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Nama : Siti Aisyah NIM
: 06081281823025 Kalkulus Peubah Banyak (Soal-soal 12.2)
Dalam soal-soal 1-16, carilah semua turunan parsial pertama masing-masing fungsi! 1. π(π₯, π¦) = (2π₯ β π¦)4 ππ§ = 4 (2π₯ β π¦)4β1 (2) = 8 (2π₯ β π¦)3 ππ₯ ππ§ = 4 (2π₯ β π¦)4β1 (β1) = β4 (2π₯ β π¦)3 ππ¦ 3.
π(π₯, π¦) =
π₯ 2 βπ¦ 2 π₯π¦
ππ§ π’β² π£ β π’π£β² 2π₯(π₯π¦) β (π₯ 2 β π¦ 2 )π¦ 2π₯ 2 π¦ β π₯ 2 π¦ + π¦ 3 π₯ 2 + π¦ 2 = = = = ππ₯ π£2 π₯2π¦2 π₯2π¦2 π₯2π¦ ππ§ π’β² π£ β π’π£β² β2π¦(π₯π¦) β (π₯ 2 β π¦ 2 )π₯ β2π₯π¦ 2 β π₯ 3 + π₯π¦ 2 π₯2 + π¦2 = = = = β ππ¦ π£2 π₯2π¦2 π₯2π¦2 π₯π¦ 2 5.
π(π₯, π¦) = π π¦ sin π₯ ππ₯ (π₯, π¦) = π π¦ cos π₯ ππ¦ (π₯, π¦) = π π¦ sin π₯
7.
π(π₯, π¦) = βπ₯ 2 β π¦ 2 1 1 1 1 ππ§ 1 1 = (π₯ 2 β π¦ 2 )2 = (π₯ 2 β π¦ 2 )2β1(2π₯) = (2π₯)(π₯ 2 β π¦ 2 )β2 = (π₯)(π₯ 2 β π¦ 2 )β2 ππ₯ 2 2 1 1 1 ππ§ 1 2 1 = (π₯ β π¦ 2 )2β1 (β2π¦) = (β2π¦)(π₯ 2 β π¦ 2 )β2 = (βπ¦)(π₯ 2 β π¦ 2 )β2 ππ¦ 2 2
9.
π(π₯, π¦) = π βπ₯π¦ ππ₯ (π₯, π¦) = βπ¦π βπ₯π¦ ππ¦ (π₯, π¦) = βπ₯π βπ₯π¦
11. π(π₯, π¦) = tanβ1(4π₯ β 7π¦) π(π₯, π¦) = π§ =
1 = cot (4π₯ β 7π¦) tan(4π₯ β 7π¦)
ππ₯ (π₯, π¦) = ππ π 2 (4π₯ β 7π¦)4 ππ₯ (π₯, π¦) = β4 πππ ππ 2 (4π₯ β 7π¦)
13. π(π₯, π¦) = π¦ cos (π₯ 2 + π¦ 2 ) ππ§ = π¦{β sin(π₯ 2 + π¦ 2 )}2π₯ = β2π₯π¦ sin(π₯ 2 + π¦ 2 ) ππ₯ ππ§ = {π¦[β sin(π₯ 2 + π¦ 2 )]2π¦} + {cos(π₯ 2 + π¦ 2 )} = β2π¦ 2 sin(π₯ 2 + π¦ 2 ) + cos(π₯ 2 + π¦ 2 ) ππ¦ 15. πΉ(π₯, π¦) = 2 sin π₯ cos π¦ ππ§ = 2 cos π¦ cos π₯ = 2 cos π₯ cos π¦ ππ₯ ππ§ = 2 sin π₯ ( β sin π¦) = β2 sin π₯ sin π¦ ππ¦ Dalam soal-soal 17-20, periksa kebenaran bahwa π 2π π 2π = ππ¦ ππ₯ ππ₯ ππ¦ 17. π(π₯, π¦) = 2π₯ 2 π¦ 3 β π₯ 3 π¦ 5 Bukti : ππ₯ = 4π₯π¦ 3 β 3π₯ 2 π¦ 5 ππ¦ = 6π₯ 2 π¦ 2 β 5π₯ 3 π¦ 4 ππ₯π¦ = 12π₯π¦ 3 β 15π₯ 2 π¦ 5 ππ¦π₯ = 12π₯π¦ 3 β 15π₯ 2 π¦ 5 (TERBUKTI) 19. π(π₯, π¦) = 3π 2π₯ cos π¦ ππ₯ (π₯, π¦) = 6π 2π₯ cos π¦ ππ₯π¦ (π₯, π¦) = 6π 2π₯ (β sin π¦) = β6 π 2π₯ sin π¦ ππ¦ (π₯, π¦) = 3π 2π₯ (β sin π¦) = β3π 2 sin π¦ ππ¦π₯ (π₯, π¦) = β6π 2π₯ sin π¦ 21. π(π₯, π¦) =
2π₯βπ¦
πβ(π₯, π¦) =
π₯π¦
, carilah πβ(3, β2)dan ππ¦ (3, β2)
(π₯π¦)(2) β (2π₯ β π¦)(π¦) (π₯π¦)Β²
πβ(π₯, π¦) =
π¦Β² π₯Β²π¦Β²
1 π₯Β² 1 πβ(3, β2) = 3Β² (π₯, π¦)(β1) β (2π₯ β π¦)(π₯) ππ¦ (π₯, π¦) = (π₯π¦)Β² πβ(π₯, π¦) =
ππ¦ (π₯, π¦) =
β2π₯Β² π₯Β²π¦Β²
ππ¦ (π₯, π¦) =
2 π¦Β²
ππ¦ (3, β2) =
2 1 =β 2 β2 2
23. Jikaπ(π₯, π¦) = tanβ1(π¦ 2 /π₯) , carilah ππ₯ (β5, β2) dan ππ¦ = (β5, β2). 25. Carilahkemiringangarissinggungpadakurvaperpotonganpermukaan36π§ = 4π₯ 2 + 9π¦ 2 dengan bidang π₯ = 3 di titik (3,2,2). π§ = π(π₯, π¦) = ππ¦ (π₯, π¦) =
π₯Β² π¦Β² + 9 4
π¦ 2
ππ¦ (3,2) = 1 27. Carilah kemiringan garis singgug pada kurva perpotongan permukaan 2π§ = 3
β9π₯ 2 + 9π¦ 2 β 36 dengan bidang π¦ = 1 di titik (2, 1, 2).
2π§ = β9π₯ 2 + 9π¦ 2 β 36 1 1 π§ = π(π₯, π¦) = ( ) (9π₯ 2 + 9π¦ 2 β 36)2 2 1
1
1
= 2 Γ 2 (9π₯ 2 + 9π¦ 2 β 36)2β1 (18π₯) 1
1
= 4 (9π₯ 2 + 9π¦ 2 β 36)β2 (18π₯) 9
1
= 2 π₯(9π₯ 2 + 9π¦ 2 β 36)β2
=
9π₯ 1
2(9π₯ 2 +9π¦ 2 β36)2 3
Maka kemiringan ππ₯ (2, 1) pada titik (2, 1, 2) 9(2)
ππ₯ (2, 1) =
1
2(9(2)2 + 9(1)2 β 36)2 18
=
1
2(36+9β36)2
=
18 6
=3 29. Volume V suatu tabung lingkaran tegak diberikan oleh π = ππ 2 β, di mana r jari-jari dan h tinggi. Jika h dipertahankan tetap pada h = 10 inci, cari laju perubahan V terhadap r ketika r = 6 inci.
Dik :
π = 6 ππππ β = 10 ππππ
π(π, β) = ππ 2 β Laju perubahan V terhadap π = ππ(π, β) ππ(π, β) = 2ππβ ππ(6,10) = 2π. 6.10 = 120π ππππ 2 31. Menurut hukum gas ideal, tekanan, suhu, dan volume gas dihubungkan oleh ππ = ππ, dengan πsuatu konstanta. Carilah laju perubahan tekanan (pon per inci kuadrat) terhadap suhu pada waktu suhu adalah 300Β°πΎ jika volume dipertahankan tetap pada 100 inci kubik.
ππ = ππ Laju perubahan tekanan (P) terhadap suhu (T) dengan suhu (T) = 300Β°πΎ dan π = 100 ππππ 3 π=
ππ π
ππ (π, π) =
π π
ππ (300,100) =
π πππ/ππππ 2 100
Suatu fungsi dua variabel yang memenuhi persamaan Laplace. π 2π π 2π + =0 ππ₯ 2 ππ¦ 2 disebut harmonik. Perlihatkan bahwa fungsi-fungsi yang didefinisikan pada soal 33 dan 34 adalah fungsi harmonik. 33. π(π₯, π¦) = π₯ 3 π¦ β π₯π¦ 3 Dikatakan fungsi harmonik jika suatu fungsi dua variable memenuhi persamaan Laplace, π 2π π 2π + =0 ππ₯ 2 ππ¦ 2 π(π₯, π¦) = π₯ 3 π¦ β π₯π¦ 3 ππ = 3π₯ 2 π¦ β π¦ 3 ππ₯ π 2π = 6π₯π¦ ππ₯ 2 ππ = π₯ 3 β 3π₯π¦ 2 ππ¦ π 2π = β6π₯π¦ ππ¦ 2 π 2π π 2π + =0 ππ₯ 2 ππ¦ 2 6π₯π¦ + (β6π₯π¦) = 0 Karena memenuhi persamaan Laplace, maka fungsi π(π₯, π¦) = π₯ 3 π¦ β π₯π¦ 3 dikatakan sebagai fungsi harmonic. 35. Jika πΉ(π₯, π¦) = 3π₯ 4 π¦ 5 β 2π₯ 2 π¦ 3 , carilah π 3 π(π₯, π¦)/ππ¦ 3 . πΉ(π₯, π¦) = 3π₯ 4 π¦ 5 β 2π₯ 2 π¦ 3 ππ = 15π₯ 4 π¦ 4 β 6π₯ 2 π¦ 2 ππ¦ π 2π = 60π₯ 4 π¦ 3 β 12π₯ 2 π¦ ππ¦ 2 π 3π = 180π₯ 4 π¦ 2 β 12π₯ 2 ππ¦ 3 37. Nyatakan yang berikut dalam cara penulisan β
π3 π
a. ππ¦π¦π¦ = ππ¦ 3
π3 π
b. ππ₯π₯π¦ = ππ¦ππ₯2 π4 π
c. ππ₯π¦π¦π¦ = ππ¦ 3 ππ₯ 39. 41. π(π₯, π¦, π§) = π βπ₯π¦π§ β ln(π₯π¦ β π§ 2 ) ππ₯ (π₯, π¦, π§) = (βπ¦π§)(π βπ₯π¦π§ ) β
(π¦) (π₯π¦ β π§ 2 )
ππ₯ (π₯, π¦, π§) = βπ¦π§π βπ₯π¦π§ β π¦(π₯π¦ β π§ 2 )β1 ππ₯ (π₯, π¦, π§) = β(π¦π§π βπ₯π¦π§ + π¦(π₯π¦ β π§ 2 )β1 )
43. 45. 47. a. Bergerak sejajar dengan sumbu y dari titik (1, 1) ke kurva level terdekat dan βπ§ 4β5 mendekati βπ¦ , diperoleh fy(1,1) = 1,25β1 = β4
b. Bergerak sejajar dengan sumbu x dari titik (-4, 2) ke kurva level terdekat dan βπ§ 1β0 2 mendekati βπ₯ , diperoleh fx(-4,2) β β2,5β(β4) = 3 c. Bergerak sejajar dengan sumbu x dari titik (-5, -2) ke kurva level terdekat dan βπ§ 1β0 2 mendekati βπ₯ , diperoleh fx(-4,-5) β β2,5β(β5) = 5 d. Bergerak sejajar dengan sumbu x dari titik (0, -2) ke kurva level terdekat dan βπ§ 1β0 8 mendekati βπ₯ , diperoleh fx(0,-2) β β19 =3 8
49. a. fy(x,y,z) = lim
π(π₯,π¦+βπ¦,π§)βπ(π₯,π¦,π§) βπ¦
βπ¦β0
b. fz(x,y,z) = lim
β(β2)
π(π₯,π¦,π§+βπ§)βπ(π₯,π¦,π§) βπ§ πΊ(π€,π₯+βπ₯,π¦,π§)βπΊ(π€,π₯,π¦,π§)
βπ§β0
c. Gy(w,x,y,z) = lim
βπ₯β0
π
d. ππ§ Ξ»(π₯, π¦ , π§ , π‘ ) = lim
βπ₯ Ξ»(π₯,π¦,π§+βπ§,π‘)βΞ»(π₯,π¦,π§,π‘) βπ§
βπ¦β0
π(ππ, π1, π2, +βπ2 ,β¦,ππ )βπ(ππ, π1, π2,β¦,ππ )
π
e. ππ π(ππ, π1, π2, β¦ , ππ ) = lim ( 2
βπ2 β0
βπ2
)