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Hint. The relation $$ad-cb=1,$$ requires that $(a,c)\ne (0,0)$, and says that the point $(d,b)$ lies on the line $$ax-cy=1,$$ which is described parametrically by $$(x,y)=t(c,a)+(a^2+c^2)^{-1/2}(a,-c).$$ Thus $$\left(\begin{matrix}a &b \\ c& d\end{matrix}\right) =\left(\begin{matrix}a &at-c(a^2+c^2)^{-1/2} \\ c& ... 4 The Lagrangian method brings conditionally stationary points to the fore, if there are any. In this example there are none, as you have found out. Now x^2-y^2=2 defines a hyperbola \gamma with apexes at (\pm\sqrt{2},0) and asymptotes y=\pm x. The function f(x,y):=x-y essentially measures the distance from the point (x,y) to the ascending ... 3 It depends on what draws you to the text, and what prompts you to post this question: Is it a matter of (mostly) enjoyment (that the prospect of digging into Loomis and Sternberg is enticing to you, so the question is more along the lines, "is it so fun/challenging that it would be a shame to miss out by not tackling it?"), ...Or a question of necessity ... 3 The point here is that for this question, you don't need to! Notice that you haven't been given an explicit parameterization of your curve, or even an explicit description of your curve? That should be a dead-giveaway that this integral is expected to be independent of path -- that is, that only the start and endpoint of your contour matter! Another sign ... 2 Let us first remember what is the meaning of each symbol: D_\nu f(x_0) is the derivative of f at x=x_0 in the \nu-direction, i.e.,$$ D_\nu f(x_0)=\lim_{h\to 0}\frac{f(x_0+hv)-f(x_0)}{h}. \tag*{(*)} $$On the other hand, we have the following the definition: The function f:U\to\mathbb R^m, where U\subset\mathbb R^d open and x_0\in U, is ... 2 L(\gamma)=L(\sigma) is obvious. The difficult thing is to prove that these suprema can be written as an integral. Nevertheless, here is why one has L(\gamma)=L(\sigma): Both L(\gamma) and L(\sigma) are the sup of the same set, namely the set of all sums of the form$$\sum_{k=1}^N |\gamma(t_k)-\gamma(t_{k-1})|$$with ... 2 Calculus is overkill: this is just geometry. Since 0 < c < 1, cx + (1-c) y \ge 0 for all (x,y) in the square. cx + (1-c) y = b, on the other hand, is a straight line that does pass through the square. Figure out where it intersects the boundary of the square (the details will depend on how b relates to c and 1-c). Your region is ... 2 I'm convinced the examples using a sine curve with decreasing amplitude work out fine, but I think calculating or bounding the arc length might be tedious. Let me suggest something more elementary. Consider the curve \gamma:[0,1]\to\mathbb R^2 given by$$ t\mapsto \left(t, \frac{1}{\lfloor 1/t\rfloor} - \left| 2t \lceil 1/t\rceil-2-\frac{1}{\lfloor ...

2

Consider the level surface $F(x,y,z)=k$. If $\alpha: \mathbb{R} \rightarrow S$ is a curve on the surface then $F(\alpha (t))=k$ for all $t$. But, the chain-rule for multivariate calculus says $\nabla F(\alpha(t)) \cdot \alpha'(t)=0$ for all $t$. Thus, for any time $t_o$ we find the tangent $\alpha'(t_o)$ is perpendicular to $\nabla F(\alpha(t_o))$. But, the ...

2

Hint: for the interior of disk you may use usual condition $\nabla f(x,y) = 0$. For the boundary of the disk you may use parametrisation $x = \cos \phi$, $y = \sin \phi$ and minimize it w.r.t. to $\phi$ (remember, it's $2\pi$-periodic, so you should investigate it only on segment $\lbrack 0, 2\pi )$ really). Or you may use Lagrange multipliers method.

2

When you are given a matrix-valued function ${\bf x}\to A({\bf x})$ and know for sure that it is the Jacobian of some vector-valued function $${\bf f}:\quad{\mathbb R}^n\to{\mathbb R}^m,\qquad {\bf x}\to{\bf f}({\bf x})$$ then it is easy to recover ${\bf f}$ from $A$. Indeed, the columns of $A$ are nothing else but the partial derivatives $${\bf ... 2 We have the system:$$\tag 1 \frac{du}{dt}=r u\left(1-\frac{u}{q}\right)-\frac{u^2}{1+u^2}$$where r, q are positive dimensionless parameters. To find the equilibrium points, we set (1) equal to zero and we see u=0 is one and the other two occur where the death rate equals the birth rate (intersection) and is given by:$$\tag 2 r \left[1 - ...

2

This result holds at least for $\alpha>0$. I will describe the method that allows to make an analysis even in the case of arbitrary dimension. Denote $\vec r=(x,y)$, $r=\|\vec r\|$. First element: let the function $f_\epsilon\in L^1_{loc}(\Bbb R^2)$ be given by $$f_\epsilon = \begin{cases}r^\alpha,&r>\epsilon,\\\epsilon^\alpha,&r\le\epsilon. ... 2 When f is differentiable at p then there exists a function r with \lim_{X\to 0}r(X)=0, such that for some h>0 one has$$f(p+X)-f(p)=df(p).X +|X|\,r(X) \qquad\bigl(0<|X|<h\bigr)\ .\tag{1}$$Let a vector V\ne0 be given and put X:=tV in (1). Then we obtain$$f(p+tV)-f(p)=t\> df(p).V +t\,|V|\>r(tV)\qquad\bigl(0<t<h'\bigr)\ ...

2

You cannot apply divergence theorem directly because $\nabla \cdot F$ has singularity at the origin. Here is an alternative way: Let $G : S^{2} \to S : (x, y, z) \to (ax, by, cz)$ be diffeomorphism. If $\omega$ denotes the 2-form we are integrating, then $$G^{*}\omega = \frac{bc}{ax} dy \wedge dz + \frac{ca}{by} dz \wedge dx + \frac{ab}{cz} dx \wedge dy$$ ...

2

I recommend that you rotate your sphere so that the plane is perpendicular to the $z$ axis. The domain of integration then becomes quite natural in cylindrical coordinates and the resulting integral is quite manageable. Since you asked for a hint, I'll leave it at that for the time being but could edit this answer to include a solution later.

1

It appears as though you're missing a norm in your original definition. The total (Frechet) derivative of a function $f:X\rightarrow Y$ is the linear operator $f^\prime(x)\in\mathscr{L}(X,Y)$ such that $$\lim_{\|h\|\rightarrow 0}\frac{\|f(x+h)-f(x)-f^\prime(x)h\|}{\|h\|}=0.$$ Now, the definition of the limit means that for any $\epsilon>0$ there exists ...

1

Extension from $\mathbb{R}^d$ to $\mathbb{R}^k$ can be done by making the function depend on first $d$ coordinates only. So we work with a problem of extending function $f\colon \mathbb{R}^m \to \mathcal{M}_{k\times d}$ into function $g_3\colon \mathbb{R}^m \to \mathcal{M}_{k\times k}$. We need only the case $m=d$, but we will not use that. First, let's go ...

1

The geometrical or physical interpretation of these integrals has to come from the people who set them up. Nevertheless, you have to be aware that $$\vec A:=\int\nolimits_\gamma \phi\ \vec{d\ell},\qquad \vec B:=\int\nolimits_\gamma \vec\psi\times \vec{d\ell}\tag{1}$$ can be considered as a shorthand notation for a limit of certain Riemann sums. By ...

1

The first part is correct. In the second, you did not correctly account for the derivative of $x_ix_j$ with respect to $x_i$, in the case $i=j$. Let's separate the two cases: \begin{align} -\frac{\partial}{\partial x_j}\left(\frac{\partial u^j}{\partial x_i}\right) &=-\frac{\partial}{\partial x_j}\left(\frac{1}{|x|}\right)+\frac{\partial}{\partial ... 1 This answer addresses part 2 which asks for the curl of \mathbf B. Your friend is correct that the curl is identically zero. The components of the curl of a vector field \mathbf{B}(\mathbf{x})=\mathbf{e}_{\rho}B_{\rho}+\mathbf{e}_{\phi}B_{\phi}+\mathbf{e}_{z}B_{z} in cylindrical coordinates are:\nabla\times\mathbf{B} = ...

1

Your idea is more or less correct, but you might want to made that rigorous. Note that $f$ is differentiable at $(0,0)$ if there is a linear map $T: \mathbb R^2 \to \mathbb R$ such that $$f(x, y) = f(0,0) + T(x, y) + o(\sqrt{x^2+ y^2})\ .$$ That is $$(x^3 + y^3)^{\frac{1}{3}} = ax + by + o(\sqrt{x^2 + y^2})\ .$$ By putting $y=0$ and $x=0$, we have ...

1

No. if you consider $$f(x, y) = \frac{x^2}{y}$$, this has no limit as $(x, y) \to (0, 0)$ basically because $y$ can go to zero really fast (like $x^3$, for instance) so in this case the limit is $\infty$, or they can both go to zero with the same "speed", so the limit is $0$. $$f(x, x^3) \to \infty$$ but $$f(x, x) \to 0$$. What the text is saying is ...

1

Yes; by definition/construction, the derivative is the best local linear approximation of the (local) change of a differentiable function. The best local linear approximation to a linear function $L$ is then $L$ itself. Strictly speaking the differential is the linear function that locally approximates the change of the function, and the derivative (or its ...

1

The function $f$ is well defined on $RP^2$. The three functions $$\phi:\quad{\mathbb R}^2\to RP^2,\qquad(x,y)\mapsto (x,y,1)/_\sim$$ and similarly $\psi$ and $\chi$ form an atlas on $RP^2$. Expressing $f$ in terms of $\phi$ we obtain $$\tilde f(x,y)=[(x,y,1)]={xy+x+y\over 1+x^2+y^2}$$ which is obviously $\in C^\infty({\mathbb R}^2)$.

1

It seems that it has to be constant. From the first coordinate of vector field you obtain that $\phi (x,y,z)$ must be of form $\phi(x,y,z) = xy\sin z + C_1(y, z)$. Now you have to compare the vector field which is generated by this potential with your initial vector field. They agree on the first component, but you have to check 2nd and 3rd yourself. It's ...

1

The equation $U(u)=V(u)$ leads to an equation $P(u)=0$ where $P$ is a polynomial of degree three. There are here cases: There are three different real roots, and therefore three steady states There are two different real roots, one simple and one doble There is one real root and two complex conjugate roots, and only one steady state. The values of $q$ ...

1

We first take the partial derivatives with respect to each variable and set them to zero: $$\frac{\partial f}{\partial x}=6x^2 + 2y^2 - 1=0$$ $$\frac{\partial f}{\partial y}=4xy - 2y=0$$ From the second equation, we have either: $y=0$ or $x=\frac{1}{2}$. Now we substitute each value into the first equation to get the corresponding variable: for $y=0$, we ...

1

The situations of differential equations in one variable and several variable are deeply different. In several variables, you have $f(x_1,\dots,x_k)=(f_1(x_1,\dots,x_k),\dots,f_m(x_1,\dots,x_k))$ and the question is about knowing the $f_i$ once one knows the partial derivatives of them (the Jacobian). Well, the world of PDEs (Partial Differential Equations) ...

1

Let $f \, : \, U \subset \mathbb{R}^{d} \, \longrightarrow \, \mathbb{R}^m$ differentiable at $x_{0}$ (with $U$ open in $\mathbb{R}^d$). The definition of differentiability at $x_{0}$ you gave is equivalent to : $$f(x_{0}+h) = f(x_{0}) + T_{x_{0}}f(h) + \Vert h \Vert \varepsilon(h)$$ (since the linear map $T$ depends on $f$ and $x_{0}$, I'd rather write ...

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