New answers tagged

0

What you are talking about is spherical cap. Yes you can find with single / double integral. There seems to be some mistake in how you have written it. You know radius of the circle at any given $z$ is $\sqrt{a^2-z^2}$. You use the knowledge of perimeter of the circle or the area of the circle at a given $z$ to make this a double or single integral. $\...


1

Collinearity only applies to straight lines, so there's a problem with your argument. However, a much easier argument is available. There are no points on the two-sheeted hyperbola with $z=0$. You have found a point on the surface with $z>0$ and also one with $z<0$. Argue that by the intermediate value theorem, any curve connecting these points ...


0

The equation of the plane tangent to $z=f(x,y)$ at point $(x_0,y_0,z_0)$ is$$z = f\left( {{x_0},{y_0}} \right) + {f_x}\left( {{x_0},{y_0}} \right)\left( {x - {x_0}} \right) + {f_y}\left( {{x_0},{y_0}} \right)\left( {y - {y_0}} \right)$$ $$f_x(x,y)=3 x^2+\frac{3 y}{x}-\log y;\;f_y(x,y)=-\frac{x}{y}+3 \log x+2 y$$ $$z=7+6(x-1)+(y-1)\to z=6x+y$$


2

This is the matrix representing projection onto $v$.


2

First solution: Here is a quick analytic solution first. Let $p_0=(x_0,y_0,z_0)$ be a point on both spheres. Then the tangent plane in $p_0$ to the one or the other sphere has the equation: $$ \begin{aligned} 2x_0(x-x_0) + 2y_0(y-y_0) + 2z_0(z-z_0) - 2a(x-x_0) &=0\ ,\\ 2x_0(x-x_0) + 2y_0(y-y_0) + 2z_0(z-z_0) - 2b(y-y_0) &=0\ , \end{aligned} $$ so ...


1

The set $D$, which is equal to$$\{(x,y)\in(0,\infty)\times[0,\infty)\mid x+y=1\}=\{(x,1-x)\mid x\in(0,1]\},$$is a closed subset of $(0,\infty)\times[0,\infty)$. In fact, it is not a closed subset of $\Bbb R^2$, but that's not relevant here.


3

Let$$f(x,y,z)=(x-a)^2+y^2+z^2\quad\text{and}\quad g(x,y,z)=x^2+(y-b)^2+z^2.$$Then your spheres are the surfaces$$\{(x,y,z)\in\Bbb R^3\mid f(x,y,z)=a^2\}\quad\text{and}\quad\{(x,y,z)\in\Bbb R^3\mid g(x,y,z)=b^2\}.$$If $p=(x_0,y_0,z_0)\in\gamma$ then the tangent plane to the first sphere at $p$ is orthogonal to$$\nabla f(x_0,y_0,z_0)=\bigl(2(x_0-a),2y_0,2z_0\...


2

The gradients are not applied properly. Let $f(x,y,z)=x^2+y^2+z^2-2ax$ and $g(x,y,z)=x^2+y^2+z^2-2by$. Then the spheres are given by $f(x,y,z)=0$ respectively $g(x,y,z)=0$. The gradients of $f$ and $g$ are orthogonal to their tangent spaces. So to verify if the tangent spaces are orthogonal, we need that for every $(x,y,z)$ on both spheres that $\langle \...


1

On your specific question about how the equation in polar form becomes $r = 3 \cos \theta$, In polar form, $x = r \cos \theta, y = r \sin \theta$ Now your curve is $(x-3/2)^2 +y^2 = 9/4 \implies x^2 + y^2 = 3x$ $r^2 cos^2 \theta + r^2 \sin^2 \theta = 3r \cos \theta$ $r^2 = 3r \cos \theta \implies r = 3 \cos \theta$ Here is a diagram showing how $r$ changes ...


1

The domain $D= \{(x,y) \in \mathbb{R}^2 \mid 1\leqslant x^2+y^2 \leqslant 4, x\leqslant0, y \geqslant0 \}$ can be easily written with polar coordinates as follows. Let $x=r\cos\theta$ and $y=r\sin\theta$. Since $1\leq x^2+y^2\leq 4$, you get that $r\in [1,2]$. Also, $x\leq 0$ and $y\geq 0$, which means we only have the part of the annulus in the second ...


1

Starting from @WA Don's answer, wa have $$9y^6+12y^5+4y^4-1024=(y-2) \left(3 y^2+8 y+16\right) \left(3 y^3+2 y^2+32\right)$$ $(3 y^2+8 y+16)$ does not show any real root nad we are left with the solutions of $$3 y^3+2 y^2+32=0$$ whidh has only one real root since $\Delta=-249856$. Using the hyperbolic method, the real root is given by $$y=-\frac{2}{9} \left(...


1

Since you are asking how to solve problems like this, I add this answer. Look carefully at the denominator: For $y=-2x$ it is not defined while for $x\neq 0$ the numerator $x^2-y^2 = x^2-(-2x)^2 = -3x^2\neq 0$. This suggests that if you move to $(0,0)$ along a curve close t0 $y=-2x$, you may push the value of the expression $\frac{{x^2 - y^2}}{{2x + y}}$ to $...


0

\begin{align} \lim_{(x,y)\to(0,0)}\frac{{(x + y)\operatorname{tg}(x^2+y^2)}}{{\sqrt{x^2 + y^2}}} &= \left(\lim_{(x,y)\to(0,0)}(x + y)\right) \left(\lim_{(x,y)\to(0,0)}\frac{{\operatorname{tg}(x^2+y^2)}}{{\sqrt{x^2 + y^2}}}\right) \\ &= 0\cdot0 = 0 \end{align} by the hint from @José Carlos Santos!


0

It's the former. A function $f:\mathbb R^n\longrightarrow\mathbb R^m$ takes elements of $\mathbb R^n$, so vectors of the form $x=(x_1,\dots,x_n)$, and maps each of them to exactly one element of $\mathbb R^m$, so vectors of the form $y=(y_1,\dots,y_m)$. Other than what I said here, there are no restrictions on how it does this. All components of the output ...


1

If you change the order of integrations you get $$\int_{0}^{1}\int_{0}^{y^5}(\pi \sin(5\pi x)/(1-x^{1/5}) dxdy=\int_{0}^{1}\int_{x^{1/5}}^{1}(\pi \sin(5\pi x)/(1-x^{1/5}) dydx.$$ This gives the integral $$\int_{0}^{1}(\pi \sin(5\pi x) dx.$$


0

It is of the form $$f(x)= (f_1(x), \dots, f_m(x)),$$ since each function $f_j$ has the domain $\mathbb R^n.$


0

Assuming that $x=(x_1,\ldots,x_n)$, then the answer is $(1)$. The functions of the form $(2)$ are a tiny subset of those of the form $(1)$.


1

Let $A$ be the set of $(x_1, x_2, ..., x_n)$ such that $$\sum x_i = c$$ and $x_i \geq 0$ for all $i$. On a point $x$ of the boundary of $A$, there is always a component $x_i$ of $x$ which is equal to zero, hence $f = 0$ on the boundary. As $f$ is continuous and $A$ is compact, $f$ has a global maximum on $A$. Denote by $x^\ast$ one such maximum. The fact ...


1

You have no control over $\frac y x$ so your answer is wrong. The expression is not defined along that lime $y=-2x$ so the limit is not even well defined. Also along the sequence $(\frac 1 n, \frac 1 {n^{2}}-\frac 2 n)$ you get a non-zero limit.


1

Here is a somewhat more geometric explanation of @Yiorgos or @Ted-Shrifin answers. $f$ satisfies a first order partial differential equation. This equation says something about a directional derivative, namely that the derivative in the direction $ (-1,1)^t $ is 0. To make things clear, we can change variables so the direction $(-1,1)^t$ is a coordinate ...


1

You drop a t taking a partial derivative. The derivation goes like this: $\begin{align} \frac {\partial f(x)} {\partial x_i} & = \int_0^1 \frac {\partial } {\partial x_i} \left( x_j v_j(tx)\right) \, \mathrm d t \\ & = \int_0^1 v_i(tx) + x_j \frac {\partial} {\partial x_i}[v_j(tx)]\, \mathrm d t = \int_0^1 v_i(tx) + x_j \frac {\partial v_j} {\...


0

You have to get $$2x = 2x$$. Since you got: $$2x = 1$$ it is not conservative. Here is a graph. Note that if you go up via the left or right and come back down the middle you wouldn't get a zero vector for a closed loop. desmos link


2

You are right. Left & right derivatives and directional derivative are different notions – they coexist in fact, also in one-dimensional spaces such as $\mathbf{R}$. The directional derivative can be defined on a generic vector- or affine space or on a differential manifold, without the existence of a norm or scalar product. It's therefore useful not ...


1

It is easier to set up your integral as $dz \, dx \, dy$ given you have one of the equations in just $x, y$. Here is how you can see it - if you take infinitely thin vertical strips from down to up between $y = 0$ to the plane, you realize you hit upon $x = 2y$ till $x = 1$ and then on it is the other plane $x + 2y + z = 2$. So you will have to break your ...


1

y goes from 0 to 0.5 And for a fixed y we know that x goes from $2y$ to $2 - 2y$ because the x values are to the right of the line $x = 2y$ and to the left of the line $x = 2 - 2y$ So we end up with: $\int_0^{0.5}\int_{2y}^{2-2y}\int_{0}^{2 - x -2y}dzdxdy$


0

The directional derivative is not the same as the limit approaching from the left or the right. If you use the negative unit vector, you're asking what the slope is if you're traveling to the right vs. to the left. You get the negative of the slope if you use the negative unit vector. But that is correct. The divide by the magnitude of v ($|v|$) is ...


1

The limit is NOT zero. It does not exist! If $x=y^2$, as $y\to 0$, then $x\to 0$ and $$\frac{x^2}{x + y^2} =\frac{y^4}{y^2 + y^2}=y^2\to 0.$$ On the other hand, along the curve $x=-y^2+y^4$, as $y\to 0$, then $x\to 0$ and $$\frac{x^2}{x + y^2} =\frac{(-y^2+y^4)^2}{-y^2+y^4 + y^2}=\frac{y^4-2y^6+y^8}{y^4}= 1-2y^2+y^4\to 1.$$ Indeed, in your first approach, $\...


0

If $f_{xy}$ and $f_{yx}$ are not required to exist, then obviously we can have a function where $f_{xy}$ is continuous but $f_{yx}$ does not exist, such as $f=|y|$. Theorem. If $f_{xy}$ and $f_{yx}$ exist everywhere, and $f_{xy}$ is continuous at $(0,0)$, then $f_{yx}$ is continuous at $(0,0)$. Proof. First, the two mixed partial derivatives are equal at $(0,...


1

Assume that $f:\textbf{R}^n\rightarrow\textbf{R}$ and $f=f(x_1,x_2,\ldots,x_n)$ is a function of $n$ variables. By saying that $x_i=x_i(\xi)$, then $C:\overline{x}=\{x_1(\xi),x_2(\xi),\ldots,x_n(\xi)\}$, $\xi\in\textbf{R}$, then $C$ is one dimentional object in $\textbf{R}^n$ and hence $C$ is a curve of $\textbf{R}^n$. Then $$ \frac{df}{d\xi}=\sum^{n}_{k=1}\...


1

All it means is you have a scalar field $f(x,y,z) = xz$ instead of a vector field. Your surface is a plane given by $2x+2y+z=2 \,$ where $(x\geq0, y\geq0, z\geq0)$. Given the integral is wrt $y, z$, rewrite your plane in the form $g(y,z) = x = 1 - y - \frac{z}{2} \,$. $\displaystyle \frac{\partial g}{\partial y} = - 1, \frac{\partial g}{\partial z} = - \frac{...


0

The planes can be rewritten as $z_1=2-x-y$ and $z_2=2x+2y+10$. Then $$z_2-z_1=3x+3y+8\,,$$ so if $-1\le x,y\le 1$, then $z_2-z_1\ge-3-3+8=2$. If $\langle x,y,z\rangle$ is in side the vertical cylinder between $y=1-x^2$ and $y=x^2-1$, clearly $-1\le x,y\le 1$, so the $z_2$ plane lies entirely above the $z_1$ plane within that cylinder: the intersection of the ...


1

For any regular curve $r(t)$, the velocity is $\dot r(t)$; this vector is clearly tangent to $r(t)$ for every $t$; the acceleration vector is evidently $\ddot r(t)$, but it is not in general tangent to $r(t)$. We may find both the tangential and normal components of $\ddot r(t)$; first, we form the unit tangent vector field $T(t)$ to $r(t)$: $T(t) = \dfrac{...


0

There is a key result called the Fubini-Tonelli Theorem (and a variety of related results) that roughly says that for any reasonable function, it does not matter what order you do the integrals in. Writing it down (actually two nested copies of it) in the $\mathbb{R}^3$ case, it becomes: For any measureable function $f$, we have $$\int |f| dxdydz = \int |f| ...


0

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\...


1

If you are familiar with differential forms the answer is simply that the volume element is a decomposable top form, i.e. the wedge product of $n$ 1-forms. In Euclidean space with the usual coordinates $x^i$ $d V= dx^1\wedge \cdots \wedge dx^n$ and if you plug in the transformation law $dx^i \mapsto \sum _j \partial x^i /\partial u ^j du^j$ you get out a ...


3

A set $C$ in $R^n$ is closed exactly iff all the Cauchy sequences in $C$ converge to an element inside $C$. You can carry a Cauchy sequence in $F(C) $ to a cauchysequence in $C$ with the preimage. This Cauchy sequence converges in $C$. Using continuity of $F$ we get that the sequence in $F(C) $ converges as well to an element of $F(C) $.


1

Consider the function, for $a<b$, $$\phi_{a,b}(t) = \begin{cases} \exp(\frac{-1}{(t - a)(b - t)})& a < t <b,\\ 0 &\text{otherwise}. \end{cases}$$ which is $C^\infty$ compactly supported. Now you can take $$\eta_{a,b}(t) = 1 - \frac{\int_{-\infty}^t \phi_{a,b}(s)~ds}{\int_{-\infty}^\infty \phi_{a,b}(s)~ds}$$ which is $C^\infty$, equal to $1$ ...


4

It's an direct consequence of the Hermite-Hadamard inequality We have with $f(x)=px^{p-1}$ a convex function , $x\geq y\geq 0$ and $p\geq 1$: $$\frac{1}{x-y}\int_{y}^{x}f(x)dx\leq \frac{f(x)+f(y)}{2}$$ Or : $$\frac{x^p-y^p}{x-y}\leq \frac{px^{p-1}+py^{p-1}}{2}$$


2

WLOG $x\ge y$ let $t=\frac{x}{y}\ge 1$ we have to prove $$f(t)=(p-1)t^p-pt^{p-1}+pt-p+1\ge 0$$ notice $$f'(t)=p(p-1)(t^{p-1}-t^{p-2})+p>0$$ thus $f(t)$ is increasing also $f(1)=0$ hance $f(t)\ge 0$ for all $x,y$ in domain given (because the case $x\le y$ can be proved analogously)


5

Wlog $x>y$. Then with $f(t):=x^p$, we have from the Mean Value Theorem that there exists some $\xi$ with $x>\xi>y$ such that $$\frac{x^p-y^p}{x-y}=\frac{f(x)-f(y)}{x-y}=f'(\xi) =p\xi^{p-1}\le p\max\{x^{p-1},y^{p-1}\}<p(x^{p-1}+y^{p-1}).$$


0

What you stated you did are the correct steps. First, differentiating each component leads to $$\frac{dr(t)}{dt} = \langle -3\sin(3t), -3\cos(3t), 9\cos(3t) - 3\sin(3t) \rangle \tag{1}\label{eq1A}$$ At $t = \frac{\pi}{4}$ means $3t = \frac{3\pi}{4}$, with $\sin(3t) = \frac{1}{\sqrt{2}}$ and $\cos(3t) = -\frac{1}{\sqrt{2}}$. Thus, \eqref{eq1A} gives $$\left\...


1

As per Divergence theorem, $\, \displaystyle \iint_{S} \vec{F} \cdot \hat{n}\,d$S $\displaystyle = \iiint_V (div \,\vec{F}) \,dV$ We have, $\, \displaystyle \vec{F} = \frac{\vec{r}}{r^2}$ $\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$ $r^2 = x^2 + y^2 + z^2$ $\displaystyle div \vec{F} = \nabla \cdot \vec{F} = \frac{\partial}{\partial x} (\frac{x}{r^2})+ \...


1

Under the assumption that $x^2+y^2=4$, you should have been able to prove that $$\frac{1}{25}\left(2x+\frac{2x}{x^2+y^2}\right)^2 + \frac{1}{9}\left(2y-\frac{2y}{x^2+y^2} \right)^2 = 1,$$so possibly you made an algebra mistake. But this would have only showed that $T$ takes the circle into the ellipse. Note that $$(\Bbb R^2 \setminus \{(0,0)\}) \setminus \{(...


0

Here's how I would prove the hard case: Step 1: Show that $\int_\alpha F\cdot n\, ds=\int_{C_r} F\cdot n\, ds$, where $C_r$ is a closed circle of radius $r$, chosen small enough so that the corresponding disk fits inside the closed region created by $\alpha$. You can show this fact that splitting up the region enclosed by $\alpha$ into two regions that avoid ...


1

Here's a picture with some more annotations: I've highlighted, in blue, an arc $c$ of the larger circle and, in red, an arc $d$ of the smaller circle, as well as drawn, in green, an arrow from the center of the smaller circle $B$ to $P$. The part of the expression you understand is essentially the position of $B$, so let's look at how to determine the ...


2

How about this method? $$\frac{2(x + y)²}{2x² + y²}-3=\frac{2x^2+4xy+2y^2-(6x^2+3y^2)}{2x^2+y^2}=-\frac{4x^2+4xy+y^2}{2x^2+y^2}=-\frac{(2x-y)^2}{2x^2+y^2}\le0..$$


2

Your notation is clumsy so let's clean up the problem and see what we are doing. You have a function $F:\mathbb{R}^2\times \mathbb{R}^2\to \mathbb{R}^2$ given by $F(x,y,u,v)=(f_1(x,y,u,v),f_2(x,y,u,v))^T$, where \begin{align} f_1(x,y,u,v)&=x^3-3xy^2-u \\ f_2(x,y,u,v)&=3x^2y-y^3-v \end{align} (The order of the $x,y$ terms and the $u,v$ terms may be ...


2

Well $2x^2+y^2> 0$ and $2(x+y)^2\ge 0$ so So $\frac{2(x + y)^2}{2x^2+ y^2} \leq 3\iff$ $2(x+y)^2 \leq3(2x^2 + y^2) \iff$ $2x^2 + 4xy +2y^2 \leq 6x^2 + 3y^2 \iff$ $0 \leq 4x^2 -4xy +y^2 \iff$ $0 \leq(2x -y)^2 \iff$ God's in his heaven and all's right in the world.


2

Let $y=l.x$; we have: $\frac{2[(l+1)x]^2}{x^2(l^2+2)}=\frac {2l^2+4l+2}{l62+2}=2+\frac{4l-2}{l^2+2}$ $\frac{4l-2}{l^2+2}<1$ because: $(l-2)^2>0$ $l^2-4l+2+2>0$ $4l-2<l^2+2$ If $l=2$ the equality holds.


3

by C-S $$(2x^2+y^2)(1+2)\ge {(\sqrt{2}x+\sqrt{2}y)}^2=2{(x+y)}^2$$


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