Помогите плиз, 15 баллов, дифференциал. Created by vipsergazyev. algebra-ru

Помогите плиз, 15 баллов, дифференциал...

vipsergazyev @vipsergazyev

July 2022 1 29 Report

Помогите плиз, 15 баллов, дифференциал

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$\boxed {\ z=f(x,y)\ \ ,\ \ \ x=\varphi (t)\ ,\ y=\psi (t)\ \ ,\ \ \dfrac{dz}{dt}=\dfrac{\partial z}{\partial x}\cdot \dfrac{dx}{dt}+\dfrac{\partial z}{\partial y}\cdot \dfrac{dz}{dy}\ }$

$1)\ \ z=x^2+xy+y^2\ ,\ \ \ x=t^2\ ,\ y=t\\\\\\\dfrac{\partial z}{\partial x}=2x+y\ \ ,\ \ \dfrac{dx}{dt}=2t\\\\\\\dfrac{\partial z}{\partial y}=x+2y\ \ ,\ \ \dfrac{dy}{dt}=1\\\\\\\dfrac{dz}{dt}=(2x+y)\cdot 2t+(x+2y)=(2t^2+t)\cdot 2t+(t^2+2t)=4t^3+3t^2+2t$

$2)\ \ z=\sqrt{x^2+y^2}\ \ ,\ \ x=sint\ ,\ y=cost\\\\\\\dfrac{\partial z}{\partial x}=\dfrac{1}{2\sqrt{x^2+y^2}}\cdot 2x=\dfrac{x}{\sqrt{x^2+y^2}}\ \ ,\ \ \ \ \ \dfrac{dx}{dt}=cost\\\\\\\dfrac{\partial z}{\partial y}=\dfrac{1}{\sqrt{x^2+y^2}}\cdot 2y=\dfrac{y}{\sqrt{x^2+y^2}}\ \ ,\ \ \ \ \dfrac{dy}{dt}=-sint\\\\\\\dfrac{dz}{dt}=\dfrac{x}{\sqrt{x^2+y^2}}\cdot cost+\dfrac{y}{\sqrt{x^2+y^2}}\cdot (-sint)=\\\\\\=\dfrac{sint\cdot cost}{\sqrt{sin^2t+cos^2t}}-\dfrac{cost\cdot sint}{\sqrt{sin^2t+cos^2t}}=0$

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