New answers tagged

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,...


0

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(-...


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: \begin{equation}E(x,y,z)=\begin{pmatrix}x\\y\\z\end{pmatrix} \tag{1}\end{equation} \begin{equation}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 ...


1

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 $f$.


1

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 ...


1

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 $y$.


3

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 $0$


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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