Arely Briggs

2022-01-25

I have the following multivariable function: $f(x,y)=x{y}^{2}$ , and I must prove it is continuous at $(1,1)$ . I have come to: $|{({(x-1)}^{2}+{(y-1)}^{2})}^{\frac{3}{2}}-1|\le {({(x-1)}^{2}+{(y-1)}^{2})}^{\frac{3}{2}}+1$ (Triangle inequality) $<{\delta}^{3}+1=\u03f5$ .

What can I do with this 1? Any tips are welcome and thank you in advance!

What can I do with this 1? Any tips are welcome and thank you in advance!

Addisyn Thompson

Beginner2022-01-26Added 16 answers

One can pose this problem by considering the graph of the function as follow.

From the definition of continuity the$\epsilon$ -neighbourhood of the image $f(1,1)=1$ determines a couple of planes $z=1+\epsilon$ and $z=1-\epsilon$

The intersections of the planes$1\pm \epsilon$ with the graph of $f$ determine two hyperbolas

$\left(\begin{array}{c}x\\ y\\ 1\pm \epsilon \end{array}\right)$

with$x{y}^{2}=1\pm \epsilon$ .

Also a little circle of radius$\delta$ , which is the border of a $\delta$ -neighbourhood centered at the preimage $(1,1)}^{\mathrm{\top}$ , when is transformed gives a closed curve $C$ viewed in the graph of $f$ .

Then there is a point in the curve$C$ which is the farest away from $p={(1,1,1)}^{\mathrm{\top}}$ , which will be determined by considering that for the curve $C$ can be parametrized as $C\left(y\right)={(\frac{1+\epsilon}{{y}^{2}},y,1+\epsilon )}^{\mathrm{\top}}$ and the distance function

$d(p,C\left(y\right))=\sqrt{{(\frac{1+\epsilon}{{y}^{2}}-1)}^{2}+{(y-1)}^{2}+{\epsilon}^{2}}.$

From the definition of continuity the

The intersections of the planes

with

Also a little circle of radius

Then there is a point in the curve

In a regression analysis, the variable that is being predicted is the "dependent variable."

?

a. Intervening variable

b. Dependent variable

c. None

d. Independent variableWhat is ${R}^{*}$ in math?

Repeated addition is called ?

A)Subtraction

B)Multiplication

C)DivisionMultiplicative inverse of 1/7 is _?

Does the series converge or diverge this $\sum n!/{n}^{n}$

Use Lagrange multipliers to find the point on a surface that is closest to a plane.

Find the point on $z=1-2{x}^{2}-{y}^{2}$ closest to $2x+3y+z=12$ using Lagrange multipliers.

I recognize $z+2{x}^{2}+{y}^{2}=1$ as my constraint but am unable to recognize the distance squared I am trying to minimize in terms of 3 variables. May someone help please.Just find the curve of intersection between ${x}^{2}+{y}^{2}+{z}^{2}=1$ and $x+y+z=0$

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The significance of partial derivative notation

If some function like $f$ depends on just one variable like $x$, we denote its derivative with respect to the variable by:

$\frac{\mathrm{d}}{\mathrm{d}x}f(x)$

Now if the function happens to depend on $n$ variables we denote its derivative with respect to the $i$th variable by:

$\frac{\mathrm{\partial}}{\mathrm{\partial}{x}_{i}}f({x}_{1},\cdots ,{x}_{i},\cdots ,{x}_{n})$

Now my question is what is the significance of this notation? I mean what will be wrong if we show "Partial derivative" of $f$ with respect to ${x}_{i}$ like this? :

$\frac{\mathrm{d}}{\mathrm{d}{x}_{i}}f({x}_{1},\cdots ,{x}_{i},\cdots ,{x}_{n})$

Does the symbol $\mathrm{\partial}$ have a significant meaning?The function $f(x,y,z)$ is a differentiable function at $(0,0,0)$ such that ${f}_{y}(0,0,0)={f}_{x}(0,0,0)=0$ and $f({t}^{2},2{t}^{2},3{t}^{2})=4{t}^{2}$ for every $t>0$. Define $u=(6/11,2/11,9/11)$, with the given about. Is it possible to calculate ${f}_{u}(1,2,3)$ or ${f}_{u}(0,0,0)$, or ${f}_{z}(0,0,0)$?

Given topological spaces ${X}_{1},{X}_{2},\dots ,{X}_{n},Y$, consider a multivariable function $f:\prod _{i=1}^{n}{X}_{i}\to Y$ such that for any $({x}_{1},{x}_{2},\dots ,{x}_{n})\in \prod _{i=1}^{n}{X}_{i}$, the functions in the family $\{x\mapsto f({x}_{1},\dots ,{x}_{i-1},x,{x}_{i+1},\dots ,{x}_{n}){\}}_{i=1}^{n}$ are all continuous. Must $f$ itself be continuous?

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Let $f:M(n,\mathbb{R})\to M(n,\mathbb{R})$ and let $f(A)=A{A}^{t}$. Then find derivative of $f$, denoted by $df$ .

So, Derivative of $f(df)$ if exists, will satisfy $limH\to 0\frac{||f(A+H)-f(A)-df(H)||}{||H||}=0$.if $F(x,y)$ and $y=f(x)$,

$\frac{dy}{dx}=-\frac{\frac{\mathrm{\partial}}{\mathrm{\partial}x}\left(F\right)}{\frac{\mathrm{\partial}}{\mathrm{\partial}y}\left(F\right)}$

1) $F(x,y)$ 𝑎𝑛𝑑 $y=f(x)$ so his means that the function $F$ is a function of one variable which is $x$

2) while we were computing 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒𝑠 we treated $y$ and $x$ as two independent variables although that $y$ changes as $x$ changes but while doing the 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒𝑠 w.r.t $x$ we treated $y$ and $x$ as two independent varaibles and considered $y$ as a constantLet $f:{\mathbb{R}}^{2}\to \mathbb{R}$ be defined as

$f(x,y)=\{\begin{array}{ll}({x}^{2}+{y}^{2})\mathrm{cos}\frac{1}{\sqrt{{x}^{2}+{y}^{2}}},& \text{for}(x,y)\ne (0,0)\\ 0,& \text{for}(x,y)=(0,0)\end{array}$

then check whether its differentiable and also whether its partial derivatives ie ${f}_{x},{f}_{y}$ are continuous at $(0,0)$. I dont know how to check the differentiability of a multivariable function as I am just beginning to learn it. For continuity of partial derivative I just checked for ${f}_{x}$ as function is symmetric in $y$ and $x$. So ${f}_{x}$ turns out to be

${f}_{x}(x,y)=2x\mathrm{cos}\left(\frac{1}{\sqrt{{x}^{2}+{y}^{2}}}\right)+\frac{x}{\sqrt{{x}^{2}+{y}^{2}}}\mathrm{sin}\left(\frac{1}{\sqrt{{x}^{2}+{y}^{2}}}\right)$

which is definitely not $0$ as $(x,y)\to (0,0)$. Same can be stated for ${f}_{y}$. But how to proceed with the first part?