘₯0 ) + 𝐹 ′ 𝑦 (π‘₯, 𝑦, 𝑧)(𝑦 − 𝑦0 )
=≫ 1(𝑧 − 0) =
1
1
(π‘₯ − 1) − (𝑦 − 𝑦0 )
3
3
1
1
=≫ 𝑧 = π‘₯ − 𝑦
3
3
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