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First, you should have said in the beginning, that you want the derivative calculated in $(0,1)$. Remember that $\mathrm{d}\left(g \circ f\right)_p = \mathrm{d}g_{f(p)} \cdot \mathrm{d}f_p$. But $\mathrm{d}g_{f(0,1)}$ and $\mathrm{d}f_{(0,1)}$ are just Jacobian matrices that you shouldn't have problem calculating. Recording, if $f:\mathbb{R}^3 \rightarrow ... 0 1) You can try to find a function which is defined by cases. If I remember well, $$f(x,y) = \begin{cases} 1, \text{if} \hspace{2pt}y = x^2 \\0, \text{if else} \end{cases}$$ may work... but you can verify yourself, for good practice. 2) Having all directional derivatives does NOT mean that it's differentiable at the given point, you would have to check the ... 0 Your jacobian matrix is correct, but you did something wrong. You're taking$f=(f_1, \ldots ,f_n)$, where$f_k\colon \mathbb R^{\color{red}n}\to \mathbb R, x\mapsto \dfrac{x_k}{|x|}$, for each$k\in \{1, \ldots ,n\}$. So the entry$(i,j)$at each point$x\in \mathbb R^{\color{red} n}$is given by$\dfrac{\partial f_i}{\partial x_j}(x)$. Your computations ... 1 So u mean if you do it like in your second case i.e $$\int \int_R{ }F(x,y)dydx$$ where R is the region bounded as 0<=x<=1, 0<=y<=1-x.thats taking points from that 2-d region R and evaluating F over these points. Then thats wrong if you were asked to integrate the function F(x,y)=x+y on the line segment. 0 For simplicity we can take the case of$R^2$. We want to go from (a,b) to (c,d) parallel to coordinate axes, but not get out of the given set. We move from (a,b) to (a,c) and test if the path (a,b)-(a,c) is in the set. If yes, we than move from (a,c) to (b,c) and test if (a,c)-(b-c) is in the set. If yes, we are done. But if (a,b)-(a,c) is not in the set, ... 0 First note that for any cube$C=[-r,r]^n\subseteq\mathbb R^n$any point$c=(c_0, c_1,\ldots c_{n-1})\in C$is polygonally connected to the center of$C$along the axes. $$(0,0,0,\ldots,0)\to(c_0,0,0, \ldots, 0)\to(c_0,c_1,0,\ldots0)\to\ldots\to(c_0,c_1,\ldots c_{n-1})$$ Let$G$be any nonempty open connected set in$\mathbb R^n$and let$a\in G$. Now set ... 0 without going into the realm of Higher Dimensional Space. You could still understand triple integral in such a way say function F is denotes Temperature, at each point in region B. Then it would be practical for us 0 You can view these excellent answers (especially the longest one, who gives you every kind of references) Obviuosly all credits go to the authors of the answers linked above :) 1 From the second equation, we have: $$x^2 y - x - 1 = 0 \implies y = \dfrac{x+1}{x^2}$$ Substitute this into the first and it reduces nicely to: $$-4x(x^2-1) = 0 \implies x = 0, \pm ~ 1$$ We toss out$x = 0$, due to division by zero. Now, solve for the two$y$values. $$(x,y)=~(−1,0),~(1,2)$$ 0 Hint: Think of them like a system of equations. Don't forget$-2x^4y+2x^3+2x^2=0$, which you can divide by$2x^2$on both sides, then solve for$y$in terms of$x$and plug it into the big mess you have. 0 You should review the rules for taking derivatives, specifically the chain rule. (Either that, or expand all squares in$F$, but this is more algebraically painful.) The basic idea is that the derivative of the square of some function$g$should be computed as$(g^2)'=2 g g'$. For example, the derivative of$(3x-7y+5)^2$with respect to$y$is ... 0 That depends on how you define vector derivative. There are generally two ways. One is applying abstract index notation, then $$\frac{d}{dx}x^T=\left(\frac{dx_i}{dx^j}\right)=(\delta_{ij})=(e_1\otimes\cdots\otimes e_n)^T$$ where$e_i$s are unit vector whose$i$th component is one and zero otherwise. Another way to look at it is to regard as directional ... 0 What sort of object can be the derivative of a vector-valued function whose values are row vectors and whose arguments are column vectors? Generally, what kind of object can be the derivative of a function whose values are members of one vector space$W$and whose arguments are members of another vector space$V$? $$f: V\to W$$ The answer is that the ... 1 Look at the definition of "tangent plane of a surface$S$at a point$p \in S$" in your textbook. As mentioned in the comments, there's a pretty good chance that it's defined to be a vector space satisfying certain conditions. If so, then it certainly contains the origin (since all vector spaces contain the origin). Intuitively, you would expect the ... 1 Let's split the solution into several steps. Suppose your surface$S$is given by the equation$F(x)=0$with$F$being, let's say,$C^1$function. The tangent space$T_xS$to the surface$S$at the point$x$is orthogonal to$\nabla F(x)$. One of equivalent definition of$T_xS$is the following: Let$\gamma:[-1,1]\to S$be a$C^1$map (in the sense of ... 0 First assume that for any sequence$x_n\in \Omega\setminus \{x_0\}$with$x_n\to x_0$, we have that $$\liminf u(x_n)>0$$ hence, by the maximum principle$u(x)>0$in$\Omega\setminus \{x_0\}$. On the other hand, if for any sequence$x_n\in \Omega\setminus \{x_0\}$with$x_n\to x_0$, we have that $$\limsup u(x_n)<0$$ then, by the maximum principle ... 2 I THINK: The author identified the tangent space$T_p\mathbb{R}^3$and$\mathbb{R}^3$itself. In this case$T_pS$is a plane passing through the origin i.e. a subspace of co-dimension one and hence the normal and$p$determine the plane. 1 The complete answer is given on pages 22-27 of my 2011 vector calculus notes. I think many good calculus text include these heuristic arguments, I found them in Thomas' calculus a few editions back. Long story short, what you should really do to understand is to prove Greene's and Stokes' Theorems, this will give you deeper insight into the nature of your ... 1 Would like someone to review this answer. let's say$x=\begin{pmatrix} x_1 \\x_2\\ \vdots \\x_n\end{pmatrix}$and$\xi =\begin{pmatrix} \xi_1 \\\xi_2\\ \vdots \\\xi_n\end{pmatrix}$then$f(x)=<x,\xi>^2 = (x_1\xi_1+x_2\xi_2+...+x_n\xi_n)^2$now, instead of actually computing$(x_1\xi_1+x_2\xi_2+...+x_n\xi_n)^2$, we can derive it as such. We can say ... 0 The range is$\mathbb{R}$if$D<0$then$f(0,\sqrt{|D|})=-|D|=D$if$D>0$then$f(1/2\sqrt{D})=D$if$D=0$take$f(0,0)=0$0 It can be disproved by the following physical experiment. Consider a large flat capacitor with the voltage$V$. Inside the capacitor, the electric field is constant and equal to$\frac{V}{d}$($d$is the distance between the plates). It's the largest possible electric field in the system. If you move a unit charge that was outside but near a plate through ... 1 Counterexample. Let $$f(x,y)={\mathcal {Im}}\log (x+iy),$$ where the complex logarithm is defined in$\mathbb C\smallsetminus (-\infty,0]$, and hence for$x<0$, we have $$\lim_{y\to 0^+} f(x,y)=\pi \quad\text{whereas}\quad \lim_{y\to 0^-} f(x,y)=-\pi.$$ If we set as$Uthe following domain $$U=\{(x,y): x^2+y^2>1\}\smallsetminus\{(x,0): ... 2 First of all, it is clearly true if m=1. Indeed, in such case f : U\to\mathbb R, and setting$$ g(t)=f\big((1-t)a+tb\big), $$then f(b)-f(a)=g(1)-g(0)=\int_0^1 g'(t)\,dt, or equivalently$$ f(b)-f(a)=\int_0^1 \frac{d}{dt} f\big((1-t)a+tb\big)\,dt= \int_0^1 \nabla f\big((1-t)a+tb\big)\cdot (b-a)\,dt, and hence \begin{align} \lvert ... 3 I doubt you can do that if U is not convex or satisfies some similar condition. You will have to replace ||b-a|| by the length of the shortest curve in U from b to a. If U is, e.g., convex, this is straightforward, since then the line from a to b is contained in U and thenf(b)-f(a) = f(a+t(b-a))|_{t=1} - f(a+t(b-a))|_{t=0} = ... 2 A point(x, y)$is called a local minimum (maximum) if there is a open ball$B$containing$(x, y)$and$f(z, w) \geq (\leq) f(x, y)$for all$(z, w)\in B$. So in your case$(c, y)$is not a minimum (maximum) for all$y$. Note that one candidate of$f$is$f(x, y) = x-c$. It is harmonic and is not constant. 1 It is easier to see it from an approximation point of view, let's expand$y(x)$around some point$x_0$using Taylor terms, then: $$y(x)= y(x_0)+\nabla y^T(x-x_0)+(x-x_0)^TH(x-x_0)+hot(x)$$ where$\nabla y$is the gradient and$H$is the Hessian matrix (second order derivative) of$y(x)$both evaluated at$x_0$, and$hot(x)$are the higher order terms. ... 0 You don't have to do integrals! Divide atmospheric pressure A = 101.3 kPa by g = 9.8 m/s2 to give the mass per unit area (kg/m2). Multiply this by the area of the earth and you're done. (Assumptions: g is a constant over the height of the atmosphere; g independent of latitude; neglect the mass of the air displaced by the volume of the land about sea ... 0 In Electrostatic,$\hat{r}/r^{2}$is the electric field produced by a charge$+1$at the origin of coordinates. The charge density associated to that charge is represented by a Dirac Delta Function. Namely,$\delta\left(\vec{r}\right) = \nabla\cdot\left[\left(\hat{r}/r^{2}\right)/\left(4\pi\right)\right]$. 1 The flux of$v$across the sphere$|x|=r$is equal to$(4\pi r^2 )( r^{-2}) = 4\pi$, i.e., is independent of$R$. It follows that the integral of divergence over every spherical shell$R_1<|x|<R_2$is zero. Since the field is rotationally invariant, so it its divergence: that is, the divergence depends on$|x|$only. Putting 1 and 2 together, we ... 2 A vector field$F|_S\colon S\to\mathbb{R}^n$is an assignment of$n$-dimensional real vectors to points in a subset of$\mathbb{R}^n$so it's really just a vector-valued function$F\colon\mathbb{R}^n\to\mathbb{R}^n$restricted to a subdomain$S\subset\mathbb{R}^n$. Note that the domain and codomain of$F$have the same dimension, but$S$can possibly be a ... 2 I'm not sure if this is the type of thing you are looking for, and it is only a partial answer. I don't know about$\operatorname{curl}$, but for$\operatorname{div}$I've always thought that this came from expectations of what we are looking for. In a physical sense it seems sensible that, given a domain$\Omega$to study, we may want to know how much ... 0 Note: not a full answer, but something which may lend a hand: Let$t=4m,\ s=4n,\ x=d,$and define $$g(m,x)=(1-x)(1-x^{2m-1}),\quad h(n,x)=(1-x^n)^2.$$ Then your function is $$f(x)=\frac{g(m,x)}{g(m,x)+h(n,x)}=\frac{1}{1+h(n,x)/g(n,x)}.$$ Then$f(x)$will increase or decrease iff$h/g$respectively decreases or increases, so to keep things easier we can ... 0 Solve for y^2: y^2 = 16 - x^2 ===> f(x,y) = 2x^2 + 3(16 - x^2) - 4x - 5 = -x^2 - 4x + 43 = -(x + 2)^2 + 47 = g(x). Consider f(x,y) = g(x) = 47 - (x + 2)^2 on [-4, 4]. Clearly: g(x) <= 47 and that max g(x) = 47 when x = -2, so y^2 = 16 - (-2)^2 = 12 , and y = 12^(1/2) and y = - 12^(1/2). Observe that g is minimized when (x + 2)^2 is maximized, and this ... 1 In the simplest case, we have curves with implicit equations like$y=f(x)$. Obviously these can be parameterised -- you just take$x=t$and$y=f(t)$. Not very interesting. Moving to two dimensions, we have curves given by implicit equations like$f(x,y)=0$. In other words we have sets of the form$S = \{(x,y) \in \mathbb{R}^2 : f(x,y)=0\}$, and it is ... 0 If your question is whether any subset of$\mathbb R^n$can be described as the graph of a function, then the answer is no .The graph of , e.g.,$S^n; n\geq 1$cannot be expressed as a continuous map$f: \mathbb R \rightarrow \mathbb R $, since this would be a continuous bijection ( when restricted to a map from the unit interval$I$into the image$S^1$), ... 1 Extension from$\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$, but we will not use that. First, let's go ... 1 1) what do you mean by explicitly? A coordinate map is a bijection on it's image, so if$v\in \alpha(U)$there exists exactly one$u\in U$such that$\alpha(u)= v$. But in general there is not way to say more. 2) The tangent space of$\mathbb{R}^m$at a point$u$in$\mathbb{R}^m$is just$\mathbb{R}^m$. The tangent space of the manifold$\alpha(U)$in$v ...

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for 1/8 of sphere volume = 1/8*4/3*pi*r^3=pi/6 by subtracting volume of cube (1*1*1) v=1-pi/6=.476 ,here the power of astroidoid approximatly =.562 but for the power of astroidoid =.5 volume is about .39

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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: $$\tag 2 r \left[1 - ... 1 In general, to find the maxima and minima of a given function, you can set the first derivatives, f_h, f_r equal to 0 and then check endpoints/places where the function is undefined. Lagrange multipliers are usually applied with a constraint, not to find global maxima and minima of a function. But for this specific function, we can see that the function ... 0 Your definition of locally Lipschitz is wrong. At least it is different from wiki. It should be like this: There is M<\infty and neighbourhood U of a that for any x,y \in U \|f(x)-f(y)\|<M\|x-y\| Now I can give you counter example. You are looking for this function:$$ f(x) = x^2 sin\left(\frac1{x^2}\right) $$It is everywhere ... 1 Here's a geometric answer that should lead to a formula after some thought. Imagine the graph of a function f(x) which changes from 0 to x_0 basically linearly as x varies from 0 to x_0^2, and then exponentially decays back down to 0 as x \to \infty. Clearly this can be made into a continuous family of functions as x_0 varies, with the limit ... 1 The geometrical or physical interpretation of these integrals has to come from the people who set them up. Nevertheless, you have to be aware that$$\vec A:=\int\nolimits_\gamma \phi\ \vec{d\ell},\qquad \vec B:=\int\nolimits_\gamma \vec\psi\times \vec{d\ell}\tag{1}$$can be considered as a shorthand notation for a limit of certain Riemann sums. By ... 1 The equation U(u)=V(u) leads to an equation P(u)=0 where P is a polynomial of degree three. There are here cases: There are three different real roots, and therefore three steady states There are two different real roots, one simple and one doble There is one real root and two complex conjugate roots, and only one steady state. The values of q ... 0 The easiest (though not the most elegant) definition of \int_\gamma \phi\, d\vec{\ell} is via coordinates: the integral is a vector with components \int_\gamma \phi\, dx, \int_\gamma \phi\, dy, \int_\gamma \phi\, dz. The same idea applies to the integral \int_\gamma \vec{\psi}\times\,d\vec{\ell}. The components of the resulting vector are ... 2 When f is differentiable at p then there exists a function r with \lim_{X\to 0}r(X)=0, such that for some h>0 one has$$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)\ ...

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This follows by applying the chain rule to the function $f\circ p$, where $p(t)=x_0+tv$. Note that $D_vf(x_0)=(f\circ p)'(0)$, this may even be the definition.

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Yes! sure! In fact if $f(x)$ and $g(x)$ are two derivable functions in $[a,b]$, and $r(x)=f(x) +g(x)$, then in [a,b] $D[r(x)] = D[f(x)]+D[g(x)]$. In fact $$D[f(x_0)] = lim_{\epsilon \to 0} \frac{f(x_0+ \epsilon)-f(x_0)}{\epsilon}$$ and $$D[g(x_0)] = lim_{\epsilon \to 0} \frac{g(x_0+ \epsilon)-g(x_0)}{\epsilon}$$ so $$D[f(x_0)] + D[g(x_0)] = lim_{\epsilon \to ... 0 For X,Y \sigma-finite measure spaces and a measurable function, f:X\times Y\to\mathbb{R}, as long as any one of the following three integrals are finite, then all three integrals, without the absolute value bars, exist and are equal.$$\int_X\left(\int_Y |f(x,y)|\,\text{d}y\right)\,\text{d}x\int_Y\left(\int_X ...

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Yes, this is true. The key fact to be proved is that for every fixed $x$, the function $g(h) = f(x+hv)$ of variable $h\in\mathbb R$ is nondecreasing. Since $f$ is continuous, it suffices to prove the above for a dense subset of $x$-values. Using Fubini's theorem, you can show that for almost every $x$, the line through $x$ in direction $v$ meets the set ...

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