Donald Johnson

2022-01-04

First of all, I'm new to multivariable calculus... in a multivariable function, by assuming that its domain is going to be $R}^{2$ and its image is going to be all real numbers, the graph of that function is defined as a subset of $R}^{3$ in which the x and y axis are going to receive the inputs, and the output is going to be in $z(x,y,f(x,y))$ ? Is that correct? Will its graph, in this example, be some kind of surface?

Edward Patten

Beginner2022-01-05Added 38 answers

Yes. If such a f(x,y) is defined for all real pairs (x,y), for each such pair corresponds a single real value, which could be taken as the z coordinate of a point. This interpretation leads to a surface, as you suppose.

Using Wolfram Alpha, for instance, you can get the plots of the following functions. Guessing the surfaces before looking them up would be an interesting exercise.

$f(x,y)=x$

$f(x,y)=y$

$f(x,y)=x+y$

$f(x,y)=x-y$

Using Wolfram Alpha, for instance, you can get the plots of the following functions. Guessing the surfaces before looking them up would be an interesting exercise.

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 $x+y+z=0$

Which equation illustrates the identity property of multiplication? A$(a+\mathrm{bi})\times c=(\mathrm{ac}+\mathrm{bci})$ B$(a+\mathrm{bi})\times 0=0$ C$(a+\mathrm{bi})\times (c+\mathrm{di})=(c+\mathrm{di})\times (a+\mathrm{bi})$ D$(a+\mathrm{bi})\times 1=(a+\mathrm{bi})$

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?

Let $x$ be an independent variable. Does the differential dx depend on $x$?(from the definition of differential for variables & multivariable functions)

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?