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1

Spelling out what is already indicated by Did: Setting $r=\sqrt{x^2+y^2}$ the equation reads $(r-3)^2+z^2=1$, a circle of radius 1 centered at $(3,0)$ in the $(r,z)$ (half) plane. Indeed in the $(x,y,z)$ coordinates it is a torus (make a drawing). For the parametrization, set $r=(3+\cos s)$ and $z=\sin s$. And then $$x=(3+\cos s) \cos t, \ \ y=(3+\cos s) \... 0 Hint for part 1: Without loss of generality you can assume \mathbf{x} = \mathbf{0}. I'll write the standard coordinates as (x_1, \dots, x_n). Since f is differentiable, and since the ball is compact, for every \epsilon > 0, there exists R > 0 sufficiently small such that for all \mathbf{y} \in B_R(0), y_1 \neq 0,$$\left| \frac{f(y_1,...

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No, you are correct if you ignore the bound on the $x$-axis. Without using transformations or such, we immediately see that you have the correct result: \begin{align}\int_{-2}^2 y \int^{1-y^2/4}_{-1+y^2} \mathrm d\,x\,\mathrm d\, y ~=&~ \tfrac 1 2\int_{-2}^2 y(4-y^2)\,\mathrm d\, y \\[1ex] =&~ 0 \end{align} [ Ever quicker: If you like, sketch ...

0

\begin{align}\mathbf x'(t) &= \frac{d}{dt}\big(r(t)\cos(\theta(t))\mathbf i + r(t)sin(\theta(t))\mathbf j\big) \\ &= \big(r'(t)\cos(\theta(t))-r(t)\omega(t)\sin(\theta(t))\big)\mathbf i + \big(r'(t)\sin(\theta(t))+r(t)\omega(t)\cos(\theta(t))\big)\mathbf j \\ &= r'(t)\big(\cos(\theta(t))\mathbf i+\sin(\theta(t))\mathbf j\big)+r(t)\omega(t)\big(-... 1 Hint: f(x,y) is uniformly continuous on any [a,b]\times [0,1]. 2 A uniform limit of real-analytic functions certainly need not be real-analytic. Any continuous function on [a,b] is a uniform limit of polynomials (and is hence the sum of a uniformly and absolutely convergent series of polynomials). I can't think of any "simple" criteria. Even uniform convergence of a sequence of functions together with uniform ... 1 "Health is determined by" is the key phrase in your question. If we can say that health is a SUM, like H = w_1 + w_2 + \ldots + w_{20} $$where w_i is the fraction of the RDA, but limited to a maximum of 1, then this is a constrained optimization problem, and pretty well adapted to standard techniques like the simplex method. If "Health" is some ... 0 E = E(x,y,z) means that E is a generic vector field whose components depend on x, y, z. Namely, being a vector field in \mathbb{R}^3, E has 3 components which depend on x, y, z. You can denote these three components however you like. For example, if you denote them as E_x, E_y, E_z, then the third notation would be the correct one. 0 Remark: As notation is quite subjective, that's how I see it. (Note that my point of view already disagrees with Don Antonio's one) If we write them using the standard notation they appear as: $$E(x,y,z)=\begin{pmatrix}x\\y\\z\end{pmatrix} \tag{1}$$ E(x,y,z)=\begin{pmatrix}x(x,y,z)\\y(x,y,z)\\z(x,y,z)\end{pmatrix} \... 0 Consider f(x,y)=y^{x-1} then F(x) defined above is \frac{y^{x}}{x}. The definition of continuous requires that the function equal it's limit at the point in question among other things. Clearly \frac{y^{x}}{x} is not defined and thus not equal to it's limit at the point x = 0. Yet f(0,y)=\frac{1}{y} has no problem for suitable y. Thus the ... 1 You have three equations from the first order conditions:$$2xy^2z^2=2\lambda x \qquad 2x^2yz^2=2\lambda y\qquad 2x^2y^2z=2\lambda z $$Suppose x=0. Then the first equation is satisfied, and the other two equations imply that \lambda=0 (since we cannot have x=y=z=0 as that does not satisfy the constraint). This gives us infinitely many solutions with ... 2 But if I take this curve (that I found simply equaling the limit to 1)$$x = \sqrt{\frac{y^2}{y-1}}$$But \varphi(y)=\left(\sqrt{\frac{y^2}{y-1}},y\right) is not a valid path to (0,0), Namely, it is only defined for y>1, so you cannot follow \varphi(y) while having y\to 0. -1 You didn't consider the case x = y = \lambda = 0, z = 1 which satisfies the nabla equations, and gives the minimum value for f of 0. 0 Green's Theorem is for closed curves! Use the vector-line integral computation namely;$$\textrm{Work} = \int_{c} \textbf{F} \cdot d \textbf{s} = \int_{t=a}^b \textbf{F}(c(t)) \cdot c'(t) \ dt$$For a handwavy reason why you use this computation is because \textbf{F}(c(t)) is interpreted as the force acting on a particle at position c(t). Then if your ... 1 How is the interior derivative a derivative? I wouldn't say it is. My background is in Clifford algebra, and that discipline's equivalent of this operation is universally referred to as a product operation, not a derivative operation. What is the geometric content of Hodge duality? Short version: you're finding the orthogonal complement of whatever ... 1 a) Im(f) is unbounded, so no absolute min/max. b) Im(f)=[0, \infty) the absolute minimum is 0 but there is no max. c) The image of a continuous real function defined on a compact set is compact so there is an absolute max and min. d) the absolute max/min doesn't exist because Im(f)=(2, \infty) 0 Continuous differentiability of the function f: \mathbb{R}^m \to \mathbb{R}^n (in terms of partial derivatives) is equivalent to existence and continuity of the map$$Df: \mathbb{R}^m \to L(\mathbb{R}^m, \mathbb{R}^n) x \to Df_x$$which takes a point to the derivative at the point. Any book on analysis on \mathbb{R}^n will have a proof of this fact.... 0 The mean value theorem is applied to the real function$$t \mapsto f(x+\sum_{i=1}^{k-1}h_ie_i+te_k),$$which is continuous since it is differentiable, as its derivative is given by the partial derivative of the function f (just apply the definition of derivative). For your edit, what he is using is the fact that \Vert h_k \Vert \leq \Vert h \Vert, and ... 0 Hint$$\sum_{i=1}^n |h_i| \le \sum_{i=1}^n \|\mathbf h\| = \|\mathbf h\|\sum_{i=1}^n 1$$2 The first question is: Under what circumstances is$$ \frac \partial {\partial\theta} \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty \bullet\bullet\bullet $$the same as$$ \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty \frac \partial {\partial\theta} \bullet\bullet\bullet \text{ ?} $$The next question is: Why is$$ \frac \partial {\partial\theta} \...

3

Let us rewrite the product rule as follows: $$(fg)'=f'g+g'f=\frac{f'}{f}fg+\frac{g'}{g}fg=\left(\frac{f'}{f}+\frac{g'}{g}\right)fg$$ Yours is just the generalization to $n$ factors, but is handled in the exact same way.

0

$$G(x,y)=H(x,y)+L(x,y)$$ $$\begin{cases} dH=\frac{\partial H}{\partial x}dx+\frac{\partial H}{\partial y}dy \\ dL=\frac{\partial L}{\partial x}dx+\frac{\partial L}{\partial y}dy \\ dG=dH+dL=\left(\frac{\partial H}{\partial x}+\frac{\partial L}{\partial x}\right)dx+\left(\frac{\partial H}{\partial y}+\frac{\partial L}{\partial y}\right)dy \end{cases}$$ $$\... 0 Let f : \mathbb R^n \to \mathbb R^n be defined by$$f (\mathrm x) := \| \mathrm x \|_2^2 \, \mathrm x = (\mathrm x^T \mathrm x) \, \mathrm x$$Hence,$$\begin{array}{rl} \mathrm d f &= (\mathrm d \mathrm x^T \mathrm x) \, \mathrm x + (\mathrm x^T \mathrm d \mathrm x) \, \mathrm x + (\mathrm x^T \mathrm x) \, \mathrm d \mathrm x\\\\ &= (\mathrm x ...

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Hint: consider $g:R^2\times R\rightarrow R^3$ defined by $g(x,y,z)=(f(x,y),z)$ shows that the rank of the differential is 3 and deduce that it is a local diffeomorphism by using the local inversion theorem, then consider the composition of $g$ with the projection (which is an open map) on $R^2$ which is $f$.

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For linear equations, the technique of separation of variables is used to find all separated solutions of the form $X_1(x_1)X_2(x_2)\cdots X_n(x_n)$. You find them all if your equation can be separated. If you make a change of variables, then you will generally find a different set of separated solutions. For example, you might separate $X(x)Y(y)$, or ...

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No. $G$ is not a function of $H$, so it doesn't make sense to talk about the derivative of $G$ with respect to $H$ in any sense. $G$ is a function of $x$ and $y$.

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For a ternary function $f(x,y,t)$, the expression $\dfrac{df}{dz}$ would usually be read as the total derivative of $f$ with respect to the exogenous argument $z$. In general, you have for this total derivative that $$\frac{df}{dz} = \frac{\partial f}{\partial x} \cdot \frac{dx}{dz} + \frac{\partial f}{\partial y} \cdot \frac{dy}{dz} + \frac{\... 4 The function f is of three variables, so you should be using$$\frac{\partial f}{\partial x},$$or f_x or f_1 to denote the partial derivative of f with respect to x. Here df/dx has no meaning because f is a function of three variables. You would only use this notation (or f') if f was a function of x alone. (You might also use f' to ... 1 It's not hard, you just need to work with piecewise definitions. Derivatives in general are a local notion, so really all that matters is you're not explicitly in the knife-edge case where t_0 = t_1 (aka the diagonal of your domain). As long as t_0 \neq t_1 you can always find a sufficiently small open nbhd around your point where you can just treat the ... 1 Consider a point with polar coordinates (r,\theta). It lies, of course, at the distance r from the origin. A change of d\theta in the value of \theta will move this point a distance r \, d\theta along the circle of radius r. (Notice the factor r; it says that the farther out you are, the bigger is the effect of a change in the angle.) The ... 1 It is wrong at very beginning. p is a function change with x and y. If only x changes and y is invariant, \frac{\partial x}{\partial p}=1 and \frac{\partial y}{\partial p}=0 because \frac{\partial x}{\partial p} = \lim_{\triangle p->0} \frac{\triangle x}{\triangle p}  If only y changes and x is invariant, \frac{\partial y}{\... 6 Your mistake is thinking in terms of "dependent" / "independent" symbols without introducing precise mathematical meaning for that. You can't say that \frac{\partial p}{\partial x}=1 implies \frac{\partial x}{\partial p}=1. p is a function of two variables (x,y), and to calculate \frac{\partial x}{\partial p}, you should introduce another variable,... 1 Your guesses are correct. Just note that \hat e_\theta increases in the anticlockwise direction. So it won't be like you have guessed, rather it will be what is given in the pdf. 4 By using \frac{\partial x}{\partial p}\neq 0 and \frac{\partial y}{\partial p}\neq 0, you make x and y depending on p ! Therefore, it has no sense to consider p(x,y). Indeed, if p(x,y) would have sense, then p would be dependent and independent of x and y, which is impossible. 1 For a point \mathbf{x}(r,\theta) = (x(r,\theta),y(r,\theta)), \hat{e_r} is defined as a length 1 vector in the direction of \frac{\partial \mathbf{x}}{\partial r}, and similarly for \hat{e_\theta}. Hence they represent the so-called 'infinitesimal' direction of change when r or \theta increases. 5 In polar coordinates you have$$\nabla g = \frac{\partial g}{\partial r} \hat r + \frac{1}{r}\frac{\partial g}{\partial \theta } \hat \theta $$where \hat r and \hat \theta are the unit orthogonal vectors at any point. So you can calculate$$\nabla g \cdot \hat r = (\frac{\partial g}{\partial r} \hat r + \frac{1}{r}\frac{\partial g}{\partial \theta } \...

0

If we visualize it, it's the reigon under a cone over that's in a sphere, like this: http://sketchtoy.com/67282842 (Sorry for the bad drawing skills) so I THINK this would be a parametrization $x=r\sin\phi \cos \theta, y=r\sin \phi \sin \theta, z=r\cos \phi, \phi\in [\pi/4,\pi], \theta\in [-\pi,\pi], r\in[0,2]$. Don't take my word for it, wait for other ...

1

If you really want to avoid spherical coordinates, you can use the decomposition: $$\int_{||x||\geq \delta}\frac{dx}{||x||^{d+1}}=\sum_{k=0}^{\infty}\int_{2^k\delta\leq ||x||<2^{k+1}\delta}\frac{dx}{||x||^{d+1}}$$ $$\leq \sum_{k=0}^{\infty}(2^k\delta)^{-d-1}m(\{x:2^k\delta\leq ||x||<2^{k+1}\delta\})$$ where $m$ is Lebesgue measure. For a very crude ...

3

I don't see why you think the coordinate transformation is troublesome... Take radial and spherical coordinates, but forget about the spherical bit... i.e. $$||x|| = r \quad \& \quad dx = \omega(\theta) r^{d-1} dr$$ where $\omega(\theta)$ and doesn't depend on $r$ (and doesn't depend on $\delta$). Thus $$\int_{ ||x|| \geq \delta} \frac{ dx}{||x||^{... 0 As Thomas pointed out just remember that the tangent plane at the point needs to be shifted to the point first because the gradient (as it is a vector) begins at the origin. Therefore, what you want is$$\nabla f(F(t_0))\cdot( (F(t_0) + F'(t_0)) - F(t_0)) = \nabla f(F(t_0))\cdot F'(t_0) = 0$$You essentially had the solution! 1 An alternate approach.$$ F(x,y)=4{x}^{2} + 3{y}^2 + \cos(2x^{2}+y^{2}) - 1$$is continuous and differentiable everywhere.$$\dfrac{\partial F}{\partial x}=8x-4x\sin(2x^{2}+y^{2})=0 $$implies that x=0 since 2-\sin(2x^{2}+y^{2})\ne0. Likewise$$\dfrac{\partial F}{\partial y}=6y-2y\sin(2x^{2}+y^{2})=0 $$implies that y=0 since 3-\sin(2x^{2}+y^{2}... 13 Call t=2x^2+y^2. Clearly t \ge 0, and your equation can be rewritten as$$2t+ \cos t + y^2=1$$Now, it is easily checked that the function 2t+ \cos t is strictly increasing (simply compute its derivative, and check that it's >0), so that we have the following inequalities$$1= 2t+ \cos t + y^2 \ge 2 \cdot 0 + \cos 0 + y^2 = 1 + y^2 \ge 1$$which ... 1 (1) Note that your integrand is periodic in all the integration variables. We will use this fact later. Let us introduce the new variables x_1= \phi_1- \phi_2, x_2 = \phi_2 -\phi_3 and x_3 = \phi_3-\phi_1. We immediately notice that x_3 = - x_1 - x_2. The goal is thus, to perform a change of variables from (\phi_1, \phi_2 , \phi_3) to (X, x_1, ... 1 At the saddle point, you need to expand up to the second order and the Taylor development essentially becomes$$f-f_0=ax^2+2bxy+cy^2,$$(a parabolic hyperboloid) which you can factor as the product of two lines. 1 Another approach that is sometimes helpful: use polar coordinates$$\begin{cases}x=r\cos t\\y=r\sin t\end{cases}\implies\frac{3x^2y^2}{x^4+y^4}=\frac{3\cos^2t\sin^2t}{\underbrace{\cos^4t+\sin^4t}_{=(\cos^2t+\sin^2t)^2-2\cos^2t\sin^2t}}==\frac{\frac34\sin^22t}{1-\frac12\sin^22t}$$Either way, it is clear that \;r\to0\implies\; the limit depends on ... 3 To show the statement is false, notice that for all x\neq 0: f(x,x)=3/2 but f(x,2x)=12/17, so the limit does not exist. 0 take the way y=x, then the limit become:$$\lim_{x\to 0}\frac{3x^2x^2}{x^4+x^4}=\lim_{x\to 0}\frac{3x^4}{2x^4}=\lim_{x\to 0}\frac{3}{2}=\frac{3}{2}$$clearly the limit is not 0 2 Hint: You are trying to prove something that is false. 0 Hint: f is positive at each point of the positive x and y axes. Added hint for the new question: If you consider f(x,0),f(0,y) for x,y>0, you'll find the one sided partial derivatives at (0,0) are 1 and 0 respectively. Is it true that [f(x,x)-f(0,0)]/x \to \nabla f(0,0)\cdot (1,1) as x\to 0^+? That would be true if f were smooth ... 2 First we parameterize our surface by two families of curves \{u=c\} and \{v=d\}: The we can see that$$\mathbf r_u = \frac{\partial \mathbf r(u_0,v_0)}{\partial u} = \lim_{h\to 0} \frac{\mathbf r(u_0+h,v_0)-\mathbf r(u_0,v_0)}{h} will be tangent to the curve $u=u_0$ at the point $(u_0,v_0)$ and thus also tangent to the surface. Likewise for \$\...

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