Let $y=f(x)=({x}_{1}^{2}+2{x}_{2},{x}_{1}{x}_{2}-3{x}_{1})$

Is the linear approximation just $f(y)=f(x)+A(y-x)$ whenever y is approximately near $x$?

Is the linear approximation just $f(y)=f(x)+A(y-x)$ whenever y is approximately near $x$?

Waldronjw
2022-07-15
Answered

Let $y=f(x)=({x}_{1}^{2}+2{x}_{2},{x}_{1}{x}_{2}-3{x}_{1})$

Is the linear approximation just $f(y)=f(x)+A(y-x)$ whenever y is approximately near $x$?

Is the linear approximation just $f(y)=f(x)+A(y-x)$ whenever y is approximately near $x$?

You can still ask an expert for help

Wade Atkinson

Answered 2022-07-16
Author has **12** answers

A differentiable function $f:{\mathbb{R}}^{2}\to {\mathbb{R}}^{2}$ can be approximated by it's derivative in the sense that given a point, say, $(1,1)$, you can write:

$f(1+\delta x,1+\delta y)\approx f(1,1)+Df(1,1)(\delta x,\delta y)$

where $Df(1,1)$ is a matrix applied to the vector $(\delta x,\delta y)$. So, yes, if $y$ is near $x$ (as ($(1+\delta x,1+\delta y)$) is near ($(1,1)$)), then you can approximate the difference $f(y)-f(x)$ by the derivative evaluated at $x$, applied to the difference $y-x$. That is:

$f(y)\approx f(x)+Df(x)(y-x)$

Compare this to the one dimensional case, where the tangent line to the graph of a differentiable function locally approximates the graph, in the sense that:

$f(y)\approx f(x)+{f}^{\prime}(x)(y-x)$

$f(1+\delta x,1+\delta y)\approx f(1,1)+Df(1,1)(\delta x,\delta y)$

where $Df(1,1)$ is a matrix applied to the vector $(\delta x,\delta y)$. So, yes, if $y$ is near $x$ (as ($(1+\delta x,1+\delta y)$) is near ($(1,1)$)), then you can approximate the difference $f(y)-f(x)$ by the derivative evaluated at $x$, applied to the difference $y-x$. That is:

$f(y)\approx f(x)+Df(x)(y-x)$

Compare this to the one dimensional case, where the tangent line to the graph of a differentiable function locally approximates the graph, in the sense that:

$f(y)\approx f(x)+{f}^{\prime}(x)(y-x)$

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Find the first partial derivatives of the function.

$z=x\mathrm{sin}\left(xy\right)$

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I am having trouble isolating the x and y to separate side in the differential equations below. Could someone give me a hint as to how to to this.

Equation 1:

$\frac{dy}{dx}-\frac{x}{y}=\frac{1}{x}$

Equation 2:

$xy\frac{dy}{dx}={y}^{2}$

Equation 1:

$\frac{dy}{dx}-\frac{x}{y}=\frac{1}{x}$

Equation 2:

$xy\frac{dy}{dx}={y}^{2}$

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In a course on partial differential equations I came through this theorem about the general solution of a first order quasi-linear partial differential equation.

1. The general solution of a first-order, quasi-linear partial differential equation

$a(x,y,u){u}_{x}+b(x,y,u){u}_{y}=c(x,y,u)$

is given by $f(\varphi ,\psi )=0$, where $f$ is an arbitrary function of $\varphi (x,y,u)$ and $\psi (x,y,u).$.

2. $\varphi =C1$ and $\psi =C2$ are solution curves of the characteristic equations

$\frac{dx}{a}=\frac{dy}{b}=\frac{du}{c}.$

Is there any geometric interpretation of both these points so that I can have a better intuitive understanding of the graphical representation of $f$,$\varphi $ and $\psi $ ?

1. The general solution of a first-order, quasi-linear partial differential equation

$a(x,y,u){u}_{x}+b(x,y,u){u}_{y}=c(x,y,u)$

is given by $f(\varphi ,\psi )=0$, where $f$ is an arbitrary function of $\varphi (x,y,u)$ and $\psi (x,y,u).$.

2. $\varphi =C1$ and $\psi =C2$ are solution curves of the characteristic equations

$\frac{dx}{a}=\frac{dy}{b}=\frac{du}{c}.$

Is there any geometric interpretation of both these points so that I can have a better intuitive understanding of the graphical representation of $f$,$\varphi $ and $\psi $ ?

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Applications of First Order Differential Equations: A dead body was found within a closed room of a house where the temperature was a constant ${24}^{\circ}C$ . At the time of discovery, the core temperature of the body was determined to be ${28}^{\circ}C$ . One hour later, a second measurement showed that the core temperature of the body was ${26}^{\circ}C$ . The core temperature of the body at the time of death is ${37}^{\circ}C$ . A. Determine how many hours elapsed before the body was found. B. What is the temperature of the body 5 hours from the death of the victim?

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Determine whether the given differential equation is exact. If it is exact, solve it.

$(x-{y}^{3}+{y}^{2}\mathrm{sin}x)dx=(3x{y}^{2}+2y\mathrm{cos}x)dy$

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Third order Cauchy-Euler differential equation

Solve:$x}^{3}y-{x}^{2}y+2xy-2y={x}^{3$

Solve: