Artikel

12.1: Fungsi-Nilai Vektor dan Keluk Ruang - Matematik


Kajian kami mengenai fungsi bernilai vektor menggabungkan idea dari pemeriksaan awal kalkulus pemboleh ubah tunggal kami dengan penerangan vektor kami dalam tiga dimensi dari bab sebelumnya. Dalam bahagian ini, kami memperluas konsep dari bab-bab sebelumnya dan juga meneliti idea-idea baru mengenai lengkung dalam ruang tiga dimensi. Definisi dan teorema ini menyokong penyampaian bahan dalam bab ini dan juga pada bab-bab yang tinggal dalam teks.

Definisi Fungsi Berharga Vektor

Langkah pertama kami dalam mengkaji kalkulus fungsi bernilai vektor adalah menentukan apa sebenarnya fungsi bernilai vektor. Kita kemudian dapat melihat grafik fungsi bernilai vektor dan melihat bagaimana mereka menentukan lengkung dalam kedua dan tiga dimensi.

Definisi: Fungsi bernilai vektor

Fungsi bernilai vektor adalah fungsi bentuk

[ vecs r (t) = f (t) , hat { mathbf {i}} + g (t) , hat { mathbf {j}} quad text {atau} quad vecs r (t) = f (t) , hat { mathbf {i}} + g (t) , hat { mathbf {j}} + h (t) , hat { mathbf { k}}, ]

di mana komponen berfungsi (f ), (g ), dan (h ), adalah fungsi nilai sebenar dari parameter (t ).

Fungsi bernilai vektor juga boleh ditulis dalam bentuk

[ vecs r (t) = ⟨f (t), , g (t)⟩ ; ; teks {atau} ; ; vecs r (t) = ⟨f (t), , g (t), , h (t)⟩. ]

Dalam kedua kes tersebut, bentuk pertama fungsi menentukan fungsi dua dimensi vektor-nilai dalam satah; bentuk kedua menerangkan fungsi bernilai tiga dimensi vektor di ruang angkasa.

Kami sering menggunakan (t ) sebagai parameter kerana (t ) dapat mewakili masa.

Parameter (t ) mungkin terletak di antara dua nombor nyata: (a≤t≤b ), atau nilainya mungkin merangkumi seluruh set nombor nyata.

Setiap fungsi komponen yang membentuk fungsi bernilai vektor mungkin mempunyai sekatan domain yang menerapkan sekatan pada nilai (t ).

Domain fungsi bernilai vektor ( vecs r ) adalah persimpangan domain fungsi komponennya, iaitu, itu adalah sekumpulan semua nilai (t ) yang ditentukan fungsi bernilai vektor .

Contoh ( PageIndex {1} ): Mencari domain fungsi bernilai vektor

Nyatakan domain fungsi bernilai vektor ( vecs r (t) = sqrt {2-t} , hat { mathbf {i}} + ln (t + 3) , hat { mathbf {j}} + e ^ t , hat { mathbf {k}} ).

Penyelesaian

Kami mempertimbangkan domain semula jadi setiap fungsi komponen. Perhatikan bahawa kami menyenaraikan domain di kedua-duanya notasi pembina set dan notasi selang.

Fungsi: Domain:
( begin {array} {ll} sqrt {2-t} & & big {, t , | , t le 2 big } quad text {atau} quad (- infty, 2 besar]
ln (t + 3) & & big {, t , | , t gt -3 big } quad text {atau} quad (-3, infty)
e ^ t & & (- infty, infty) end {array} )

Domain ( vecs r ) adalah persimpangan domain ini, jadi mesti mengandungi semua nilai (t ) yang berfungsi di ketiga-tiganya, tetapi tidak ada nilai (t ) yang tidak berfungsi di mana-mana salah satu fungsi ini.

Oleh itu, domain ( vecs r ) adalah: ( teks {D} _ { vecs r}: besar {, t , | , -3 lt t le 2 besar } ) atau ((-3, 2 besar] ).

Perhatikan bahawa hanya satu bentuk domain ( vecs r ) yang perlu diberikan. Yang pertama, ( besar {, t , | , -3 lt t le 2 besar } ), berada di notasi pembina set, sementara yang kedua, ((-3, 2 besar] ), masuk notasi selang.

Contoh ( PageIndex {2} ): Menilai Fungsi Berharga Vektor dan Menentukan Domain

Untuk setiap fungsi bernilai vektor berikut, nilai ( vecs r (0) ), ( vecs r ( frac { pi} {2}) ), dan ( vecs r ( frac {2 pi} {3}) ). Adakah fungsi ini mempunyai batasan domain?

  1. ( vecs r (t) = 4 cos t , hat { mathbf {i}} + 3 sin t , hat { mathbf {j}} )
  2. ( vecs r (t) = 3 tan t , hat { mathbf {i}} + 4 sec t , hat { mathbf {j}} + 5t , hat { mathbf { k}} )

Penyelesaian

  1. Untuk mengira setiap nilai fungsi, ganti nilai (t ) yang sesuai dengan fungsi:

    start {align *} vecs r (0) ; & = 4 cos (0) hat { mathbf {i}} + 3 sin (0) hat { mathbf {j}} [4pt] & = 4 hat { mathbf {i}} +0 hat { mathbf {j}} = 4 hat { mathbf {i}} [4pt] vecs r kiri ( frac { pi} {2} kanan) ; & = 4 cos kiri ( frac {π} {2} kanan) hat { mathbf {i}} + 3 sin kiri ( frac {π} {2} kanan) topi { mathbf {j}} [4pt] & = 0 hat { mathbf {i}} + 3 hat { mathbf {j}} = 3 hat { mathbf {j}} [4pt] vecs r kiri ( frac {2 pi} {3} kanan) ; & = 4 cos kiri ( frac {2π} {3} kanan) hat { mathbf {i}} + 3 sin kiri ( frac {2π} {3} kanan) topi { mathbf {j}} [4pt] & = 4 kiri (- tfrac {1} {2} kanan) hat { mathbf {i}} + 3 kiri ( tfrac { sqrt {3} } {2} kanan) hat { mathbf {j}} = - 2 hat { mathbf {i}} + tfrac {3 sqrt {3}} {2} hat { mathbf {j} } end {align *}

    Untuk menentukan sama ada fungsi ini mempunyai sekatan domain, pertimbangkan fungsi komponen secara berasingan. Fungsi komponen pertama adalah (f (t) = 4 cos t ) dan fungsi komponen kedua adalah (g (t) = 3 sin t ). Kedua-dua fungsi ini tidak mempunyai batasan domain, jadi domain ( vecs r (t) = 4 cos t , hat { mathbf {i}} + 3 sin t , hat { mathbf { j}} ) adalah semua nombor nyata.
  2. Untuk mengira setiap nilai fungsi, ganti nilai yang sesuai dengan t ke dalam fungsi: [ start {align *} vecs r (0) ; & = 3 tan (0) hat { mathbf {i}} + 4 saat (0) hat { mathbf {j}} + 5 (0) hat { mathbf {k}} [ 4pt] & = 0 hat { mathbf {i}} + 4j + 0 hat { mathbf {k}} = 4 hat { mathbf {j}} [4pt] vecs r kiri ( frac { pi} {2} kanan) ; & = 3 tan kiri ( frac { pi} {2} kanan) hat { mathbf {i}} + 4 saat kiri ( frac { pi} {2} kanan) topi { mathbf {j}} + 5 kiri ( frac { pi} {2} kanan) hat { mathbf {k}}, , teks {yang tidak ada} [4pt] vecs r kiri ( frac {2 pi} {3} kanan) ; & = 3 tan kiri ( frac {2 pi} {3} kanan) hat { mathbf {i}} + 4 saat kiri ( frac {2 pi} {3} kanan) hat { mathbf {j}} + 5 kiri ( frac {2 pi} {3} kanan) hat { mathbf {k}} [4pt] & = 3 (- sqrt {3 }) hat { mathbf {i}} + 4 (−2) hat { mathbf {j}} + frac {10π} {3} hat { mathbf {k}} [4pt] & = (- 3 sqrt {3}) hat { mathbf {i}} - 8 hat { mathbf {j}} + frac {10π} {3} hat { mathbf {k}} akhir {align *} ] Untuk menentukan sama ada fungsi ini mempunyai sekatan domain, pertimbangkan fungsi komponen secara berasingan. Fungsi komponen pertama adalah (f (t) = 3 tan t ), fungsi komponen kedua adalah (g (t) = 4 sec t ), dan fungsi komponen ketiga adalah (h (t) = 5t ). Dua fungsi pertama tidak didefinisikan untuk gandaan ganjil ( frac { pi} {2} ), jadi fungsi tidak ditentukan untuk gandaan ganjil ( frac { pi} {2} ). Oleh itu, [ text {D} _ { vecs r} = Big {t , | , t ≠ frac {(2n + 1) pi} {2} Big }, nonumber ] di mana (n ) adalah bilangan bulat.

Latihan ( PageIndex {1} )

Untuk fungsi bernilai vektor ( vecs r (t) = (t ^ 2−3t) , hat { mathbf {i}} + (4t + 1) , hat { mathbf {j}} ), menilai ( vecs r (0), , vecs r (1) ), dan ( vecs r (−4) ). Adakah fungsi ini mempunyai sekatan domain?

Petunjuk

Ganti nilai (t ) yang sesuai ke dalam fungsi.

Jawapan:

( vecs r (0) = hat { mathbf {j}}, , vecs r (1) = - 2 hat { mathbf {i}} + 5 hat { mathbf {j}} , , vecs r (−4) = 28 hat { mathbf {i}} - 15 hat { mathbf {j}} )

Domain ( vecs r (t) = (t ^ 2−3t) hat { mathbf {i}} + (4t + 1) hat { mathbf {j}} ) adalah semua nombor nyata.

Contoh ( PageIndex {1} ) menggambarkan konsep penting. Domain fungsi bernilai vektor terdiri daripada nombor nyata. Domain boleh berupa semua nombor nyata atau subset nombor nyata. Julat fungsi bernilai vektor terdiri daripada vektor. Setiap nombor nyata dalam domain fungsi bernilai vektor dipetakan ke vektor dua atau tiga dimensi.

Grafik Fungsi Berharga Vektor

Ingat bahawa vektor satah terdiri daripada dua kuantiti: arah dan magnitud. Diberikan sebarang titik di dalam pesawat (the titik awal, jika kita bergerak ke arah tertentu untuk jarak tertentu, kita sampai pada titik kedua. Ini mewakili titik terminal vektor. Kami mengira komponen vektor dengan mengurangkan koordinat titik awal dari koordinat titik terminal.

Vektor dianggap berada di kedudukan standard jika titik awal terletak di tempat asal. Semasa membuat grafik fungsi bernilai vektor, kita biasanya memetakan vektor dalam domain fungsi dalam kedudukan standard, kerana melakukannya menjamin keunikan grafik. Konvensyen ini berlaku untuk grafik fungsi bernilai vektor tiga dimensi juga. Grafik fungsi nilai vektor borang

[ vecs r (t) = f (t) , hat { mathbf {i}} + g (t) , hat { mathbf {j}} bukan nombor ]

terdiri daripada set semua titik ((f (t), , g (t)) ), dan jalan yang dilacaknya disebut keluk satah. Grafik fungsi nilai vektor borang

[ vecs r (t) = f (t) , hat { mathbf {i}} + g (t) , hat { mathbf {j}} + h (t) , hat { mathbf {k}} bukan nombor ]

terdiri daripada set semua titik ((f (t), , g (t), , h (t)) ), dan jalan yang dilacaknya disebut keluk ruang. Sebarang perwakilan lengkung satah atau lengkung ruang menggunakan fungsi bernilai vektor disebut a parameterisasi vektor keluk.

Setiap lengkung satah dan keluk ruang mempunyai orientasi, ditunjukkan oleh anak panah yang ditarik pada lengkung, yang menunjukkan arah gerakan di sepanjang lekukan ketika nilai parameter (t ) meningkat.

Contoh ( PageIndex {3} ): Membuat Grafik Fungsi Berharga Vektor

Buat grafik setiap fungsi bernilai vektor berikut:

  1. Keluk satah yang ditunjukkan oleh ( vecs r (t) = 4 cos t , hat { mathbf {i}} + 3 sin t , hat { mathbf {j}} ), ( 0≤t≤2 pi )
  2. Keluk satah yang diwakili oleh ( vecs r (t) = 4 cos (t ^ 3) , hat { mathbf {i}} + 3 sin (t ^ 3) , hat { mathbf { j}} ), (0≤t≤ sqrt [3] {2 pi} )
  3. Keluk ruang yang diwakili oleh ( vecs r (t) = 4 cos t , hat { mathbf {i}} + 4 sin t , hat { mathbf {j}} + t , topi { mathbf {k}} ), (0≤t≤4 pi )

Penyelesaian

1. Seperti mana-mana graf, kita mulakan dengan jadual nilai. Kami kemudian membuat grafik setiap vektor di lajur kedua jadual dalam kedudukan standard dan menyambungkan titik terminal setiap vektor untuk membentuk lengkung (Gambar ( PageIndex {1} )). Lengkung ini ternyata menjadi elips yang berpusat pada asal.

Jadual ( PageIndex {1} ): Jadual Nilai untuk ( vecs r (t) = 4 cos t , hat { mathbf {i}} + 3 sin t , hat { mathbf {j}} ), (0≤t≤2 pi )
(t ) ( vecs r (t) ) (t ) ( vecs r (t) )
(0) (4 hat { mathbf {i}} ) ( pi ) (- 4 topi { mathbf {i}} )
( dfrac { pi} {4} ) (2 sqrt {2} hat { mathbf {i}} + frac {3 sqrt {2}} {2} hat { mathbf {j}} ) ( dfrac {5 pi} {4} ) (- 2 sqrt {2} hat { mathbf {i}} - frac {3 sqrt {2}} {2} hat { mathbf {j}} )
( dfrac { pi} {2} ) ( mathrm {3 hat { mathbf {j}}} ) ( dfrac {3 pi} {2} ) ( mathrm {-3 hat { mathbf {j}}} )
( dfrac {3 pi} {4} ) (-2 sqrt {2} hat { mathbf {i}} + frac {3 sqrt {2}} {2} hat { mathbf {j}} ) ( dfrac {7 pi} {4} ) (2 sqrt {2} hat { mathbf {i}} - frac {3 sqrt {2}} {2} hat { mathbf {j}} )
(2 pi ) (4 hat { mathbf {i}} )

2. Jadual nilai untuk ( vecs r (t) = 4 cos (t ^ 3) , hat { mathbf {i}} + 3 sin (t ^ 3) , hat { mathbf {j}} ), (0≤t≤ sqrt [3] {2 pi} ) adalah seperti berikut:

Jadual Nilai untuk ( vecs r (t) = 4 cos (t ^ 3) , hat { mathbf {i}} + 3 sin (t ^ 3) , hat { mathbf {j}} ) , (0≤t≤ sqrt [3] {2 pi} )
(t ) ( vecs r (t) ) (t ) ( vecs r (t) )
(0) ( mathrm {4 hat { mathbf {i}}} ) ( displaystyle sqrt [3] { pi} ) ( mathrm {-4 hat { mathbf {i}}} )
( displaystyle sqrt [3] { dfrac { pi} {4}} ) ( mathrm {2 sqrt {2} hat { mathbf {i}} + frac {3 sqrt {2}} {2} hat { mathbf {j}}} ) ( displaystyle sqrt [3] { dfrac {5 pi} {4}} ) ( mathrm {-2 sqrt {2} hat { mathbf {i}} - frac {3 sqrt {2}} {2} hat { mathbf {j}}} )
( displaystyle sqrt [3] { dfrac { pi} {2}} ) ( mathrm {3 hat { mathbf {j}}} ) ( displaystyle sqrt [3] { dfrac {3 pi} {2}} ) ( mathrm {-3 hat { mathbf {j}}} )
( displaystyle sqrt [3] { dfrac {3 pi} {4}} ) ( mathrm {-2 sqrt {2} hat { mathbf {i}} + frac {3 sqrt {2}} {2} hat { mathbf {j}}} ) ( displaystyle sqrt [3] { dfrac {7 pi} {4}} ) ( mathrm {2 sqrt {2} hat { mathbf {i}} - frac {3 sqrt {2}} {2} hat { mathbf {j}}} )
( displaystyle sqrt [3] {2 pi} ) ( mathrm {4 hat { mathbf {i}}} )

Grafik lengkung ini juga elips berpusat pada asal.

3. Kami menjalani prosedur yang sama untuk fungsi vektor tiga dimensi.

Jadual Nilai untuk ( mathrm {r (t) = 4 cos t hat { mathbf {i}} + 4 sin t hat { mathbf {j}} + t hat { mathbf {k }}} ), ( mathrm {0≤t≤4 pi} )
(t ) ( vecs r (t) ) (t ) ( vecs r (t) )
( mathrm {0} ) ( mathrm {4 hat { mathbf {i}}} ) ( mathrm { pi} ) ( mathrm {-4 hat { mathbf {i}}} + pi hat { mathbf {k}} )
( dfrac { pi} {4} ) ( mathrm {2 sqrt {2} hat { mathbf {i}} + 2 sqrt {2} hat { mathbf {j}} + frac { pi} {4} hat { mathbf {k}}} ) ( dfrac {5 pi} {4} ) ( mathrm {-2 sqrt {2} hat { mathbf {i}} - 2 sqrt {2} hat { mathbf {j}} + frac {5 pi} {4} hat { mathbf {k}}} )
( dfrac { pi} {2} ) ( mathrm {4 hat { mathbf {j}} + frac { pi} {2} hat { mathbf {k}}} ) ( dfrac {3 pi} {2} ) ( mathrm {-4 hat { mathbf {j}} + frac {3 pi} {2} hat { mathbf {k}}} )
( dfrac {3 pi} {4} ) ( mathrm {-2 sqrt {2} hat { mathbf {i}} + 2 sqrt {2} hat { mathbf {j}} + frac {3 pi} {4} hat { mathbf {k}}} ) ( dfrac {7 pi} {4} ) ( mathrm {2 sqrt {2} hat { mathbf {i}} - 2 sqrt {2} hat { mathbf {j}} + frac {7 pi} {4} hat { mathbf {k}}} )
( mathrm {2 pi} ) ( mathrm {4 hat { mathbf {j}} + 2 pi hat { mathbf {k}}} )

Nilai kemudian diulang, kecuali fakta bahawa pekali ( hat { mathbf {k}} ) sentiasa meningkat ( ( PageIndex {3} )). Lengkung ini dipanggil heliks. Perhatikan bahawa jika komponen ( hat { mathbf {k}} ) dihapuskan, maka fungsi tersebut menjadi ( vecs r (t) = 4 cos t hat { mathbf {i}} + 4 sin t hat { mathbf {j}} ), yang merupakan bulatan jejari 4 yang berpusat pada asal.

Anda mungkin menyedari bahawa grafik di bahagian a. dan b. sama. Ini berlaku kerana fungsi menggambarkan lengkung b adalah apa yang disebut reparameterization fungsi yang menggambarkan lengkung a. Sebenarnya, mana-mana lengkung mempunyai bilangan reparameterisasi yang tidak terhingga; sebagai contoh, kita boleh mengganti (t ) dengan (2t ) di salah satu daripada tiga lengkung sebelumnya tanpa mengubah bentuk lengkung. Selang yang ditentukan (t ) mungkin berubah, tetapi itu sahaja. Kami kembali kepada idea ini kemudian dalam bab ini ketika kita mengkaji parameterisasi panjang lengkok. Seperti disebutkan, nama bentuk lengkung grafik di ( PageIndex {3} ) adalah heliks. Lengkung menyerupai mata air, dengan keratan rentas bulat menghadap ke bawah di sepanjang paksi (z ). Ada kemungkinan heliks juga berbentuk elips dalam penampang. Contohnya, fungsi bernilai vektor ( vecs r (t) = 4 cos t , hat { mathbf {i}} + 3 sin t , hat { mathbf {j}} + t , hat { mathbf {k}} ) menerangkan heliks elips. Unjuran heliks ini ke dalam satah (xy ) adalah elips. Terakhir, anak panah dalam grafik heliks ini menunjukkan orientasi lengkung ketika (t ) maju dari (0 ) ke (4π ).

Latihan ( PageIndex {2} )

Buat grafik fungsi bernilai vektor ( vecs r (t) = t , hat { mathbf {i}} + t ^ 3 , hat { mathbf {j}} ).

Petunjuk

Mulakan dengan membuat jadual nilai, kemudian graf titik yang ditunjukkan oleh vektor untuk setiap nilai (t ).

Jawapan

Pada ketika ini, anda mungkin dapat melihat persamaan antara fungsi bernilai vektor dan lengkung parameter. Sesungguhnya, diberikan fungsi bernilai vektor ( vecs r (t) = f (t) , hat { mathbf {i}} + g (t) , hat { mathbf {j}} ) kita boleh menentukan (x = f (t) ) dan (y = g (t) ). Grafik fungsi parameter kemudian akan setuju dengan grafik fungsi bernilai vektor, kecuali bahawa graf fungsi vektor-nilai akan dikesan oleh vektor dan bukan hanya sekumpulan titik. Oleh kerana kita dapat membuat parameter lengkung yang didefinisikan oleh fungsi (y = f (x) ), dimungkinkan juga untuk mewakili lengkung satah sewenang-wenangnya oleh fungsi bernilai vektor.

Mencari Fungsi Berharga Vektor untuk Mengesan Graf Fungsi (y = f (x) )

Seperti yang anda lihat dalam contoh di atas, fungsi bernilai vektor mengesan lengkung di satah atau di angkasa. Bagaimana jika kita ingin menulis fungsi bernilai vektor yang mengesan grafik lengkung tertentu dalam satah (xy ) -?

Graf fungsi apa yang dikesan oleh fungsi bernilai vektor dalam Latihan ( PageIndex {2} ) di atas: ( vecs r (t) = t , hat { mathbf {i}} + t ^ 3 , hat { mathbf {j}} )? Nampaknya grafik (y = x ^ 3 ), bukan?

Mengingat apa yang baru saja dinyatakan mengenai komponen fungsi bernilai vektor yang sesuai dengan persamaan parametik dari lengkung parameter, kita melihat bahawa di sini kita mempunyai:

[ start {align *} x & = t y & = t ^ 3 end {align *} bukan nombor ]

Oleh kerana (x = t ), kita dapat mengganti (t ) dalam persamaan (y = t ^ 3 ) dengan (x ), memberi kita fungsi: (y = x ^ 3 ) .

Oleh itu, kami betul dalam tekaan kami.

Bagaimana kita dapat menulis fungsi bernilai vektor untuk menelusuri grafik fungsi, (y = f (x) )?

Terdapat dua orientasi yang perlu dipertimbangkan: kiri ke kanan dan kanan ke kiri.

Mengesan Fungsi dari Kiri ke Kanan:

Untuk mengesan grafik (y = f (x) ) dari kiri ke kanan, gunakan: ( vecs r (t) = t , hat { mathbf {i}} + f (t ) , hat { mathbf {j}} )

Perhatikan bahawa yang penting di sini adalah menjadikan komponen (x ) menjadi fungsi yang semakin meningkat. Sebarang fungsi yang meningkat akan berfungsi. Kita boleh menggunakan (x = t ^ 3 ), misalnya. Tetapi kita perlu ingat untuk mengganti (x ) dalam fungsi (f (x) ) dengan ungkapan ini (t ^ 3 ), memberi kita (y = f (t ^ 3) ) . Ini bermaksud fungsi (y = f (x) ) juga dapat diukur dari kiri ke kanan oleh fungsi bernilai vektor: ( vecs r (t) = t ^ 3 , hat { mathbf {i}} + f (t ^ 3) , hat { mathbf {j}} )

Mengesan Fungsi dari Kanan ke Kiri:

Untuk mengesan grafik (y = f (x) ) dari kanan ke kiri, gunakan: ( vecs r (t) = -t , hat { mathbf {i}} + f ( -t) , hat { mathbf {j}} )

Sekali lagi perhatikan bahawa kita dapat menggunakan fungsi penurunan (t ) untuk komponen (x ) dan memperoleh fungsi bernilai vektor yang menelusuri grafik (y = f (x) ) dari kanan ke -ditinggalkan. Menggunakan (x = -t ) adalah fungsi penurunan paling mudah yang boleh kita pilih.

Contoh ( PageIndex {4} ): Mencari fungsi bernilai vektor untuk mengesan grafik (y = f (x) )

Tentukan fungsi bernilai vektor yang akan mengesan grafik (y = cos x ) dari kiri ke kanan, dan yang lain untuk mengesannya dari kanan ke kiri.

Penyelesaian

Kiri-ke-kanan: ( vecs r (t) = t , hat { mathbf {i}} + cos t , hat { mathbf {j}} )

Kanan ke kiri: ( vecs r (t) = -t , hat { mathbf {i}} + cos (-t) , hat { mathbf {j}} )

Mencari Fungsi Berharga Vektor untuk Mengesan Graf Persamaan di (x ) dan (y ) dan Naib Versa

Bagaimana jika kita ingin mencari fungsi bernilai vektor untuk mengesan grafik bulatan, elips, atau hiperbola, berdasarkan persamaan tersiratnya?

Perhatikan bahawa dalam Contoh ( PageIndex {3} ), fungsi bernilai vektor ( vecs r (t) = 4 cos t , hat { mathbf {i}} + 3 sin t , hat { mathbf {j}} ) mengesan graf elips ( frac {x ^ 2} {16} + frac {y ^ 2} {9} = 1 ).

Dalam fungsi bernilai vektor ini kita melihat bahawa: [x = 4 cos t quad text {dan} quad y = 3 sin t ]

Apa yang kita perlukan sekarang adalah cara untuk mengubahnya menjadi persamaan tersirat yang melibatkan (x ) dan (y ). Untuk mencapai ini, ingatlah identiti Pythagoras, [ cos ^ 2 t + sin ^ 2 t = 1 ].

Sekarang yang perlu kita lakukan ialah menyelesaikan persamaan di atas untuk ( cos t ) dan ( sin t ) dan kita boleh mengganti identiti ini untuk mendapatkan persamaan di (x ) dan (y ) .

Oleh itu: [ cos t = frac {x} {4} quad text {dan} quad sin t = frac {y} {3} ]

Mengganti identiti memberi kita: [ kiri ( frac {x} {4} kanan) ^ 2 + kiri ( frac {y} {3} kanan) ^ 2 = 1 ]

Menyederhanakan persamaan tersirat ini memberi kita persamaan tersirat elips dalam Contoh ( PageIndex {3} ) yang kita tulis di atas:

[ frac {x ^ 2} {16} + frac {y ^ 2} {9} = 1 ]

Untuk pergi ke arah lain dan mencari fungsi bernilai vektor yang mengesan elips, kita hanya perlu mengambil langkah-langkah ini ke arah yang berlawanan!

Contoh ( PageIndex {5} ): Menulis fungsi bernilai vektor untuk bulatan, elips, atau Hyperbola tertentu

Tulis fungsi bernilai vektor yang mengesan setiap lengkung tersirat berikut:

a. Elips: ( frac {x ^ 2} {16} + frac {y ^ 2} {9} = 1 )

b. Bulatan: (x ^ 2 + y ^ 2 = 4 )

c. Hiperbola: ( frac {x ^ 2} {25} - frac {y ^ 2} {16} = 1 )

Penyelesaian

a. Mari kita gunakan proses yang ditunjukkan di atas secara terbalik. Pertama, mari tulis semula persamaan tersirat sehingga menunjukkan jumlah kuantiti kuadrat sama dengan satu.

[ kiri ( frac {x} {4} kanan) ^ 2 + kiri ( frac {y} {3} kanan) ^ 2 = 1 angka ]

Sekarang kita memerlukan identiti yang kita gunakan di atas, ( cos ^ 2 t + sin ^ 2 t = 1 ).

Menyamakan bahagian yang kuasa dua (perhatikan sebenarnya kita mempunyai pilihan di sini untuk membuat ( cos t ) dan mana yang harus dibuat ( sin t )), kita mendapat:

[ frac {x} {4} = cos t quad text {dan} quad frac {y} {3} = sin t nonumber ]

Sekarang kita hanya perlu menyelesaikan untuk (x ) dan (y ).

[x = 4 cos t quad text {dan} quad y = 3 sin t nonumber ]

Kita sekarang boleh menulis fungsi bernilai vektor yang mengesan elips ini: ( vecs r (t) = 4 cos t , hat { mathbf {i}} + 3 sin t , hat { mathbf {j}} )

Perhatikan bahawa kita juga boleh menulis ( vecs r (t) = 4 sin t , hat { mathbf {i}} + 3 cos t , hat { mathbf {j}} ), kerana kita boleh memilih untuk menukar ( sin t ) dan ( cos t ) di atas. Ia akan mengesan elips yang sama, tetapi dengan arah yang berlawanan.

b. Menjejaki bulatan cukup mudah, tidak memerlukan proses yang kami tunjukkan di atas, walaupun pada mulanya masih bermanfaat. Ingat bahawa semua vektor pada bulatan unit dapat ditunjukkan dalam bentuk: ( vecs v = cos theta , hat { mathbf {i}} + sin theta , hat { mathbf { j}} ).

Oleh itu, fungsi bernilai vektor, ( vecs r (t) = cos t , hat { mathbf {i}} + sin t , hat { mathbf {j}} ), akan mengesan keluar bulatan unit dengan persamaan, (x ^ 2 + y ^ 2 = 1 ).

Untuk mendapatkan bulatan jejari (2 ) yang berpusat pada asal (yang merupakan grafik (x ^ 2 + y ^ 2 = 4 )), kita hanya perlu mengalikan fungsi vektor ini dengan skalar faktor (2 ).

Oleh itu, fungsi bernilai vektor yang akan mengesan lingkaran ini adalah: ( vecs r (t) = 2 cos t , hat { mathbf {i}} + 2 sin t , hat { mathbf {j}} ).

Perhatikan lagi bahawa kemungkinan lain adalah: ( vecs r (t) = 2 sin t , hat { mathbf {i}} + 2 cos t , hat { mathbf {j}} ). Ini akan mengesan lingkaran yang sama, tetapi dengan arah yang berlawanan.

Untuk menggunakan teknik di atas, anda mulakan dengan membahagikan setiap istilah dalam persamaan dengan segiempat jejari, di sini 4, sehingga meletakkan persamaan bulatan dalam "bentuk elips". Langkah selebihnya mengikut corak yang ditunjukkan di bahagian a.

c. Untuk mengesan hiperbola bentuk ( frac {x ^ 2} {a ^ 2} - frac {y ^ 2} {b ^ 2} = 1 ) atau ( frac {y ^ 2} { a ^ 2} - frac {x ^ 2} {b ^ 2} = 1 ), kita perlu mencari identiti trigonometri yang menunjukkan perbezaan dua kotak sama dengan 1. Sekiranya anda belum mempunyai identiti seperti itu, kita dapat memperolehnya dari identiti Pythagoras yang digunakan di atas. Itu dia,

[ cos ^ 2 t + sin ^ 2 t = 1 bukan nombor ]

Membahagi setiap istilah dengan ( cos ^ 2 t )

[ frac { cos ^ 2 t} { cos ^ 2 t} + frac { sin ^ 2 t} { cos ^ 2 t} = frac {1} { cos ^ 2 t} angka ]

hasil,

[1+ tan ^ 2 t = sec ^ 2 t bukan nombor ]

Menulis semula persamaan ini memberi kita identiti yang kita perlukan:

[ sec ^ 2 t - tan ^ 2 t = 1 bukan nombor ]

Sekarang, persamaan hiperbola ini adalah:

[ frac {x ^ 2} {25} - frac {y ^ 2} {16} = 1 bukan nombor ]

Tulis semula sebelah kiri untuk menunjukkan kuantiti yang kuasa dua:

[ kiri ( frac {x} {5} kanan) ^ 2 - kiri ( frac {y} {4} kanan) ^ 2 = 1 angka ]

Kita kemudian dapat menyamakan ungkapan kuadrat dengan istilah yang sesuai:

[ frac {x} {5} = sec t quad text {dan} quad frac {y} {4} = tan t bukan nombor ]

Selesaikan untuk (x ) dan (y ), kami mempunyai:

[x = 5 sec t quad text {dan} quad y = 4 tan t nonumber ]

Jadi fungsi bernilai vektor yang akan mengesan hiperbola [ frac {x ^ 2} {25} - frac {y ^ 2} {16} = 1 ] adalah [ vecs r (t) = 5 sec t , hat { mathbf {i}} + 4 tan t , hat { mathbf {j}}. nombor ]

Parameterisasi Jalan Piecewise

Ada kalanya perlu untuk membuat parameter jalan yang terdiri daripada potongan lengkung yang berbeza. Laluan sepotong ini mungkin terbuka atau membentuk sempadan kawasan tertutup seperti contoh yang ditunjukkan dalam Rajah ( PageIndex {4} ). Selain menentukan fungsi bernilai vektor untuk mengesan setiap potongan secara terpisah, dengan orientasi yang ditunjukkan, kita juga perlu menentukan rentang nilai yang sesuai untuk parameter (t ).

Perhatikan bahawa ada banyak cara untuk membuat parameter satu bahagian, jadi ada banyak cara yang betul untuk memaramatisasi jalan dengan cara ini.

Contoh ( PageIndex {6} ): Parameterisasi jalan sepotong

Tentukan parameterisasi sekeping jalan yang ditunjukkan dalam Rajah ( PageIndex {4} ), bermula dengan (t = 0 ) dan teruskan setiap bahagian.

Penyelesaian

Tugas pertama kami adalah mengenal pasti tiga kepingan di jalan sepotong ini.

Perhatikan bagaimana kami melabelnya secara berurutan sebagai ( vecs r_1 ), ( vecs r_2 ), dan ( vecs r_3 ). Sekarang kita perlu mengenal pasti fungsi untuk masing-masing dan menulis fungsi bernilai vektor yang sesuai dengan orientasi yang betul (kiri-ke-kanan atau kanan-ke-kiri).

Menentukan ( vecs r_1 ): Persamaan fungsi linear dalam bahagian ini ialah (y = x ).

Oleh kerana berorientasikan dari kiri ke kanan antara (t = 1 ) dan (t = 4 ), kita dapat menulis:

[ vecs r_ {1a} (t) = t , hat { mathbf {i}} + t , hat { mathbf {j}} quad text {for} quad 1 le t le 4 bukan nombor ]

Sekiranya kita ingin memulakan karya ini di (t = 0 ), kita hanya perlu menggeser nilai (t ) satu unit ke kiri. Salah satu cara untuk melakukannya adalah dengan menulis ( vecs r_ {1a} ) dengan sebutan (t_1 ) dan bukan (t ) untuk menjadikan terjemahan lebih mudah dilihat.

Oleh itu, kami mempunyai ( vecs r_ {1a} (t_1) = t_1 , hat { mathbf {i}} + t_1 , hat { mathbf {j}} ) untuk (1 le t_1 le 4 ).

Rajah ( PageIndex {4} ): Jalan sepotong tertutup

Dengan mengurangkan (1 ) dari setiap bahagian julat nilai parameter ini, kita mempunyai: (0 le t_1 - 1 le 3 ).

Sekarang kita membiarkan (t = t_1 - 1 ). Selesaikan untuk (t_1 ), kami memperoleh: (t_1 = t + 1 ).

Menggantikan (t_1 ) dengan ungkapan (t + 1 ) secara efektif akan mengubah julat nilai parameter satu unit ke kiri.

Oleh itu, bermula dengan (t = 0 ), kita mempunyai: [ vecs r_1 (t) = (t + 1) , hat { mathbf {i}} + (t + 1) , hat { mathbf {j}} quad text {for} quad 0 le t le 3 nonumber ]

Periksa semula bahawa fungsi bernilai vektor ini akan mengesan segmen ini ke arah yang betul sebelum pergi ke (r_2 ).

Menentukan ( vecs r_2 ): Bahagian ini mempunyai label yang menunjukkan fungsi yang grafnya ditelusuri. Sekiranya ia berorientasi dari kiri ke kanan, kita akan mempunyai:

[ text {Kiri-ke-kanan:} quad vecs r_ {2a} (t) = t , hat { mathbf {i}} + kiri (2 sqrt { frac {4-t } {3}} + 4 kanan) , hat { mathbf {j}} quad text {for} quad 1 le t le 4 nonumber ]

Tetapi kerana kita memerlukannya berorientasi dari kanan ke kiri, kita perlu mengganti (t ) dengan (- t ) dalam fungsi dan kita perlu membahagikan jurang ketaksamaan dengan -1 untuk mendapatkan yang sesuai julat. Oleh itu, kami memperoleh:

[ vecs r_ {2b} (t) = -t , hat { mathbf {i}} + kiri (2 sqrt { frac {4 - (- t)} {3}} + 4 kanan) , hat { mathbf {j}} quad text {for} quad -4 le t le -1 nonumber ]

Pastikan ia berfungsi!

Sekarang kami ingin agar karya ini bermula pada (t = 3 ) sejurus selepas yang pertama selesai. Sekali lagi mari kita menjadikannya lebih mudah dilihat dengan menulis (r_ {2b} ) dari segi (t_2 ).

[ vecs r_ {2b} (t_2) = -t_2 , hat { mathbf {i}} + kiri (2 sqrt { frac {4 - (- t_2)} {3}} + 4 kanan) , hat { mathbf {j}} quad text {for} quad -4 le t_2 le -1 nonumber ]

Untuk memaksa (r_2 ) untuk memulakan dengan (t = 3 ) dan bukan (t = -4 ), kita perlu menambahkan (7 ) pada setiap bahagian ketaksamaan. Ini menghasilkan: (3 le t_2 + 7 le 6 ).

Mari (t = t_2 + 7 ). Kemudian menyelesaikan untuk (- t_2 ) (kerana inilah yang perlu kita ganti di (r_ {2b} )), kita mempunyai: (t_2 = 7-t ).

Menggantikan (- t_2 ) dengan ( kiri (7-t kanan) ) di ( vecs r_ {2b} ), kami memperoleh:

[ vecs r_ {2} (t) = (7-t) , hat { mathbf {i}} + kiri (2 sqrt { frac {4- (7-t)} {3} } +4 kanan) , hat { mathbf {j}} quad text {for} quad 3 le t le 6 nonumber ]

Ini boleh digabungkan dengan hasil awal kami untuk (r_1 ) untuk menulis fungsi vektor-nilai yang ditentukan mengikut kepingan yang mengesan dua bahagian pertama, bermula pada (t = 0 ):

[ vecs r (t) = mulakan {kes}
(t + 1) , hat { mathbf {i}} + (t + 1) , hat { mathbf {j}}, & 0 le t le 3
(7-t) , hat { mathbf {i}} + kiri (2 sqrt { frac {t - 3} {3}} + 4 kanan) , hat { mathbf {j} }, & 3 lt le 6
end {case} bukan nombor ]

Perhatikan bahawa satu modifikasi kecil dilakukan pada julat kedua sehingga ketika (t = 3 ), tidak ada kekeliruan tentang bahagian mana yang akan dinilai.

Menentukan ( vecs r_3 ): Untuk menentukan bahagian terakhir ini, kita perlu berfikir sedikit berbeza. Ini kerana ia adalah segmen menegak, yang tidak dapat ditunjukkan dengan fungsi bentuk, (y = f (x) ). Perhatikan bahawa ia dapat ditunjukkan oleh fungsi bentuk (x = f (y) ). Membiarkan (y = t ), kita dapat menulis (x = f (t) ) dan menulis parameterisasi dalam meningkatkan nilai (y ) (bawah-ke-atas), kita akan mendapat: ( vecs r (t) = f (t) , hat { mathbf {i}} + t , hat { mathbf {j}} ).

Persamaan garis ini ialah (x = 1 ). Oleh itu, jika kita ingin membuat parameter segmen ini dengan orientasi ke atas (peningkatan nilai (y )), kita mempunyai:

[ vecs r_ {3a} (t) = 1 , hat { mathbf {i}} + t , hat { mathbf {j}} quad text {for} quad 1 le t le 6 bukan nombor ]

Tetapi kerana kita ingin menggunakan orientasi ke bawah (penurunan nilai (y )), kita perlu menggunakan fungsi penurunan (t ) untuk (y ). Seperti sebelumnya, kes yang paling mudah adalah menggunakan (y = -t ). Kemudian, dalam kes umum, kita akan mengesan fungsi (x = f (y) ) dalam arah ke bawah dengan ( vecs r (t) = f (-t) , hat { mathbf { i}} - t , hat { mathbf {j}} ).

Sekiranya (r_3 ), ini memberi kita:

[ vecs r_ {3b} (t) = 1 , hat { mathbf {i}} - t , hat { mathbf {j}} quad text {for} quad -6 le t le -1 bukan nombor ]

Perhatikan bahawa kerana (x = 1, , f (-t) = 1 ), iaitu, ia tidak mengubah komponen pertama kerana ia tetap dan bukan fungsi variabel parameter (t ).

Juga perhatikan bahawa sejak kita meniadakan (t ), kita juga harus meniadakan julatnya, membaginya dengan (- 1 ).

Seperti di atas, untuk memudahkan terjemahan, kami akan mengganti (t ) dengan (t_3 ), memberi kami:

[ vecs r_ {3b} (t) = 1 , hat { mathbf {i}} - t_3 , hat { mathbf {j}} quad text {for} quad -6 le t_3 le -1 bukan nombor ]

Sekarang, kami ingin bahagian terakhir ini bermula di (t = 6 ) di mana bahagian kedua yang kami bentuk di atas berhenti. Kita melihat bahawa kita perlu menambahkan (12 ) pada julat paramater (t ) untuk mencapainya, memberi kita julat baru (6 le t_3 + 12 le 11 ).

Mari (t = t_3 + 12 ). Kemudian menyelesaikan untuk (- t_3 ) (kerana inilah yang perlu kita ganti di (r_ {3b} )), kita mempunyai: (t_3 = 12-t ).

Menggantikan (- t_3 ) dengan ( kiri (12-t kanan) ) di ( vecs r_ {3b} ), kami memperoleh:

[ vecs r_ {3} (t) = 1 , hat { mathbf {i}} + (12 - t) , hat { mathbf {j}} quad text {for} quad 6 le t_3 le 11 bukan nombor ]

Periksa bahawa ini masih mengesan segmen menegak ini dari atas ke bawah.

Kita sekarang boleh menyatakan jawapan terakhir sebagai fungsi bernilai vektor tunggal yang ditentukan secara berturut-turut yang mengesan keseluruhan jalan ini, bermula ketika (t = 0 ).

[ vecs r (t) = mulakan {kes}
(t + 1) , hat { mathbf {i}} + (t + 1) , hat { mathbf {j}}, & 0 le t le 3
(7-t) , hat { mathbf {i}} + kiri (2 sqrt { frac {t - 3} {3}} + 4 kanan) , hat { mathbf {j} }, & 3 lt le 6
1 , hat { mathbf {i}} + (12 - t) , hat { mathbf {j}} & 6 lt t_3 le 11
end {case} bukan nombor ]

Pastikan anda mengesahkan bahawa fungsi bernilai vektor tunggal ini benar-benar mengesan keseluruhan jalan!

Had dan Kesinambungan Fungsi Berharga Vektor

Kami sekarang melihat had fungsi bernilai vektor. Ini penting untuk difahami untuk mengkaji kalkulus fungsi bernilai vektor.

Definisi: had fungsi bernilai vektor

Fungsi bernilai vektor ( vecs r ) menghampiri had ( vecs L ) ketika (t ) menghampiri (a ), bertulis

[ lim limit_ {t to a} vecs r (t) = vecs L, ]

disediakan

[ lim had_ {t ke a} besar | vecs r (t) - vecs L besar | = 0. ]

Dalam contoh berikut, kami menunjukkan cara mengira had fungsi bernilai vektor.

Contoh ( PageIndex {7} ): Menilai Had Fungsi Berfungsi Vektor

Untuk setiap fungsi bernilai vektor berikut, hitung ( lim limit_ {t to 3} vecs r (t) ) untuk

  1. ( vecs r (t) = (t ^ 2−3t + 4) hat { mathbf {i}} + (4t + 3) hat { mathbf {j}} )
  2. ( vecs r (t) = frac {2t − 4} {t + 1} hat { mathbf {i}} + frac {t} {t ^ 2 + 1} hat { mathbf {j }} + (4t − 3) hat { mathbf {k}} )

Penyelesaian

  1. Gunakan Persamaan ref {Th1} dan gantikan nilai (t = 3 ) ke dalam dua ungkapan komponen:

[ egin{align*} lim limits_{t o 3} vecs r(t) ; & = lim limits_{t o 3} left[(t^2−3t+4) hat{mathbf{i}} + (4t+3) hat{mathbf{j}} ight] [5pt] & = left[lim limits_{t o 3} (t^2−3t+4) ight]hat{mathbf{i}}+left[lim limits_{t o 3} (4t+3) ight] hat{mathbf{j}} [5pt] & = 4 hat{mathbf{i}}+15 hat{mathbf{j}} end{align*}]

  1. Use Equation ef{Th2} and substitute the value (t=3) into the three component expressions:

[ egin{align*} lim limits_{t o 3} vecs r(t) ; & = lim limits_{t o 3}left(dfrac{2t−4}{t+1}hat{mathbf{i}}+dfrac{t}{t^2+1}hat{mathbf{j}}+(4t−3) hat{mathbf{k}} ight) [5pt] & = left[lim limits_{t o 3} left(dfrac{2t−4}{t+1} ight) ight]hat{mathbf{i}}+left[lim limits_{t o 3} left(dfrac{t}{t^2+1} ight) ight] hat{mathbf{j}} +left[lim limits_{t o 3} (4t−3) ight] hat{mathbf{k}} [5pt] & = frac{1}{2} hat{mathbf{i}}+ frac{3}{10}hat{mathbf{j}}+9 hat{mathbf{k}} end{align*}]

Latihan ( PageIndex {3} )

Calculate (lim limits_{t o 2} vecs r(t)) for the function (vecs r(t) = sqrt{t^2 + 3t - 1},hat{mathbf{i}}−(4t-3),hat{mathbf{j}}− sin frac{(t+1)pi}{2},hat{mathbf{k}})

Petunjuk

Use Equation ef{Th2} from the preceding theorem.

Jawapan

[lim limits_{t o 2} vecs r(t) = 3hat{mathbf{i}}−5hat{mathbf{j}}+hat{mathbf{k}}]

Now that we know how to calculate the limit of a vector-valued function, we can define continuity at a point for such a function.

Definitions

Let (f), (g), and (h) be functions of (t). Then, the vector-valued function (vecs r(t)=f(t) hat{mathbf{i}}+g(t)hat{mathbf{j}}) is continuous at point (t=a) if the following three conditions hold:

  1. (vecs r(a)) exists
  2. (lim limits_{t o a} vecs r(t)) exists
  3. (lim limits_{t o a} vecs r(t) = vecs r(a))

Similarly, the vector-valued function (vecs r(t)=f(t) hat{mathbf{i}}+g(t)hat{mathbf{j}}+h(t)hat{mathbf{k}}) is continuous at point (t=a) if the following three conditions hold:

  1. (vecs r(a)) exists
  2. (lim limits_{t o a} vecs r(t)) exists
  3. (lim limits_{t o a} vecs r(t) = vecs r(a))

Ringkasan

  • A vector-valued function is a function of the form (vecs r(t)=f(t) hat{mathbf{i}}+ g(t) hat{mathbf{j}}) or (vecs r(t)=f(t) hat{mathbf{i}}+g(t) hat{mathbf{j}}+h(t) hat{mathbf{k}}), where the component functions (f), (g), and (h) are real-valued functions of the parameter (t).
  • The graph of a vector-valued function of the form (vecs r(t)=f(t) hat{mathbf{i}}+g(t) hat{mathbf{j}}) is called a plane curve. The graph of a vector-valued function of the form (vecs r(t)=f(t)hat{mathbf{i}}+g(t)hat{mathbf{j}}+h(t) hat{mathbf{k}}) is called a space curve.
  • It is possible to represent an arbitrary plane curve by a vector-valued function.
  • To calculate the limit of a vector-valued function, calculate the limits of the component functions separately.

Key Equations

  • Vector-valued function
    (vecs r(t)=f(t) hat{mathbf{i}}+g(t) hat{mathbf{j}}) or (vecs r(t)=f(t) hat{mathbf{i}}+g(t) hat{mathbf{j}}+h(t) hat{mathbf{k}}),or (vecs r(t)=⟨f(t),g(t)⟩) or (vecs r(t)=⟨f(t),g(t),h(t)⟩)

  • Limit of a vector-valued function
    (lim limits_{t o a} vecs r(t) = [lim limits_{t o a} f(t)] hat{mathbf{i}} + [lim limits_{t o a} g(t)] hat{mathbf{j}}) or (lim limits_{t o a} vecs r(t) = [lim limits_{t o a} f(t)] hat{mathbf{i}} + [lim limits_{t o a} g(t)] hat{mathbf{j}} + [lim limits_{t o a} h(t)] hat{mathbf{k}})

Glosari

component functions
the component functions of the vector-valued function (vecs r(t)=f(t)hat{mathbf{i}}+g(t)hat{mathbf{j}}) are (f(t)) and (g(t)), and the component functions of the vector-valued function (vecs r(t)=f(t)hat{mathbf{i}}+g(t)hat{mathbf{j}}+h(t)hat{mathbf{k}}) are (f(t)), (g(t)) and (h(t))
helix
a three-dimensional curve in the shape of a spiral
limit of a vector-valued function
a vector-valued function (vecs r(t)) has a limit (vecs L) as (t) approaches (a) if (lim limits{t o a} left| vecs r(t) - vecs L ight| = 0)
plane curve
the set of ordered pairs ((f(t),g(t))) together with their defining parametric equations (x=f(t)) and (y=g(t))
reparameterization
an alternative parameterization of a given vector-valued function
space curve
the set of ordered triples ((f(t),g(t),h(t))) together with their defining parametric equations (x=f(t)), (y=g(t)) and (z=h(t))
vector parameterization
any representation of a plane or space curve using a vector-valued function
vector-valued function
a function of the form (vecs r(t)=f(t)hat{mathbf{i}}+g(t)hat{mathbf{j}}) or (vecs r(t)=f(t)hat{mathbf{i}}+g(t)hat{mathbf{j}}+h(t)hat{mathbf{k}}),where the component functions (f), (g), and (h) are real-valued functions of the parameter (t).

Contributors

  • Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.

  • Edited by Paul Seeburger (Monroe Community College)
  • Paul Seeburger created Example (PageIndex{1}), Exercise (PageIndex{1}), and the subsections titled: Finding a Vector-Valued Function to Trace out the Graph of a Function (y = f(x)), Finding a Vector-Valued Function to Trace out the Graph of an Equation in (x) and (y) and Vice Versa, and Parameterizing a Piecewise Path.

12.1: Vector-Valued Functions and Space Curves - Mathematics

Math 211 Calculus IIIA
Instructor:
S. Levkov
Text: Thomas and Finney, Calculus, 9th edition, Addison Wesley
Masa:
Monday, Wednesday 6:00 9:00 PM.
Office Hours:
Monday, Wednesday 5:00 6:00 PM. Office: 303 ECE.
Telefon:
973 596 5621. E-mail: [email protected]

Minggu 1
Sec 9.4 Parametrization of Plane Curves - HW p741 #1,7-10,12-15,20,23,26-28
Sec 9.5 Calculus with Parametrized Curves - HW p749 #1-3,5,13,15,17,23,33,38

Sec 10.1 Vectors in the Plane - HW p 794 #1-4,7,9,11,13,15,17-19,21,24,26,30,40,41
Sec 10.2 Vectors in Space - HW p 804 #1,5,7,9,12, 15,19,20,27,33,49

Minggu ke-2
Sec 10.3 Dot Products - HW p8l2 #1,3,5,13,16,20,22,35,39,41,49,51,53,59
Sec 10.4 Cross Products - HW p820 #1,3,9,10,17,29,35,39,42

Sec 10.5 Lines and Planes in Space - HW p827

Minggu ke-3
Sec 11.1 Vector-Valued Functions and Space Curves - HW p865

#1,7,10,13,16,21,22,27,32,35,39,43,45, and read 57

Sec 11.3 Arc Length and the Unit Tangent Vector - HW p880 #1,6,8,15

Sec 12.1 Functions of Several Variables - HW p9l4 #3,6,11,25,31,32,41,45
Sec 12.2 Limits and Continuity - HW p921 #1-4,9,11,15,22,28,35

Minggu ke-4
Sec 12.3 Partial Derivatives - HW p931

#2,4,8,14,16,23,26,30,37,43,45,49,50,53,57,59,65,69
Sec 12.5 The Chain Rule - HW p950 #3,7,15,19,27,32,35,43,47

Sec 12.6 Partial Derivatives with Constrained Variables- HW p956 #3,7,11 and then 9
MIDTERM

Minggu ke-5
Sec 12.7 Directional Derivatives, Gradient Vectors and Tangent Planes - HW p967

#3,7,15,19,25,27,34,39,45,49,51,55,58 AND p942 #1,5,25,27

(Note that HW includes Linearization of Sec 12.4)

Sec 12.8 Extreme Values and Saddle Points - HW p975 #3,15,20,33,37,41,43,50,52

Sec 12.9 Lagrange Multipliers - HW p987 #3,7,10,15,17,25,32,35,37,42

Minggu ke-6
Sec 13.1 Double Integrals - HW p1010 #1,2,6,13,18,22,27,31,36,43,47,51

Sec 14.1 Line Integrals - HW p1065 #5,11,12,18,20,25,26
Sec 14.2 Vector Fields, Work, Circulation and Flux - HW p1074

Minggu ke-7
Sec 14.3 Path Independence, Potential Functions and Conservative Fields

- HW p1O83 #2,4,10,14,16,19,23,27,29,31
Sec 14.4 Green's Theorem in the Plane - HW p1093 #1,6,9,13,16,19,21,31,34
Sec 14.8 The Divergence Theorem and a Unified Theory - HW p1132 #3,5,6,8,13,15,26


2 Answers 2

The difference is that a parametrization has some extra properties. A vector valued function is a map $f:Usubsetmathbb R^m o Vsubsetmathbb R^n$

And parametric equations for a [portion of a] submanifold $M$ in Euclidean space (it's rare to parametrize things other than manifolds) is a map $varphi:Usubsetmathbb R^m o Msubsetmathbb R^n$ Where:

  • $U$ is open
  • $varphi$ is a homeomorphism onto its image
  • $operatornameDvarphi = m$ everywhere

What we could say then, is that a parametrization is always in the form of a vector valued function, but conversely, we use vector valued functions with nice properties to parametrize varieties.


Evaluating and Graphing Vector-Valued Functions

y (b) Figure 12.1.1: Sketching the graph of a vector-valued function.

Evaluating a vector-valued function at a specific value of t is straightforward simply evaluate each component function at that value of t . For instance, if r → ⁢ ( t ) = ⟨ t 2 , t 2 + t - 1 ⟩ , then r → ⁢ ( - 2 ) = ⟨ 4 , 1 ⟩ . We can sketch this vector, as is done in Figure 12.1.1 (a). Plotting lots of vectors is cumbersome, though, so generally we do not sketch the whole vector but just the terminal point. The graph of a vector-valued function is the set of all terminal points of r → ⁢ ( t ) , where the initial point of each vector is always the origin. In Figure 12.1.1 (b) we sketch the graph of r → we can indicate individual points on the graph with their respective vector, as shown.

Vector-valued functions are closely related to parametric equations of graphs. While in both methods we plot points ( x ⁢ ( t ) , y ⁢ ( t ) ) or ( x ⁢ ( t ) , y ⁢ ( t ) , z ⁢ ( t ) ) to produce a graph, in the context of vector-valued functions each such point represents a vector. The implications of this will be more fully realized in the next section as we apply calculus ideas to these functions.

Watch the video:
Domain of a Vector-Valued Function from https://youtu.be/Djtttm0C7zA


12.1: Vector-Valued Functions and Space Curves - Mathematics

The topics on this page are vector functions and space curves.

A function whose domain is a set of real numbers and whose range is a subset of 2-space (or called plane), or 3-space is called a vector-valued function of a real variable. For example, the line through a point P parallel to a nonzero vector U is the range of the vector-valued function r given by

Each t corresponds to a point, which can be thinked as a vector initiating from the origin and tip pointing at the point, in the straight line. See the graph below. Usually a vector-valued function in 2-space (resp. 3-space) has two component functions (resp. three component functions). Such as the staight line vector function in 2-space and 3-space can be written as

where point (a, b) (or (a, b, c)) is a point that the line passes through and (u1, u2) (or (u1, u2, u3)) is a vector parallel to the line. The graph above is of the vector function of the straight line r (t) = t i + (-0.6t + 2) j = 2 j + t( i - 0.6 j ). Second expression shows that r passes through point (0, 2) and is parallel to vector (1,-0.6).

If the point (x, y, z) revolves around z-axis at a distance a from it and simultaneously moves parallel to z-axis in such a way that its z-component is proportional to the angle of revolution, the resulting path is called circular helix . If t denotes the angle of revolution, we have

The uaual operations of vectors can be applied to combined two vector functions or to combine a vector function with a real-valued function. If U and V are vector-valued functions, and if f is a real-valued function, all having a common domain, we define new functions U + V, fU, and U dot V by the equations


Math 215 Examples

A vector-valued function is a function that outputs a vector rather than a single number. Often we will be working with functions whose outputs are vectors in three-dimensional space. Such a function can be written as [ vec r(t) = langle x(t), y(t), z(t) angle ] where (f(t)), (g(t)), and (h(t)) are usual functions whose outputs are single numbers (sometimes called scalar functions in contrast with vector-valued functions).

Space Curves

A vector-valued function (vec r(t)) whose values are three-dimensional functions traces out a space curve, a curve in three-dimensional space.

For example, [ vec r(t) = leftlangle t, frac<18>,frac <3> ight angle ] traces out a space curve called a twisted cubic:

Vector-valued functions don't always trace out smooth curves. For example, [ vec r(t) = leftlangle frac<30>, frac<10>, frac <5> ight angle ] has a sharp corner at the point (vec r(0) = langle 0, 0, 0 angle):

Different Parameterizations

Note that two different vector-valued functions may trace out the same curve. We say that two such vector-valued functions corresponding to the same curves are different parameterizations of the same curve.

Given any vector-valued function (vec r(t)), it is easy to construct different parameterizations of the same curve: just use (vec s(t) = vec r(at)) for any non-zero constant (a).

For example, suppose we have [egin vec r(t) &= langle (2+sin<3t>)cos, (2+sin<3t>)sin, cos <3t> angle vec s(t) &= langle (2+sin<6t>)cos<2t>,(2+sin<6t>)sin<2t>, cos <6t> angle. akhir] Notice that (vec s(t) = vec r(2t)), so as observed above, both these vector-valued functions will trace out the same space curve (a curve called a toroidal spiral):

Illustrated Example

Find a vector-valued function that traces out the curve defined as the intersection of the paraboloid (z=x^2+y^2) with the plane (y=x).

Worked Solution

Since every equation in this problem is a function of (x), we can use (x) itself as our parameter. In other words, in our eventual parameterization [ vec r(t) = langle x(t), y(t), z(t) angle, ] we can set (x(t) = t).

Since our curve lies entirely in the plane (y=x), it follows that (y(t) = x(t)) in our parameterization, or (y(t) = t). Since the curve lies on the paraboloid (z = x^2 + y^2), we similarly have [ z(t) = x(t)^2 + y(t)^2 = 2t^2. ]

Thus our final parameterization of this curve is [ vec r(t) = langle t, t, 2t^2 angle. ]

Visualizing the Example

The animation below shows the paraboloid (z=x^2+y^2) with (vec r(t) = langle t, t, 2t^2) tracing out its intersection with the plane (y=x):

Further Questions

  1. Find a parameterization of the curve in the example that traces out the curve half as fast.
  2. Find a parameterization of the curve in the example that traces out the curve backwards.
  3. What curve does (vec r(t) = langle t, -t, 2t^2 angle) trace out?
  4. In the animation in Key Concepts with two different parameterizations, which of (vec r(t)) and (vec s(t)) is represented by the blue arrows? Which is represented by the red arrows?

Using the Mathematica Demo

All graphics on this page were generated by the Mathematica notebook 13_1SpaceCurves.nb.

This notebook generates images and animations like those on this page for any curve.

As an exercise, use the notebook to provide a visual demonstration illustrating your answers to Questions 1-3.

Investigate other vector-valued funtions (vec r(t)). Can you find one that traces out a circle? A helix? Can you find one with a sharp corner somewhere other than the origin? Experiment with functions that yield sharp corners can you figure out what causes them?


CalcPlot3D¶

A useful tool for graphing vector functions and other kinds of 3D objects. Although this applet was created for use in calculus classes, it is useful to us as well. Use the following procedure to graph a vector function in CalcPlot3D.

  1. Erase the default shape that appears, by unchecking the box next to Function 1 and clicking the Graph button immediately above it.
  2. Add a parametric curve by clicking the Graph menu and choosing Add a Space Curve.
  3. In the three blanks provided, enter the x , y , and z components of the vector function, using t as the parameter. The default bounds for t (from -10 to 10 ) may be sensible for your function, but you can change them.
  4. Click Graph (on the popup window into which you typed the parametric equations).
  5. Click and drag to view from different angles.

A powerful mathematics tool that you can use on your own computer or on the web. Here is a link to a webpage that evaluates Sage code and shows you the result immediately. Type in code like the following to graph a vector function. (Replace the three components of the vector function with any three vector function components.) The 0 and 2pi are the bounds on t .

To see that example plotted, click here.


Kandungan

Vector fields on subsets of Euclidean space Edit

Given a subset S dalam R n , a vector field is represented by a vector-valued function V: SR n in standard Cartesian coordinates (x1, …, xn) . If each component of V is continuous, then V is a continuous vector field, and more generally V ialah C k vector field if each component of V is k times continuously differentiable.

A vector field can be visualized as assigning a vector to individual points within an n-dimensional space. [1]

Given two C k -vector fields V , W defined on S and a real-valued C k -function f defined on S , the two operations scalar multiplication and vector addition

define the module of C k -vector fields over the ring of C k -functions where the multiplication of the functions is defined pointwise (therefore, it is commutative with the multiplicative identity being fid(p) := 1 ).

Coordinate transformation law Edit

In physics, a vector is additionally distinguished by how its coordinates change when one measures the same vector with respect to a different background coordinate system. The transformation properties of vectors distinguish a vector as a geometrically distinct entity from a simple list of scalars, or from a covector.

Thus, suppose that (x1. xn) is a choice of Cartesian coordinates, in terms of which the components of the vector V adalah

and suppose that (y1. yn) adalah n functions of the xi defining a different coordinate system. Then the components of the vector V in the new coordinates are required to satisfy the transformation law

Such a transformation law is called contravariant. A similar transformation law characterizes vector fields in physics: specifically, a vector field is a specification of n functions in each coordinate system subject to the transformation law (1) relating the different coordinate systems.

Vector fields are thus contrasted with scalar fields, which associate a number or scalar to every point in space, and are also contrasted with simple lists of scalar fields, which do not transform under coordinate changes.

Vector fields on manifolds Edit

If the manifold M is smooth or analytic—that is, the change of coordinates is smooth (analytic)—then one can make sense of the notion of smooth (analytic) vector fields. The collection of all smooth vector fields on a smooth manifold M is often denoted by Γ ( T M ) or C ∞ ( M , T M ) (M,TM)> (especially when thinking of vector fields as sections) the collection of all smooth vector fields is also denoted by X ( M ) >(M)> (a fraktur "X").

  • A vector field for the movement of air on Earth will associate for every point on the surface of the Earth a vector with the wind speed and direction for that point. This can be drawn using arrows to represent the wind the length (magnitude) of the arrow will be an indication of the wind speed. A "high" on the usual barometric pressure map would then act as a source (arrows pointing away), and a "low" would be a sink (arrows pointing towards), since air tends to move from high pressure areas to low pressure areas. field of a moving fluid. In this case, a velocity vector is associated to each point in the fluid. are 3 types of lines that can be made from (time-dependent) vector fields. They are:
    . The fieldlines can be revealed using small iron filings. allow us to use a given set of initial and boundary conditions to deduce, for every point in Euclidean space, a magnitude and direction for the force experienced by a charged test particle at that point the resulting vector field is the electromagnetic field.
  • A gravitational field generated by any massive object is also a vector field. For example, the gravitational field vectors for a spherically symmetric body would all point towards the sphere's center with the magnitude of the vectors reducing as radial distance from the body increases.

Gradient field in euclidean spaces Edit

Vector fields can be constructed out of scalar fields using the gradient operator (denoted by the del: ∇). [4]

A vector field V defined on an open set S is called a bidang kecerunan or a bidang konservatif if there exists a real-valued function (a scalar field) f pada S such that

The associated flow is called the gradient flow , and is used in the method of gradient descent.

The path integral along any closed curve γ (γ(0) = γ(1)) in a conservative field is zero:

Central field in euclidean spaces Edit

A C ∞ -vector field over R n <0>is called a central field if

where O(n, R) is the orthogonal group. We say central fields are invariant under orthogonal transformations around 0.

The point 0 is called the center of the field.

Since orthogonal transformations are actually rotations and reflections, the invariance conditions mean that vectors of a central field are always directed towards, or away from, 0 this is an alternate (and simpler) definition. A central field is always a gradient field, since defining it on one semiaxis and integrating gives an antigradient.

Line integral Edit

A common technique in physics is to integrate a vector field along a curve, also called determining its line integral. Intuitively this is summing up all vector components in line with the tangents to the curve, expressed as their scalar products. For example, given a particle in a force field (e.g. gravitation), where each vector at some point in space represents the force acting there on the particle, the line integral along a certain path is the work done on the particle, when it travels along this path. Intuitively, it is the sum of the scalar products of the force vector and the small tangent vector in each point along the curve.

The line integral is constructed analogously to the Riemann integral and it exists if the curve is rectifiable (has finite length) and the vector field is continuous.

Given a vector field V and a curve γ , parametrized by t in [a, b] (where a and b are real numbers), the line integral is defined as

Divergence Edit

The divergence of a vector field on Euclidean space is a function (or scalar field). In three-dimensions, the divergence is defined by

with the obvious generalization to arbitrary dimensions. The divergence at a point represents the degree to which a small volume around the point is a source or a sink for the vector flow, a result which is made precise by the divergence theorem.

The divergence can also be defined on a Riemannian manifold, that is, a manifold with a Riemannian metric that measures the length of vectors.

Curl in three dimensions Edit

The curl is an operation which takes a vector field and produces another vector field. The curl is defined only in three dimensions, but some properties of the curl can be captured in higher dimensions with the exterior derivative. In three dimensions, it is defined by

The curl measures the density of the angular momentum of the vector flow at a point, that is, the amount to which the flow circulates around a fixed axis. This intuitive description is made precise by Stokes' theorem.

Index of a vector field Edit

The index of a vector field is an integer that helps to describe the behaviour of a vector field around an isolated zero (i.e., an isolated singularity of the field). In the plane, the index takes the value -1 at a saddle singularity but +1 at a source or sink singularity.

Let the dimension of the manifold on which the vector field is defined be n. Take a small sphere S around the zero so that no other zeros lie in the interior of S. A map from this sphere to a unit sphere of dimensions n − 1 can be constructed by dividing each vector on this sphere by its length to form a unit length vector, which is a point on the unit sphere S n-1 . This defines a continuous map from S to S n-1 . The index of the vector field at the point is the degree of this map. It can be shown that this integer does not depend on the choice of S, and therefore depends only on the vector field itself.

The index of the vector field as a whole is defined when it has just a finite number of zeroes. In this case, all zeroes are isolated, and the index of the vector field is defined to be the sum of the indices at all zeroes.

The index is not defined at any non-singular point (i.e., a point where the vector is non-zero). it is equal to +1 around a source, and more generally equal to (−1) k around a saddle that has k contracting dimensions and n-k expanding dimensions. For an ordinary (2-dimensional) sphere in three-dimensional space, it can be shown that the index of any vector field on the sphere must be 2. This shows that every such vector field must have a zero. This implies the hairy ball theorem, which states that if a vector in R 3 is assigned to each point of the unit sphere S 2 in a continuous manner, then it is impossible to "comb the hairs flat", i.e., to choose the vectors in a continuous way such that they are all non-zero and tangent to S 2 .

For a vector field on a compact manifold with a finite number of zeroes, the Poincaré-Hopf theorem states that the index of the vector field is equal to the Euler characteristic of the manifold.

Michael Faraday, in his concept of lines of force, emphasized that the field itself should be an object of study, which it has become throughout physics in the form of field theory.

In addition to the magnetic field, other phenomena that were modeled by Faraday include the electrical field and light field.

Consider the flow of a fluid through a region of space. At any given time, any point of the fluid has a particular velocity associated with it thus there is a vector field associated to any flow. The converse is also true: it is possible to associate a flow to a vector field having that vector field as its velocity.

Given a vector field V defined on S, one defines curves γ(t) on S such that for each t in an interval Saya

By the Picard–Lindelöf theorem, if V is Lipschitz continuous there is a unique C 1 -curve γx for each point x dalam S so that, for some ε > 0,

The curves γx are called integral curves atau trajectories (or less commonly, flow lines) of the vector field V and partition S into equivalence classes. It is not always possible to extend the interval (−ε, +ε) to the whole real number line. The flow may for example reach the edge of S in a finite time. In two or three dimensions one can visualize the vector field as giving rise to a flow on S. If we drop a particle into this flow at a point p it will move along the curve γp in the flow depending on the initial point p. Sekiranya p is a stationary point of V (i.e., the vector field is equal to the zero vector at the point p), then the particle will remain at p.

Given a smooth function between manifolds, f : MN, the derivative is an induced map on tangent bundles, f* : TMTN. Given vector fields V : MTM dan W : NTN, we say that W adalah f-related to V if the equation Wf = fV holds.

Sekiranya Vi adalah f-related to Wi, i = 1, 2, then the Lie bracket [V1, V2] is f-related to [W1, W2].

Replacing vectors by p-vectors (pth exterior power of vectors) yields p-vector fields taking the dual space and exterior powers yields differential k-forms, and combining these yields general tensor fields.

Algebraically, vector fields can be characterized as derivations of the algebra of smooth functions on the manifold, which leads to defining a vector field on a commutative algebra as a derivation on the algebra, which is developed in the theory of differential calculus over commutative algebras.


Silibus

If at any time during this semester you feel ill, in the interest of your own health and safety as well as the health and safety of your instructors and classmates, you are encouraged not to attend face-to-face class meetings or events. Please review the steps outlined below that you should follow to ensure your absence for illness will be excused. These steps also apply to not participating in synchronous online class meetings if you feel too ill to do so and missing specified assignment due dates in asynchronous online classes because of illness.

1. If you are ill and think the symptoms might be COVID-19-related:
a) Call Student Health Services at 806.743.2848 or your health care provider.
b) Self-report as soon as possible using the Dean of Students COVID-19 webpage. This website has specific directions about how to upload documentation from a medical provider and what will happen if your illness renders you unable to participate in classes for more than one week.
c) If your illness is determined to be COVID-19-related, all remaining documentation and communication will be handled through the Office of the Dean of Students, including notification of your instructors of the period of time you may be absent from and may return to classes.
d) If your illness is determined not to be COVID-19-related, please follow steps 2.a-d below.

2. If you are ill and can attribute your symptoms to something other than COVID-19:
a) If your illness renders you unable to attend face-to-face classes, participate in synchronous online classes, or miss specified assignment due dates in asynchronous online classes, you are encouraged to visit with either Student Health Services at 806.743.2848 or your health care provider. Note that Student Health Services and your own and other health care providers may arrange virtual visits.
b) During the health provider visit, request a "return to school" note.
c) E-mail the instructor a picture of that note.
d) Return to class by the next class period after the date indicated on your note.

Following the steps outlined above helps to keep your instructors informed about your absences and ensures your absence or missing an assignment due date because of illness will be marked excused. You will still be responsible to complete within a week of returning to class any assignments, quizzes, or exams you miss because of illness.

This is a distance class, all the students enrolled in this class should be highly responsible in managing their schedule. This course moves very fast. If you fall behind, even by one section, you may not be able to catch up, since each section generally depends very heavily on the ones before. A student enrolled in this class has to be capable to read and understand the textbook. If in the past you struggled in self-lecturing mathematics, then this is not the class for you and it is highly recommended you switch to a face-to-face class. The instructor expects for the student to read each section of the textbook, watch the videos and read the class-notes before attempting to solve the homework problems. When asking for help you need to show all your work, by typing it on the email (better) or by attaching a scanned copy of your work. When asking for help for a WebWork problem it is recommended you use the button email to the instructor at the bottom of the screen, otherwise you may not get any answer.


3 Jawapan 3

As you seem to have worked out for yourself, you can just write the Taylor series for each component of $f$ separately so I guess the remaining issue is how to write this down neatly.

Your "something" here is the second derivative of $f$, which is the third-order tensor comprised of the Hessian matrices $H_1,ldots,H_n$ so you need to use a notation that can deal with this kind of object. If you're comfortable with the Einstein summation convention, then you can write $f_i( heta) = f_i( heta_0) + A_ ( theta_0) ( theta - theta_0) _j + frac 1 2 H_( theta ') ( theta - theta_0) _j ( theta- theta_0) _k, $ di mana $ H_ = frac < separa ^ 2 f_i> < separa x_k separa x_j>. $

Sebagai alternatif, anda boleh menggunakan sesuatu seperti $ f ( theta) = f ( theta_0) + A ( theta_0) ( theta - theta_0) + frac 1 2 H ( theta ') ( theta- theta_0 , theta- theta_0) $ di mana $ H $ ditafsirkan sebagai bentuk bilinear pada $ mathbb R ^ m $ mengambil nilai dalam $ mathbb R ^ n. $ Anda boleh menulis ini menggunakan pendaraban matriks seperti dalam jawapan Mostafa, tetapi ini sedikit tidak standard dan anda harus jelas bahawa "vektor" $ A $ dan "matriks" anda $ H $ sebenarnya bernilai $ mathbb R ^ n $ -valu.


Tonton videonya: Part 1 Matematik Tambahan Tingkatan 4 Bab 8 Vektor KSSM (Disember 2021).