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III. TURUNAN DALAM RUANG BERDIMENSI nIII.1 Fungsi dengan Dua Peubah atau LebihSampai kini telah ada dua jenis fungsi yang muncul. Pertama, diberikan contoh f(x) = x2, yang mengaitkan bilangan riil lain f(x). Kita sebutf(x0 ini sebagai fungsi bernilai riil dari perubah riil. Jenis fungsi yang kedua, diilustrasikan oleh f(x) = (x3-x), mengaitkan bilangan riil x dengan vekto f(x). Fungsi itu kita sebut fungsi bernilai vektor dari peubah riil.Perhatian kita sekarang berpidah kesatu fungsi bernilai rill dari dua peubah riil; yakni fungsifyangmenandakansetiappasanganterurut (x,y)dalamhimpunanDpadabidang dengan bilangan riil (tunggal) f (x,y). Sebagai contoh :1) f (x,y) = x2 + 3y22) g (x,y) = 2xy Perhatiakan bahwa f (-1,4) = (-1)2 + 3(4)2 = 49 dan g (-1,4) = 2(-1) 4= -4Himpunan D disebutdaerah asal(domain)fungsi. Jika daerah asal fungsi tidak diperinci, kita ambil D berupa daerah asal alami (natural domain), yakni himpunan semua titik (x,y) = x2 + 3y2, daerah mulanya adalah seluruh bidang; untuk g (x,y) = 2x y , daerah asal mulanya adalah{(x,y): R, y0}.Daerahhasil(range) suatufungsi adalahhimpunannilai-nilainya. Jikaz= f(x,y), kitasebutxdanypeubahbebas(independent variable)danz peubah tak bebas.(dependent variable)Semuayang telah kami nyatakan,dapat meluas menjadi fungsi real dari tiga peubah reall (ataubahkannpeubahreal) kitabebasmenggunakanfungsi-fungsi yangdemikiantanpa komentar lebih lanjut.1Contoh 1Dalam bidang xy, buatlah grafik daerah asal mula untuk ( )2 22) 1 (, + y xx yy x fPenyelesaian : agar aturan ini bermakna, kita harus mengeluarkan {(x,y): y > x2} dan titik (0,1). Daerah asal yang dihasilkan dapat anda lihat dalam gambar 1. (gambar dapat dilihat pada buku paket halaman 249)Contoh 2 :Sketsalah peta kontur untuk z = f (x,y) = 2 24 9 3631y x

Penyelesaian : Misalkanz= 2 24 9 3631y x dan perhatikan bahwa z 0Jika kedua ruas kita kuadratkandan menyederhanakannya diperoleh persamaan

( )2 2 24 9 3691y x z

36 9 4 92 2 2 + + z y xYang dikenal sebagai persamaan dari sebuah elipsoid . Grafik dari fungsi ini merupakan setengah bagian atas elipsoid (gambar 4 hal 249) Untuk mengambar dengan mudah kita gunakan perangkat lunak(software: Maple dan Mathematica)Contoh 3: Sketsa grafik dari( )2 2, x y y x f z PenyelesaianGrafik berbentuk paraboloid ( gambar 5 hal 249)2Grafik Komputer : Beberapa perangkat lunak(software),termasuk Maple dan Mathematica,mampu menghasilkan grafik-grafik berdimensi tiga yang rumit dengan mudah.(belumselesailihat hal 249 250) 3 Equation.3 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TURUNANPARSIALAndaikanf adalahfungsi denganduapeubahxdany. Jikaydijagaagar tetap konstan,misalnya oy y maka( )oy x f , adalah fungsi dengan peubah tunggalx.Turunannyadiox x disebutTurunanparsialfterhadapxdi( )o oy x ,dandinyatakan sebagai ( )o o xy x f , jadi

Dengan cara serupa turunan parsial f terhadapydi( )o oy x ,dinyatakan dengan( )o o yy x f , dan dirumuskan dengan

Daripada menghitung ( )o o xy x f ,dan ( )o o yy x f ,secara langsung dari rumus didalam kotak diatas,biasanya kita dapat menentukan ( ) y x fx,da dengan mengunakan aturan aturanstandar turunan, kemudian kita memsubtitusikano oy y dan x x .Contoh 1:Tentukan( ) 2 , 1xf dan( ) 2 , 1yfjika ( )3 23 , y y x y x f + Penyelesaian :Untukmenentukan( ) y x fx,kitaanggapysebagai konstantadanmendiferensialkannya terhadap x, menghasilkan

( ) 0 2 , + xy y x fx 4( ) [ ] ( ) ( )( ) ( ) ( )( ) ( )( ) ( )2 2 2 22 2 2 22 2 2 22 2 2 2sin 2 cos .sin . 2 . cos2 . sin cossin sinxy x xy y xxzxy x y xy xx xy xyxxy xxxxy xyxxxz+ + ++( ) [ ] ( ) ( )( ) ( ) ( )( ) ( )( ) ( )2 2 2 22 2 2 22 2 2 22 2 2 2sin 2 cos .sin . 2 . cos2 . sin cossin sinxy x xy y xxzxy x y xy xx xy xyxxy xxxxy xyxxxz+ + ++( ) [ ]( ) ( )( )( )2 32 22 2 22 2cos . 22 . coscossinxy y xyzxy xy xxyyxy xxyyxyz jadi( ) 2 , 1xf= 2.1.2= 4Dengancara yangsama, kita anggap xsebagai konstanta dan menndiferensialkanya terhadap y, dan menghasilkan ( )2 29 , y x y x fy+ jadi ( ) 2 , 1yf = 37 2 . 9 12 2 +Jika ( ) y x f z , kita menggunakan notasi-notasi alternatif berikut( )( )x y x fxzy x fx,, ( )( )y y x fyzy x fy,,( )( )o oy x o o xxzy x f,| , ( )( )o oy x o o yyzy x f,| ,Simbol adalahtanda turunan parsialSimbol simbolydanx mempersentasikanoperator-operator linear,samaseperti operator-operatordxddan Dx padaturunan.Contoh 2: Jika ( )2 2sin xy x z tentukan yzdanxzPenyelesaian 5INTERPRESTASI GEOMETRIK DAN INTERPRESTASI FISIS.Tinjaulah permukaan yang mempunyai persamaan ( ) y x f z , bidang oy y memotong permukaanini pada kurvabidang QPR (gambar1 hal 256)Dannilai dari( )o o xy x f ,adalahkemiringan garis singgungterhadapkurvaini di ( ) ( )o o o oy x f y x P , , ,.Demikian pula, bidangox x memotong permukaan tersebutdi kurva bidangLPM(gambar2hal 257)dan ( )o o yy x f ,adalahkemiringangarissinggung dengan kurva ini di P.Turunan-turunan parsial juga dapat diinterprestasikan sebagai laju (sesaat)perubahan.Andaikan sebuah senarbiolaterbentang diantaratitik A dan B, dan begetar pada bidang xz (gambar3 hal 257) menunjukkanposisi senar tersebut pada waktu t.Jika s = f(x,t) menyatakan tinggi senar di titik Pdengan absis x pada waktu t, maka xz adalah kemiringansenar di P, dantzadalahlajuwaktudari perubahanketinggiandari Pdi sepanjang garisvertikalyang diketahui . Dengan kata lain, tz adalah kecepatanvertikal 6dari PContoh 3:Permukaan ( )2 22 9 , y x y x f z dan bidangy = 1 berpotongan dalamsebuah kurva seperti pada gambar 1. Tentukan persamaan persamaanparametrikuntuk garis singgungdi ) 2 , 1 , 2 ( Penyelesaian ( ) ( ) ( ) x y x y x fx4 2 321,212 2 Sehingga( ) 2 1 , 2 xfbilanganini adalahkemiringangaarissinggungterhadap kurvadi) 2 , 1 , 2 (:dalamhal ini1 / 2 adalahrasiokenaikanterhadappenambahandi sepanjang garis singgung tersebut .Hasilnya adalah bahwa garis ini mempunyai bilangan-bilangan arah( ) 2 , 0 , 1 dankarena garis ini melalui ) 2 , 1 , 2 ( maka

t z y x 2 2 1 1 2 + Menghsilkan persamana-persamaan parametrik yang diinginkan.Turunan Parsial yang Lebih TinggiKarena turunan parsial darisuatu fungsi x dan y, secara umu adalh sebuah fungsilain dari dua peubahyang sama tersebut, maka turunan tersebut dapat dideferensialkansecara parsial terhadap xdan y, menghasilkan empat buahturunanparsial kedua(secondpartialderivative) dari f.722xfxfxfxx ,`

.|

22yfyfyfyy

,`

.|( )x y fxfyf fy x xy ,`

.| 2

( )y x fyfxf fxy yx

,`

.| 2Contoh 4:Tentukan empat turunan parsial dari ( ) ( )2 3/ sin , y x y x xe y x fy+ Penyelesaian( )2 23 cos1, y xyxye y x fyx+

,`

.| ( )2 23 cos1, y xyxye y x fyy+

,`

.| ( )226 sin1, xyyxyy x fxx+

,`

.|( )332 cos2sin , xyxyxyxyxxe y x fyyy+

,`

.|

,`

.|+ ( ) y xyxy yxyxe y x fyxy22 36 cos1sin , +

,`

.|+

,`

.| ( ) y xyxy yxyxe y x fyyx22 36 cos1sin , +

,`

.|+

,`

.| Dariyx xtf f yang sebenarnya merupakan kasus untuk fungsidengan dua peubahyangDijumpai pada bab pertama.Turunan turunan parsial ketga atau lebih dapat didefinisikan secara analogis, dan notasi penulisan serupa.Jadijika f fungsi dengan dua peubah x dan y , maka turunan parsial 8ketigaf diperolehdenganmendeferensialkanf secaraparsial. Pertamaterhadapxdan kemudian dua kaliterhadap y , dapat dinyatakandengan : xyyfx yfx y fy xfy y ]]]

]]]

,`

.|22 2Secara keseluruhan terdapat delapanturunan parsial ketiga.Fungsi dengan lebih dari Dua PeubahMisalkanf adalah fungsidengan tiga peubah x,y dan z. Turunanparsial f terhadap x di (x,y,z) dinyatakan dengan ( ) z y x fx, , atau( )xz y x f , , dan didefinisikansebagai( )( ) ( )xz y x f z y x x fz y x fxx + , , , ,lim , ,0Jadi,( ) z y x fx, ,dapat diperoleh dengan melakukan y dan z sebagai konstanta mendeferensialkan terhaderhadap yap x.Turunan turunan parsialterhadap y dan zdidefinisikan secara analogis. Turunan turunan parsialdarifungsi dengan empat peubahatau lebih dapat didefinisikandengan cara yang sama. Turunan turunan parsial, sepertixyz xyf dan fyang menggunakan pendeferensialanterhadaplebihdari satupeubahdisebutTurunanparsial campuran.(mixedpartial derivative)Contoh 5:Jika ( ) zx yz xy z y x f 3 2 , , + + tentukan z y xf dan f f ,..92 2 2 2 2 2 2 2 24 22 2y x w y x w y x we wxz e xzwe zx w x wT+ + + + + +

,`

.|

,`

.| Penyelesaian Untuk memperoleh xf kita anggap y dan z sebagai konstanta, dan mendeferensialkan terhadappeubah x. Jadi( ) z y z y x fx3 , , + ( ) z x z y x fy2 , , + ( ) x y z y x fz3 2 , , + Contoh 6. Jika ( )2 2 2, , ,y x we z z y x w T+ + tentukan seluruh turnan parsial pertama dan 22 2 2, ,zTw x Tx wT Penyelesaian :2 2 2 2 2 22y x w y x we wz e zw wT+ + + +

,`

.|22 2 2 2 2 22y x w y x we xz e zx xT+ + + +

,`

.|2 2 2 2 2 22y x w y x we yz e zy yT+ + + +

,`

.|2 2 2 2 2 2y x w y x we e zz zT+ + + +

,`

.|Turunan parsial lainnya102 2 2 2 2 2 2 2 24 22 2y x w y x w y x we wxz e wzxe zw x w x T+ + + + + +

,`

.|

,`

.| ( ) 02 2 2 2 2 22222

,`

.|+ + + + y x w y x weze zz zTSoal1. Tentukan seluruh turunan parsial pertama dari setiap fungsi berikuta. ( ) ( )42 , y x y x f b. ( ) x e y x fysin , c. ( )xyx y x g ,2. Buktukanbahwa 02 2 + y x fx y f dari ( )5 3 3 22 , y x y x y x f 3. Jika ( ) ( ) ( ) 2 , 3 2 , 32, y xF dan F tentukanxy y xy x F4. Berikandalambentuk limit turunan-turunan parsialberikuta. ( ) z y x fy, ,c. ( ) z y x w Gx, , ,b.( ) z y x fz, , d. ( ) t z y xz,. ,. ,. 11