или
25x2 = 5 2 y .
Это – каноническое уравнение параболического цилиндра в прямоугольной системе координат (O, e1, e2 , e3 ).
Поскольку
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4 5 0 3 5 |
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0 1 |
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−1 5 |
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3 5 |
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−4 (5 2 ) |
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−4 (5 2 ) |
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11 25 |
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4 5 3 (5 2 ) |
3 (5 2 ) |
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каноническая система координат (O, e1, e2 , e3 ) |
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имеет начало |
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4 т |
O = −1, |
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и направляющие векторы e1 |
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Задачи для самостоятельного решения
Приведите уравнения к каноническому виду при помощи перехода к новой прямоугольной системе координат и выясните расположение относительно исходной прямоугольной системы координат следующих поверхностей второго порядка:
а) 2x2 + 2 y2 +3z2 + 4xy + 2xz + 2 yz − 2x − 2 y −16z +53 = 0 ;
б) xy + xz + yz + 2 2x + 2 6 y − 2 3z −100 = 0 ;
в) x2 + 4 y2 + z2 + 4xy − 2xz − 4 yz − 2x −6 y + 6z = 0 ;
г) 4x2 −6 y2 + 4z2 − 4xz −8y − 4z −3 = 0 ;
д) 5x2 + 7 y2 + 6z2 + 4xz + 4 yz +10x +14 y +8z − 6 = 0 ;
е) x2 +5y2 + z2 + 2xy + 6xz + 2 yz − 2x = 0 ;
ж) −8x2 + 4 y2 + 4z2 + 4xy + 4xz −10 yz +12x + 24 y + 24z −9 = 0 ; з) −18x2 −32z2 +40xy −48xz −30 yz +20x−100 y−140z −250 =0 ; и) 2x2 + y2 + 2z2 − 2xy − 2 yz + x − 4 y −3z + 2 = 0 ;
к) 72 x2 + 72 y2 +8z2 −5xy − 4xz − 4 yz −8x −8y − 4z +36 = 0 ; л) 17x2 +17 y2 +11z2 −16xy +8xz −8yz −2x+34 y−44z +57 =0 ; м) 11x2 +2 y2 +5z2 +4xy −16xz +20 yz +22x−68y−34z +62=0 ;
н) 4x2 + y2 + 4z2 − 4xy −8xz + 4 yz −14x + y +8z +30 = 0 ; о) −xy + xz + yz + 2x + 2 y − 2z = 0 ;
п) y2 + 2xy + 4xz + 2 yz − 4x − 2 y = 0 ;
р) 3x2 +12 y2 + 27z2 −12xy +18xz −36 yz −3x + 6 y −9z −18 = 0 .
1.13. z =
5. ИНДИВИДУАЛЬНЫЕ ЗАДАНИЯ ДЛЯ ТИПОВОГО РАСЧЕТА
Задача 1
Найти область определения функции двух переменных (дать геометрическое истолкование).
1.1. z = |
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ln(x + y) . |
1.2. z = ln |
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1.3. z = ln |
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1.4. z = ln |
x −3 |
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y −5 |
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1.5. z = ln( y −sin x) . |
1.6. z = log y (x2 |
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+ y 2 −9) . |
1.7. z = |
1 |
arcsin |
x + y |
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1.8. z = |
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1.9. z = |
ln(x2 y) |
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y − x |
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1.10.z = ln(− x − y).
1.11.z = ln x + ln sin y .
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1.12. z = |
ln x |
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1x + (x2 + y 2 −1)(4 − x2 − y 2 ) .
1.14. z = 1 − x2 + y 2 −1 .
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1.15. z = |
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1.16. z = |
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1.17. z = (x + |
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1.18. z = |
1 − (x2 + y)2 ln y . |
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1.19. z = |
1 −x3 + ln( y2 |
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1.20. z = |
cos x − y . |
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1.21. z = arcsin(x + y) + |
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1.22. z = arccos |
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1.23. z = |
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1.24.z = sin(x2 + y 2 ) + arcsin(x − y) .
1.25.z = arcsin xy2 + arccos(1 − x) .
1.26.z = sin x cos y .
1.27.z = sinxy .
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1.28. |
z = |
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1.29. z = |
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1.30. z = ln x ln y .
Задача 2
Найти частные производные ∂∂xz , ∂∂yz от функции z = z(x, y) .
2.1. z = ln(x + x2 + y 2 ) .
2.2.z = ln( x + y2 ) .
2.3.z = ln(1 + x) ln(1 + y3 ) .
2.4. z = |
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2.5. z = xy2 ln(x2 + y) . |
2.6. z = ln(x5 + ln y) . |
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z = (1 + log y x)3 . |
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2.8. |
z = ln(sin x + cos y) . |
2.9. |
z = ln(3 |
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2.11. |
z = |
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2.12. |
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2.13. |
z = (x sin y + y cos x)2 . |
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2.15. z = |
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2.16. z = 3 |
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195 |
2.17. z = cos xx +− yy .
2.18. z = arcsin x2 − y 2 . x2 + y 2
2.19.z = −ln cos y .
x
2.20.z = (sin x)cos y .
2.21.z = x sin( x + y 2 ) .
2.22.z = arctg x y .
2.23.z = cos x2 .
x+ y
2.24. z = arcsin |
x |
x2 + y 2 . |
2.25.z = cos xy sin xy .
2.26.z = arctg 1 +xyx2 .
y
2.27.z = e x2 −y2 .
2.28.z = arccos ln x .
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2.29.z = arctg 1x−+xyy .
2.30.z = arcsin yx2 + arccos(1 − y) .
Задача 3
Вычислить производные сложных функций.
3.1. z = x sin y + y cos x, где x = |
u |
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y = u |
3v2 ; |
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3.2. z = e4 xy , где x = cos(1 −t), |
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3.3. z = x2 |
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3.4. z = (x + y3 ) ex2 +y2 , где x = cos(t 2 ), |
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3.5. z = arcsin |
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3.6. z = tg(x + 2x2 − y), где x = |
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3.7. z = |
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где x = v −u 2 v, |
y = u + v2u ; |
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∂u |
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3.8. z = arctg |
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∂z , |
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3.9. z = |
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y = t sin t ; |
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3.10. z = arcsin(x − y), где x = ln( |
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3.11. z = arctg |
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3.12. z = ln( |
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x ; ∂z |
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3.13. z = ln( e 2 x + e 6 y ), где y = x |
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3.29. z = sin 2 x + cos2 y , |
где x =π ln t , y = lnπ t ; |
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u + v2 |
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3.30. |
z = |
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x = |
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∂z |
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∂z |
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u |
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Задача 4.
Найти вторые частные производные указанных функций. Убедиться в том, что z"xy = z"yx .
4.1.z = ex2 −y 2
4.2.z = ctg(x + y)
4.3.z = tg(x / y)
4.4.z = cos(xy2 )
4.5.z = sin(x2 − y)
4.6.z = arctg(x + y)
4.7.z = arcsin(x − y)
4.8.z = arccos(2x + y)
4.9.z = arcctg(x − 3y)
4.10.z = ln(3x2 − 2 y2 )
4.11.z = e2x2 +y2
4.12.z = ctg( y / x)