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olabilir."]},"beginTime":0,"endTime":94,"href":"/video/preview/9867457546522917982?parent-reqid=1765307968614385-10793074216298608906-balancer-l7leveler-kubr-yp-vla-21-BAL&text=Strong+Calculus+II&t=0&ask_summarization=1"},{"index":1,"title":"Vektör Alanlarının Çizgi İntegrali Formülü","list":{"type":"unordered","items":["Vektör alanının çizgi integralinin sorulduğunu, çizgi integralinin içinde büyük F harfi ve dot product (iç çarpım) işaretiyle çarpım gördüğümüzde anlarız.","Vektör alanının çizgi integrali formülü: ∫_c F·dr'dir.","Bu integral hesaplanırken, r parametresi t'ye göre parametrize edilir ve integral t'ye göre alınır."]},"beginTime":94,"endTime":388,"href":"/video/preview/9867457546522917982?parent-reqid=1765307968614385-10793074216298608906-balancer-l7leveler-kubr-yp-vla-21-BAL&text=Strong+Calculus+II&t=94&ask_summarization=1"},{"index":2,"title":"Vektör Alanlarının Çizgi İntegrali Hesaplama Adımları","list":{"type":"unordered","items":["Vektör alanının çizgi integralini hesaplamak için önce f(x,y,z) içindeki x, y, z değerleri r parametresine göre değiştirilir.","İkinci adım olarak r'nin t'ye göre türevi alınır.","Üçüncü adım olarak f(r) ile r'nin türevinin iç çarpımı yapılır ve vektörsel durumdan kurtulunur.","Son adım olarak elde edilen ifade t'ye göre integral alınır ve sınırlar belirlenir."]},"beginTime":388,"endTime":439,"href":"/video/preview/9867457546522917982?parent-reqid=1765307968614385-10793074216298608906-balancer-l7leveler-kubr-yp-vla-21-BAL&text=Strong+Calculus+II&t=388&ask_summarization=1"},{"index":3,"title":"Örnek Çözüm","list":{"type":"unordered","items":["Örnek olarak f(x,y,z) = 8x²yz + 5zj - 4xyk vektör alanı ve r(t) = t i + t²j + t³k eğrisi verilmiştir.","İlk adım olarak f(r) = 8t⁴i + 5t³j - 4t³k olarak hesaplanır.","İkinci adım olarak r'(t) = i + 2j + 3k olarak bulunur.","Üçüncü adım olarak f(r)·r'(t) = 8t⁷ + 10t⁴ - 12t⁵ olarak hesaplanır.","Dördüncü adım olarak integral ∫₀¹ (8t⁷ + 10t⁴ - 12t⁵) dt alınır ve sonuç 1 olarak bulunur."]},"beginTime":439,"endTime":825,"href":"/video/preview/9867457546522917982?parent-reqid=1765307968614385-10793074216298608906-balancer-l7leveler-kubr-yp-vla-21-BAL&text=Strong+Calculus+II&t=439&ask_summarization=1"},{"index":4,"title":"Parametreze İşlemi Gerektiren Örnek","list":{"type":"unordered","items":["İkinci örnek olarak f(x,y,z) = x k vektör alanı ve doğru parçası verilmiştir.","Doğru parçasının parametrize edilmesi için x = -1 + 4t, y = 2 - 2t, z = t olarak bulunur.","r(t) = -1 + 4t i + 2 - 2t j + t k olarak hesaplanır.","Bu örnekte parametreze işlemi gereklidir çünkü r parametresi doğrudan verilmemiştir."]},"beginTime":825,"endTime":1025,"href":"/video/preview/9867457546522917982?parent-reqid=1765307968614385-10793074216298608906-balancer-l7leveler-kubr-yp-vla-21-BAL&text=Strong+Calculus+II&t=825&ask_summarization=1"},{"index":5,"title":"Vektör Alanının Çizgi İntegrali Hesaplama","list":{"type":"unordered","items":["Vektör alanının çizgi integrali hesaplanırken, r'nin türevi t olarak belirleniyor.","f(r(t)) ifadesi hesaplanırken, x yerine -1+4t, y yerine 2-2t, z yerine t değerleri yerleştiriliyor.","r(t) ve r'(t) vektörleri iç çarpım yapılarak 18t²-6t sonucu elde ediliyor."]},"beginTime":1031,"endTime":1205,"href":"/video/preview/9867457546522917982?parent-reqid=1765307968614385-10793074216298608906-balancer-l7leveler-kubr-yp-vla-21-BAL&text=Strong+Calculus+II&t=1031&ask_summarization=1"},{"index":6,"title":"İntegral Hesaplama ve Sonuç","list":{"type":"unordered","items":["18t²-6t ifadesi 0'dan 1'e kadar t'ye göre integral alınıyor.","İntegral hesaplandığında sonuç 3 olarak bulunuyor.","Bu video, vektör alanlarının çizgi integrallerinin nasıl hesaplandığını gösteriyor."]},"beginTime":1205,"endTime":1255,"href":"/video/preview/9867457546522917982?parent-reqid=1765307968614385-10793074216298608906-balancer-l7leveler-kubr-yp-vla-21-BAL&text=Strong+Calculus+II&t=1205&ask_summarization=1"},{"index":7,"title":"Gelecek Konular","list":{"type":"unordered","items":["Bir sonraki videoda vektör alanı konsorveti (korumacı vektör alanı) konusu ele alınacak.","Korumacı vektör alanının çizgi integrali hesaplamaları için potansiyel fonksiyonun yardımıyla özel bir formül kullanılacak.","Bu konu, hocaların sormayı en çok sevdikleri bölümlerden biri olarak belirtiliyor."]},"beginTime":1255,"endTime":1298,"href":"/video/preview/9867457546522917982?parent-reqid=1765307968614385-10793074216298608906-balancer-l7leveler-kubr-yp-vla-21-BAL&text=Strong+Calculus+II&t=1255&ask_summarization=1"}],"linkTemplate":"/video/preview/9867457546522917982?parent-reqid=1765307968614385-10793074216298608906-balancer-l7leveler-kubr-yp-vla-21-BAL&text=Strong+Calculus+II&t=%%timestamp%%&ask_summarization=1"},"isAdultDoc":false,"relatedParams":{"text":"Calculus-II : Vektör Alanlarının Çizgi İntegralini Hesaplama","related_orig_text":"Strong Calculus 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üniversite matematiği derslerinden calculus-II dersine ait "Calculus-2 İngilizce Final Sınav Örneği-1" videosudur. Hazırlayan: Kemal Duran (Matematik Öğretmeni)...","preview":{"posterSrc":"//avatars.mds.yandex.net/get-vthumb/3784843/bf9e8eecc746be298fe790bfbd3f22d7/564x318_1","videoSrc":"https://video-preview.s3.yandex.net/65_bBAIAAAA.mp4","videoType":"video/mp4"},"target":"_self","position":"1","reqid":"1765307968614385-10793074216298608906-balancer-l7leveler-kubr-yp-vla-21-BAL","summary":{"isFull":true,"fullTextUrl":"/video/result?ask_summarization=1&numdoc=1&noreask=1&nomisspell=1&parent-reqid=1765307968614385-10793074216298608906-balancer-l7leveler-kubr-yp-vla-21-BAL&text=videoid:7193410124373256789","teaser":[{"list":{"type":"unordered","items":["Bu video, bir eğitmen tarafından sunulan Calculus II dersindeki İngilizce sınav sorularının ve çift katlı integral konusunun çözümünü içeren bir eğitim içeriğidir.","Video, dört ana bölümden oluşmaktadır: İlk bölümde çok değişkenli fonksiyonların yönlü türevi ve Lagrange çarpanları yöntemiyle maksimum-minimum değerlerin bulunması ele alınmaktadır. İkinci bölümde ise lokal minimum, maksimum ve sedle point bulma konuları işlenmektedir. Üçüncü ve dördüncü bölümlerde ise çift katlı integralin hesaplanması, alan ve hacim problemlerinin çözümü, integral sınırlarının belirlenmesi ve polar koordinatların kullanımı detaylı olarak anlatılmaktadır.","Videoda her bir soru adım adım çözülmekte, gerekli formüller hatırlatılmakta ve hesaplamalar detaylı olarak gösterilmektedir. Özellikle birim çember üzerindeki çift katlı integralin çözümü ve integral alma sırasının değiştirilmesi gibi konulara ağırlık verilmektedir. Video, üniversite sınavlarına hazırlanan öğrenciler için faydalı bir kaynak niteliğindedir."]},"endTime":3103,"title":"Calculus II İngilizce Sınav Soruları ve Çift Katlı İntegral Çözümleri","beginTime":0}],"fullResult":[{"index":0,"title":"Calculus II İngilizce Final Sınav Örneği","list":{"type":"unordered","items":["Bu videoda Calculus II dersinden İngilizce sorulardan oluşan bir sınav örneğinin çözümü gösterilecektir.","İlk soruda, f(x,y) = x(x²+y²) fonksiyonunun (1,1) noktasındaki u vektörü yönündeki yönlü türevi hesaplanacaktır.","Yönlü türev formülü Df(a,b) = ∇f(a,b) · (u/|u|) olarak hesaplanır, burada ∇f(a,b) fonksiyonun gradyanı, u vektörün birim vektörüdür."]},"beginTime":0,"endTime":110,"href":"/video/preview/7193410124373256789?parent-reqid=1765307968614385-10793074216298608906-balancer-l7leveler-kubr-yp-vla-21-BAL&text=Strong+Calculus+II&t=0&ask_summarization=1"},{"index":1,"title":"Yönlü Türev Hesaplama","list":{"type":"unordered","items":["Fonksiyonun gradyanı ∇f = (f_x, f_y) olarak hesaplanır, burada f_x = y + (2x²+2y²)/(2(x²+y²)) ve f_y = (2y)/(x²+y²) olur.","(1,1) noktasındaki gradyan vektörü ∇f(1,1) = (1+ln2, 1) olarak bulunur.","U vektörünün birim vektörü (3/5, 4/5) olarak hesaplanır ve yönlü türev (1+ln2)·(3/5) + 1·(4/5) = 7/5 + 3/5·ln2 olarak bulunur."]},"beginTime":110,"endTime":398,"href":"/video/preview/7193410124373256789?parent-reqid=1765307968614385-10793074216298608906-balancer-l7leveler-kubr-yp-vla-21-BAL&text=Strong+Calculus+II&t=110&ask_summarization=1"},{"index":2,"title":"Lagrange Çarpanları ile Maksimum ve Minimum Bulma","list":{"type":"unordered","items":["İkinci soruda, f(x,y) = x+y² fonksiyonunun x²+y²+xy=1 kısıtlaması altında mutlak maksimum ve minimum değerleri Lagrange çarpanları yöntemiyle bulunacaktır.","Kısıtlama denklemi g(x,y) = x²+y²+xy-1 şeklinde yazılarak, ∇f = λ∇g eşitliği kurulur.","Eşitlikten x=0 ve x=-2y ilişkileri elde edilir ve bunlar kısıtlama denkleminde yerine konularak kritik noktalar bulunur."]},"beginTime":398,"endTime":822,"href":"/video/preview/7193410124373256789?parent-reqid=1765307968614385-10793074216298608906-balancer-l7leveler-kubr-yp-vla-21-BAL&text=Strong+Calculus+II&t=398&ask_summarization=1"},{"index":3,"title":"Kritik Noktalar ve Sonuç","list":{"type":"unordered","items":["Kritik noktalar (0,1), (0,-1), (-2/√3, 1/√3) ve (2/√3, -1/√3) olarak bulunur.","Bu noktalar f(x,y) fonksiyonuna yerleştirilerek değerleri hesaplanır.","Sonuç olarak, tüm kritik noktaların f(x,y) değerleri 1 olarak bulunur."]},"beginTime":822,"endTime":850,"href":"/video/preview/7193410124373256789?parent-reqid=1765307968614385-10793074216298608906-balancer-l7leveler-kubr-yp-vla-21-BAL&text=Strong+Calculus+II&t=822&ask_summarization=1"},{"index":4,"title":"Lagrange Multiplier Probleminin Çözümü","list":{"type":"unordered","items":["Lagrange Multiplier problemi çözülerek absolute maximum değeri 1 ve eksi 1 noktalarında, absolute minimum değeri eksi 1/3 olan noktada bulunmuştur.","Sorunun çözümü detaylı olarak gösterilmiştir ve işlem hatası olup olmadığını belirtmek için yorumlar kısmında düzeltme yapılacağı belirtilmiştir."]},"beginTime":859,"endTime":984,"href":"/video/preview/7193410124373256789?parent-reqid=1765307968614385-10793074216298608906-balancer-l7leveler-kubr-yp-vla-21-BAL&text=Strong+Calculus+II&t=859&ask_summarization=1"},{"index":5,"title":"Lokal Minimum, Maksimum ve Sedle Point Bulma","list":{"type":"unordered","items":["f(x,y) = 4 - 9x² - 2xy - y² fonksiyonunun lokal minimum, maksimum ve sedle point'lerini bulmak için gradient f alınıp sıfıra eşitlenmiştir.","Fonksiyonun x ve y'ye göre kısmi türevleri alınarak dört aday kritik nokta bulunmuştur.","Kritik noktaların sınıflandırılması için Dxy = fxx * fyy - (fxy)² formülü kullanılmıştır."]},"beginTime":984,"endTime":1251,"href":"/video/preview/7193410124373256789?parent-reqid=1765307968614385-10793074216298608906-balancer-l7leveler-kubr-yp-vla-21-BAL&text=Strong+Calculus+II&t=984&ask_summarization=1"},{"index":6,"title":"Kritik Noktaların Sınıflandırılması","list":{"type":"unordered","items":["(0,√2) ve (0,-√2) noktaları için Dxy = -32 sonucu elde edilerek sedle point olduğu belirlenmiştir.","(2/3,0) noktasında Dxy = 32 ve fxx = -16 olduğundan lokal maksimum noktası olduğu tespit edilmiştir.","(-2/3,0) noktasında da Dxy = 32 ve fxx = 16 olduğundan lokal minimum noktası olduğu belirlenmiştir."]},"beginTime":1251,"endTime":1605,"href":"/video/preview/7193410124373256789?parent-reqid=1765307968614385-10793074216298608906-balancer-l7leveler-kubr-yp-vla-21-BAL&text=Strong+Calculus+II&t=1251&ask_summarization=1"},{"index":7,"title":"Çift Katlı İntegral ile Alan Hesaplama","list":{"type":"unordered","items":["y = x² parabolü ve y = x + 2 doğrusu tarafından kapatılan R bölgesinin alanını çift katlı integral kullanarak bulmak istenmektedir.","Çift katlı integralin sınırları R bölgesinden elde edilir ve alan bulmak için integralin içerisine 1 yazılmalıdır."]},"beginTime":1605,"endTime":1659,"href":"/video/preview/7193410124373256789?parent-reqid=1765307968614385-10793074216298608906-balancer-l7leveler-kubr-yp-vla-21-BAL&text=Strong+Calculus+II&t=1605&ask_summarization=1"},{"index":8,"title":"Çift Katlı İntegral ile Alan Hesaplama","list":{"type":"unordered","items":["Bir çift katlı integralde, fonksiyon içine 1 yazıldığında, belirtilen bölgenin alanı hesaplanır.","Verilen problemde, y=x² ve y=2x-2 doğruları arasındaki kesişme noktaları x=-1 ve x=2 olarak bulunur.","Alan hesaplamak için çift katlı integral sınırları belirlenir: x sınırı -1'den 2'ye, y sınırı x²'den x+2'ye kadar değişir."]},"beginTime":1669,"endTime":1814,"href":"/video/preview/7193410124373256789?parent-reqid=1765307968614385-10793074216298608906-balancer-l7leveler-kubr-yp-vla-21-BAL&text=Strong+Calculus+II&t=1669&ask_summarization=1"},{"index":9,"title":"İntegral Hesaplama","list":{"type":"unordered","items":["Alan hesaplaması için çift katlı integral yazılır: ∫∫ dA = ∫∫(x+2-x²) dx dy.","İntegral hesaplandığında alan 9/2 olarak bulunur.","İntegral hesaplamasında önce y'ye göre, sonra x'e göre integral alınır."]},"beginTime":1814,"endTime":1922,"href":"/video/preview/7193410124373256789?parent-reqid=1765307968614385-10793074216298608906-balancer-l7leveler-kubr-yp-vla-21-BAL&text=Strong+Calculus+II&t=1814&ask_summarization=1"},{"index":10,"title":"Hacim Hesaplama Problemi","list":{"type":"unordered","items":["Bir yüzeyin altında ve belirli bir bölge üzerindeki hacim hesaplanmak isteniyor.","Z=16-x²-y² yüzeyi ve y=2√x, y=4x-2 doğruları ile x ekseninin sınırladığı bölge inceleniyor.","Hacim hesaplaması için çift katlı integral kullanılacak ve sınırlar belirlenecek."]},"beginTime":1922,"endTime":2095,"href":"/video/preview/7193410124373256789?parent-reqid=1765307968614385-10793074216298608906-balancer-l7leveler-kubr-yp-vla-21-BAL&text=Strong+Calculus+II&t=1922&ask_summarization=1"},{"index":11,"title":"Grafik Çizimi ve Sınırların Belirlenmesi","list":{"type":"unordered","items":["Fonksiyonların grafiklerini çizmek, alan ve hacim hesaplamalarında çok önemlidir.","y=2√x ve y=4x-2 doğrularının kesişme noktası x=1, y=2'dir.","İntegral sınırları belirlerken, sayı sınırı y'ye, fonksiyon sınırı x'e verildiğinde daha kolay hesaplama yapılabilir."]},"beginTime":2095,"endTime":2351,"href":"/video/preview/7193410124373256789?parent-reqid=1765307968614385-10793074216298608906-balancer-l7leveler-kubr-yp-vla-21-BAL&text=Strong+Calculus+II&t=2095&ask_summarization=1"},{"index":12,"title":"İntegral Oluşturma","list":{"type":"unordered","items":["Hacim hesaplaması için çift katlı integral oluşturulur: ∫∫(16-x²-y²) dx dy.","İntegral sınırları 0'dan 2'ye, y²/4'ten (y+2)/4'e kadar değişir.","İntegral hesaplaması için önce x'e göre, sonra y'e göre integral alınacak."]},"beginTime":2351,"endTime":2443,"href":"/video/preview/7193410124373256789?parent-reqid=1765307968614385-10793074216298608906-balancer-l7leveler-kubr-yp-vla-21-BAL&text=Strong+Calculus+II&t=2351&ask_summarization=1"},{"index":13,"title":"Çift Katlı İntegralde İntegral Alma Sırasını Değiştirme","list":{"type":"unordered","items":["Çift katlı integralde integral alma sırasını değiştirmek için integral alma bölgesinin grafiğini çizmek gerekir.","İntegralde y sınırları x²'den 2x, x sınırları ise 0'dan 2'ye kadar verilmiştir.","İntegral alma bölgesi, y=x² ve y=2x fonksiyonlarının kesiştiği noktalar arasında belirlenir ve bu noktalar 0 ve 2'dir."]},"beginTime":2457,"endTime":2590,"href":"/video/preview/7193410124373256789?parent-reqid=1765307968614385-10793074216298608906-balancer-l7leveler-kubr-yp-vla-21-BAL&text=Strong+Calculus+II&t=2457&ask_summarization=1"},{"index":14,"title":"İntegral Alma Sırasını Değiştirme İşlemi","list":{"type":"unordered","items":["İntegral alma sırasını değiştirmek için önce y sınırları 0'dan 4'e, sonra x sınırları 2'den √y'ye kadar belirlenir.","Sol sınır fonksiyonu x=y/2, sağ sınır fonksiyonu ise x=√y olarak bulunur.","İntegral alma sırası değiştirildiğinde, integralin içerisinde hiçbir değişiklik olmaz ve sonuç aynı olacaktır."]},"beginTime":2590,"endTime":2753,"href":"/video/preview/7193410124373256789?parent-reqid=1765307968614385-10793074216298608906-balancer-l7leveler-kubr-yp-vla-21-BAL&text=Strong+Calculus+II&t=2590&ask_summarization=1"},{"index":15,"title":"Sınavın Son Sorusu ve Çözümü","list":{"type":"unordered","items":["Son soruda birim çember (merkezi orijin, yarıçapı 1 olan çember) üzerindeki çift katlı integral hesaplanmaktadır.","İntegral alma bölgesi çember olduğunda polar koordinatlar kullanılmalıdır.","r sınırları 0'dan 1'e, θ sınırları 0'dan 2π'ye kadar belirlenir ve dx dy yerine rdrdθ yazılır."]},"beginTime":2753,"endTime":2890,"href":"/video/preview/7193410124373256789?parent-reqid=1765307968614385-10793074216298608906-balancer-l7leveler-kubr-yp-vla-21-BAL&text=Strong+Calculus+II&t=2753&ask_summarization=1"},{"index":16,"title":"İntegralin Hesaplanması","list":{"type":"unordered","items":["İntegralde x yerine r cosθ, y yerine r sinθ yazılır ve integral polar koordinatlara dönüştürülür.","İntegralin hesaplanmasında değişken değiştirme yöntemi kullanılır ve sonucu -2π olarak buluruz.","Çift katlı integral hacim hesaplar, bazen negatif sonuçlar verebilir, bu durumda hacim için mutlak değeri almak 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Ayrıca, Fransız matematikçilerin dizileri sıfırdan başlatma alışkanlığı gibi kültürel farklılıklar da açıklanmaktadır."]},"endTime":640,"title":"Dizilerin Matematiksel Tanımı ve Özellikleri","beginTime":0}],"fullResult":[{"index":0,"title":"Dizinin Tanımı","list":{"type":"unordered","items":["Dizi, İngilizcesi sequence olan, fonksiyon kavramının bir alt kümesi olan basit bir kavramdır.","Dizi, girdileri özel tanımlanmış fonksiyonlardır.","Dizilerde girdiler sayma sayılar kümesi (pozitif doğal sayılar) olmalıdır ve hiçbir girdi formülü tanımsız yapmamalıdır."]},"beginTime":1,"endTime":139,"href":"/video/preview/11780236931523443076?parent-reqid=1765307968614385-10793074216298608906-balancer-l7leveler-kubr-yp-vla-21-BAL&text=Strong+Calculus+II&t=1&ask_summarization=1"},{"index":1,"title":"Dizinin Özellikleri","list":{"type":"unordered","items":["Dizilerde girdi genellikle \"n\" harfi ile gösterilir ve 1'den başlayıp sonsuza kadar gider.","Dizilerde her terim için \"n\" değeri direkt 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