Aptarimas:Matematika/Kryptinė išvestinė

gali buti neteisingi arba beprasmiai

• Pavyzdys. Apskaičiuoti išvestinę funkcijos $u={\sqrt {100-x^{2}-y^{2}-z^{2}}}$ taške M(1; 2; 2) kriptimi vektoriaus ${\vec {MM_{1}}},$ kur $M_{1}$ - taškas su koordinatėmis (3; 4; 5).

Sprendimas. Rasime vienetinį vektorių ${\vec {l}},$ turinti duotąją kryptį:

${\vec {MM_{1}}}=(3-1;\,4-2;\;5-2)=(2;\,2;\;3)=2i+2j+3k;$

$\|{\vec {MM_{1}}}\|={\sqrt {2^{2}+2^{2}+3^{2}}}={\sqrt {4+4+9}}={\sqrt {17}}=4.123105626;$

${\frac {\vec {MM_{1}}}{\|{\vec {MM_{1}}}\|}}={\frac {2i+2j+3k}{\sqrt {17}}}={\frac {2}{\sqrt {17}}}i+{\frac {2}{\sqrt {17}}}j+{\frac {3}{\sqrt {17}}}k,$

Iš kur $\cos \alpha ={\frac {2}{\sqrt {17}}},\;\cos \beta ={\frac {2}{\sqrt {17}}},\;\cos \gamma ={\frac {3}{\sqrt {17}}}.$ Apskaičiuosime dalines išvestines funkcijos taške M(1; 2; 2):

$f_{x}'(x,\;y,\;z)={\frac {1}{2}}\cdot {\frac {-2x}{\sqrt {100-x^{2}-y^{2}-z^{2}}}}=-{\frac {x}{\sqrt {100-x^{2}-y^{2}-z^{2}}}},$

$f_{y}'(x,\;y,\;z)={\frac {1}{2}}\cdot {\frac {-2y}{\sqrt {100-x^{2}-y^{2}-z^{2}}}}=-{\frac {y}{\sqrt {100-x^{2}-y^{2}-z^{2}}}},$

$f_{z}'(x,\;y,\;z)={\frac {1}{2}}\cdot {\frac {-2z}{\sqrt {100-x^{2}-y^{2}-z^{2}}}}=-{\frac {z}{\sqrt {100-x^{2}-y^{2}-z^{2}}}},$

iš kur

$f_{x}'(1,\;2,\;2)=-{\frac {x}{\sqrt {100-x^{2}-y^{2}-z^{2}}}}=-{\frac {1}{\sqrt {100-1^{2}-2^{2}-2^{2}}}}=-{\frac {1}{\sqrt {100-1-4-4}}}=-{\frac {1}{\sqrt {91}}},$

$f_{y}'(1,\;2,\;2)=-{\frac {y}{\sqrt {100-x^{2}-y^{2}-z^{2}}}}=-{\frac {2}{\sqrt {100-1^{2}-2^{2}-2^{2}}}}=-{\frac {2}{\sqrt {100-1-4-4}}}=-{\frac {2}{\sqrt {91}}},$

$f_{z}'(1,\;2,\;2)=-{\frac {z}{\sqrt {100-x^{2}-y^{2}-z^{2}}}}=-{\frac {2}{\sqrt {100-1^{2}-2^{2}-2^{2}}}}=-{\frac {2}{\sqrt {100-1-4-4}}}=-{\frac {2}{\sqrt {91}}},$

Pagal formulę ${\frac {\partial z}{\partial l}}={\frac {\partial z}{\partial x}}\cos \alpha +{\frac {\partial z}{\partial y}}\cos \beta$ gausime

${\frac {\partial z}{\partial l}}=-{\frac {1}{\sqrt {91}}}\cdot {\frac {2}{\sqrt {17}}}-{\frac {2}{\sqrt {91}}}\cdot {\frac {2}{\sqrt {17}}}-{\frac {3}{\sqrt {91}}}\cdot {\frac {2}{\sqrt {17}}}={\frac {-2-4-6}{\sqrt {91\cdot 17}}}=-{\frac {12}{\sqrt {1547}}}=-0.305095701.$

• Pavyzdys. Apskaičiuoti išvestinę funkcijos $u={\sqrt {100-x^{2}-y^{2}-z^{2}}}$ taške M(1; 2; 2) kriptimi vektoriaus ${\vec {MM_{1}}},$ kur $M_{1}$ - taškas su koordinatėmis (4; 3; 5).

Sprendimas. Rasime vienetinį vektorių ${\vec {l}},$ turinti duotąją kryptį:

${\vec {MM_{1}}}=(4-1;\,3-2;\;5-2)=(3;\,1;\;3)=3i+1j+3k;$

$\|{\vec {MM_{1}}}\|={\sqrt {3^{2}+1^{2}+3^{2}}}={\sqrt {9+1+9}}={\sqrt {19}}=4.358898944;$

${\frac {\vec {MM_{1}}}{\|{\vec {MM_{1}}}\|}}={\frac {3i+1j+3k}{\sqrt {19}}}={\frac {3}{\sqrt {19}}}i+{\frac {1}{\sqrt {19}}}j+{\frac {3}{\sqrt {19}}}k,$

Iš kur $\cos \alpha ={\frac {3}{\sqrt {19}}},\;\cos \beta ={\frac {1}{\sqrt {19}}},\;\cos \gamma ={\frac {3}{\sqrt {19}}}.$ Apskaičiuosime dalines išvestines funkcijos taške M(1; 2; 2):

$f_{x}'(x,\;y,\;z)={\frac {1}{2}}\cdot {\frac {-2x}{\sqrt {100-x^{2}-y^{2}-z^{2}}}}=-{\frac {x}{\sqrt {100-x^{2}-y^{2}-z^{2}}}},$

$f_{y}'(x,\;y,\;z)={\frac {1}{2}}\cdot {\frac {-2y}{\sqrt {100-x^{2}-y^{2}-z^{2}}}}=-{\frac {y}{\sqrt {100-x^{2}-y^{2}-z^{2}}}},$

$f_{z}'(x,\;y,\;z)={\frac {1}{2}}\cdot {\frac {-2z}{\sqrt {100-x^{2}-y^{2}-z^{2}}}}=-{\frac {z}{\sqrt {100-x^{2}-y^{2}-z^{2}}}},$

iš kur

$f_{x}'(1,\;2,\;2)=-{\frac {x}{\sqrt {100-x^{2}-y^{2}-z^{2}}}}=-{\frac {1}{\sqrt {100-1^{2}-2^{2}-2^{2}}}}=-{\frac {1}{\sqrt {100-1-4-4}}}=-{\frac {1}{\sqrt {91}}},$

$f_{y}'(1,\;2,\;2)=-{\frac {y}{\sqrt {100-x^{2}-y^{2}-z^{2}}}}=-{\frac {2}{\sqrt {100-1^{2}-2^{2}-2^{2}}}}=-{\frac {2}{\sqrt {100-1-4-4}}}=-{\frac {2}{\sqrt {91}}},$

$f_{z}'(1,\;2,\;2)=-{\frac {z}{\sqrt {100-x^{2}-y^{2}-z^{2}}}}=-{\frac {2}{\sqrt {100-1^{2}-2^{2}-2^{2}}}}=-{\frac {2}{\sqrt {100-1-4-4}}}=-{\frac {2}{\sqrt {91}}},$

Pagal formulę ${\frac {\partial z}{\partial l}}={\frac {\partial z}{\partial x}}\cos \alpha +{\frac {\partial z}{\partial y}}\cos \beta$ gausime

${\frac {\partial z}{\partial l}}=-{\frac {1}{\sqrt {91}}}\cdot {\frac {3}{\sqrt {19}}}-{\frac {2}{\sqrt {91}}}\cdot {\frac {1}{\sqrt {19}}}-{\frac {3}{\sqrt {91}}}\cdot {\frac {3}{\sqrt {19}}}={\frac {-3-2-9}{\sqrt {91\cdot 19}}}=-{\frac {14}{\sqrt {1729}}}=-0.336690249.$