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üniversite matematiği derslerinden calculus-II dersine ait "Calculus-2 İngilizce Final Sınav Örneği-1" videosudur. 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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=1765308054756475-15830990373066760636-balancer-l7leveler-kubr-yp-klg-76-BAL&text=Quick+Calculus&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=1765308054756475-15830990373066760636-balancer-l7leveler-kubr-yp-klg-76-BAL&text=Quick+Calculus&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=1765308054756475-15830990373066760636-balancer-l7leveler-kubr-yp-klg-76-BAL&text=Quick+Calculus&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=1765308054756475-15830990373066760636-balancer-l7leveler-kubr-yp-klg-76-BAL&text=Quick+Calculus&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=1765308054756475-15830990373066760636-balancer-l7leveler-kubr-yp-klg-76-BAL&text=Quick+Calculus&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 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∫∫(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=1765308054756475-15830990373066760636-balancer-l7leveler-kubr-yp-klg-76-BAL&text=Quick+Calculus&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 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