Question

Помогите помогите пожалуйста!!!!!!!!

Answer 1

Ответ:

$\sqrt{(3,02)^2-(2,004)^2+11}\ \ \ \Rightarrow \\\\a)\ \ z=f(x,y)=\sqrt{x^2-y^2+11}\ \ ,\ \ a=3,02\ ,\ \ b=2,004\\\\x_0=3\ ,\ \ y_0=2\\\\b)\ \ \Delta x=a-x_0=3,02-3=0,2\ \ ,\ \ \ \Delta y=b-y_0=2,004-2=0,004\\\\f(a,b)\approx f(x_0,y_0)+dz\\\\dz=\dfrac{\partial z}{\partial x}\cdot dx+\dfrac{\partial z}{\partial y}\cdot dy\\\\\dfrac{\partial z}{\partial x}=\dfrac{2x}{2\sqrt{x^2-y^2+11}}=\dfrac{x}{\sqrt{x^2-y^2+11}}\ \ ,\ \ \dfrac{\partial z}{\partial x}\Big|_{(x_0,y_0)}=\dfrac{3}{\sqrt{9-4+11}}=\dfrac{3}{4}$

$\dfrac{\partial z}{\partial y}=\dfrac{-y}{\sqrt{x^2-y^2+11}}\ \ ,\ \ \ \dfrac{\partial z}{\partial y}\Big|_{(3,2)}=\dfrac{-2}{4}=-\dfrac{1}{2}\\\\\\dz=z'_{x}(3,2)\cdot \Delta x+z'_{y}(3,2)\cdot \Delta y=\dfrac{3}{4}\cdot 0,02-\dfrac{1}{2}\cdot 0,004=0,015-0,002=0,013\\\\f(x_0;y_0)=f(3;2)=\sqrt{3^2-2^2+11}=\sqrt{16}=4\\\\f(a;b)=f(3,02\ ;\ 2,004)\approx f(3;2)+dz=4+0,013=\boxed{\ 4,013\ }$