Question

Oblicz pochodną kierunkową daje naj.

Answer 1

Odpowiedź:

[tex]f(x,y)=(x^{2} -y)\sqrt{x^{2} +y^{2} } \qquad P(0,1)\quad \vec v=[2,3]\\f'_{\vec v_k}=grad(f(P))\circ \vec v_k\\\displaystyle f'_x=(x^{2} -y)'\sqrt{x^{2} +y^{2} } +(\sqrt{x^{2} +y^{2} } )'(x^{2} -y)\\f'_x=2x\sqrt{x^{2} +y^{2} } +\frac{2x}{2\sqrt{x^{2} +y^{2} } } (x^{2} -y)\\f'_x(0,1)=2\cdot0\sqrt{0+1} +0=0\\f'_y=-1\sqrt{x^{2} +y^{2} } +\frac{2y}{2\sqrt{x^{2} +y^{2} } } (x^{2} -y)\\f'_x(0,1)=-1\sqrt{0+1} +\frac{1}{\sqrt{0+1} } (0-1)=-1-1=-2\\grad f(P)=[0,-2]\\[/tex]

[tex]|\vec v|=\sqrt{2^2+3^2} =\sqrt{13} \\\displaystyle \vec v_k=\left[\frac{2}{\sqrt{13} } ,\frac{3}{\sqrt{13} } \right]\\f'_{\vec v_k}=[0,-2]\circ\left[\frac{2}{\sqrt{13} } ,\frac{3}{\sqrt{13} } \right]=0\cdot\frac{2}{\sqrt{13} } -2\cdot\frac{3}{\sqrt{13} } =-\frac{6}{\sqrt{13} } =\underline{-\frac{6\sqrt{13} }{13}}[/tex]