garnentas3m

2022-01-04

Draw the gradient $\mathrm{\nabla}f$ and level curves for $f(x,y)={x}^{2}-y$ at a few common locations. For the point where the level curve $f(x,y)=1$ meets the tangent line, write an equation $(\sqrt{2},1)$.

Shannon Hodgkinson

Beginner2022-01-05Added 34 answers

Step 1

If

Step 2

The gradient of a multivariable function

To find the gradient at a particular point

temnimam2

Beginner2022-01-06Added 36 answers

Second part:

The equation of the tangent line to the function$y=f\left(x\right)$ at $x={x}_{0}$ is given by $y=f\left({x}_{0}\right)+{f}^{\prime}\left({x}_{0}\right)(x-{x}_{0})$ . In this problem the $f(x,y)=1$ , since $f(x,y)={x}^{2}-y$ the equation of curve in the x-y plane becomes ${x}^{2}-y=1$ or $y={x}^{2}-1$ .

To find the equation of the tangent line to the curve$y={x}^{2}-1$ at $(\sqrt{2},1)$ , find the value of $f\left(x\right)={x}^{2}-1$ and its derivative at $x=\sqrt{2}$ and then substitute obtained values in $y=f\left({x}_{0}\right)+{f}^{\prime}\left({x}_{0}\right)(x-{x}_{0})$

$f\left(x\right)={x}^{2}-1$

$f\left(\sqrt{2}\right)={\left(\sqrt{2}\right)}^{2}-1$

$=1$

${f}^{\prime}\left(x\right)=2x$

$f}^{\prime}\left(\sqrt{2}\right)=2\sqrt{2$

$y=f(\sqrt{2}+{f}^{\prime}\left(\sqrt{2}\right)(x-\sqrt{2})$

$=1+2\sqrt{2}(x-\sqrt{2})$

$=2\sqrt{2}x-2\sqrt{2}\cdot \sqrt{2}+1$

$=2\sqrt{2}x-3$

The equation of the tangent line to the function

To find the equation of the tangent line to the curve

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

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a. Intervening variable

b. Dependent variable

c. None

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A)Subtraction

B)Multiplication

C)DivisionMultiplicative inverse of 1/7 is _?

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

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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?