Recent questions in Multivariable calculus

Multivariable calculusAnswered question

zookeeper1930r8k 2023-03-29

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

a. Intervening variable

b. Dependent variable

c. None

d. Independent variable

Multivariable calculusAnswered question

Bradyn Cherry 2022-12-26

Repeated addition is called ?

A)Subtraction

B)Multiplication

C)Division

A)Subtraction

B)Multiplication

C)Division

Multivariable calculusAnswered question

klepnin4wv 2022-12-19

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

Multivariable calculusAnswered question

Leandro Acosta 2022-12-18

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.

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.

Multivariable calculusAnswered question

Will Osborn 2022-12-04

Just find the curve of intersection between ${x}^{2}+{y}^{2}+{z}^{2}=1$ and $x+y+z=0$

Multivariable calculusAnswered question

Jaden Bird 2022-11-30

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$(a+\mathrm{bi})\times 1=(a+\mathrm{bi})$

Multivariable calculusAnswered question

lascieflYr 2022-11-30

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?

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?

Multivariable calculusAnswered question

Kierra Griffith 2022-11-22

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 ${f}_{z}(0,0,0)$?

Multivariable calculusAnswered question

jorgejasso85xvx 2022-11-19

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?

Multivariable calculusAnswered question

Jenny Roberson 2022-11-18

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

Multivariable calculusAnswered question

Alberto Calhoun 2022-11-18

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 $limH\to 0\frac{||f(A+H)-f(A)-df(H)||}{||H||}=0$.

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

Multivariable calculusAnswered question

Nicholas Hunter 2022-11-17

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 constant

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

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 constant

Multivariable calculusAnswered question

Laila Murphy 2022-11-17

Let $f:{\mathbb{R}}^{2}\to \mathbb{R}$ 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?

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

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?

Multivariable calculusAnswered question

Sophie Marks 2022-11-16

How "messy" can a multivariable function be?

$f(x,y)=\frac{2xy}{{x}^{2}+{y}^{2}}\phantom{\rule{1em}{0ex}}f(0,0)=0$

$f(x,y)=\frac{2xy}{{x}^{2}+{y}^{2}}\phantom{\rule{1em}{0ex}}f(0,0)=0$

Multivariable calculusAnswered question

Alberto Calhoun 2022-11-13

Many mathematical texts define a multivariable function $f$ in the following way

$f:=f(x,y)$

However, if we focus on the fact that a function is really a binary relation on two sets, (say the real numbers), the definition would be as follows

$f:{\mathbb{R}}^{2}\to \mathbb{R}$

This seems to imply that the domain of the function is a set of ordered pairs of the form $(x,y)$.

The set $\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{p}\mathrm{h}(f)\subset {\mathbb{R}}^{2}\times \mathbb{R}$, would then comprise ordered pairs of the form

$\{(({x}_{0},{y}_{0}),a),(({x}_{1},{y}_{1}),b),\dots \}$

In line with this, does it not follow that the correction notation for $f$ should be be

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

$f:=f(x,y)$

However, if we focus on the fact that a function is really a binary relation on two sets, (say the real numbers), the definition would be as follows

$f:{\mathbb{R}}^{2}\to \mathbb{R}$

This seems to imply that the domain of the function is a set of ordered pairs of the form $(x,y)$.

The set $\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{p}\mathrm{h}(f)\subset {\mathbb{R}}^{2}\times \mathbb{R}$, would then comprise ordered pairs of the form

$\{(({x}_{0},{y}_{0}),a),(({x}_{1},{y}_{1}),b),\dots \}$

In line with this, does it not follow that the correction notation for $f$ should be be

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

Multivariable calculusAnswered question

Siemensueqw 2022-11-11

What is the definition of a path along a multivariable function?

$\begin{array}{r}f(x,y)=\{\begin{array}{ll}\frac{{x}^{2\alpha}}{{x}^{2}+{y}^{2}},& \text{if (}x\text{,}y\text{)}\ne \text{(0,0)}\\ 0& \text{if (}x\text{,}y\text{) = (0,0)}\end{array}\end{array}$

Long story short, we ended up switching to polar coordinates and simplifying down to:

$\underset{r\to 0}{lim}{r}^{2\alpha -2}\cdot {\mathrm{cos}}^{2\alpha}\theta $

Besides the fact that we can now analyze different cases based on the value of $\alpha $, we came to the conclusion that since the point is approached at by infinitely different "paths" (depending on the value of $\theta $), the limit doesn't even exist in the first place.

$\begin{array}{r}f(x,y)=\{\begin{array}{ll}\frac{{x}^{2\alpha}}{{x}^{2}+{y}^{2}},& \text{if (}x\text{,}y\text{)}\ne \text{(0,0)}\\ 0& \text{if (}x\text{,}y\text{) = (0,0)}\end{array}\end{array}$

Long story short, we ended up switching to polar coordinates and simplifying down to:

$\underset{r\to 0}{lim}{r}^{2\alpha -2}\cdot {\mathrm{cos}}^{2\alpha}\theta $

Besides the fact that we can now analyze different cases based on the value of $\alpha $, we came to the conclusion that since the point is approached at by infinitely different "paths" (depending on the value of $\theta $), the limit doesn't even exist in the first place.

Multivariable calculusAnswered question

Jaiden Elliott 2022-11-10

Why $f(x,y)=\{\begin{array}{l}\frac{x{y}^{2}}{{x}^{2}+{y}^{2}}{\textstyle \text{for}}(x,y)\ne (0,0)\\ 0{\textstyle \text{for}}(x,y)=(0,0)\end{array}$

is continuous and

$f(x)=\{\begin{array}{l}2{\textstyle \text{for}}0=x10\\ 5{\textstyle \text{for}}x=10\end{array}$

seem to be discontinuity?

is continuous and

$f(x)=\{\begin{array}{l}2{\textstyle \text{for}}0=x10\\ 5{\textstyle \text{for}}x=10\end{array}$

seem to be discontinuity?

As you start exploring calculus and analysis, you will encounter multivariable calculus equations that are self-explanatory as well because all of them will contain at least two questions related to each variable being involved. See our multivariable calculus examples to receive more help and information regarding how these are used. The answers to solving these multivariable calculus questions should be based on finding the deterministic behavior. These are used in engineering and those fields where the parametric equations solver will provide you an optimal control of time dynamic systems. As an interesting subject, applying at least one equation in practice will keep you inspired!