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0

Use the chain rule to obtain \begin{align*} 0\leq g'(t)=\nabla f(\cos(t)/t,\sin(t)/t)\cdot\left(-\frac{\cos (t)}{t^2}-\frac{\sin (t)}{t},\frac{\cos (t)}{t}-\frac{\sin (t)}{t^2}\right). \end{align*} Now, let $t\to0$.


1

We can also use a tweaked version of the Weierstrass approximation theorem: Given $f$ continuous on $[a,b],$ there are polynomials $p_m \to f$ uniformly on $[a,b]$ such that $p_m(a) = f(a), p_m(b)=f(b)$ for all $m.$ (For the proof, see * below.) Given this, suppose $\gamma:[0,1] \to U$ is a path from $x$ to $y$ in $U\subset \mathbb {R}^n.$ We can then ...


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Given any two points in your domain, pick a continuous path between them. Because the closed interval $I$ is compact, you can cover this path with finitely many balls; use this to construct a piecewise linear path between the two points. (With enough care in this step and the previous you can ensure that the path is actually injective.) Now smooth the ...


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If you already proved the particular case when the curve is a graph, then you can use the implicit function theorem! By the theorem, every $C^1$ curve is (locally) a graph, either of the form $(x, f(x))$ or of the form $(f(y),y)$. So, your proof can be expanded as follows: Take a curve $(x(t),y(t))$, where $t\in[a,b]$. For every point $t_0\in [a,b]$, ...


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$$\begin{align} (\delta_{im}\delta_{kl} - \delta_{il}\delta_{km})\frac{\partial f_m}{\partial x_l}f_k&=\frac{\partial f_i}{\partial x_k}f_k-\frac{\partial f_k}{\partial x_i}f_k \\ &\ne \frac{\partial}{\partial x_k}f_if_k - \frac{\partial}{\partial x_i}f_k^2 \end{align}$$ which was the incorrect result. It is straightforward to see that ...


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If $\dim S_1 = \dim S_2 = 2$ and $S_1 \subset S_2$, then you have $S_1 = S_2$. To see this, suppose $ \{ u_1,u_2\}$ is a basis of $S_1$. If $S_1 \neq S_2$ then there exists $u_3 \in S_2 \setminus S_1$, and the collection $\{u_1,u_2,u_3\}$ is linearly independent (otherwise $u_3 \in S_1$). Hence $\dim S_2 \ge 3$, a contradiction.


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Let $\def\v#1{{\bf#1}}\v w\in W_2$. Then since $S_2$ is a basis for $W_2$ we have $\v w=\lambda\v v_1+\mu\v v_2$ for some scalars $\lambda,\mu$; from what you are given, $\v v_1=\alpha_1\v u_1+\alpha_2\v u_2$ and $\v v_2=\beta_1\v u_1+\beta_2\v u_2$; so $\v w=(\lambda\alpha_1+\mu\beta_1)\v u_1+(\lambda\alpha_2+\mu\beta_2)\v u_2$, and this is in $W_1$. ...


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You don't need, At first, every vector of $S_2$ is linear combination of $S_1$, then $v_1,v_2$ are linear combinatios of $u_1,u_2$, so $span(S_2)\subset span(S_1)$ Then exists some $\alpha_1, \alpha_2, \beta_1, \beta_2$ such: $v_1=\alpha_1u_1+\alpha_2u_2$ and $v_2=\beta_1u_1+ \beta_2u_2$ Then solving the sistem for $u_1, u_2$, you can prove that their are ...


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For some reason you made the derivative operator act on both $f$'s. The calculation should go as follows, $$ \left((\nabla \times \vec{f} ) \times \vec{f} \right)_k = \epsilon_{ijk} ( \nabla \times \vec{f} )_i f_j $$ $$ = \epsilon_{ijk} \epsilon_{\alpha \beta i} \left( \partial_\alpha f_\beta \right) f_j $$ $$ = \epsilon_{ijk} \epsilon_{ i \alpha ...


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It's a bit confusing keeping the two $f$'s together. But your work looks correct up till the end. I think the correct expression should be $$ (\nabla \times f)\times f=(f \cdot\nabla)f-\frac12 \nabla(f\cdot f) $$


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i think you can explain what you are doing in words. you have not made clear that you understand what linear dependence of vectors mean. true, you are showing some row reduction, but why and how does it relate to the question. one way to do is to explain how you determine linear independence by supposing $$a(v_1+v_2-v_3) + b(2v_1+2v_3) + c(-v_1 + v_2 - ...


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You can find if a set is linearly dependent or independent by calculating the determinant. If it's zero, then the set is dependent. If it's non-zero, then the set is independent.


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If one understands why you came up with the original matrix and how the linear dependence follows from your fast reasoning and writings derived from that matrix, then it's fine. However, to make a proof that is unquestionable, you would probably need more detail. However, if it makes you feel any better, if I were a professor I would be inclined to give you ...


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$$\int_{0}^{\pi} \int_{0}^{2\cos(\Theta)} rsin(\Theta) rcos(\Theta)\ r \ dr d\Theta$$ this is how you would do it basically


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Well the answer is not simply "no" because it could be the case. The answer to if it guarantees that its a subspace however, is indeed no. Consider the basic example using $\mathbb{R^3}$. Consider the space W consisting of vectors of the form $v=(x,y,z)$ such that all $x,y$ and $z$ are integers. So of course the zero vector is in W say $v_1,v_2 \in W$ ...


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Here's an example: Let $V=\Bbb R^2$ and $W=\{(n,m):n,m\in\Bbb Z\}$. $W$ is certainly closed under addition, contains $(0,0)$, but it is not closed under scalr multiplication. For example, $\frac12(1,1)\notin W$.


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Uniqueness for Laplace's equation on a domain $D$ (with, say, piecewise-differentiable boundary $\partial D$) with, say, Dirichlet boundary conditions $u=f$ on $\partial D$ is easy: suppose there are two such solutions, $u$ and $v$. Take $$ -\nabla^2 (u-v) = 0, $$ multiply by $u-v$ and integrate over $D$, and use the divergence theorem: $$ 0 = -\int_D ...


0

Some comments first. For the usual spherical coordinates: (1) $\theta$ is measured as the angle with respect to the origin in the $xy$ plane. On the unit sphere, when you let $\theta$ be a constant value, e.g. $\theta = \frac{\pi}{4}$, then you will get a curve that is a longitude of the sphere. This is half of a circle joining the north and south pole. ...


2

The $6$ different orders of integration can be worked out straight-forwardly: x-y-z This is what you did first. The bounds are $$\int_0^1 \int_{x^2}^{\sqrt x} \int_0^{x+y}$$ x-z-y Again $x$ is independent, so $x\in[0,1]$. Now we bind $z\in [0, x+y]$. For this, $y$ is at most $\sqrt x$, so $z\in [0,x+\sqrt x]$. This results in $y \ge z-x$ in addition to ...


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Both false. 1) Consider the complement of a two-point set. 2) Let $C$ be a semicircle.


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You need to parameterize the surface. In this case the surface $S$ is the graph of $z = 1 - x - y$ over the triangle $T$ described $x \ge 0$, $y \ge 0$, $x + y \le 1$. If you use the simplest parameterization $r : T \to S$ given by $$r(x,y) = (x,y,1-x-y)$$ you get $r_x = (1,0,-1)$ and $r_y = (0,1,-1)$ so that $$r_x \times r_y = (-1,1,-1)$$ which points ...


2

So, we have $f(x,y)=3y-y^3-3yx^2$. Hence, for critical points $(x,y)$, we have, $$\begin{cases}\dfrac{\partial f}{\partial x}=-6xy=0\\ \dfrac{\partial f}{\partial y}=3-3y^2-3x^2=0\end{cases}$$ First equation implies that either $x=0$ or $y=0$ or both $x,y=0$. Now, if $x=y=0$, then the second equation is not satisfied, so $(0,0)$ is not a critical point. ...


1

The directional derivative of a scalar function $f(\mathbf{x}) = f(x_1, x_2, \ldots, x_n)$ along a vector $\mathbf{v}$ is, by definition $$\nabla_{\mathbf{v}}{f}(\mathbf{x}) = \lim_{h \rightarrow 0}{\frac{f(\mathbf{x} + h\mathbf{v}) - f(\mathbf{x})}{h}}\hbox{.}$$ In case when $\nabla{f}(\mathbf{x})$ exists, $\nabla_{\mathbf{v}}{f}(\mathbf{x}) = \nabla ...


0

Note that your function $$ det: \mathbb{R}^{n \times n } \rightarrow \mathbb{R} $$ Thus its frechet derivative $Ddet : \mathbb{R}^{n \times n } \rightarrow \mathit{L} (\mathbb{R}^{n \times n }, \mathbb{R} ) $, where $\mathit{L} (\mathbb{R}^{n \times n }, \mathbb{R} ) $ is the vector space of all linear maps(here it is forms) from $\mathbb{R}^{n \times ...


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The Frechet derivative $dF(x)$ of a function $F$ defined on a Banach space $X$ is defined as a (continous) linear functional on $X$, i.e. a mapping $dF(x):X\to\mathbb{R}$, or, more precisely $dF(x)\in X^*$ ($dF(x)$ is an element in the dual space of $X$). In your case you have $\det:\mathbb{R}^{n\times n}\to \mathbb{R}$, i.e. $X = \mathbb{R}^{n\times n}$. ...


1

If $$F=F_1\frac{\partial}{\partial x}+F_2\frac{\partial}{\partial y}$$ then $$\int_CF\times da = k \int_C(F_1\, dy - F_2 \, dx).$$ Now apply Green's theorem.


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I think the easiest way to follow through with the calculations in a case like this is to use the well-known vector calculus identities, $\nabla \cdot (fX) = \nabla f \cdot X + f \nabla \cdot X, \tag{1}$ $\nabla \times (fX) = \nabla f \times X + f \nabla \times X, \tag{2}$ where $f$ is a scalar function and $X$ is a vector field, both assumed to be ...


0

No, stipulating that the supports of the functions $\varphi\in\Phi$ be Jordan-measurable is not needed. It is sufficient for the existence of the integral $\int \phi\cdot |f|$ that $f$ be locally bounded and almost everywhere continuous and the support of $\varphi$ be compact. This is because $\varphi(x)=0$ at the boundary of $\textrm{Supp}\,\varphi$, so ...


1

I find your notation a little confusing so I'll restate this with the notation I'm used to: $$\frac{\partial f}{\partial t} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial t}$$ Like you say, this is still in terms of $x$ and $y$ at least in some of it, like $\frac{\partial f}{\partial ...


2

You multiply the function by the position vector and then just calculate the derivatives... [\begin{array}{l} \vec r = \left( {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right)\\ f(x,y,z) = \left( {{x^2} + {y^2}} \right)\log (1 - z) = {x^2}\log (1 - z) + {y^2}\log (1 - z)\\ f(x,y,z)\vec r = \vec F = \left( {\begin{array}{*{20}{c}} {{x^3}\log (1 - z) + ...


1

Remember that if $\bf F$ is a vector field which domain is simply connected, then a potential will exist if and only if $\nabla \times {\bf F} = {\bf 0}$. So if you have a integral $$\int_C {\bf F} \cdot {\rm d}{\bf r}$$ and $\bf F$ admits a potential $f$, then you can find $f$ and use the FTC. You would have to analyze what is easier: finding the potential ...


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Hint: For 2, you have two line integrals; one for each scalar component of the 2-D vector. Integrate the components separately. Each integral is zero. For 3, you have a tensor $\vec x d\vec x$ that has $4$ components. They are $$-\cos t \sin t$$ $$\cos^2t$$ $$-\sin^2t$$ $$\cos t \sin t$$ For 4, note that $ds=d\vec x/||d\vec x||=dt$ and ||\vec ...


1

Since the vector field $\langle x_2^2 \cos(x_1), 2x_2(1 + \sin(x_1))\rangle$ is the gradient of the scalar field $\phi(x_1,x_2) = x_2^2(1 + \sin(x_1))$, by the fundamental theorem of line integrals, your integral is zero.


0

Stop torturing yourself with parametrizations and use Green's theorem: you'll get $0$ in just a few seconds.


1

It's like in polar coordinates, where for the plane we have $0<r<\infty,0<\theta<2\pi$ or $-\infty<r<\infty,0<\theta<\pi$


3

Hint Before changing coordinates, factor $$e^{-(x^2 + y^2 + z^2)} = e^{-(x^2 + y^2)}e^{-z^2}$$ and separate the integral in $z$: $$\iiint_R z e^{-(x^2 + y^2 + z^2)} dV = \int_0^1 ze^{-z^2} dz \iint_{{\Bbb R}^2} e^{-(x^2 + y^2)} dx\,dy .$$


0

Hint: Write $$ \| f(x_2)-f(x_1)-Df(x_0)(x_2-x_1)\| \leq \| f(x_2)-f(x_1)-Df(x_1)(x_2-x_1)\| +\| (Df(x_1)-Df(x_0))(x_2-x_1)\|. $$ Then both terms can be made small using the continuity of the derivative.


1

As far as continuity: Since $x^2$ and $y$ are in additive relationship, independently of the path, if both tend to $0$ the limit is zero. So the function is continuous at $(0,0)$. I don't understand why you divided by $x$, though... Considering the definition of the directional derivative we need a vector of absolute value one: $$\overline ...


1

$g(x,0)= 0 \implies g_x(0,0) = 0, g(0,y) = y \implies g_y(0,0) = 1.$ So if $Dg(0,0)$ exists, it must be the linear transformation $(x,y)\to y.$ The question is then: Is it true that $$g(x,y) = g(0,0) + y + o((x^2+y^2)^{1/2})$$ as $(x,y) \to (0,0)$? Yes! But I'll leave that one for you. (It will require some thought.) Given the above, the directional ...


1

Another way to proceed is to transform variables to spherical coordinates. To that end, let $x=r \sin \theta \cos \phi$, $y=r\sin \theta \sin \phi$, and $z=r\cos \theta$, where $0\le \theta \le \pi$ and $0\le \phi \le 2\pi$. Then, the limit that is in question is $$\lim_{r\to 0}\left(\frac{\sin r^2}{r(|\sin \theta \cos \phi|+|\sin \theta \sin ...


1

How about $(x,y,z)=(\cos t,\sin t,-\sin t\cos t)$ Check the derivative is never the zero vector.


1

Write $\frac{\sin(x^2+y^2+z^2)}{|x|+|y|+|z|}=\frac{\sin(x^2+y^2+z^2)}{x^2+y^2+z^2}\frac{\sqrt{x^2+y^2+z^2}}{|x|+|y|+|z|}\sqrt{x^2+y^2+z^2}$. The first factor tends to $1$, the second is bounded and the last tends to zero. To answer your question in your comment: $$x^2+y^2+z^2\leq x^2+y^2+z^2+2|x||y|+2|x||z|+2|y||z|=(|x|+|y|+|z|)^2$$ Therefore ...


1

Here is a generalization of your approach for the one-dimensional case to all cases, making heavy use of the axiom of choice. For convenience, set $\eta(x):= f(x) - f(a) - D_a f(x-a)$. Define $R \colon X \to L(X, Y)$ by choosing at each $x \in X$ an $R(x) \in L(X, \operatorname{span}\{\eta(x)\}) \subset L(X,Y)$ such that $R(x)(x-a) = \eta(x)$ and ...


0

A quick-and-dirty mnemonic is that "by cancellation of differentials", you "should" have $$\frac{dy}{dx} = \frac{\frac{\partial G}{\partial x}}{\frac{\partial G}{\partial y}}$$ You have this, but with an additional minus sign, and with division replaced by a matrix inverse. This mnemonic was part of how I was taught this rule in thermodynamics, where it is ...


0

I always had trouble remembering the implicit function theorem as well. It helped me to formulate it this way:¹ If for banach spaces $E$, $F$ a function $f \colon E \oplus F → F$ is $C^k$ such that $f(0) = 0$ and the tangential of $f∘i_F \colon F → F $ is invertible at $0$ for the inclusion $i_F \colon F → E \oplus F,~x ↦ 0 \oplus x$, … then there ...


1

$\newcommand{\Reals}{\mathbf{R}}$In case it's a helpful mnemonic, the implicit function theorem (in this setting) is essentially a non-linear version of Gaussian elimination from linear algebra. Specifically, let $m$ and $k$ be positive integers, and put $n = m + k$. Suppose $A$ is an $m \times n$ matrix, and let $F:\Reals^{n} \to \Reals^{m}$ be ...


1

I have written an answer to explain what is going on, but a little mistake made me loose the LaTeX original file. Fortunately, I have it on PDF ; I include it as a picture : Take care that your $F$ is transformed in $f$ from the $7$-th line! I add a picture which summarizes what I wrote : Hope it will be helpful!


0

As far as I can understand, the question implies that: For uniform areal density the center of gravity is at center of inner circle. Suggest a function for areal density such that the unbalanced mass center of gravity is forced out of the inner circle. Try out some odd functions like: $$ \rho (y) = a + b y , a > b , \rho(y) = c + d y^3, c>0, ...



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