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I think I have understood an idea by Dominik, a kind user of the site whom I thank again. Let $\rho\in C^{k}(A)$, $k\ge 1$, where $A$ is an open set such that $D\subset A$ and $\forall\mathbf{x}\notin D\quad\rho(\mathbf{x})=0$. Without loss of generality we can assume $A=\mathbb{R}^3$ because $\rho$ can be extended to such a function. Then the function ...

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If you want an $\varepsilon$-argument: Let $\varepsilon > 0$; let $(x',y') \in \mathbb{R}^{2}$ such that $x'^{2} + y'^{2} < 1$; then $$|\sqrt{1-x^{2}-y^{2}} - \sqrt{1- x'^{2}-y'^{2}}| \leq \frac{|x-x'||x+x'| + |y-y'||y+y'|}{\sqrt{1-x^{2}-y^{2}} + \sqrt{1-x'^{2}-y'^{2}}} < \frac{|x-x'||x+x'| + |y-y'||y+y'|}{\sqrt{1-x'^{2}-y'^{2}}}.$$ If $|x-x'|, ... 0 A direct proof: Denote our ball by$B$. We observe that$f(B)=[0,1]$. Let$0 \leq a < b \leq 1$. Define$I = (a,b)$(open interval) and consider$A = f^{-1}\left( I \right) $. Observe that $$A = \left\{x \in \mathbb{R}^2 \: \middle| \: \sqrt{1- a} < \left| x \right| < \sqrt{1- b} \right\}$$ which is open. Thus the preimage of an open interval (and ... -1 Yeah! But if you just want to, you can also use the other method which is Linear Differential equation, where in the standard solution goes like this: dx/dy+xP(y)=Q(y)--> if eq. is linear in x; dy/dx+yP(x)=Q(x)--> if eq. is linear in y. since the given problem is linear in x, use dx/dy+xP(y)=Q(y) where P(y)=1 and Q(y)=e^(−y)sec^(2)y and IF (Integrating ... 0 A function$f:\mathbb R^n \to \mathbb R^n$is called differentiable at$0\in \mathbb R^n$if there is a linear operator$\nabla f(0) : \mathbb R^n \to \mathbb R^n$so that $$\lim_{h\to 0}\frac{|f(0 +h) - f(0) - \nabla f(0)\cdot h|}{|h|} = 0,$$ in this case we have $$\lim_{h\to 0}\frac{|f(h) - \nabla f(0)\cdot h|}{|h|} = 0,$$ Now fix any$x\neq 0$. ... 0 Seeing as differentiation usually defines the slope of a curve/function.. I don't see how you could make a single point be differentiable, sorry. 0 It is simpler if you compute the composition explicitly. Compute$\phi(x)=f(f(x)) = x$. Then$\phi'(x) = I$. 0 The$i$th component function of f is $$f_{i}(x_{1},...,x_{n})=\frac{x_{i}}{||x||^{2}}=\frac{x_{i}}{{x_{1}^{2}+...+x_{n}^{2}}}$$ You need to take all$n$derivatives of this function, for each$i$. At this point, it's just partial differentiation and quotient rule. For example, $$\frac{\partial f_{i}}{\partial ... 2 Since I don't know how to post drawings of vectors you'll have to do the drawing for me. Draw a vector with any orientation you like and call it \mathbf N. Consider the point at the tail of the vector and call it \mathbf p. Consider all vectors of all lengths whose tails are at p and are perpendicular to \mathbf N. Can you see that this set of ... 1 The answer ultimately depends on what level of rigor the situation calls for. In a typical multivariable calculus course, for instance, you may usually take for granted that polynomials (in any number of variables) and f(z) = \ln(z) are both continuous functions on their domains, whereas this may or may not be the case in a course on foundational real ... 3 Your function is a composition of two functions, f:\mathbb{R}^2 \to \mathbb{R} defined by f(x,y) = 1 + 3x + 4y + x^2 + y^4 and h: \mathbb{R}_{> 0} \to \mathbb{R} defined by h(z) = \ln z. You first need to show that near (0,0), your composition of functions is defined (this is not hard, because f(x,y) is close to 1 for x,y close to 0. Then ... 0 Partial answer: The Laplacian measures how a function changes “on average” as you move away from a given point. It’s rotationally invariant, so \Delta u=0 describes a property of a function on the Euclidean plane that doesn’t depend on the choice of Euclidean coordinates. 1 The normal to a tangent plane is, up to sign, unique, if you are working in a codimension 1 case. So if you can show that the planes have the same normal or the normal differs only by a nonzero (positive or negative) factor, then the planes coincide. If two normal vectors are linearly independent (as in your example) the planes do not coincide. 2 Note the or in the book's definition of a surface. The author defines a surface as a graph of a map f \colon \mathbf R^2 \to \mathbf R\def\R{\mathbf R}, given by$$ z = f(x,y) $$or y = g(x,z) or x = h(y,z). In each of this three cases, the tangent plane has a different form, its one of the three equations you give above. Let's look at the first one. ... 2 Let f=f(\boldsymbol x,\boldsymbol u)=f(x_1,...,n_n,u_1,...,u_r). Then$$\nabla f=\left(\frac{\partial f}{\partial x_1},...,\frac{\partial f}{\partial x_n},\frac{\partial f}{\partial u_1},...,\frac{\partial f}{\partial u_r} \right),\nabla _{\boldsymbol x}f=\left(\frac{\partial f}{\partial x_1},...,\frac{\partial f}{\partial x_n} \right),$$and ... 1 If f is homogeneous of degree 1, then f(tx) = t f(x) for t \ge 0. Deriving this with respect to t (and using the chain rule for derivatives) we get$$\sum _i \frac {\partial f} {\partial x_i} \frac {\Bbb d (t x_i)} {\Bbb d t} = f(x)$$and evaluating at t=0 gives$$\nabla f (0) \cdot x = f(x) .$$(This is a particular case of Euler's theorem ... 2 You need to be careful with any notation since often authors define and use symbols in different manner. Although in this particular case, my mathematical experience tells me that most common are: 1. "Gradient's variable": following Ruben Tobar's answer, it may be notation providing information with respect to which variables the whole gradient should be ... 0 f(f(x))=\frac{\frac{x}{\lvert x\rvert^2}}{\left\lvert\frac{x}{\lvert x\rvert^2}\right\rvert^2}=\frac{x}{\lvert x\rvert ^2}\times\frac{\lvert x^2\rvert^2}{| x|^2}=x. Hence f(x)=f^{-1}(x) 0 You have to show f\circ f(x) = x, and this is quite simple: f(f(x)) = f\left(\dfrac{x}{||x||^2}\right)=\dfrac{\dfrac{x}{||x||^2}}{\left|\dfrac{x}{||x||^2}\right|^2}= \dfrac{\dfrac{x}{||x||^2}}{\dfrac{||x||^2}{||x||^4}}= \dfrac{x}{||x||^4}\cdot ||x||^4 = x. Done. 1 We can see it directly from definition: Let x \neq 0; let y := f(x); then$$ f(y) = \frac{y}{|y|^{2}} = \frac{x/|x|^{2}}{1} = \frac{x}{|x|^{2}}. $$1 In general, solving systems of polynomials is a very hard problem—so hard that two whole fields, algebraic geometry and commutative algebra, have grown up around it. If you want to solve an arbitrary system of polynomials, you're probably going to need heavy-duty tools like Gröbner bases. Your system of polynomials, however, is far from arbitrary. For one ... 1 hint: Subtract any two of the equations: z(x-y) = 0 = y(x-z) = x(y-z). Can you take it from here? 2 The definition of the partial derivative of x^y with respect to x is$$\frac{\partial }{\partial x} x^y = \lim_{h\to 0} \frac{(x+h)^y - x^y}{h}.$$In this way, we may simply treat y as a constant when differentiating. So,$$\frac{\partial }{\partial x} x^y = yx^{y-1}$$for the same reason that$$\frac{d }{d x} x^n = nx^{n-1}.$$0 It should be \frac{d}{{dx}}\left( {{x^y}} \right) = y{x^{y - 1}} 1 On the path x=t,y=0 we have$$f(x,y)=f(t,0)=0\qquad \text{i.e.}\quad f(x,y)\to 0$$On the \color{red}{\text{ellipse}} x=\sin t, y=\frac{\cos t}{\sqrt{2}} we have (x,y)\to (1,0) as t\to \frac{\pi}{2}^+$$f(x,y)=\frac{\frac{\cos t}{\sqrt{2}}}{\sqrt{1-\sin^2 t-\frac{1}{2}\cos^2 t}}=\frac{\frac{\cos t}{\sqrt{2}}}{\sqrt{\cos^2 t-\frac{1}{2}\cos^2 ... 1 The second one appears to belong to a family of surfaces defined by Banchoff based on the Chmutov surfaces of order$n$. Explicitly, the implicit equation is given by $$3+8(x^4+y^4+z^4) = 8(x^2+y^2+z^2).$$ The first one is not known to me. It appears to have a tetrahedral symmetry. 0 draw a graph of a sector of circle with centre$O$, centre angle$x$. we will find that: $$\sin(x)<x \implies \sin(x)/x < 1$$ and $$x < \tan(x) = \sin(x)/\cos(x)$$ then we have $$\cos(x) < \sin(x)/x < 1.$$ By squeeze theorem,$\lim \sin(x)/x =1$as$x$tends to$0^+$. 0 Your problem can be fixed. That means that one should take instead the function$F(x,y,z)=f(x,y)-z$and tell that $${\rm grad}F=({\rm grad f},-1)=(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},-1)$$ This triplete is perpendicular to the tangents$(1,0,\frac{\partial f}{\partial x})$and$(0,1,\frac{\partial f}{\partial y})$of the level ... 1 The tangent plane to your surface in the point$(a,b,f(a,b))$is generated by$(1,0,\frac{\partial f}{\partial x}(a,b))$and$(0,1,\frac{\partial f}{\partial y}(a,b))$, so a normal vector to the surface in$(a,b,f(a,b))$is$(\frac{\partial f}{\partial x}(a,b)),\frac{\partial f}{\partial y}(a,b)),-1)$, which is not the gradient of your$f$,$(\frac{\partial ...

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The way to approach these types of problems is to first find the boundaries. We have $$x^y=1\implies y\log x=0$$ Hence the boundary is defined by the lines $y=0$ and $x=1$ (and don't forget we also have $x>0$). This gives you $4$ regions to check, and you can test each region simply by plugging in a point of the region to see if it satisfies the ...

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By your definition, it is. Empty set always satisfies $\forall x \in \emptyset ~P(x)$ for any $P$.

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An optimal value of f(x,y) subject to the constraint g(x,y)=c depends on the value of c. If you vary c the optimal value of f and its associated $\lambda$ will vary. For a given c let P(c) be the point at which f is optimal, and f(P(c)) the optimal value, and $\lambda(c)$ the associated multiplier. How does f(P(c)) change as c changes? The chain rule gives ...

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Since there are three variables so it will be differentiated by product rule i.e. $\frac{dz}{dx}7e^{7z}= yz+ \frac{dy}{dx}xz+ \frac{dz}{dx}xy$

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As you have stated, along $y=0$ the limit is $0$. Along $x = y^2$, $\frac{xy^2}{x^2+y^4} = \frac{y^4}{y^4 + y^4} \to \frac{1}{2}$, so the limit does not exist.

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$x=y/2=(z-1)/8$. We can also write $(x,y,z) =(0,0,1)+(1,2,8)t .$ The line is parallel to the line that pases through $(0,0,0)$ and through $(1,2,8)$.

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Hint: $\lvert g(x,y) \rvert$ is bounded by a function of $y$ alone.

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This answer assumes $f:\mathbb{R}^n\to \mathbb{R}$. For convenience I'll denote by $DF$ the total derivative or the Jacobian of $F$ (which in your question you denoted by "$F'$" which is just not convenient, but I'll try to include your notation as well). From your notation I assume that the gradient $\nabla$ is a column vector and $\nabla^2$ is the Hessian. ...

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Take $x=y$, then $\lim=\lim_{x\to0}\frac{x^8}{(x^4+x^2)^3}=0$ Take $x=y^2$, then $\lim=\lim_{y\to0}\frac{y^{12}}{8y^12}=\frac{1}{8}$ Thus, the limit does not exist. QED.

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Let $\bar{x} \in B(\bar{a})$ and $\bar{y} \in \mathbb{R}^n$. Note that $B(\bar{a})$ is an open set. So we can find an $r>0$ such that $B(\bar{x},r) \subseteq B(\bar{a})$. Hence, we can find a $\theta >0$ such that $\bar{x}+h\bar{y} \in B(\bar{x},r),$ for all $h \in [-\theta,\theta]$. (I request you to make a diagram to convince yourself about this ...

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You can use integration by parts, or you can use the hyperbolic substitution $t = \sqrt{2}\sinh(u)$. I'll use the latter. With $t = \sqrt{2}\sinh(u)$, we have $\sqrt{2 + t^2} = \sqrt{2}\cosh(u)$ and $dt = \sqrt{2}\cosh(u)\, du$, so $$\int_0^\pi \sqrt{2+t^2}\, dt = \int_0^{\sin^{-1}(\pi/\sqrt{2})} 2\cosh^2(u)\, du.$$ Using the identity $2\cosh^2(u) = 1 + ... 0 Your integrals are equal to $$1/2\,\pi \,\sqrt {2+{\pi }^{2}}-\ln \left( 2 \right) +\ln \left( \sqrt {2}\pi +\sqrt {2\,{\pi }^{2}+4} \right)$$ 0 A conservative vector field is if it is the gradient of some scalar field. I.e.$F = \nabla S$. What you need to do is to show that the gradient of a scalar field can not have a non-zero curl. Some hints on the way: Use definition of conservative field above. Use definition of curl Try and put them together and see what happens. 0 This is a question from homework of a course that is running. The professor doesn't say that he allows asking questions here so please don't answer the question if you know it. For the one who asked the problem, please stop posting the questions in the homework here. 2 Recall that$x^2+y^2\geqslant 2xy$for any choice of$x,y$, so that $$\left|\frac{4xy^2}{x^2+y^2}\right|\leqslant \left|\frac{4xy^2}{2xy}\right| = |2y|$$ 3 Substitute$x=r \sin \theta$and$y=r \cos \theta$. $$\lim_{r \to 0} 4 r^3 \cos \theta \sin^2 \theta=0$$ 2 hint:$ 0 \leq \left|\dfrac{4xy^2}{x^2+y^2}\right| \leq |4x|$0 Let$g(x)=x^2$for$x\geqslant0$with$g(x)=2x^2$for$x<0$, and set$f(x,y)=g(x)+g(y)-2x-2y$for all$x,y\in\Bbb R$. Then$f$is smooth at its minimum, where$f(x,y)=-2$at$(1,1)$, but its second derivative is undefined along the$x$and$y$axes. 0 I think this formula will help you:$(a\cdot b)'= a'\cdot b+ a\cdot b'$. And we know$T\cdot T=1$, a constant. What can we say about it? 3 I claim that the limit is equal to zero, if$\alpha>2$, and does not exist, if$\alpha\le2$. Assume first that$\alpha>2$. We can then use the AM-GM inequality,$a+b\ge2\sqrt{ab}$, valid for all positive numbers$a,b$, to give the denominator a lower bound $$x^4+y^2\ge 2\sqrt{x^4y^2}=2x^2|y|.$$ Using this we get$$... 1 This isn’t the correct generalization of the Mean Value Theorem to$\mathbb R^n$-valued functions with$n>1$. Let’s look at just the$\mathbb R\to\mathbb R^n$case. You can apply the MVT to each component$f_j$of$f$individually, as you’re doing, to find a$z_j$such that$f_j(b)-f_j(a)=f_j'(z_j)(b-a)\$, but the problem you run into is that there’s no ...

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