Kryptinė išvestinė nusako funkcijos pasikeitimo greitį. Pavyzdžiui, jei yra apverstas paraboloidas
z
=
100
−
x
2
−
y
2
{\displaystyle z=100-x^{2}-y^{2}}
(
x
1
;
y
1
)
=
(
0
;
0
)
{\displaystyle (x_{1};y_{1})=(0;0)}
z lygus 0 (funkcija neturi kitimo greičio). Tai reiškia, kad lėtai didėja (arba mažėja) z reikšmė, kai x ir y yra maži, o kai x ir y reikšmės yra didesnės tuomet toks pat x ir y pasiketimas (pavyzdžiui dydžiu 0,1) įtakoja didelį reikšmės z pasikeitimą. Įstatykime, pavyzdžiui, pasikeitimą x ir y reikšmių dydžiu 1, vienu atveju, kai x =2 ir y =2 ir kitu atveju, kai x =4, y =4. Taigi
z
1
=
(
100
−
2
2
−
2
2
)
−
(
100
−
(
2
+
1
)
2
−
(
2
+
1
)
2
)
=
(
100
−
4
−
4
)
−
(
100
−
9
−
9
)
=
92
−
82
=
10
;
{\displaystyle z_{1}=(100-2^{2}-2^{2})-(100-(2+1)^{2}-(2+1)^{2})=(100-4-4)-(100-9-9)=92-82=10;}
z
2
=
(
100
−
4
2
−
4
2
)
−
(
100
−
(
4
+
1
)
2
−
(
4
+
1
)
2
)
=
(
100
−
16
−
16
)
−
(
100
−
25
−
25
)
=
68
−
50
=
18.
{\displaystyle z_{2}=(100-4^{2}-4^{2})-(100-(4+1)^{2}-(4+1)^{2})=(100-16-16)-(100-25-25)=68-50=18.}
Matome, kad su didesniomis x ir y reikšmėmis gaunamas ir didesnis z reikšmės pasikeitimas, pridėjus
Δ
x
{\displaystyle \Delta x}
Δ
y
{\displaystyle \Delta y}
x ir y atitinkamai.
Randame gradientą funkcijos
z
=
100
−
x
2
−
y
2
{\displaystyle z=100-x^{2}-y^{2}}
grad
z
=
−
2
x
i
−
2
y
j
;
{\displaystyle {\text{grad}}z=-2x\mathbf {i} -2y\mathbf {j} ;}
grad
z
(
2
;
2
)
=
−
2
⋅
2
i
−
2
⋅
2
j
=
−
4
i
−
4
j
;
{\displaystyle {\text{grad}}z(2;2)=-2\cdot 2\mathbf {i} -2\cdot 2\mathbf {j} =-4\mathbf {i} -4\mathbf {j} ;}
grad
z
(
4
;
4
)
=
−
2
⋅
4
i
−
2
⋅
4
j
=
−
8
i
−
8
j
.
{\displaystyle {\text{grad}}z(4;4)=-2\cdot 4\mathbf {i} -2\cdot 4\mathbf {j} =-8\mathbf {i} -8\mathbf {j} .}
Taške N(4; 4) funkcijos
z
=
100
−
x
2
−
y
2
{\displaystyle z=100-x^{2}-y^{2}}
Kryptinė išvestinė nusako funkcijos kitimo greitį tik tam tikra kryptimi, tarsi, jei vektorius, eis tik y kryptimi, t. y.
cos
α
=
0
{\displaystyle \cos \alpha =0}
cos
β
=
1
{\displaystyle \cos \beta =1}
z
=
100
−
x
2
−
y
2
{\displaystyle z=100-x^{2}-y^{2}}
y reikšmės.
Raskime, kada funkcijos kitimo greitis bus didžiausias ar vektoriaus
a
→
=
0
i
+
1
j
{\displaystyle {\vec {a}}=0\mathbf {i} +1\mathbf {j} }
b
→
=
1
i
+
1
j
{\displaystyle {\vec {b}}=1\mathbf {i} +1\mathbf {j} }
b ortas yra
b
→
∘
=
1
i
+
1
j
1
2
+
1
2
=
1
i
+
1
j
2
=
1
2
i
+
1
2
j
.
{\displaystyle {\vec {b}}^{\circ }={\frac {1\mathbf {i} +1\mathbf {j} }{\sqrt {1^{2}+1^{2}}}}={\frac {1\mathbf {i} +1\mathbf {j} }{\sqrt {2}}}={\frac {1}{\sqrt {2}}}\mathbf {i} +{\frac {1}{\sqrt {2}}}\mathbf {j} .}
O vektoriaus a ortas yra toks pat kaip vektorius a . Taigi, gauname kryptinių išvestinių dydžius, vektoriaus a ir vektoriaus b kryptimis.
∂
z
(
2
;
2
)
∂
l
=
a
→
∘
⋅
grad
z
(
2
;
2
)
=
0
⋅
(
−
4
)
+
1
⋅
(
−
4
)
=
−
4
;
{\displaystyle {\frac {\partial z(2;2)}{\partial l}}={\vec {a}}^{\circ }\cdot {\text{grad}}z(2;2)=0\cdot (-4)+1\cdot (-4)=-4;}
∂
z
(
2
;
2
)
∂
l
=
b
→
∘
⋅
grad
z
(
2
;
2
)
=
1
2
⋅
(
−
4
)
+
1
2
⋅
(
−
4
)
=
−
8
2
=
−
5.65685425.
{\displaystyle {\frac {\partial z(2;2)}{\partial l}}={\vec {b}}^{\circ }\cdot {\text{grad}}z(2;2)={\frac {1}{\sqrt {2}}}\cdot (-4)+{\frac {1}{\sqrt {2}}}\cdot (-4)={\frac {-8}{\sqrt {2}}}=-5.65685425.}
∂
z
(
4
;
4
)
∂
l
=
a
→
∘
⋅
grad
z
(
4
;
4
)
=
0
⋅
(
−
8
)
+
1
⋅
(
−
8
)
=
−
8
;
{\displaystyle {\frac {\partial z(4;4)}{\partial l}}={\vec {a}}^{\circ }\cdot {\text{grad}}z(4;4)=0\cdot (-8)+1\cdot (-8)=-8;}
∂
z
(
4
;
4
)
∂
l
=
b
→
∘
⋅
grad
z
(
4
;
4
)
=
1
2
⋅
(
−
8
)
+
1
2
⋅
(
−
8
)
=
−
16
2
=
−
11.3137085.
{\displaystyle {\frac {\partial z(4;4)}{\partial l}}={\vec {b}}^{\circ }\cdot {\text{grad}}z(4;4)={\frac {1}{\sqrt {2}}}\cdot (-8)+{\frac {1}{\sqrt {2}}}\cdot (-8)={\frac {-16}{\sqrt {2}}}=-11.3137085.}
Funkcijos kitimo greitis, šiame pavyzdyje (taške M(2; 2) ir taške N(4; 4)) yra didžiausias vektoriaus (b ) kryptimi, kuris su Ox ašimi sudaro 45 laipsnių kampą.
Galime matyti, kad funkcijos
z
=
100
−
x
2
−
y
2
{\displaystyle z=100-x^{2}-y^{2}}
M arba N ) ir koordinačių pradžios taškas O (0; 0). Apskaičiuokime atkarpų OM ir ON ilgius.
O
M
=
2
2
+
2
2
=
4
+
4
=
8
=
2
2
.
{\displaystyle OM={\sqrt {2^{2}+2^{2}}}={\sqrt {4+4}}={\sqrt {8}}=2{\sqrt {2}}.}
O
N
=
4
2
+
4
2
=
16
+
16
=
32
=
4
2
.
{\displaystyle ON={\sqrt {4^{2}+4^{2}}}={\sqrt {16+16}}={\sqrt {32}}=4{\sqrt {2}}.}
Parinkime taškus
E
(
0
;
2
2
)
{\displaystyle E(0;2{\sqrt {2}})}
F
(
0
;
4
2
)
.
{\displaystyle F(0;4{\sqrt {2}}).}
z
=
100
−
x
2
−
y
2
{\displaystyle z=100-x^{2}-y^{2}}
a
→
=
0
i
+
1
j
{\displaystyle {\vec {a}}=0\mathbf {i} +1\mathbf {j} }
grad
z
(
0
;
2
2
)
=
−
2
⋅
0
i
−
2
⋅
2
2
j
=
0
i
−
4
2
j
;
{\displaystyle {\text{grad}}z(0;2{\sqrt {2}})=-2\cdot 0\mathbf {i} -2\cdot 2{\sqrt {2}}\mathbf {j} =0\mathbf {i} -4{\sqrt {2}}\mathbf {j} ;}
grad
z
(
0
;
4
2
)
=
−
2
⋅
0
i
−
2
⋅
4
2
j
=
0
i
−
8
2
j
.
{\displaystyle {\text{grad}}z(0;4{\sqrt {2}})=-2\cdot 0\mathbf {i} -2\cdot 4{\sqrt {2}}\mathbf {j} =0\mathbf {i} -8{\sqrt {2}}\mathbf {j} .}
∂
z
(
0
;
2
2
)
∂
l
=
a
→
∘
⋅
grad
z
(
2
;
2
2
)
=
0
⋅
0
+
1
⋅
(
−
4
2
)
=
−
4
2
=
−
5.656854249
;
{\displaystyle {\frac {\partial z(0;2{\sqrt {2}})}{\partial l}}={\vec {a}}^{\circ }\cdot {\text{grad}}z(2;2{\sqrt {2}})=0\cdot 0+1\cdot (-4{\sqrt {2}})=-4{\sqrt {2}}=-5.656854249;}
∂
z
(
0
;
4
2
)
∂
l
=
a
→
∘
⋅
grad
z
(
2
;
4
2
)
=
0
⋅
0
+
1
⋅
(
−
8
2
)
=
−
8
2
=
−
11.3137085.
{\displaystyle {\frac {\partial z(0;4{\sqrt {2}})}{\partial l}}={\vec {a}}^{\circ }\cdot {\text{grad}}z(2;4{\sqrt {2}})=0\cdot 0+1\cdot (-8{\sqrt {2}})=-8{\sqrt {2}}=-11.3137085.}
Kaip matome, jei spindulys r plokštumoje xOy vienodas iki bet kokio taško, tai ir funkcijos
z
=
100
−
x
2
−
y
2
{\displaystyle z=100-x^{2}-y^{2}}
Pavyzdis . Rasime funkcijos
u
=
x
3
y
3
z
3
{\displaystyle u=x^{3}y^{3}z^{3}}
M
0
{\displaystyle M_{0}}
M
0
M
1
→
{\displaystyle {\vec {M_{0}M_{1}}}}
M
1
(
3
;
2
;
4
)
{\displaystyle M_{1}(3;2;4)}
Sprendimas . Randame
M
0
M
1
→
=
(
3
−
2
;
2
−
(
−
1
)
;
4
−
3
)
=
(
1
;
3
;
1
)
,
M
0
M
1
→
=
1
2
+
3
2
+
1
2
=
1
+
9
+
1
=
11
,
{\displaystyle {\vec {M_{0}M_{1}}}=(3-2;\;2-(-1);\;4-3)=(1;\;3;\;1),\quad {\vec {M_{0}M_{1}}}={\sqrt {1^{2}+3^{2}+1^{2}}}={\sqrt {1+9+1}}={\sqrt {11}},}
cos
α
=
1
11
,
cos
β
=
3
11
,
cos
γ
=
1
11
.
{\displaystyle \cos \alpha ={\frac {1}{\sqrt {11}}},\quad \cos \beta ={\frac {3}{\sqrt {11}}},\quad \cos \gamma ={\frac {1}{\sqrt {11}}}.}
cos
2
α
+
cos
2
β
+
cos
2
γ
=
1
11
+
9
11
+
1
11
=
11
11
=
1.
{\displaystyle \cos ^{2}\alpha +\cos ^{2}\beta +\cos ^{2}\gamma ={\frac {1}{11}}+{\frac {9}{11}}+{\frac {1}{11}}={\frac {11}{11}}=1.}
Toliau randame išvestines
∂
u
∂
x
=
3
x
2
y
3
z
3
,
∂
u
∂
y
=
3
x
3
y
2
z
3
,
∂
u
∂
x
=
3
x
3
y
3
z
2
{\displaystyle {\frac {\partial u}{\partial x}}=3x^{2}y^{3}z^{3},\quad {\frac {\partial u}{\partial y}}=3x^{3}y^{2}z^{3},\quad {\frac {\partial u}{\partial x}}=3x^{3}y^{3}z^{2}}
M
0
{\displaystyle M_{0}}
∂
u
∂
x
|
M
0
=
(
3
x
2
y
3
z
3
)
|
M
0
=
3
⋅
2
2
⋅
(
−
1
)
3
⋅
3
3
=
−
3
⋅
4
⋅
27
=
−
324
,
{\displaystyle {\frac {\partial u}{\partial x}}|_{M_{0}}=(3x^{2}y^{3}z^{3})|_{M_{0}}=3\cdot 2^{2}\cdot (-1)^{3}\cdot 3^{3}=-3\cdot 4\cdot 27=-324,}
∂
u
∂
y
|
M
0
=
(
3
x
3
y
2
z
3
)
|
M
0
=
3
⋅
2
3
⋅
(
−
1
)
2
⋅
3
3
=
3
⋅
8
⋅
27
=
648
,
{\displaystyle {\frac {\partial u}{\partial y}}|_{M_{0}}=(3x^{3}y^{2}z^{3})|_{M_{0}}=3\cdot 2^{3}\cdot (-1)^{2}\cdot 3^{3}=3\cdot 8\cdot 27=648,}
∂
u
∂
z
|
M
0
=
(
3
x
3
y
3
z
2
)
|
M
0
=
3
⋅
2
3
⋅
(
−
1
)
3
⋅
3
2
=
−
3
⋅
8
⋅
9
=
−
216.
{\displaystyle {\frac {\partial u}{\partial z}}|_{M_{0}}=(3x^{3}y^{3}z^{2})|_{M_{0}}=3\cdot 2^{3}\cdot (-1)^{3}\cdot 3^{2}=-3\cdot 8\cdot 9=-216.}
Įrašę šias išvvestinių ir krypties kosinusų reikšmes į formulę, gauname:
∂
u
∂
l
=
∂
u
∂
x
cos
α
+
∂
u
∂
y
cos
β
+
∂
u
∂
z
cos
γ
=
−
324
⋅
1
11
+
648
⋅
3
11
−
216
⋅
1
11
=
{\displaystyle {\frac {\partial u}{\partial l}}={\frac {\partial u}{\partial x}}\cos \alpha +{\frac {\partial u}{\partial y}}\cos \beta +{\frac {\partial u}{\partial z}}\cos \gamma =-324\cdot {\frac {1}{\sqrt {11}}}+648\cdot {\frac {3}{\sqrt {11}}}-216\cdot {\frac {1}{\sqrt {11}}}=}
=
−
324
⋅
1
11
+
1944
11
−
216
⋅
1
11
=
1944
−
324
−
216
11
=
1404
11
=
423.3219278.
{\displaystyle =-324\cdot {\frac {1}{\sqrt {11}}}+{\frac {1944}{\sqrt {11}}}-216\cdot {\frac {1}{\sqrt {11}}}={\frac {1944-324-216}{\sqrt {11}}}={\frac {1404}{\sqrt {11}}}=423.3219278.}
Pavyzdys. Apskaičiuoti išvestinę funkcijos
z
=
x
2
+
y
2
x
{\displaystyle z=x^{2}+y^{2}x}
M (1; 2) kriptimi vektoriaus
M
M
1
→
,
{\displaystyle {\vec {MM_{1}}},}
M
1
{\displaystyle M_{1}}
Sprendimas . Rasime vienetinį vektorių
l
→
,
{\displaystyle {\vec {l}},}
M
M
1
→
=
(
3
−
1
;
0
−
2
)
=
(
2
;
−
2
)
=
2
i
→
−
2
j
→
;
{\displaystyle {\vec {MM_{1}}}=(3-1;\,0-2)=(2;\,-2)=2{\vec {i}}-2{\vec {j}};}
‖
M
M
1
→
‖
=
2
2
+
(
−
2
)
2
=
4
+
4
=
8
=
2
2
=
2.828427125
;
{\displaystyle \|{\vec {MM_{1}}}\|={\sqrt {2^{2}+(-2)^{2}}}={\sqrt {4+4}}={\sqrt {8}}=2{\sqrt {2}}=2.828427125;}
M
M
1
→
‖
M
M
1
→
‖
=
2
i
→
−
2
j
→
2
2
=
1
2
i
→
−
1
2
j
→
,
{\displaystyle {\frac {\vec {MM_{1}}}{\|{\vec {MM_{1}}}\|}}={\frac {2{\vec {i}}-2{\vec {j}}}{2{\sqrt {2}}}}={\frac {1}{\sqrt {2}}}{\vec {i}}-{\frac {1}{\sqrt {2}}}{\vec {j}},}
Iš kur
cos
α
=
1
2
,
cos
β
=
−
1
2
.
{\displaystyle \cos \alpha ={\frac {1}{\sqrt {2}}},\;\cos \beta =-{\frac {1}{\sqrt {2}}}.}
M (1; 2):
f
x
′
(
x
,
y
)
=
2
x
+
y
2
,
f
y
′
(
x
,
y
)
=
2
x
y
,
{\displaystyle f_{x}'(x,\;y)=2x+y^{2},\;f_{y}'(x,\;y)=2xy,}
f
x
′
(
1
;
2
)
=
2
⋅
1
+
2
2
=
2
+
4
=
6
,
f
y
′
(
1
;
2
)
=
2
⋅
1
⋅
2
=
4.
{\displaystyle f_{x}'(1;\;2)=2\cdot 1+2^{2}=2+4=6,\;f_{y}'(1;\;2)=2\cdot 1\cdot 2=4.}
Pagal formulę
∂
z
∂
l
=
∂
z
∂
x
cos
α
+
∂
z
∂
y
cos
β
{\displaystyle {\frac {\partial z}{\partial l}}={\frac {\partial z}{\partial x}}\cos \alpha +{\frac {\partial z}{\partial y}}\cos \beta }
∂
z
∂
l
=
6
⋅
1
2
−
4
⋅
1
2
=
6
−
4
2
=
2
2
=
1.414213562.
{\displaystyle {\frac {\partial z}{\partial l}}=6\cdot {\frac {1}{\sqrt {2}}}-4\cdot {\frac {1}{\sqrt {2}}}={\frac {6-4}{\sqrt {2}}}={\frac {2}{\sqrt {2}}}=1.414213562.}
Pavyzdys . Duota funkcija
u
=
x
2
+
y
2
+
z
2
.
{\displaystyle u=x^{2}+y^{2}+z^{2}.}
∂
u
∂
s
{\displaystyle {\frac {\partial u}{\partial s}}}
M (1; 1; 1):a) kryptimi vektoriaus
S
1
=
2
i
+
j
+
3
k
;
{\displaystyle {\mathsf {S_{1}}}=2{\mathsf {i}}+{\mathsf {j}}+3{\mathsf {k}};}
b) kryptimi vektoriaus
S
2
=
i
+
j
+
k
.
{\displaystyle {\mathsf {S_{2}}}={\mathsf {i}}+{\mathsf {j}}+{\mathsf {k}}.}
Sprendimas .a) Randame nukreipiančius kosinusus vektoriaus
S
1
:
{\displaystyle {\mathsf {S_{1}}}:}
cos
α
=
2
2
2
+
1
2
+
3
2
=
2
4
+
1
+
9
=
2
14
,
{\displaystyle \cos \alpha ={\frac {2}{\sqrt {2^{2}+1^{2}+3^{2}}}}={\frac {2}{\sqrt {4+1+9}}}={\frac {2}{\sqrt {14}}},}
cos
β
=
1
2
2
+
1
2
+
3
2
=
1
4
+
1
+
9
=
1
14
,
{\displaystyle \cos \beta ={\frac {1}{\sqrt {2^{2}+1^{2}+3^{2}}}}={\frac {1}{\sqrt {4+1+9}}}={\frac {1}{\sqrt {14}}},}
cos
γ
=
3
2
2
+
1
2
+
3
2
=
3
4
+
1
+
9
=
3
14
.
{\displaystyle \cos \gamma ={\frac {3}{\sqrt {2^{2}+1^{2}+3^{2}}}}={\frac {3}{\sqrt {4+1+9}}}={\frac {3}{\sqrt {14}}}.}
Todėl,
∂
u
∂
s
1
=
∂
u
∂
x
2
14
+
∂
u
∂
y
1
14
+
∂
u
∂
z
3
14
.
{\displaystyle {\frac {\partial u}{\partial s_{1}}}={\frac {\partial u}{\partial x}}{\frac {2}{\sqrt {14}}}+{\frac {\partial u}{\partial y}}{\frac {1}{\sqrt {14}}}+{\frac {\partial u}{\partial z}}{\frac {3}{\sqrt {14}}}.}
Dalinės išvestinės taške M(1; 1; 1) bus
∂
u
∂
x
=
2
x
,
∂
u
∂
y
=
2
y
,
∂
u
∂
z
=
2
z
;
{\displaystyle {\frac {\partial u}{\partial x}}=2x,\quad {\frac {\partial u}{\partial y}}=2y,\quad {\frac {\partial u}{\partial z}}=2z;}
(
∂
u
∂
x
)
M
=
2
⋅
1
=
2
,
(
∂
u
∂
y
)
M
=
2
,
(
∂
u
∂
z
)
M
=
2.
{\displaystyle \left({\frac {\partial u}{\partial x}}\right)_{M}=2\cdot 1=2,\quad \left({\frac {\partial u}{\partial y}}\right)_{M}=2,\quad \left({\frac {\partial u}{\partial z}}\right)_{M}=2.}
Taigi,
∂
u
∂
s
1
=
2
⋅
2
14
+
2
⋅
1
14
+
2
⋅
3
14
=
12
14
=
3.207134903.
{\displaystyle {\frac {\partial u}{\partial s_{1}}}=2\cdot {\frac {2}{\sqrt {14}}}+2\cdot {\frac {1}{\sqrt {14}}}+2\cdot {\frac {3}{\sqrt {14}}}={\frac {12}{\sqrt {14}}}=3.207134903.}
b) Randame nukreipiančius kosinusus vektoriaus
S
2
:
{\displaystyle {\mathsf {S_{2}}}:}
cos
α
=
1
1
2
+
1
2
+
1
2
=
1
3
,
{\displaystyle \cos \alpha ={\frac {1}{\sqrt {1^{2}+1^{2}+1^{2}}}}={\frac {1}{\sqrt {3}}},}
cos
β
=
1
1
2
+
1
2
+
1
2
=
1
3
,
{\displaystyle \cos \beta ={\frac {1}{\sqrt {1^{2}+1^{2}+1^{2}}}}={\frac {1}{\sqrt {3}}},}
cos
γ
=
1
1
2
+
1
2
+
1
2
=
1
3
.
{\displaystyle \cos \gamma ={\frac {1}{\sqrt {1^{2}+1^{2}+1^{2}}}}={\frac {1}{\sqrt {3}}}.}
Todėl,
∂
u
∂
s
2
=
2
⋅
1
3
+
2
⋅
1
3
+
2
⋅
1
3
=
6
3
=
2
3
=
3.464101615.
{\displaystyle {\frac {\partial u}{\partial s_{2}}}=2\cdot {\frac {1}{\sqrt {3}}}+2\cdot {\frac {1}{\sqrt {3}}}+2\cdot {\frac {1}{\sqrt {3}}}={\frac {6}{\sqrt {3}}}=2{\sqrt {3}}=3.464101615.}
Pastebėsime, kad
2
3
>
12
14
.
{\displaystyle 2{\sqrt {3}}>{\frac {12}{\sqrt {14}}}.}
Taške A(1; 2; 3) apskaičiuokite lauko
u
(
x
;
y
;
z
)
=
z
2
+
2
arctan
(
x
−
y
)
{\displaystyle u(x;y;z)=z^{2}+2\arctan(x-y)}
∂
u
∂
x
=
2
1
+
(
x
−
y
)
2
;
∂
u
∂
y
=
−
2
1
+
(
x
−
y
)
2
;
∂
u
∂
z
=
2
z
;
{\displaystyle {\frac {\partial u}{\partial x}}={\frac {2}{1+(x-y)^{2}}};\;{\frac {\partial u}{\partial y}}={\frac {-2}{1+(x-y)^{2}}};\;{\frac {\partial u}{\partial z}}=2z;}
g
r
a
d
u
|
A
=
2
1
+
(
1
−
2
)
2
i
−
2
1
+
(
1
−
2
)
2
j
+
2
⋅
3
k
=
i
−
j
+
6
k
.
{\displaystyle gradu|_{A}={\frac {2}{1+(1-2)^{2}}}i-{\frac {2}{1+(1-2)^{2}}}j+2\cdot 3k=i-j+6k.}
|
A
B
|
=
(
−
3
)
2
+
(
−
2
)
2
+
(
−
2
)
2
=
17
;
{\displaystyle |AB|={\sqrt {(-3)^{2}+(-2)^{2}+(-2)^{2}}}={\sqrt {17}};}
cos
α
=
−
3
17
,
cos
β
=
−
2
17
,
cos
γ
=
−
2
17
.
{\displaystyle \cos \alpha ={\frac {-3}{\sqrt {17}}},\;\cos \beta ={\frac {-2}{\sqrt {17}}},\;\cos \gamma ={\frac {-2}{\sqrt {17}}}.}
kryptinė išvestinė:
∂
u
∂
s
|
A
=
1
⋅
−
3
17
−
1
⋅
−
2
17
+
6
⋅
−
2
17
=
−
13
17
.
{\displaystyle {\frac {\partial u}{\partial s}}|_{A}=1\cdot {\frac {-3}{\sqrt {17}}}-1\cdot {\frac {-2}{\sqrt {17}}}+6\cdot {\frac {-2}{\sqrt {17}}}=-{\frac {13}{\sqrt {17}}}.}
Pavyzdys . Reikia surasti išvestinę funkcijos
z
=
y
y
−
x
,
{\displaystyle z={\frac {y}{y-x}},}
Ox , taške M(1; 3). Kitaip tariant, reikia surasti kryptinę išvestinę funkcijos
z
=
y
y
−
x
{\displaystyle z={\frac {y}{y-x}}}
Ox ašimi sudaro 60 laipsnių kampą.Sprendimas .
x
2
=
cos
(
60
o
)
=
cos
π
3
=
1
2
=
0.5
,
{\displaystyle x_{2}=\cos(60^{o})=\cos {\frac {\pi }{3}}={\frac {1}{2}}=0.5,}
y
2
=
sin
(
60
o
)
=
sin
π
3
=
3
2
=
0.866025403.
{\displaystyle y_{2}=\sin(60^{o})=\sin {\frac {\pi }{3}}={\frac {\sqrt {3}}{2}}=0.866025403.}
Krypties vektorius yra
e
→
=
1
2
i
+
3
2
j
=
(
1
2
;
3
2
)
.
{\displaystyle {\vec {e}}={\frac {1}{2}}i+{\frac {\sqrt {3}}{2}}j=({\frac {1}{2}};\;{\frac {\sqrt {3}}{2}}).}
Randame dalines išvestines:
z
x
′
=
−
−
y
(
y
−
x
)
2
=
y
(
y
−
x
)
2
;
{\displaystyle z_{x}'=-{\frac {-y}{(y-x)^{2}}}={\frac {y}{(y-x)^{2}}};}
z
y
′
=
(
y
y
−
x
)
′
=
y
′
(
y
−
x
)
−
y
(
y
−
x
)
′
(
y
−
x
)
2
=
1
⋅
(
y
−
x
)
−
y
⋅
1
(
y
−
x
)
2
=
−
x
(
y
−
x
)
2
.
{\displaystyle z_{y}'=({\frac {y}{y-x}})'={\frac {y'(y-x)-y(y-x)'}{(y-x)^{2}}}={\frac {1\cdot (y-x)-y\cdot 1}{(y-x)^{2}}}=-{\frac {x}{(y-x)^{2}}}.}
Dalinių išvestinių reikšmės taške M(1; 3) yra:
∂
z
∂
x
|
M
=
3
(
3
−
1
)
2
=
3
2
2
=
3
4
,
{\displaystyle {\frac {\partial z}{\partial x}}|_{M}={\frac {3}{(3-1)^{2}}}={\frac {3}{2^{2}}}={\frac {3}{4}},}
∂
z
∂
y
|
M
=
−
1
(
3
−
1
)
2
=
−
1
2
2
=
−
1
4
.
{\displaystyle {\frac {\partial z}{\partial y}}|_{M}=-{\frac {1}{(3-1)^{2}}}=-{\frac {1}{2^{2}}}=-{\frac {1}{4}}.}
grad
z
(
1
;
3
)
=
(
3
4
;
−
1
4
)
.
{\displaystyle {\text{grad}}z(1;3)=\left({\frac {3}{4}};-{\frac {1}{4}}\right).}
Kryptinės išvestinės reikšmė yra:
∂
z
(
1
;
3
)
∂
l
=
e
→
⋅
grad
z
(
1
;
3
)
=
1
2
⋅
3
4
+
3
2
⋅
(
−
1
4
)
=
3
−
3
8
.
{\displaystyle {\frac {\partial z(1;3)}{\partial l}}={\vec {e}}\cdot {\text{grad}}z(1;3)={\frac {1}{2}}\cdot {\frac {3}{4}}+{\frac {\sqrt {3}}{2}}\cdot \left(-{\frac {1}{4}}\right)={\frac {3-{\sqrt {3}}}{8}}.}
Pavyzdys. Apskaičiuoti išvestinę funkcijos
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
M (1; 2) kriptimi vektoriaus
M
M
1
→
,
{\displaystyle {\vec {MM_{1}}},}
M
1
{\displaystyle M_{1}}
Sprendimas . Rasime vienetinį vektorių
l
→
,
{\displaystyle {\vec {l}},}
M
M
1
→
=
(
3
−
1
;
4
−
2
)
=
(
2
;
2
)
=
2
i
+
2
j
;
{\displaystyle {\vec {MM_{1}}}=(3-1;\,4-2)=(2;\,2)=2i+2j;}
‖
M
M
1
→
‖
=
2
2
+
2
2
=
4
+
4
=
8
=
2
2
=
2.828427125
;
{\displaystyle \|{\vec {MM_{1}}}\|={\sqrt {2^{2}+2^{2}}}={\sqrt {4+4}}={\sqrt {8}}=2{\sqrt {2}}=2.828427125;}
M
M
1
→
‖
M
M
1
→
‖
=
2
i
+
2
j
2
2
=
1
2
i
+
1
2
j
,
{\displaystyle {\frac {\vec {MM_{1}}}{\|{\vec {MM_{1}}}\|}}={\frac {2i+2j}{2{\sqrt {2}}}}={\frac {1}{\sqrt {2}}}i+{\frac {1}{\sqrt {2}}}j,}
Iš kur
cos
α
=
1
2
,
cos
β
=
1
2
.
{\displaystyle \cos \alpha ={\frac {1}{\sqrt {2}}},\;\cos \beta ={\frac {1}{\sqrt {2}}}.}
M (1; 2):
f
x
′
(
x
,
y
)
=
2
x
,
f
y
′
(
x
,
y
)
=
2
y
,
{\displaystyle f_{x}'(x,\;y)=2x,\;f_{y}'(x,\;y)=2y,}
f
x
′
(
1
;
2
)
=
2
⋅
1
=
2
,
f
y
′
(
1
;
2
)
=
2
⋅
2
=
4.
{\displaystyle f_{x}'(1;\;2)=2\cdot 1=2,\;f_{y}'(1;\;2)=2\cdot 2=4.}
Pagal formulę
∂
z
∂
l
=
∂
z
∂
x
cos
α
+
∂
z
∂
y
cos
β
{\displaystyle {\frac {\partial z}{\partial l}}={\frac {\partial z}{\partial x}}\cos \alpha +{\frac {\partial z}{\partial y}}\cos \beta }
∂
z
∂
l
=
2
⋅
1
2
+
4
⋅
1
2
=
2
+
4
2
=
6
2
=
4.242640687.
{\displaystyle {\frac {\partial z}{\partial l}}=2\cdot {\frac {1}{\sqrt {2}}}+4\cdot {\frac {1}{\sqrt {2}}}={\frac {2+4}{\sqrt {2}}}={\frac {6}{\sqrt {2}}}=4.242640687.}
Pavyzdys. Apskaičiuoti išvestinę funkcijos
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
M (1; 2) kriptimi vektoriaus
M
M
1
→
,
{\displaystyle {\vec {MM_{1}}},}
M
1
{\displaystyle M_{1}}
Sprendimas . Rasime vienetinį vektorių
l
→
,
{\displaystyle {\vec {l}},}
M
M
1
→
=
(
4
−
1
;
3
−
2
)
=
(
3
;
1
)
=
3
i
+
1
j
;
{\displaystyle {\vec {MM_{1}}}=(4-1;\,3-2)=(3;\,1)=3i+1j;}
‖
M
M
1
→
‖
=
3
2
+
1
2
=
9
+
1
=
10
=
3.16227766
;
{\displaystyle \|{\vec {MM_{1}}}\|={\sqrt {3^{2}+1^{2}}}={\sqrt {9+1}}={\sqrt {10}}=3.16227766;}
M
M
1
→
‖
M
M
1
→
‖
=
3
i
+
1
j
10
=
3
10
i
+
1
10
j
,
{\displaystyle {\frac {\vec {MM_{1}}}{\|{\vec {MM_{1}}}\|}}={\frac {3i+1j}{\sqrt {10}}}={\frac {3}{\sqrt {10}}}i+{\frac {1}{\sqrt {10}}}j,}
Iš kur
cos
α
=
3
10
,
cos
β
=
1
10
.
{\displaystyle \cos \alpha ={\frac {3}{\sqrt {10}}},\;\cos \beta ={\frac {1}{\sqrt {10}}}.}
M (1; 2):
f
x
′
(
x
,
y
)
=
2
x
,
f
y
′
(
x
,
y
)
=
2
y
,
{\displaystyle f_{x}'(x,\;y)=2x,\;f_{y}'(x,\;y)=2y,}
f
x
′
(
1
;
2
)
=
2
⋅
1
=
2
,
f
y
′
(
1
;
2
)
=
2
⋅
2
=
4.
{\displaystyle f_{x}'(1;\;2)=2\cdot 1=2,\;f_{y}'(1;\;2)=2\cdot 2=4.}
Pagal formulę
∂
z
∂
l
=
∂
z
∂
x
cos
α
+
∂
z
∂
y
cos
β
{\displaystyle {\frac {\partial z}{\partial l}}={\frac {\partial z}{\partial x}}\cos \alpha +{\frac {\partial z}{\partial y}}\cos \beta }
∂
z
∂
l
=
2
⋅
3
10
+
4
⋅
1
10
=
6
+
4
10
=
10
10
=
10
=
3.16227766.
{\displaystyle {\frac {\partial z}{\partial l}}=2\cdot {\frac {3}{\sqrt {10}}}+4\cdot {\frac {1}{\sqrt {10}}}={\frac {6+4}{\sqrt {10}}}={\frac {10}{\sqrt {10}}}={\sqrt {10}}=3.16227766.}
Pavyzdys. Apskaičiuoti išvestinę funkcijos
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
M (1; 2) kriptimi vektoriaus
M
M
1
→
,
{\displaystyle {\vec {MM_{1}}},}
M
1
{\displaystyle M_{1}}
Sprendimas . Rasime vienetinį vektorių
l
→
,
{\displaystyle {\vec {l}},}
M
M
1
→
=
(
6
−
1
;
8
−
2
)
=
(
5
;
6
)
=
5
i
+
6
j
;
{\displaystyle {\vec {MM_{1}}}=(6-1;\,8-2)=(5;\,6)=5i+6j;}
‖
M
M
1
→
‖
=
5
2
+
6
2
=
25
+
36
=
61
=
7.810249676
;
{\displaystyle \|{\vec {MM_{1}}}\|={\sqrt {5^{2}+6^{2}}}={\sqrt {25+36}}={\sqrt {61}}=7.810249676;}
M
M
1
→
‖
M
M
1
→
‖
=
5
i
+
6
j
61
=
5
61
i
+
6
61
j
,
{\displaystyle {\frac {\vec {MM_{1}}}{\|{\vec {MM_{1}}}\|}}={\frac {5i+6j}{\sqrt {61}}}={\frac {5}{\sqrt {61}}}i+{\frac {6}{\sqrt {61}}}j,}
Iš kur
cos
α
=
5
61
,
cos
β
=
6
61
.
{\displaystyle \cos \alpha ={\frac {5}{\sqrt {61}}},\;\cos \beta ={\frac {6}{\sqrt {61}}}.}
M (1; 2):
f
x
′
(
x
,
y
)
=
2
x
,
f
y
′
(
x
,
y
)
=
2
y
,
{\displaystyle f_{x}'(x,\;y)=2x,\;f_{y}'(x,\;y)=2y,}
f
x
′
(
1
;
2
)
=
2
⋅
1
=
2
,
f
y
′
(
1
;
2
)
=
2
⋅
2
=
4.
{\displaystyle f_{x}'(1;\;2)=2\cdot 1=2,\;f_{y}'(1;\;2)=2\cdot 2=4.}
Pagal formulę
∂
z
∂
l
=
∂
z
∂
x
cos
α
+
∂
z
∂
y
cos
β
{\displaystyle {\frac {\partial z}{\partial l}}={\frac {\partial z}{\partial x}}\cos \alpha +{\frac {\partial z}{\partial y}}\cos \beta }
∂
z
∂
l
=
2
⋅
5
61
+
4
⋅
6
61
=
10
+
24
61
=
34
61
=
4.353253918.
{\displaystyle {\frac {\partial z}{\partial l}}=2\cdot {\frac {5}{\sqrt {61}}}+4\cdot {\frac {6}{\sqrt {61}}}={\frac {10+24}{\sqrt {61}}}={\frac {34}{\sqrt {61}}}=4.353253918.}
Pavyzdys. Apskaičiuoti išvestinę funkcijos
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
M (1; 2) kriptimi vektoriaus
M
M
1
→
,
{\displaystyle {\vec {MM_{1}}},}
M
1
{\displaystyle M_{1}}
Sprendimas . Rasime vienetinį vektorių
l
→
,
{\displaystyle {\vec {l}},}
M
M
1
→
=
(
8
−
1
;
6
−
2
)
=
(
7
;
4
)
=
7
i
+
4
j
;
{\displaystyle {\vec {MM_{1}}}=(8-1;\,6-2)=(7;\,4)=7i+4j;}
‖
M
M
1
→
‖
=
7
2
+
4
2
=
49
+
16
=
65
=
8.062257748
;
{\displaystyle \|{\vec {MM_{1}}}\|={\sqrt {7^{2}+4^{2}}}={\sqrt {49+16}}={\sqrt {65}}=8.062257748;}
M
M
1
→
‖
M
M
1
→
‖
=
7
i
+
4
j
65
=
7
65
i
+
4
65
j
,
{\displaystyle {\frac {\vec {MM_{1}}}{\|{\vec {MM_{1}}}\|}}={\frac {7i+4j}{\sqrt {65}}}={\frac {7}{\sqrt {65}}}i+{\frac {4}{\sqrt {65}}}j,}
Iš kur
cos
α
=
7
65
,
cos
β
=
4
65
.
{\displaystyle \cos \alpha ={\frac {7}{\sqrt {65}}},\;\cos \beta ={\frac {4}{\sqrt {65}}}.}
M (1; 2):
f
x
′
(
x
,
y
)
=
2
x
,
f
y
′
(
x
,
y
)
=
2
y
,
{\displaystyle f_{x}'(x,\;y)=2x,\;f_{y}'(x,\;y)=2y,}
f
x
′
(
1
;
2
)
=
2
⋅
1
=
2
,
f
y
′
(
1
;
2
)
=
2
⋅
2
=
4.
{\displaystyle f_{x}'(1;\;2)=2\cdot 1=2,\;f_{y}'(1;\;2)=2\cdot 2=4.}
Pagal formulę
∂
z
∂
l
=
∂
z
∂
x
cos
α
+
∂
z
∂
y
cos
β
{\displaystyle {\frac {\partial z}{\partial l}}={\frac {\partial z}{\partial x}}\cos \alpha +{\frac {\partial z}{\partial y}}\cos \beta }
∂
z
∂
l
=
2
⋅
7
65
+
4
⋅
4
65
=
14
+
16
65
=
30
65
=
3.721042038.
{\displaystyle {\frac {\partial z}{\partial l}}=2\cdot {\frac {7}{\sqrt {65}}}+4\cdot {\frac {4}{\sqrt {65}}}={\frac {14+16}{\sqrt {65}}}={\frac {30}{\sqrt {65}}}=3.721042038.}
Pavyzdys. Apskaičiuoti išvestinę funkcijos
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
M (1; 2) kriptimi vektoriaus
M
M
1
→
,
{\displaystyle {\vec {MM_{1}}},}
M
1
{\displaystyle M_{1}}
Sprendimas . Rasime vienetinį vektorių
l
→
,
{\displaystyle {\vec {l}},}
M
M
1
→
=
(
3
−
1
;
6
−
2
)
=
(
2
;
4
)
=
2
i
+
4
j
;
{\displaystyle {\vec {MM_{1}}}=(3-1;\,6-2)=(2;\,4)=2i+4j;}
‖
M
M
1
→
‖
=
2
2
+
4
2
=
4
+
16
=
20
=
2
5
=
4.472135955
;
{\displaystyle \|{\vec {MM_{1}}}\|={\sqrt {2^{2}+4^{2}}}={\sqrt {4+16}}={\sqrt {20}}=2{\sqrt {5}}=4.472135955;}
M
M
1
→
‖
M
M
1
→
‖
=
2
i
+
4
j
2
5
=
1
5
i
+
2
5
j
,
{\displaystyle {\frac {\vec {MM_{1}}}{\|{\vec {MM_{1}}}\|}}={\frac {2i+4j}{2{\sqrt {5}}}}={\frac {1}{\sqrt {5}}}i+{\frac {2}{\sqrt {5}}}j,}
Iš kur
cos
α
=
1
5
,
cos
β
=
2
5
.
{\displaystyle \cos \alpha ={\frac {1}{\sqrt {5}}},\;\cos \beta ={\frac {2}{\sqrt {5}}}.}
M (1; 2):
f
x
′
(
x
,
y
)
=
2
x
,
f
y
′
(
x
,
y
)
=
2
y
,
{\displaystyle f_{x}'(x,\;y)=2x,\;f_{y}'(x,\;y)=2y,}
f
x
′
(
1
;
2
)
=
2
⋅
1
=
2
,
f
y
′
(
1
;
2
)
=
2
⋅
2
=
4.
{\displaystyle f_{x}'(1;\;2)=2\cdot 1=2,\;f_{y}'(1;\;2)=2\cdot 2=4.}
Pagal formulę
∂
z
∂
l
=
∂
z
∂
x
cos
α
+
∂
z
∂
y
cos
β
{\displaystyle {\frac {\partial z}{\partial l}}={\frac {\partial z}{\partial x}}\cos \alpha +{\frac {\partial z}{\partial y}}\cos \beta }
∂
z
∂
l
=
2
⋅
1
5
+
4
⋅
2
5
=
2
+
8
5
=
10
5
=
2
5
=
4.472135955.
{\displaystyle {\frac {\partial z}{\partial l}}=2\cdot {\frac {1}{\sqrt {5}}}+4\cdot {\frac {2}{\sqrt {5}}}={\frac {2+8}{\sqrt {5}}}={\frac {10}{\sqrt {5}}}=2{\sqrt {5}}=4.472135955.}
Pavyzdys. Apskaičiuoti išvestinę funkcijos
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
M (1; 2) kriptimi vektoriaus
M
M
1
→
,
{\displaystyle {\vec {MM_{1}}},}
M
1
{\displaystyle M_{1}}
Sprendimas . Rasime vienetinį vektorių
l
→
,
{\displaystyle {\vec {l}},}
M
M
1
→
=
(
1
−
1
;
6
−
2
)
=
(
0
;
4
)
=
0
⋅
i
+
4
⋅
j
;
{\displaystyle {\vec {MM_{1}}}=(1-1;\,6-2)=(0;\,4)=0\cdot \mathbf {i} +4\cdot \mathbf {j} ;}
‖
M
M
1
→
‖
=
0
2
+
4
2
=
0
+
16
=
4
;
{\displaystyle \|{\vec {MM_{1}}}\|={\sqrt {0^{2}+4^{2}}}={\sqrt {0+16}}=4;}
M
M
1
→
‖
M
M
1
→
‖
=
0
i
+
4
j
4
=
0
i
+
1
j
,
{\displaystyle {\frac {\vec {MM_{1}}}{\|{\vec {MM_{1}}}\|}}={\frac {0i+4j}{4}}=0i+1j,}
Iš kur
cos
α
=
0
,
cos
β
=
1.
{\displaystyle \cos \alpha =0,\;\cos \beta =1.}
M (1; 2):
f
x
′
(
x
,
y
)
=
2
x
,
f
y
′
(
x
,
y
)
=
2
y
,
{\displaystyle f_{x}'(x,\;y)=2x,\;f_{y}'(x,\;y)=2y,}
f
x
′
(
1
;
2
)
=
2
⋅
1
=
2
,
f
y
′
(
1
;
2
)
=
2
⋅
2
=
4.
{\displaystyle f_{x}'(1;\;2)=2\cdot 1=2,\;f_{y}'(1;\;2)=2\cdot 2=4.}
Pagal formulę
∂
z
∂
l
=
∂
z
∂
x
cos
α
+
∂
z
∂
y
cos
β
{\displaystyle {\frac {\partial z}{\partial l}}={\frac {\partial z}{\partial x}}\cos \alpha +{\frac {\partial z}{\partial y}}\cos \beta }
∂
z
∂
l
=
2
⋅
0
+
4
⋅
1
=
4.
{\displaystyle {\frac {\partial z}{\partial l}}=2\cdot 0+4\cdot 1=4.}
Pavyzdys. Apskaičiuoti išvestinę funkcijos
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
M (1; 2) kriptimi vektoriaus
M
M
1
→
,
{\displaystyle {\vec {MM_{1}}},}
M
1
{\displaystyle M_{1}}
Sprendimas . Rasime vienetinį vektorių
l
→
,
{\displaystyle {\vec {l}},}
M
M
1
→
=
(
3
−
1
;
2
−
2
)
=
(
2
;
0
)
=
2
⋅
i
+
0
⋅
j
;
{\displaystyle {\vec {MM_{1}}}=(3-1;\,2-2)=(2;\,0)=2\cdot \mathbf {i} +0\cdot \mathbf {j} ;}
‖
M
M
1
→
‖
=
2
2
+
0
2
=
4
+
0
=
2
;
{\displaystyle \|{\vec {MM_{1}}}\|={\sqrt {2^{2}+0^{2}}}={\sqrt {4+0}}=2;}
M
M
1
→
‖
M
M
1
→
‖
=
2
i
+
0
j
2
=
1
i
+
0
j
,
{\displaystyle {\frac {\vec {MM_{1}}}{\|{\vec {MM_{1}}}\|}}={\frac {2i+0j}{2}}=1i+0j,}
Iš kur
cos
α
=
1
,
cos
β
=
0.
{\displaystyle \cos \alpha =1,\;\cos \beta =0.}
M (1; 2):
f
x
′
(
x
,
y
)
=
2
x
,
f
y
′
(
x
,
y
)
=
2
y
,
{\displaystyle f_{x}'(x,\;y)=2x,\;f_{y}'(x,\;y)=2y,}
f
x
′
(
1
;
2
)
=
2
⋅
1
=
2
,
f
y
′
(
1
;
2
)
=
2
⋅
2
=
4.
{\displaystyle f_{x}'(1;\;2)=2\cdot 1=2,\;f_{y}'(1;\;2)=2\cdot 2=4.}
Pagal formulę
∂
z
∂
l
=
∂
z
∂
x
cos
α
+
∂
z
∂
y
cos
β
{\displaystyle {\frac {\partial z}{\partial l}}={\frac {\partial z}{\partial x}}\cos \alpha +{\frac {\partial z}{\partial y}}\cos \beta }
∂
z
∂
l
=
2
⋅
1
+
4
⋅
0
=
2.
{\displaystyle {\frac {\partial z}{\partial l}}=2\cdot 1+4\cdot 0=2.}
Vaizdas:Gradient182183t.jpg 182 ir 183. Nustatyti gradientą funkcijos
z
=
x
2
2
+
y
2
3
{\displaystyle z={\frac {x^{2}}{2}}+{\frac {y^{2}}{3}}}
M(2; 4) . Ir apskaičiuoti kryptinę išvestinę, vektoriaus
a
→
=
3
i
+
4
j
{\displaystyle {\vec {a}}=3{\mathsf {i}}+4{\mathsf {j}}}
Sprendimas . Čia
∂
z
∂
x
=
x
|
M
=
2
,
∂
z
∂
y
=
2
3
y
|
M
=
8
3
.
{\displaystyle {\frac {\partial z}{\partial x}}=x|_{M}=2,\quad {\frac {\partial z}{\partial y}}={\frac {2}{3}}y|_{M}={\frac {8}{3}}.}
Todėl
grad
z
=
2
i
+
8
3
j
.
{\displaystyle {\text{grad}}\,z=2{\mathsf {i}}+{\frac {8}{3}}{\mathsf {j}}.}
Lygtis linijos lygio (pav. 183), praeinančios per duotą tašką, bus
x
2
2
+
y
2
3
=
22
3
,
{\displaystyle {\frac {x^{2}}{2}}+{\frac {y^{2}}{3}}={\frac {22}{3}},}
nes įstačius taško M reikšmes gauname:
x
2
2
+
y
2
3
=
2
2
2
+
4
2
3
=
2
+
16
3
=
2
⋅
3
+
16
3
=
22
3
.
{\displaystyle {\frac {x^{2}}{2}}+{\frac {y^{2}}{3}}={\frac {2^{2}}{2}}+{\frac {4^{2}}{3}}=2+{\frac {16}{3}}={\frac {2\cdot 3+16}{3}}={\frac {22}{3}}.}
y
=
3
(
22
3
−
x
2
2
)
.
{\displaystyle y={\sqrt {3\left({\frac {22}{3}}-{\frac {x^{2}}{2}}\right)}}.}
Randame vektoriaus a ortą:
a
→
∘
=
3
i
+
4
j
3
2
+
4
2
=
3
i
+
4
j
9
+
16
=
3
i
+
4
j
25
=
3
i
+
4
j
5
=
3
5
i
+
4
5
j
.
{\displaystyle {\vec {a}}^{\circ }={\frac {3{\mathsf {i}}+4{\mathsf {j}}}{\sqrt {3^{2}+4^{2}}}}={\frac {3{\mathsf {i}}+4{\mathsf {j}}}{\sqrt {9+16}}}={\frac {3{\mathsf {i}}+4{\mathsf {j}}}{\sqrt {25}}}={\frac {3{\mathsf {i}}+4{\mathsf {j}}}{5}}={\frac {3}{5}}{\mathsf {i}}+{\frac {4}{5}}{\mathsf {j}}.}
Randame kryptinės išvestinės reikšmę, taške M(2; 4) , vektoriaus
a
→
=
3
i
+
4
j
{\displaystyle {\vec {a}}=3{\mathsf {i}}+4{\mathsf {j}}}
∂
z
∂
l
|
M
=
∂
z
(
2
;
4
)
∂
l
=
∇
z
(
2
;
4
)
⋅
a
→
∘
=
2
⋅
3
5
+
8
3
⋅
4
5
=
6
5
+
32
15
=
18
15
+
32
15
=
18
+
32
15
=
50
15
=
10
3
=
3.3
(
3
)
.
{\displaystyle {\frac {\partial z}{\partial l}}|_{M}={\frac {\partial z(2;4)}{\partial l}}=\nabla z(2;4)\cdot {\vec {a}}^{\circ }=2\cdot {\frac {3}{5}}+{\frac {8}{3}}\cdot {\frac {4}{5}}={\frac {6}{5}}+{\frac {32}{15}}={\frac {18}{15}}+{\frac {32}{15}}={\frac {18+32}{15}}={\frac {50}{15}}={\frac {10}{3}}=3.3(3).}
Vaizdas:Gradient182183t.jpg 182 ir 183. a) Nustatyti funkcijos
z
=
x
2
2
+
y
2
3
{\displaystyle z={\frac {x^{2}}{2}}+{\frac {y^{2}}{3}}}
M(2; 4) ; b) surasti vektorių, kurio kryptimi, kryptinės išvestinės reikšmė taške M(2; 4) yra didžiausia ir apskaičiuoti kryptinės išvestinės reikšmę to vektoriaus kryptimi taške M(2; 4) . Sprendimas .a) Didžiausia funkcijos kitimo kryptis yra nusakoma vektoriumi
grad
z
.
{\displaystyle {\text{grad}}\,z.}
∂
z
∂
x
=
x
|
M
=
2
,
∂
z
∂
y
=
2
3
y
|
M
=
8
3
.
{\displaystyle {\frac {\partial z}{\partial x}}=x|_{M}=2,\quad {\frac {\partial z}{\partial y}}={\frac {2}{3}}y|_{M}={\frac {8}{3}}.}
Todėl
grad
z
=
2
i
+
8
3
j
.
{\displaystyle {\text{grad}}\,z=2{\mathsf {i}}+{\frac {8}{3}}{\mathsf {j}}.}
b) Kryptinės išvestinės reikšmė taške M(2; 4) yra didžiausia vektoriaus
grad
z
=
2
i
+
8
3
j
{\displaystyle {\text{grad}}\,z=2{\mathsf {i}}+{\frac {8}{3}}{\mathsf {j}}}
Randame vektoriaus
grad
z
=
∇
z
=
2
i
+
8
3
j
{\displaystyle {\text{grad}}\,z=\nabla z=2{\mathsf {i}}+{\frac {8}{3}}{\mathsf {j}}}
∇
∘
z
(
2
;
4
)
=
2
i
+
8
3
j
2
2
+
(
8
3
)
2
=
2
i
+
8
3
j
4
+
64
9
=
2
i
+
8
3
j
4
⋅
9
+
64
9
=
2
i
+
8
3
j
36
+
64
9
=
2
i
+
8
3
j
100
9
=
2
i
+
8
3
j
10
3
=
2
⋅
3
10
i
+
8
3
3
10
j
=
6
10
i
+
8
10
j
=
3
5
i
+
4
5
j
.
{\displaystyle \nabla ^{\circ }z(2;4)={\frac {2{\mathsf {i}}+{\frac {8}{3}}{\mathsf {j}}}{\sqrt {2^{2}+({\frac {8}{3}})^{2}}}}={\frac {2{\mathsf {i}}+{\frac {8}{3}}{\mathsf {j}}}{\sqrt {4+{\frac {64}{9}}}}}={\frac {2{\mathsf {i}}+{\frac {8}{3}}{\mathsf {j}}}{\sqrt {\frac {4\cdot 9+64}{9}}}}={\frac {2{\mathsf {i}}+{\frac {8}{3}}{\mathsf {j}}}{\sqrt {\frac {36+64}{9}}}}={\frac {2{\mathsf {i}}+{\frac {8}{3}}{\mathsf {j}}}{\sqrt {\frac {100}{9}}}}={\frac {2{\mathsf {i}}+{\frac {8}{3}}{\mathsf {j}}}{\frac {10}{3}}}=2\cdot {\frac {3}{10}}{\mathsf {i}}+{\frac {8}{3}}{\frac {3}{10}}{\mathsf {j}}={\frac {6}{10}}{\mathsf {i}}+{\frac {8}{10}}{\mathsf {j}}={\frac {3}{5}}{\mathsf {i}}+{\frac {4}{5}}{\mathsf {j}}.}
Randame kryptinės išvestinės reikšmę, taške M(2; 4) , vektoriaus
grad
z
=
2
i
+
8
3
j
{\displaystyle {\text{grad}}\,z=2{\mathsf {i}}+{\frac {8}{3}}{\mathsf {j}}}
∂
z
∂
l
|
M
=
∂
z
(
2
;
4
)
∂
l
=
∇
z
(
2
;
4
)
⋅
∇
∘
z
(
2
;
4
)
=
2
⋅
3
5
+
8
3
⋅
4
5
=
6
5
+
32
15
=
18
15
+
32
15
=
18
+
32
15
=
50
15
=
10
3
=
3.3
(
3
)
.
{\displaystyle {\frac {\partial z}{\partial l}}|_{M}={\frac {\partial z(2;4)}{\partial l}}=\nabla z(2;4)\cdot \nabla ^{\circ }z(2;4)=2\cdot {\frac {3}{5}}+{\frac {8}{3}}\cdot {\frac {4}{5}}={\frac {6}{5}}+{\frac {32}{15}}={\frac {18}{15}}+{\frac {32}{15}}={\frac {18+32}{15}}={\frac {50}{15}}={\frac {10}{3}}=3.3(3).}
Vaizdas:Gradient182183t.jpg 182 ir 183. Nustatyti gradientą funkcijos
z
=
x
2
2
+
y
2
3
{\displaystyle z={\frac {x^{2}}{2}}+{\frac {y^{2}}{3}}}
M(2; 4) . Ir apskaičiuoti kryptinę išvestinę, vektoriaus
a
→
=
4
i
+
3
j
{\displaystyle {\vec {a}}=4{\mathsf {i}}+3{\mathsf {j}}}
Sprendimas . Čia
∂
z
∂
x
=
x
|
M
=
2
,
∂
z
∂
y
=
2
3
y
|
M
=
8
3
.
{\displaystyle {\frac {\partial z}{\partial x}}=x|_{M}=2,\quad {\frac {\partial z}{\partial y}}={\frac {2}{3}}y|_{M}={\frac {8}{3}}.}
Todėl
grad
z
=
2
i
+
8
3
j
.
{\displaystyle {\text{grad}}\,z=2{\mathsf {i}}+{\frac {8}{3}}{\mathsf {j}}.}
Lygtis linijos lygio (pav. 183), praeinančios per duotą tašką, bus
x
2
2
+
y
2
3
=
22
3
,
{\displaystyle {\frac {x^{2}}{2}}+{\frac {y^{2}}{3}}={\frac {22}{3}},}
nes įstačius taško M reikšmes gauname:
x
2
2
+
y
2
3
=
2
2
2
+
4
2
3
=
2
+
16
3
=
2
⋅
3
+
16
3
=
22
3
.
{\displaystyle {\frac {x^{2}}{2}}+{\frac {y^{2}}{3}}={\frac {2^{2}}{2}}+{\frac {4^{2}}{3}}=2+{\frac {16}{3}}={\frac {2\cdot 3+16}{3}}={\frac {22}{3}}.}
y
=
3
(
22
3
−
x
2
2
)
.
{\displaystyle y={\sqrt {3\left({\frac {22}{3}}-{\frac {x^{2}}{2}}\right)}}.}
Randame vektoriaus a ortą:
a
→
∘
=
4
i
+
3
j
4
2
+
3
2
=
4
i
+
3
j
16
+
9
=
4
i
+
3
j
25
=
4
i
+
3
j
5
=
4
5
i
+
3
5
j
.
{\displaystyle {\vec {a}}^{\circ }={\frac {4{\mathsf {i}}+3{\mathsf {j}}}{\sqrt {4^{2}+3^{2}}}}={\frac {4{\mathsf {i}}+3{\mathsf {j}}}{\sqrt {16+9}}}={\frac {4{\mathsf {i}}+3{\mathsf {j}}}{\sqrt {25}}}={\frac {4{\mathsf {i}}+3{\mathsf {j}}}{5}}={\frac {4}{5}}{\mathsf {i}}+{\frac {3}{5}}{\mathsf {j}}.}
Randame kryptinės išvestinės reikšmę, taške M(2; 4) , vektoriaus
a
→
=
4
i
+
3
j
{\displaystyle {\vec {a}}=4{\mathsf {i}}+3{\mathsf {j}}}
∂
z
∂
l
|
M
=
∂
z
(
2
;
4
)
∂
l
=
∇
z
(
2
;
4
)
⋅
a
→
∘
=
2
⋅
4
5
+
8
3
⋅
3
5
=
8
5
+
24
15
=
24
15
+
24
15
=
24
+
24
15
=
48
15
=
16
5
=
3.2.
{\displaystyle {\frac {\partial z}{\partial l}}|_{M}={\frac {\partial z(2;4)}{\partial l}}=\nabla z(2;4)\cdot {\vec {a}}^{\circ }=2\cdot {\frac {4}{5}}+{\frac {8}{3}}\cdot {\frac {3}{5}}={\frac {8}{5}}+{\frac {24}{15}}={\frac {24}{15}}+{\frac {24}{15}}={\frac {24+24}{15}}={\frac {48}{15}}={\frac {16}{5}}=3.2.}
Vaizdas:Gradient182183t.jpg 182 ir 183. Nustatyti gradientą funkcijos
z
=
x
2
2
+
y
2
3
{\displaystyle z={\frac {x^{2}}{2}}+{\frac {y^{2}}{3}}}
M(2; 4) . Ir apskaičiuoti kryptinę išvestinę, vektoriaus
a
→
=
3
i
+
3
j
{\displaystyle {\vec {a}}=3{\mathsf {i}}+3{\mathsf {j}}}
Sprendimas . Čia
∂
z
∂
x
=
x
|
M
=
2
,
∂
z
∂
y
=
2
3
y
|
M
=
8
3
.
{\displaystyle {\frac {\partial z}{\partial x}}=x|_{M}=2,\quad {\frac {\partial z}{\partial y}}={\frac {2}{3}}y|_{M}={\frac {8}{3}}.}
Todėl
grad
z
=
2
i
+
8
3
j
.
{\displaystyle {\text{grad}}\,z=2{\mathsf {i}}+{\frac {8}{3}}{\mathsf {j}}.}
Randame vektoriaus a ortą:
a
→
∘
=
3
i
+
3
j
3
2
+
3
2
=
3
i
+
3
j
9
+
9
=
3
i
+
3
j
18
=
3
i
+
3
j
3
2
=
1
2
i
+
1
2
j
.
{\displaystyle {\vec {a}}^{\circ }={\frac {3{\mathsf {i}}+3{\mathsf {j}}}{\sqrt {3^{2}+3^{2}}}}={\frac {3{\mathsf {i}}+3{\mathsf {j}}}{\sqrt {9+9}}}={\frac {3{\mathsf {i}}+3{\mathsf {j}}}{\sqrt {18}}}={\frac {3{\mathsf {i}}+3{\mathsf {j}}}{3{\sqrt {2}}}}={\frac {1}{\sqrt {2}}}{\mathsf {i}}+{\frac {1}{\sqrt {2}}}{\mathsf {j}}.}
Randame kryptinės išvestinės reikšmę, taške M(2; 4) , vektoriaus
a
→
=
3
i
+
3
j
{\displaystyle {\vec {a}}=3{\mathsf {i}}+3{\mathsf {j}}}
∂
z
∂
l
|
M
=
∂
z
(
2
;
4
)
∂
l
=
∇
z
(
2
;
4
)
⋅
a
→
∘
=
2
⋅
1
2
+
8
3
⋅
1
2
=
2
2
+
8
3
2
=
6
3
2
+
8
3
2
=
6
+
8
3
2
=
14
3
2
=
7
2
3
=
3.299831646.
{\displaystyle {\frac {\partial z}{\partial l}}|_{M}={\frac {\partial z(2;4)}{\partial l}}=\nabla z(2;4)\cdot {\vec {a}}^{\circ }=2\cdot {\frac {1}{\sqrt {2}}}+{\frac {8}{3}}\cdot {\frac {1}{\sqrt {2}}}={\frac {2}{\sqrt {2}}}+{\frac {8}{3{\sqrt {2}}}}={\frac {6}{3{\sqrt {2}}}}+{\frac {8}{3{\sqrt {2}}}}={\frac {6+8}{3{\sqrt {2}}}}={\frac {14}{3{\sqrt {2}}}}={\frac {7{\sqrt {2}}}{3}}=3.299831646.}
Kryptinė išvestinė didžiausia Gradiento kryptimi. 
