Aptarimas:Matematika/Gradientas

Ryšiai kaip kryptinių išvestinių skaičiavime

• Pavyzdis. Rasime funkcijos $u=x^{3}y^{3}z^{3}$ kryptinę išvestinę taške $M_{0}$(2; -1; 3) vektoriaus ${\vec {M_{0}M_{1}}}$ kryptimi, kai $M_{1}(3;2;4)$.

Sprendimas. Randame ${\vec {M_{0}M_{1}}}=(3-2;\;2-(-1);\;4-3)=(1;\;3;\;1),\quad {\vec {M_{0}M_{1}}}={\sqrt {1^{2}+3^{2}+1^{2}}}={\sqrt {1+9+1}}={\sqrt {11}},$ tuomet $\cos \alpha ={\frac {1}{\sqrt {11}}},\quad \cos \beta ={\frac {3}{\sqrt {11}}},\quad \cos \gamma ={\frac {1}{\sqrt {11}}}.$

$\cos ^{2}\alpha +\cos ^{2}\beta +\cos ^{2}\gamma ={\frac {1}{11}}+{\frac {9}{11}}+{\frac {1}{11}}={\frac {11}{11}}=1.$

Toliau randame išvestines ${\frac {\partial u}{\partial x}}=3x^{2}y^{3}z^{3},\quad {\frac {\partial u}{\partial y}}=3x^{3}y^{2}z^{3},\quad {\frac {\partial u}{\partial x}}=3x^{3}y^{3}z^{2}$ ir apskaičiuojame jų reikšmes taške $M_{0}$:

${\frac {\partial u}{\partial x}}|_{M_{0}}=(3x^{2}y^{3}z^{3})|_{M_{0}}=3\cdot 2^{2}\cdot (-1)^{3}\cdot 3^{3}=-3\cdot 4\cdot 27=-324,$

${\frac {\partial u}{\partial y}}|_{M_{0}}=(3x^{3}y^{2}z^{3})|_{M_{0}}=3\cdot 2^{3}\cdot (-1)^{2}\cdot 3^{3}=3\cdot 8\cdot 27=648,$

${\frac {\partial u}{\partial z}}|_{M_{0}}=(3x^{3}y^{3}z^{2})|_{M_{0}}=3\cdot 2^{3}\cdot (-1)^{3}\cdot 3^{2}=-3\cdot 8\cdot 9=-216.$

Įrašę šias išvvestinių ir krypties kosinusų reikšmes į formulę, gauname:

${\frac {\partial u}{\partial l}}={\frac {\partial u}{\partial x}}\cos \alpha +{\frac {\partial u}{\partial y}}\cos \beta +{\frac {\partial u}{\partial z}}\cos \gamma =-324\cdot {\frac {1}{\sqrt {11}}}+648\cdot {\frac {3}{\sqrt {11}}}-216\cdot {\frac {1}{\sqrt {11}}}=$

$=-324\cdot {\frac {1}{\sqrt {11}}}+{\frac {1944}{\sqrt {11}}}-216\cdot {\frac {1}{\sqrt {11}}}={\frac {1944-324-216}{\sqrt {11}}}={\frac {1404}{\sqrt {11}}}=423.3219278.$

Irelevant

Vaizdas:Gradient182183t.jpg

182 ir 183.

• Nustatyti gradientą funkcijos $u={\frac {x^{2}}{2}}+{\frac {y^{2}}{3}}$ (pav. 182) taške M(2, 4). Funkcija yra elipsinis paraboloidas.

Sprendimas. Čia

${\frac {\partial u}{\partial x}}=x|_{M}=2,\quad {\frac {\partial u}{\partial y}}={\frac {2}{3}}y|_{M}={\frac {8}{3}}.$

Todėl

${\text{grad}}\,u=2{\mathsf {i}}+{\frac {8}{3}}{\mathsf {j}}.$

Lygtis linijos lygio (pav. 183), praeinančios per duotą tašką, bus

${\frac {x^{2}}{2}}+{\frac {y^{2}}{3}}={\frac {22}{3}},$

nes įstačius taško M reikšmes gauname:

${\frac {x^{2}}{2}}+{\frac {y^{2}}{3}}={\frac {2^{2}}{2}}+{\frac {4^{2}}{3}}=2+{\frac {16}{3}}={\frac {2\cdot 3+16}{3}}={\frac {22}{3}}.$

$y={\sqrt {3\left({\frac {22}{3}}-{\frac {x^{2}}{2}}\right)}}.$

Dar randame:

$\|{\text{grad }}\,u\|_{M}={\sqrt {2^{2}+\left({\frac {8}{3}}\right)^{2}}}={\sqrt {4+{\frac {64}{9}}}}={\sqrt {\frac {36+64}{9}}}={\sqrt {\frac {100}{9}}}={\frac {10}{3}}=3.3(3);$

$\cos \alpha ={\frac {{\frac {\partial u}{\partial x}}|_{M}}{\|{\text{grad }}\,u\|_{M}}}={\frac {2}{\frac {10}{3}}}={\frac {6}{10}}=0.6;$

$\cos \beta ={\frac {{\frac {\partial u}{\partial y}}|_{M}}{\|{\text{grad }}\,u\|_{M}}}={\frac {\frac {8}{3}}{\frac {10}{3}}}={\frac {8}{10}}=0.8.$

• Nustatyti gradientą funkcijos $u={\frac {x^{2}}{3}}+{\frac {y^{2}}{3}}$ taške $M(2;\;3{\sqrt {2}}).$ Funkcija u yra paraboloidas.

Sprendimas. Čia

${\frac {\partial u}{\partial x}}={\frac {2x}{3}}|_{M}={\frac {4}{3}},\quad {\frac {\partial u}{\partial y}}={\frac {2}{3}}y|_{M}={\frac {2}{3}}\cdot 3{\sqrt {2}}=2{\sqrt {2}}.$

Todėl

${\text{grad}}\,u={\frac {4}{3}}{\mathsf {i}}+2{\sqrt {2}}{\mathsf {j}}.$

Lygtis linijos lygio, praeinančios per duotą tašką, bus

${\frac {x^{2}}{2}}+{\frac {y^{2}}{3}}={\frac {70}{9}},$

nes įstačius taško M reikšmes gauname:

${\frac {x^{2}}{3}}+{\frac {y^{2}}{3}}={\frac {\left({\frac {4}{3}}\right)^{2}}{3}}+{\frac {(3{\sqrt {2}})^{2}}{3}}={\frac {\frac {16}{3}}{3}}+{\frac {({\sqrt {18}})^{2}}{3}}={\frac {16}{9}}+{\frac {18}{3}}={\frac {16}{9}}+{\frac {18\cdot 3}{3\cdot 3}}={\frac {16+54}{9}}={\frac {70}{9}}=7.777777778.$

$y={\sqrt {3\left({\frac {70}{9}}-{\frac {x^{2}}{3}}\right)}}={\sqrt {{\frac {70}{3}}-x^{2}}}.$

Dar randame:

$\|{\text{grad }}\,u\|_{M}={\sqrt {\left({\frac {4}{3}}\right)^{2}+(2{\sqrt {2}})^{2}}}={\sqrt {{\frac {16}{9}}+8}}={\sqrt {\frac {16+72}{9}}}={\sqrt {\frac {88}{9}}}={\frac {2{\sqrt {22}}}{3}}=3.12694384;$

$\cos \alpha ={\frac {{\frac {\partial u}{\partial x}}|_{M}}{\|{\text{grad }}\,u\|_{M}}}={\frac {\frac {4}{3}}{{\frac {2}{3}}\cdot {\sqrt {22}}}}={\frac {2}{\sqrt {22}}}=0.426401432;$

$\cos \beta ={\frac {{\frac {\partial u}{\partial y}}|_{M}}{\|{\text{grad }}\,u\|_{M}}}={\frac {2{\sqrt {2}}}{{\frac {2}{3}}\cdot {\sqrt {22}}}}={\frac {3{\sqrt {2}}}{\sqrt {22}}}={\frac {3}{\sqrt {11}}}=0.904534033.$

$\cos ^{2}\alpha +\cos ^{2}\beta =\left({\frac {2}{\sqrt {22}}}\right)^{2}+\left({\frac {3}{\sqrt {11}}}\right)^{2}={\frac {4}{22}}+{\frac {9}{11}}={\frac {2}{11}}+{\frac {9}{11}}={\frac {11}{11}}=1.$