Soal dan Pembahasan Analisis Vektor (Diferensiasi, Gradien, Divergensi dan Curl) PDF

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SOAL DAN PEMBAHASAN ANALISIS VEKTOR BAB GRADIEN, DIVERGENSI DAN CURL 1. Jika 𝜙 = 2𝑥𝑧 4 − 𝑥 2 𝑦 carilaah ∇𝜙 dan |∇𝜙| pada titik (2, −2,1) Jawab: 𝜕 𝜕 𝜕 𝜕(2𝑥𝑧 4−𝑥 2 𝑦) 𝜕(2𝑥𝑧 4 −𝑥 2 𝑦) 𝜕(2𝑥𝑧 4−𝑥 2 𝑦) ∇𝜙 = (𝜕𝑥 𝑖 + 𝜕𝑦 𝑗 + 𝜕𝑧 𝑘) (2𝑥𝑧 4 − 𝑥 2 𝑦) = 𝑖+ 𝑗+ 𝑘 𝜕𝑥 𝜕𝑦 𝜕𝑧 4 2 3 = (2𝑧 − 2𝑥𝑦)𝑖 − 𝑥 𝑗 + 8𝑥𝑧 𝑘 Pada titi...

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SOAL DAN PEMBAHASAN ANALISIS VEKTOR BAB GRADIEN, DIVERGENSI DAN CURL 1.

Jika 𝜙 = 2𝑥𝑧 4 − 𝑥 2 𝑦 carilaah ∇𝜙 dan |∇𝜙| pada titik (2, −2,1) Jawab: ∇𝜙

2.

𝜕

𝜕𝑥

𝑖+

4

𝜕

𝜕𝑦

𝑗+

𝜕

𝜕𝑧

𝑘) (2𝑥𝑧 4 − 𝑥 2 𝑦) = 2

3

𝜕(2𝑥𝑧 4−𝑥 2 𝑦) 𝜕𝑥

𝑖+

𝜕(2𝑥𝑧 4 −𝑥 2 𝑦) 𝜕𝑦

𝑗+

𝜕(2𝑥𝑧 4−𝑥 2 𝑦) 𝜕𝑧

𝑘

= (2𝑧 − 2𝑥𝑦)𝑖 − 𝑥 𝑗 + 8𝑥𝑧 𝑘 Pada titik (2, −2,1) maka: ∇𝜙 = (2𝑧 4 − 2𝑥𝑦)𝑖 − 𝑥 2 𝑗 + 8𝑥𝑧 3 𝑘 = (2.14 − 2.2. −2)𝑖 − 22 𝑗 + (8.2.13 )𝑘 = 10𝑖 − 4𝑗 + 16𝑘 Sehingga: |∇𝜙| = √102 + (−4)2 + 162 = √100 + 16 + 256 = √372 = 2√93

Jika 𝐴 = 2𝑥 2 𝑖 − 3𝑦𝑧𝑗 + 𝑥𝑧 2 𝑘 dan 𝜙 = 2𝑧 − 𝑥 3 𝑦. Carilah 𝐴 ∙ ∇𝜙 dan 𝐴 × ∇𝜙 pada titik (1, −1,1) Jawab: ∇𝜙

Maka: 𝐴 ∙ ∇𝜙

3.

=(

=(

𝜕

𝜕𝑥

𝑖+

𝜕

𝜕𝑦

𝑗+

𝜕

𝜕𝑧

𝑘) (2𝑧 − 𝑥 3 𝑦) =

= −3𝑥 2 𝑦𝑖 − 𝑥 3 𝑗 + 2𝑘

𝜕(2𝑧−𝑥 3𝑦) 𝜕𝑥

𝑖+

𝜕(2𝑧−𝑥 3𝑦)

𝑗+

𝜕𝑦

𝜕(2𝑧−𝑥 3𝑦) 𝜕𝑧

𝑘

= (2𝑥 2 𝑖 − 3𝑦𝑧𝑗 + 𝑥𝑧 2 𝑘) ∙ (−3𝑥 2 𝑦𝑖 − 𝑥 3 𝑗 + 2𝑘) = 2𝑥 2 (−3𝑥 2 𝑦) + (−3𝑦𝑧)(−𝑥 3 ) + (𝑥𝑧 2 )2 = −6𝑥 4 𝑦 + 3𝑥 3 𝑦𝑧 + 2𝑥𝑧 2 Diketahuai pada titik (1, −1,1) maka: 𝐴 ∙ ∇𝜙 = −6𝑥 4 𝑦 + 3𝑥 3 𝑦𝑧 + 2𝑥𝑧 2 = −6(14 )(−1) + 3(13 )(−1)(1) + 2(1)(12 ) = 5 Selanjutnya, 𝑖 𝑗 𝑘 2 −3𝑦𝑧 𝑥𝑧 2 | = 𝑖(−6𝑦𝑧 + 𝑥 4 𝑧 2 ) − 𝑗(4𝑥 2 + 3𝑥 3 𝑦𝑧 2 ) + 𝑘(−2𝑥 5 − 9𝑥 2 𝑦 2 𝑧) 𝐴 × ∇𝜙 = | 2𝑥 −3𝑥 2 𝑦 −𝑥 3 2 Diketahuai pada titik (1, −1,1) maka: 𝐴 × ∇𝜙 = 𝑖((−6(−1)(1) + (14 )(14 )) − 𝑗((4(12 ) + 3(13 )(−1)(12 )) + 𝑘((−2(15 ) − 9(12 )(−12 )(1)) = 𝑖 (6 + 1) − 𝑗(4 − 3) + 𝑘(−2 − 9) = 7𝑖 − 𝑗 − 11𝑘 𝑦

Jika 𝐹 = 𝑥 2 𝑧 + 𝑒 𝑥 dan 𝐺 = 2𝑧 2 𝑦 − 𝑥𝑦 2 . Carilah ∇(𝐹 + 𝐺) dan ∇(𝐹𝐺) pada titik (1,0, −2) Jawab: ∇𝐹 ∇𝐺

Maka:

=(

𝜕

𝜕𝑥

𝑖+

= (2𝑥𝑧 − =(

𝜕

𝑖+ 𝜕𝑥

𝜕

𝜕𝑦 𝑦

𝑥2 𝜕

𝜕𝑦

𝑗+ 𝑦

𝜕

𝜕𝑧

𝑦 𝑥

𝑘) (𝑥 2 𝑧 + 𝑒 ) = 𝑦

1

𝑒 𝑥 ) 𝑖 + ( 𝑒 𝑥 ) 𝑗 + (𝑥 2 )𝑘

𝑗+

𝜕

𝜕𝑧

𝑥

𝑘) (2𝑥 2 𝑦 − 𝑧𝑦 2 ) =

= −𝑦 2 𝑖 + 2𝑧 2 𝑗 − 2𝑥𝑦𝑗 + 4𝑧𝑦𝑘

∇(𝐹 + 𝐺) = ∇𝐹 + ∇𝐺 = (2𝑥𝑧 − 𝑦

𝑦

𝜕(𝑥 2𝑧+𝑒 𝑥 )

𝑦 𝑥

2]

𝑦

𝑥

𝑦

1

𝜕𝑥

𝑖+

𝜕(2𝑧 2 𝑦−𝑥𝑦 2) 𝜕𝑥

𝑦

𝜕(𝑥 2𝑧+𝑒 𝑥 ) 𝜕𝑦

𝑖+

𝑗+

𝜕(2𝑧 2𝑦−𝑥𝑦 2 ) 𝜕𝑦

𝑦

𝜕(𝑥 2𝑧+𝑒 𝑥 ) 𝜕𝑧

𝑗+

𝑘

𝜕(2𝑧 2𝑦−𝑥𝑦 2 ) 𝜕𝑧

𝑘

𝑦

𝑥 𝑥 ( 2 )𝑘 − 𝑦 2 𝑖 + 2𝑧 2 𝑗 − 2𝑥𝑦𝑗 + 4𝑧𝑦𝑘 2 𝑒 )𝑖 + ( 𝑒 )𝑗 + 𝑥

1

𝑦 𝑥

𝑥

= [2𝑥𝑧 − 2 𝑒 − 𝑦 𝑖 + [𝑥 𝑒 + (2𝑧 2 − 2𝑥𝑦)] 𝑗 + (𝑥 2 + 4𝑧𝑦)𝑘 𝑥 Diketahui pada titik (1,0, −2) maka: ∇(𝐹 + 𝐺) = (−4 − 0 − 0)𝑖 + (1 + 8 − 0)𝑗 + (1 + 0)𝑘 = −4𝑖 + 9𝑗 + 𝑘 Selanjutnya: ∇(𝐹𝐺) = 𝐹∇𝐺 + 𝐺∇𝐹 𝑦 𝑦 𝑦 1 𝑦 = (𝑥 2 𝑧 + 𝑒 𝑥 ) (−𝑦2 𝑖 + 2𝑧2 𝑗 − 2𝑥𝑦𝑗 + 4𝑧𝑦𝑘) + (2𝑧 2 𝑦 − 𝑥𝑦 2 )((2𝑥𝑧 − 2 𝑒 𝑥 ) 𝑖 + ( 𝑒 𝑥 ) 𝑗 + (𝑥 2 )𝑘) 𝑥 𝑥 Diketahui pada titik (1,0, −2) maka:

Nafida Hetty Marhaeni

∇(𝐹𝐺) 4.

= (−2 + 1)(0𝑖 + 8𝑗 − 0𝑗 + 0𝑘) + (0)(−4𝑖 − 0𝑖 + 𝑗 + 𝑘) = −1(8𝑗) = −8𝑗

Carilah ∇|𝑟|3 Jawab: ∇|𝑟 |3

=(

=

𝜕

𝜕𝑥

𝑖+

𝜕

𝜕𝑦

𝑗+

3 𝜕(𝑥 2+𝑦 2 +𝑧 2 )2

𝜕𝑥

3

𝜕

3

𝑘) (√𝑥 2 + 𝑦 2 + 𝑧 2 ) = ( 𝜕𝑧 3 𝜕(𝑥 2+𝑦 2 +𝑧 2 )2

𝑖+

𝜕𝑦 1 2

3

𝜕

𝜕𝑥

3 𝜕(𝑥 2+𝑦 2 +𝑧 2 )2

𝑗+

𝜕𝑧

𝑖+ 𝑘

𝜕

𝜕𝑦

𝑗+

1

𝜕

𝜕𝑧

3

𝑘) (𝑥 2 + 𝑦 2 + 𝑧 2 )2

3

1

= (2𝑥 )(𝑥 2 + 𝑦 2 + 𝑧 2 ) 𝑖 + (2𝑦)(𝑥 2 + 𝑦 2 + 𝑧 2 )2 𝑗 + (2𝑧)(𝑥 2 + 𝑦 2 + 𝑧 2 )2 𝑘 2

2

2

= 3𝑥(𝑥 + 𝑦 + 𝑧

5.

1

2 )2 1

2

2

2

𝑖 + 3𝑦(𝑥 + 𝑦 + 𝑧

= 3(𝑥 2 + 𝑦 2 + 𝑧 2 )2 (𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘) = 3(𝑥 + 𝑦 + 𝑧)(𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘) = 3𝑟𝑟⃗

1

2 )2

2

2

1

𝑗 + 3𝑧(𝑥 + 𝑦 2 + 𝑧 2 )2 𝑘

6

Hitunglah ∇ (3𝑟 2 − 4√𝑟 + 3 ) √𝑟

Jawab: 6 (3𝑟 2 − 4√𝑟 + 3 ) √𝑟

1 2

)2

− 4(𝑥 + 𝑦 + 𝑧 ) + 6 (𝑥 + 𝑦 + 𝑧 )

=(

𝜕𝑥

= 3(𝑥 + 𝑦 + 𝑧 1 2

6

= 3(𝑥 + 𝑦 + 𝑧)2 − 4√𝑥 + 𝑦 + 𝑧 + 3 𝑥+𝑦+𝑧 1 − 3

1 − 3

1 2

√ 1 − 3

1

1

= (3𝑥 2 − 4𝑥 + 6𝑥 ) + (3𝑦 2 − 4𝑦 + 6𝑦 ) + (3𝑧 2 − 4𝑧 2 + 6𝑧 −3 ) 6

∇ (3𝑟 2 − 4√𝑟 + 3 ) √𝑟

1

𝜕

𝑖+

𝜕

𝜕𝑦

4

𝑗+

𝜕

𝜕𝑧

1 1 1 −1 −1 −1 𝑘) [(3𝑥2 − 4𝑥2 + 6𝑥 3 ) + (3𝑦2 − 4𝑦2 + 6𝑦 3 ) + (3𝑧2 − 4𝑧2 + 6𝑧 3 )]

1

4

1

4

= (6𝑥 − 2𝑥 −2 − 2𝑥 −3 ) 𝑖 + (6𝑦 − 2𝑦 −2 − 2𝑦 −3 ) 𝑗 + (6𝑧 − 2𝑧 −2 − 2𝑧 −3 ) 𝑘 1

1

1

4

4

4

= 6𝑥𝑖 + 6𝑦𝑗 + 6𝑧𝑘 − 2𝑥 −2 𝑖 − 2𝑦 −2 𝑗 − 2𝑧 −2 𝑘 − 2𝑥 −3 𝑖 − 2𝑦 −3 𝑗 − 2𝑧 −3 𝑘 3

3

3

7

7

7

= (𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘) (6 − 2𝑥 −2 − 2𝑦 −2 − 2𝑧 −2 − 2𝑥 −3 𝑖 − 2𝑦 −3 𝑗 − 2𝑧 −3 𝑘) 7

3

6.

= 𝑟⃗ (6 − 2𝑟 −2 − 2𝑟 −3 )

Carilah ∇𝜓 dimana 𝜓 = (𝑥 2 + 𝑦 2 + 𝑧 2 )𝑒 −√𝑥 2 2 2 Jawab: Note: 𝑒 −√𝑥 +𝑦 +𝑧 = 𝑒 −|𝑟| ∇𝜓

=(

𝜕

𝜕𝑥

= (2𝑥𝑒 −√𝑥 +𝑦 +𝑧 2

2

2

𝑖+

𝜕

𝜕𝑦

− 𝑒 −√𝑥

𝑗+

2 +𝑦2 +𝑧 2

𝜕

𝜕𝑧

2 +𝑦 2 +𝑧 2

𝑘) ((𝑥 2 + 𝑦 2 + 𝑧 2 )𝑒 −√𝑥

(𝑥 2 + 𝑦 2 + 𝑧 2)) 𝑖 + (2𝑦𝑒 −√𝑥

2 +𝑦2 +𝑧 2

2 +𝑦 2+𝑧 2

− 𝑒 −√𝑥

2 +𝑦2 +𝑧 2

)

(𝑥 2 + 𝑦 2 + 𝑧 2 )) 𝑗 + (2𝑧𝑒 −√𝑥

2 +𝑦2 +𝑧 2

− 𝑒 −√𝑥

2 +𝑦2 +𝑧 2

(𝑥 2 + 𝑦 2 + 𝑧 2 )) 𝑘

= (2𝑥𝑒 −|𝑟| − 𝑒 −|𝑟| (𝑥 2 + 𝑦 2 + 𝑧 2 )) 𝑖 + (2𝑦𝑒 −|𝑟| − 𝑒 −|𝑟| (𝑥 2 + 𝑦 2 + 𝑧 2 )) 𝑗 + (2𝑧𝑒 −|𝑟| − 𝑒 −|𝑟| (𝑥 2 + 𝑦 2 + 𝑧 2 )) 𝑘

7.

= 𝑒 −|𝑟| (2𝑥 − (𝑥 2 + 𝑦 2 + 𝑧 2 ))𝑖 + 𝑒 −|𝑟| (2𝑦 − (𝑥 2 + 𝑦 2 + 𝑧 2 ))𝑗 + 𝑒 −|𝑟| (2𝑧 − (𝑥 2 + 𝑦 2 + 𝑧 2 ))𝑘 = 𝑒 −|𝑟| (2𝑥 − (𝑥 2 + 𝑦 2 + 𝑧 2 ))𝑖 + (2𝑦 − (𝑥 2 + 𝑦 2 + 𝑧 2 ))𝑗 + (2𝑧 − (𝑥 2 + 𝑦 2 + 𝑧 2 ))𝑘 = 𝑒 −|𝑟| [2𝑥𝑖 + 2𝑦𝑗 + 2𝑧𝑘 − (𝑥 2 + 𝑦 2 + 𝑧 2 )𝑖 − (𝑥 2 + 𝑦 2 + 𝑧 2 )𝑗 − (𝑥 2 + 𝑦 2 + 𝑧 2 )𝑘] =𝑒 −|𝑟| (2(𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘) − (𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘)(𝑥 + 𝑦 + 𝑧)) = 𝑒 −|𝑟| ((𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘))(2 − (𝑥 + 𝑦 + 𝑧)) = 𝑒 −|𝑟| 𝒓(2 − 𝑟)

Jika ∇𝜙 = 2𝑥𝑦𝑧 3 𝑖 + 𝑥 2 𝑧 3 𝑗 + 3𝑥 2 𝑦𝑧 2 𝑘, carilah 𝜙(𝑥, 𝑦, 𝑥) jika 𝜙(1, −2,2) = 4 Jawab: 𝜕 𝜕 𝜕 ∇𝜙 = ( 𝑖 + 𝜕𝑦 𝑗 + 𝜕𝑧 𝑘) 𝜙 = 2𝑥𝑦𝑧 3 𝑖 + 𝑥 2 𝑧 3 𝑗 + 3𝑥 2 𝑦𝑧 2 𝑘 𝜕𝑥

Nafida Hetty Marhaeni

Maka:

(2𝑥𝑦𝑧 3 )𝜕𝑥 + (𝑥 2 𝑧 3 )𝜕𝑦 + (3𝑥 2 𝑦𝑧 2 )𝜕𝑧 = 0 Akan dicari solusi persamaan tersebut. 𝜕𝜙(𝑥, 𝑦, 𝑥 ) = ∫(2𝑥𝑦𝑧 3 )𝜕𝑥 + 𝜔(𝑦, 𝑧) = 𝑥 2 𝑦𝑧 3 + 𝜔(𝑦, 𝑧) 𝜕𝑥 𝜕𝜙(𝑥,𝑦,𝑥) Samakan 𝜕𝑦 = 𝑥 2 𝑧 3 untuk memperoleh 𝜔(𝑦, 𝑧) 𝜕(𝑥 2 𝑦𝑧 3 + 𝜔(𝑦, 𝑧)) = 𝑥 2𝑧 3 𝜕𝑦 𝑥 2 𝑧 3 + 𝜔′ (𝑦, 𝑧) = 𝑥 2 𝑧 3 𝜔′ (𝑦, 𝑧) = 0 ′( Integralkan 𝜔 𝑦, 𝑧) untuk memperoleh 𝜔 (𝑦, 𝑧)

∫ 𝜔′ (𝑦, 𝑧) = ∫ 0 𝜕𝑦 = 𝐶1 + 𝜔(𝑧)

Diperoleh: 𝜙(𝑥, 𝑦, 𝑥 ) = 𝑥 2 𝑦𝑧 3 + 𝐶1 + 𝜔(𝑧) 𝜕𝜙(𝑥,𝑦,𝑥) Samakan 𝜕𝑧 = 3𝑥 2 𝑦𝑧 2 untuk memperoleh 𝜔(𝑧)

𝜕(𝑥 2 𝑦𝑧 3 + 𝐶1 + 𝜔 (𝑧)) = 3𝑥 2 𝑦𝑧 2 𝜕𝑧 3𝑥 2 𝑦𝑧 2 + 𝜔′ (𝑧) = 3𝑥 2 𝑦𝑧 2 𝜔 ′ (𝑧 ) = 0 Integralkan 𝜔′ (𝑧) untuk memperoleh 𝜔(𝑧) ∫ 𝜔′ (𝑧) = ∫ 0 𝜕𝑧 = 𝐶2

Diperoleh: 𝜙(𝑥, 𝑦, 𝑥 ) = 𝑥 2 𝑦𝑧 3 + 𝐶1 + 𝐶2 (jumlahan 𝐶1 + 𝐶2 = 𝐶) maka: 𝜙(𝑥, 𝑦, 𝑥 ) = 𝑥 2 𝑦𝑧 3 + 𝐶 Diberikan syarat awal 𝜙(1, −2,2) = 4 maka: 12 (−2)23 + 𝐶 = 4 −16 + 𝐶 = 4 𝐶 = 20 2 3 ( ) Jadi, 𝜙 𝑥, 𝑦, 𝑥 = 𝑥 𝑦𝑧 + 20 8.

Jika ∇𝜓 = (𝑦 2 − 2𝑥𝑦𝑧 3 )𝑖 + (3 + 2𝑥𝑦 − 𝑥 2 𝑧 3 )𝑗 + (6𝑧 3 − 3𝑥 2 𝑦𝑧 2 )𝑘. Carilah 𝜓 Jawab: 𝜕 𝜕 𝜕 ∇𝜓 = (𝜕𝑥 𝑖 + 𝑗 + 𝑘) 𝜓 = (𝑦 2 − 2𝑥𝑦𝑧 3 )𝑖 + (3 + 2𝑥𝑦 − 𝑥 2 𝑧 3 )𝑗 + (6𝑧 3 − 3𝑥 2 𝑦𝑧 2 )𝑘 𝜕𝑦

Maka:

𝜕𝑧

(𝑦 2 − 2𝑥𝑦𝑧 3 )𝜕𝑥 + (3 + 2𝑥𝑦 − 𝑥 2 𝑧 3 )𝜕𝑦 + (6𝑧 3 − 3𝑥 2 𝑦𝑧 2 )𝜕𝑧 = 0 Akan dicari solusi persamaan tersebut. 𝜕𝜓(𝑥, 𝑦, 𝑥 ) = ∫(𝑦 2 − 2𝑥𝑦𝑧 3 )𝜕𝑥 + 𝜔(𝑦, 𝑧) = 𝑥𝑦 2 − 𝑥 2 𝑦𝑧 3 + 𝜔(𝑦, 𝑧) 𝜕𝑥 𝜕𝜓(𝑥,𝑦,𝑥) = (3 + 2𝑥𝑦 − 𝑥 2 𝑧 3 ) untuk memperoleh 𝜔(𝑦, 𝑧) Samakan 𝜕𝑦

𝜕(𝑥𝑦 2 − 𝑥 2 𝑦𝑧 3 + 𝜔(𝑦, 𝑧)) = (3 + 2𝑥𝑦 − 𝑥 2 𝑧 3 ) 𝜕𝑦 2𝑥𝑦 − 𝑥 2 𝑧 3 + 𝜔′ (𝑦, 𝑧) = (3 + 2𝑥𝑦 − 𝑥 2 𝑧 3 ) 𝜔′ (𝑦, 𝑧) = 3 Integralkan 𝜔′ (𝑦, 𝑧) untuk memperoleh 𝜔 (𝑦, 𝑧) ∫ 𝜔′ (𝑦, 𝑧) = ∫ 3 𝜕𝑦 = 3𝑦 + 𝜔(𝑧)

Diperoleh: 𝜓(𝑥, 𝑦, 𝑥 ) = 𝑥𝑦 2 − 𝑥 2 𝑦𝑧 3 + 3𝑦 + 𝜔(𝑧)

Nafida Hetty Marhaeni

9.

𝜕𝜓(𝑥,𝑦,𝑥)

= 6𝑧 3 − 3𝑥 2 𝑦𝑧 2 untuk memperoleh 𝜔(𝑧) 𝜕 (𝑥𝑦 2 − 𝑥 2 𝑦𝑧 3 + 3𝑦 + 𝜔(𝑧)) = 6𝑧 3 − 3𝑥 2 𝑦𝑧 2 𝜕𝑧 −3𝑥 2 𝑦𝑧 2 + 𝜔′ (𝑧) = 6𝑧 3 − 3𝑥 2 𝑦𝑧 2 𝜔′ (𝑧) = 6𝑧 3 ′( ) Integralkan 𝜔 𝑧 untuk memperoleh 𝜔(𝑧) 3 ∫ 𝜔′ (𝑧) = ∫ 6𝑧 3 𝜕𝑧 = 𝑧 4 + 𝐶 2 3 Diperoleh: 𝜓(𝑥, 𝑦, 𝑥 ) = 𝑥𝑦 2 − 𝑥 2 𝑦𝑧 3 + 3𝑦 + 𝑧 4 + 𝐶 2 Samakan

𝜕𝑧

Jika 𝐴 = 𝑥 2 𝑧𝑖 + 𝑦𝑧 3 𝑗 − 3𝑥𝑦𝑘, 𝐵 = 𝑦 2 𝑖 − 𝑦𝑧𝑗 + 2𝑥𝑘 dan 𝜙 = 2𝑥 2 + 𝑦𝑧 maka carilah: (a) 𝐴 ∙ (∇𝜙); (b) (𝐴 ∙ ∇)𝜙; (c) (𝐴 ∙ ∇)𝐵; (d) 𝐵(𝐴 ∙ ∇); (e) (∇ ∙ 𝐴)𝐵 Jawab: 𝜕 𝜕 𝜕 ∇𝜙 = ( 𝑖 + 𝑗 + 𝑘) (2𝑥 2 + 𝑦𝑧) = 4𝑥𝑖 + 𝑧𝑗 + 𝑦𝑘 𝐴∙∇

∇∙𝐴

𝜕𝑥 ( 2

𝜕𝑦

𝜕𝑧

3

= 𝑥 𝑧𝑖 + 𝑦𝑧 𝑗 − 3𝑥𝑦𝑘 ) ∙ (

=(

Sehingga, (a) 𝐴 ∙ (∇𝜙) (b) (𝐴 ∙ ∇)𝜙

𝜕

𝑖+ 𝜕𝑥

𝜕

𝜕𝑦

𝑗+

𝜕

𝜕𝑧

𝜕

𝜕𝑥

𝑖+

𝜕

𝜕𝑦 3

𝑗+

𝜕

𝜕𝑧

𝑘) = 𝑥 2 𝑧

𝜕

𝜕𝑥

+ 𝑦𝑧 3

𝑘) ∙ (𝑥 2 𝑧𝑖 + 𝑦𝑧 𝑗 − 3𝑥𝑦𝑘) = 2𝑥𝑧 + 𝑧 3

𝜕

𝜕𝑦

− 3𝑥𝑦

𝜕

𝜕𝑧

= (𝑥 2 𝑧𝑖 + 𝑦𝑧 3 𝑗 − 3𝑥𝑦𝑘) ∙ (4𝑥𝑖 + 𝑧𝑗 + 𝑦𝑘) = 4𝑥 3 𝑧 + 𝑦𝑧 4 − 3𝑥𝑦 2 𝜕 𝜕 𝜕 = (𝑥 2 𝑧 + 𝑦𝑧 3 𝜕𝑦 − 3𝑥𝑦 𝜕𝑧) (2𝑥 2 + 𝑦𝑧) = 𝑥 2 𝑧(4𝑥 ) + 𝑦𝑧 3 (𝑧) − 3𝑥𝑦(𝑦) 𝜕𝑥

3

= 4𝑥 𝑧 + 𝑦𝑧 4 − 3𝑥𝑦 2 𝜕 𝜕 𝜕 = (𝑥 2 𝑧 + 𝑦𝑧 3 𝜕𝑦 − 3𝑥𝑦 𝜕𝑧) (𝑦 2 𝑖 − 𝑦𝑧𝑗 + 2𝑥𝑘)

(c) (𝐴 ∙ ∇)𝐵

𝜕𝑥

= 𝑥 2 𝑧(2𝑘) + 𝑦𝑧 3 (2𝑦𝑖 − 𝑧𝑗) − 3𝑥𝑦(−𝑦𝑗) = 2𝑥 2 𝑧𝑘 + 2𝑦 2 𝑧 3 𝑖 − 𝑦𝑧 4 𝑗 + 3𝑥𝑦 2 𝑗 = 2𝑦 2 𝑧 3 𝑖 + (3𝑥𝑦 2 − 𝑦𝑧 4 )𝑗 + 2𝑥 2 𝑧𝑘 𝜕 𝜕 = (𝑦 2 𝑖 − 𝑦𝑧𝑗 + 2𝑥𝑘) (𝑥 2 𝑧 + 𝑦𝑧 3 − 3𝑥𝑦

(d) 𝐵(𝐴 ∙ ∇)

= (𝑥 2 𝑦 2 𝑧𝑖 − 𝑥 2 𝑦𝑧 2 𝑗 + 2𝑥 2 𝑧𝑘)

𝜕

𝜕𝑥

𝜕𝑥

𝜕𝑦

𝜕

𝜕𝑧

)

+ (𝑦 3 𝑧 3 𝑖 − 𝑦 2 𝑧 4 𝑗 + 2𝑥𝑦𝑧 3 𝑘)

𝜕

𝜕𝑦

+ (−3𝑥𝑦 3 𝑖 + 3𝑥𝑦 2 𝑧𝑗 − 6𝑥 2 𝑦𝑘)

= (2𝑥𝑧 + 𝑧 3 )(𝑦 2 𝑖 − 𝑦𝑧𝑗 + 2𝑥𝑘 ) = (2𝑥𝑦 2 𝑧 + 𝑦 2 𝑧 3 )𝑖 − (2𝑥𝑦𝑧 2 + 𝑦𝑧 4 )𝑗 + (4𝑥 2 𝑧 + 2𝑥𝑧 3 )𝑘

(e) (∇ ∙ 𝐴)𝐵

10. Jika 𝐴 = 𝑦𝑧 2 𝑖 − 3𝑥𝑧 2 𝑗 + 2𝑥𝑦𝑧𝑘, 𝐵 = 3𝑥𝑖 + 4𝑧𝑗 − 𝑥𝑦𝑘 dan 𝜙 = 𝑥𝑦𝑧 maka carilah: (a) 𝐴 × (∇𝜙); (b) (𝐴 × ∇)𝜙; (c) (∇ × 𝐴) × 𝐵; (d) 𝐵 ∙ ∇ × 𝐴 Jawab: 𝜕 𝜕 𝜕 ∇𝜙 = ( 𝑖 + 𝑗 + 𝑘) (𝑥𝑦𝑧) = 𝑦𝑧𝑖 + 𝑥𝑧𝑗 + 𝑥𝑦𝑘 𝐴×∇

(∇ × 𝐴)

Sehingga:

𝜕𝑦

𝜕𝑥

𝑖 2 𝑦𝑧 =[ 𝜕

𝜕𝑥

== [

𝑖

𝜕

𝜕𝑦

𝜕

𝜕𝑥 2

𝑦𝑧

𝜕𝑧

𝑗 −3𝑥𝑧 2

𝑖

𝑘 2𝑥𝑦𝑧] = 𝑖 (−3𝑥𝑧 2 𝜕

𝜕𝑧

𝑗

𝜕

𝜕𝑦

−3𝑥𝑧2 𝑗

𝑘 𝜕

𝜕𝑧

𝜕

𝜕𝑧

− 2𝑥𝑦𝑧

𝜕

𝜕𝑦

) − 𝑗 (𝑦𝑧 2

𝜕

𝜕𝑧

− 2𝑥𝑦𝑧

𝜕

𝜕𝑥

) + 𝑘 (𝑦𝑧 2

𝜕

𝜕𝑦

+ 3𝑥𝑧 2

𝜕

𝜕𝑥

𝜕

𝜕𝑧

)

] = 𝑖(2𝑥𝑧 + 6𝑥𝑧) − 𝑗(2𝑦𝑧 − 2𝑦𝑧) + 𝑘(−3𝑧2 − 𝑧2 ) = 8𝑥𝑧𝑖 − 4𝑧2 𝑘

2𝑥𝑦𝑧

𝑘

(a) 𝐴 × (∇𝜙) = [𝑦𝑧2 −3𝑥𝑧2 2𝑥𝑦𝑧] = 𝑖(−3𝑥2 𝑦𝑧2 − 2𝑥2 𝑦𝑧2 ) − 𝑗(𝑥𝑦2 𝑧2 − 2𝑥𝑦2 𝑧2 ) + 𝑘(𝑥𝑦𝑧3 + 3𝑥𝑦𝑧3 ) 𝑦𝑧 𝑥𝑧 𝑥𝑦 2 2 2 2 = −5𝑥 𝑦𝑧 𝑖 + 𝑥𝑦 𝑧 𝑗 + 4𝑥𝑦𝑧 3 𝑘

Nafida Hetty Marhaeni

(b) (𝐴 × ∇)𝜙

= [𝑖 (−3𝑥𝑧2

𝜕

𝜕

− 2𝑥𝑦𝑧 ) − 𝑗 (𝑦𝑧2 𝜕𝑦

𝜕𝑧

𝜕

𝜕𝑧

𝜕

𝜕

− 2𝑥𝑦𝑧 ) + 𝑘 (𝑦𝑧2 𝜕𝑥

𝜕𝑦

+ 3𝑥𝑧2

𝜕

𝜕𝑥

)] 𝑥𝑦𝑧

= (−3𝑥𝑧 2 𝑥𝑦 − 2𝑥𝑦𝑧𝑥𝑧)𝑖 − (𝑦𝑧 2 𝑥𝑦 − 2𝑥𝑦𝑧𝑦𝑧)𝑗 + (𝑦𝑧 2 𝑥𝑧 + 3𝑥𝑧 2 𝑦𝑧)𝑘 = (−3𝑥 2 𝑦𝑧 2 − 2𝑥 2 𝑦𝑧 2 )𝑖 − (𝑥𝑦2 𝑧2 − 2𝑥𝑦2 𝑧2 )𝑗 + (𝑥𝑦𝑧3 + 3𝑥𝑦𝑧3 )𝑘 = −5𝑥 2 𝑦𝑧 2 𝑖 + 𝑥𝑦 2 𝑧 2 𝑗 + 4𝑥𝑦𝑧 3 𝑘 Note: 𝑨 × (𝛁𝝓) = (𝑨 × 𝛁)𝝓

𝑖 (c) (∇ × 𝐴) × 𝐵 = [8𝑥𝑧

𝑗 0

𝑘 −4𝑧2 ] = 𝑖(16𝑧3 ) − 𝑗(−8𝑥2 𝑦𝑧 + 12𝑥𝑧2 ) + 𝑘(32𝑥𝑧2 )

3𝑥 4𝑧 −𝑥𝑦 = 16𝑧 3 𝑖 + (8𝑥 2 𝑦𝑧 − 12𝑥𝑧 2 )𝑗 + 32𝑥𝑧 2 𝑘 (d) 𝐵 ∙ ∇ × 𝐴 = (3𝑥𝑖 + 4𝑧𝑗 − 𝑥𝑦𝑘) ∙ (8𝑥𝑧𝑖 − 4𝑧2 𝑘) = 24𝑥2 𝑧 + 4𝑥𝑦𝑧2 2

11. Jika 𝑹 = 𝑒 −𝑡 𝑥 𝒊 + ln|−𝑡 3 + 3𝑡|𝒋 + cos 𝑡 𝒌, carilah: (a) Jawab:

𝑑𝑹 𝑑𝑡

; (b)

𝑑2 𝑹 𝑑𝑡

𝑑2 𝑹

𝑑𝑹

| | ; (d)| 2 | 2 ; (c) 𝑑𝑡 𝑑𝑡

Ingat: 𝑦 = 𝑢𝑣 → 𝑦 ′ = 𝑢′ 𝑣 + 𝑣′𝑢 dan 𝑦 =

𝑢 𝑣

→ 𝑦′ =

𝑢 ′𝑣−𝑣 ′𝑢 𝑣2

2

𝑦 = 𝑒 𝑓(𝑥) → 𝑦 ′ = 𝑓 ′ (𝑥 )𝑒 𝑓(𝑥) → 𝑦 ′′ = 𝑓 ′′ (𝑥 )𝑒 𝑓(𝑥) + 𝑓 ′ (𝑥 )𝑒 𝑓(𝑥) 𝑓 ′ (𝑥 ) = 𝑓 ′′ 𝑒 𝑓(𝑥) + (𝑓 ′ (𝑥 )) 𝑒 𝑓(𝑥) 𝑦 = 𝑙𝑛 𝑓 (𝑥 ) → 𝑦 ′ = 𝑢 = 𝑒 −𝑡

2𝑥

𝑓 ′ (𝑥) 𝑓(𝑥)

→ 𝑦 ′′ =

→ 𝑢′ = −2𝑡𝑥𝑒 −𝑡

b. c. d.

𝑑𝑹 𝑑𝑡

2

= −2𝑡𝑥𝑒 −𝑡 𝑥 𝑖 +

𝑑2 𝑹 𝑑𝑡 2

𝑑𝑹

= (−2𝑥𝑒 −𝑡

2𝑥

−𝑡 2 +3𝑡

−2𝑡+3

−𝑡 2 +3𝑡

𝑑2 𝑹

|

𝑑𝑡 2

→ 𝑣 ′′ =

2

2𝑥

+ 4𝑡 2 𝑥 2 𝑒 −𝑡

−2𝑡 2 +6𝑡−9

𝑗 − sin 𝑡𝑘

(−𝑡 2 +3𝑡) 2

−2𝑡 2 +6𝑡−9

2

2

| = √(−2𝑥𝑒 −𝑡

2𝑥

−2𝑡+3 2

−𝑡 2 +3𝑡

) + (− sin 𝑡) 2 = √4𝑡 2 𝑥 2 𝑒 −2𝑡

2

2

−2𝑡 2 +6𝑡−9 2

2

Jawab:

b. c. d.

𝜕𝑨 𝜕𝑥

𝜕𝑨 𝜕𝑦

2𝑥

+ 4𝑡 2 𝑥 2 𝑒 −𝑡 𝑥 ) + ( (−𝑡 2 +3𝑡)2 ) + (− cos 𝑡) 2

12. Jika 𝑨 = 5 sin 𝑥𝑦 𝒊 + 3𝑒 𝑦𝑥 𝒋 − (3𝑥 2 + 2𝑥𝑦 3 )𝒌 carilah: (a) a.

2𝑥

+ 4𝑡 2 𝑥 2 𝑒 −𝑡 𝑥 )𝑖 + ( (−𝑡 2 +3𝑡)2 ) 𝑗 − cos 𝑡 𝑘

2 | | = √(−2𝑡𝑥𝑒 −𝑡 𝑥 ) + (

𝑑𝑡

(𝑓(𝑥))

→ 𝑢′′ = −2𝑥𝑒 −𝑡

−2𝑡+3

𝑣 = ln|−𝑡 2 + 3𝑡| → 𝑣 ′ = a.

2𝑥

𝑓′′ (𝑥)𝑓(𝑥)+𝑓′ (𝑥)𝑓′(𝑥)

𝜕𝑨 𝜕𝑥

; (b)

𝜕𝑨 𝜕𝑦

+

; (c)

4𝑡 2 −12𝑡+9

𝑡 4 −6𝑡 3 +9𝑡 2

𝜕2𝑨

𝜕𝑥𝜕𝑦

; (d)

+ sin2 𝑡

𝜕 2𝑨

𝜕𝑦𝜕𝑥

2

= 5𝑦 cos 𝑥𝑦 𝑖 + 6𝑥𝑦𝑒 𝑦𝑥 𝑗 − (6𝑥 + 2𝑦 3 )𝑘 2

= 5𝑥 cos 𝑥𝑦 𝑖 + 3𝑥 2 𝑒 𝑦𝑥 𝑗 − 6𝑥𝑦 2 𝑘

𝜕 2𝑨

𝜕𝑥𝜕𝑦 𝜕 2𝑨

𝜕𝑦𝜕𝑥

= =

2

𝜕(5𝑥 cos 𝑥𝑦𝑖+3𝑥 2 𝑒 𝑦𝑥 𝑗−6𝑥𝑦 2𝑘) 𝜕𝑥

2

𝜕(5𝑦 cos 𝑥𝑦𝑖+6𝑥𝑦𝑒 𝑦𝑥 𝑗−(6𝑥+2𝑦 3 )𝑘) 𝜕𝑦

2

2

= (5 cos 𝑥𝑦 − 5𝑥𝑦 sin 𝑥𝑦)𝑖 + (6𝑥𝑒 𝑦𝑥 + 6𝑥 3 𝑦𝑒 𝑦𝑥 𝑗 − 6𝑦 2 𝑘 2

2

= (5 cos 𝑥𝑦 − 5𝑥𝑦 sin 𝑥𝑦)𝑖 + (6𝑥𝑒 𝑦𝑥 + 6𝑥 3 𝑦𝑒 𝑦𝑥 )𝑗 − 6𝑦𝑘

Nafida Hetty Marhaeni

13. Jika 𝑨 = 3𝑥𝑦 2 𝑧𝒊 + 2𝑥𝑦𝑧 3 𝒋 + 4𝑥 2 𝑦 3 𝑧𝒌 dan 𝑩 = −6𝑥𝑦𝑧𝒊 + 3𝑥𝑦 3 𝒋 − 2𝑧𝒌, carilah (1, −2,0)

𝜕2

𝜕𝑥𝜕𝑦

(𝑨 × 𝑩) pada titik

Jawab: 𝐴×B 𝜕

𝜕𝑦

𝑖 𝑗 2 3 3𝑥𝑦 𝑧 2𝑥𝑦𝑧 [ = −6𝑥𝑦𝑧 3𝑥𝑦 3

𝑘 4𝑥 2 𝑦 3 𝑧] = 𝑖(−4𝑥𝑦𝑧 4 − 12𝑥 3 𝑦 6 𝑧) − 𝑗(−6𝑥𝑦 2 𝑧 2 + 24𝑥 3 𝑦 4 𝑧2 ) + 𝑘(9𝑥 2 𝑦 5 𝑧 + 12𝑥 2 𝑦 2 𝑧4 ) −2𝑧

(𝑨 × 𝑩) = 𝑖 (−4𝑥𝑧 4 − 72𝑥 3 𝑦 5 𝑧) − 𝑗(−12𝑥𝑦𝑧 2 + 96𝑥 3 𝑦 3 𝑧 2 ) + 𝑘(45𝑥 2 𝑦 4 𝑧 + 24𝑥 2 𝑦𝑧 4 )

𝜕2

𝜕𝑥𝜕𝑦

𝜕

𝜕

(𝑨 × 𝑩) = ( (𝑨 × 𝑩)) = 𝑖 (−4𝑧 4 − 216𝑥 2 𝑦 5 𝑧) − 𝑗(−12𝑦𝑧 2 + 288𝑥 2 𝑦 3 𝑧 2 ) + 𝑘(90𝑥𝑦 4 𝑧 + 48𝑥𝑦𝑧 4 ) 𝜕𝑥 𝜕𝑦

14. Jika ∇𝜁 = (3𝑦 2 − 8𝑥𝑦𝑧 3 )𝒊 + (6𝑥𝑦 − 4𝑥 2 𝑧 3 )𝒋 − 12𝑥 2 𝑦𝑧 2 𝒌. Carilah 𝜁(𝑥, 𝑦, 𝑧) jika 𝜁 (1, −2,2) = 15. Jawab: 𝜕 ∇𝜁 = ( 𝑖 + 𝜕𝑥 Maka:

𝜕

𝜕𝑦

𝑗+

𝜕

𝜕𝑧

𝑘) 𝜁 = (3𝑦 2 − 8𝑥𝑦𝑧 3 )𝒊 + (6𝑥𝑦 − 4𝑥 2 𝑧 3 )𝒋 − 12𝑥 2 𝑦𝑧 2 𝒌

(3𝑦 2 − 8𝑥𝑦𝑧 3 )𝜕𝑥 + (6𝑥𝑦 − 4𝑥 2 𝑧 3 )𝜕𝑦 + (−12𝑥 2 𝑦𝑧 2 )𝜕𝑧 = 0 Akan dicari solusi persamaan tersebut. 𝜕𝜁(𝑥, 𝑦, 𝑥 ) = ∫(3𝑦 2 − 8𝑥𝑦𝑧 3 )𝜕𝑥 + 𝜔(𝑦, 𝑧) = 3𝑥𝑦 2 − 4𝑥 2 𝑦𝑧 3 + 𝜔(𝑦, 𝑧) 𝜕𝑥 𝜕𝜁(𝑥,𝑦,𝑥) = 6𝑥𝑦 − 4𝑥 2 𝑧 3 untuk memperoleh 𝜔(𝑦, 𝑧) Samakan 𝜕𝑦

𝜕(3𝑥𝑦 2 − 4𝑥 2 𝑦𝑧 3 ) = 6𝑥𝑦 − 4𝑥 2 𝑧 3 𝜕𝑦 6𝑥𝑦 − 4𝑥 2 𝑧 3 + 𝜔′ (𝑦, 𝑧) = 6𝑥𝑦 − 4𝑥 2 𝑧 3 𝜔′ (𝑦, 𝑧) = 0 ′( Integralkan 𝜔 𝑦, 𝑧) untuk memperoleh 𝜔 (𝑦, 𝑧) ∫ 𝜔′ (𝑦, 𝑧) = ∫ 0 𝜕𝑦 = 𝐶1 + 𝜔(𝑧)

Diperoleh: 𝜁 (𝑥, 𝑦, 𝑥 ) = 3𝑥𝑦 2 − 4𝑥 2 𝑦𝑧 3 + 𝐶1 + 𝜔(𝑧) 𝜕𝜁(𝑥,𝑦,𝑥) Samakan 𝜕𝑧 = −12𝑥 2 𝑦𝑧 2 untuk memperoleh 𝜔(𝑧) 𝜕(3𝑥𝑦 2 − 4𝑥 2 𝑦𝑧 3 + 𝐶1 + 𝜔(𝑧)) = −12𝑥 2 𝑦𝑧 2 𝜕𝑧 −12𝑥 2 𝑦𝑧 2 + 𝜔′ (𝑧) = −12𝑥 2 𝑦𝑧 2 𝜔 ′ (𝑧 ) = 0 Integralkan 𝜔′ (𝑧) untuk memperoleh 𝜔(𝑧) ∫ 𝜔′ (𝑧) = ∫ 0 𝜕𝑧 = 𝐶2

Diperoleh: 𝜁 (𝑥, 𝑦, 𝑥 ) = 3𝑥𝑦 2 − 4𝑥 2 𝑦𝑧 3 + 𝐶1 + 𝐶2 (jumlahan 𝐶1 + 𝐶2 = 𝐶) maka: 𝜁 (𝑥, 𝑦, 𝑥 ) = 3𝑥𝑦 2 − 4𝑥 2 𝑦𝑧 3 + 𝐶 Diberikan syarat awal 𝜁 (1, −2,2) = 15 maka: 3(1)(−2)2 − 4(1)2 (−2)(2)3 + 𝐶 = 15 12 + 64 + 𝐶 = 15 76 + 𝐶 = 15 𝐶 = −61 Jadi, 𝜁 (𝑥, 𝑦, 𝑥 ) = 3𝑥𝑦 2 − 4𝑥 2 𝑦𝑧 3 − 61

Nafida Hetty Marhaeni

15. Diketahui 𝜆 = −3𝑥 2 𝑦 3 𝑧 4 ; 𝑨 = 6𝑥𝑦 2 𝒊 + 𝑥 3 𝑦𝑧 2 𝒌, dan 𝑩 = 3𝑥𝑦𝒊 − 2𝑥 3 𝑦𝑧 2 𝒋. Carilah: a. ∇ × (𝜆𝑩)

b. ∇ ∙ (∇ × 𝑨)

c. ∇2 𝜆 = ∇ ∙ ∇𝜆 dimana ∇2 =

𝜕2

𝜕𝑥 2

+

𝜕2

𝜕𝑦 2

+

𝜕2

𝜕𝑧 2

menyatakan operator Laplacian.

Jawab: 𝜆𝐵

= (−3𝑥 2 𝑦 3 𝑧 4 )( 3𝑥𝑦𝒊 − 2𝑥 3 𝑦𝑧 2 𝒋) = −9𝑥 3 𝑦 4 𝑧 4 𝒊 + 6𝑥 5 𝑦 4 𝑧 6 𝒋

∇𝜆

6𝑥𝑦2 0 𝜕 𝜕 =( 𝑖+ 𝑗+

∇×A

a.

𝑖

𝑗

𝜕 2 2 3 2 𝜕𝑧 ] = 𝑖(𝑥 𝑧 ) − 𝑗(3𝑥 𝑦𝑧 ) − 𝑘(12𝑥𝑦) 𝑥3 𝑦𝑧2

𝜕𝑦

𝜕𝑥

∇ × (𝜆𝑩)

𝑘

𝜕 𝜕𝑦

𝜕 = [ 𝜕𝑥

𝜕

𝜕𝑧

𝑖

𝜕

=[

𝑘) (−3𝑥 2 𝑦 3 𝑧 4 ) = −6𝑥𝑦 3 𝑧 4 𝑖 − 9𝑥 2 𝑦 2 𝑧 4 𝑗 − 12𝑥 2 𝑦 3 𝑧 3 𝑗

𝑘

𝜕

𝜕𝑥 3 4 4

𝜕

𝜕𝑧 ]

𝜕𝑦 5 4 6

−9𝑥 𝑦 𝑧

6𝑥 𝑦 𝑧

0

= 𝑖(−36𝑥 5 𝑦 4 𝑧 5 ) − 𝑗(36𝑥 3 𝑦 4 𝑧 3 ) + 𝑘(30𝑥 4 𝑦 4 𝑧 6 + 36𝑥 3 𝑦 3 𝑧 4 )

d. ∇ ∙ (∇ × 𝑨) = (

𝜕

𝜕𝑥

𝑖+

𝜕

𝑗+

𝜕𝑦

e. ∇2 𝜆 = ∇ ∙ ∇𝜆 dimana ∇2 = 𝜕

𝜕

𝜕

𝜕

𝜕𝑧

𝜕2

𝑘) ∙ (𝑥 3 𝑧 2 𝑖 − 3𝑥 2 𝑦𝑧 2 𝑗 − 12𝑥𝑦𝑘) = 3𝑥 2 𝑧 − 3𝑥 2 𝑧 2 = 3𝑥 2 𝑧(1 − 𝑧)

+ 𝜕𝑥 2

𝜕2

𝜕𝑦

2 +

𝜕2

𝜕𝑧 2

menyatakan operator Laplacian.

∇ ∙ ∇𝜆 = (𝜕𝑥 𝑖 + 𝜕𝑦 𝑗 + 𝜕𝑧 𝑘) ∙ (−6𝑥𝑦 3 𝑧 4 𝑖 − 9𝑥 2 𝑦 2 𝑧 4 𝑗 − 12𝑥 2 𝑦 3 𝑧 3 ) = −6𝑦 3 𝑧 4 − 18𝑥 2 𝑦𝑧 4 − 36𝑥 2 𝑦 3 𝑧 2

Akan ditunjukkan ∇2 𝜆 = ∇ ∙ ∇𝜆 Perhatikan bahwa: ∇2 𝜆

=(

𝜕2

𝜕𝑥 2

𝜕

+

𝜕2

𝜕𝑦 2

+

𝜕2

𝜕𝑧 2 𝜕

) (−3𝑥 2 𝑦 3 𝑧 4 ) = 𝜕

𝜕

𝜕𝑥

𝜕

( (−3𝑥 2 𝑦 3 𝑧 4 )) + 𝜕𝑥

𝜕

𝜕𝑦

𝜕