# Tag Info

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The lower bound is the region in the $xy$-plane, where $z=0$. The upper bound is the surface $z = 1+x+y$.

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Following @Dan’s comment, a reasonable way to approach this kind of problem is to look for a coordinate transformation that straightens out the boundary of the region and turns it into a square or triangle. Trying $u=x^2y$ looks promising since its level curves form two edges of $D$ and it appears as a factor in the integrand. From the given inequalities, ...

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if $b\ge \frac 12$, then there isn't enough room to move about. $a^2 x^2, b^2 y^2, b^2 z^2,$ are all greater than $0.$ $b^2 y^2 + b^2 z^2 > yz.$ And the only solution is $(0,0,0).$ If $|b| < \frac 12$, then you get a double cone.

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Suppose we have a function $f : \mathbb{R}^n \to \mathbb{R}$. We now introduce function $g : \mathbb{R}^{n+1} \to \mathbb{R}$, defined by $$g (x,z) = f (x) - z$$ Differentiating, we get a vector field $\nabla g : \mathbb{R}^{n+1} \to \mathbb{R}^{n+1}$ $$\nabla g (x,z) = \begin{bmatrix} \nabla_x f (x)\\-1\end{bmatrix}$$ If we pick a point on the surface ...

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1) The substitution is used obtain the final expression in terms of an error function. Notice that they set $h=0$ after this substitution, and in that case $\eta=x/(2\sqrt{k\tau})$ is the square root of the exponent, i.e. $\eta^2=x^2/(4k\tau)$. 2) It looks like a typo in the book

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How you interpret the gradient depends on how you represent the surface. For example, for a surface written in the form $z=f(x,y)$, the gradient is more like the slope interpretation. More specifically, if you take the dot product of the gradient at a point $(x,y)$ and a small displacement vector $\Delta r$ then you will approximately get the change in $z$ ...

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Stereographic projection is conformal and you can also show the gauss map is conformal for a minimal surface. This is not very hard; simply assume $<dN_p(t_1),dN_p(t_2)>=\lambda(p)<t_1,t_2> \forall t_1,t_2 \in T_pS$ and then take the basis of $T_pS$ consisting of the principal directions of the gauss map. A quick bit of algebra will show that ...

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(Assuming this is what you are looking for.) For example, $$\frac{d}{dx}f(x,y)=\frac{d}{dx}\int_y^x e^{t^2}dt=e^{x^2}$$ by the Fundamental Theorem of Calculus. Finding the other partial is very similar.

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Fix a point outside the cylinder, without loss of generality at $\mathbf r=(a,0,0)$. By symmetry, the magnetic field at $\mathbf r$ has only a $y$ component, and since $k$ only has a $z$ component, the only relevant component of $\mathbf r-\mathbf x$ is the $x$ component. The $y$ component of the left-hand side is \begin{align} \frac1{2\pi ...

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First if $0\le y\le2$ then we are going to need to integrate (3-y) or we will get a negative number. To find the limits of y, I suggest you draw a line that is parallel to the y-axis. How would you represent the endpoints of this segment? $x^2 + y^2 = 4\\ y^2 = 4-x^2\\ y = \sqrt{4-x^2}$ $\int_0^{2} \int_0^{\sqrt{4-x^2}} (3-y) dydx$ Now, if you want to ...

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Given that you are working with a cynlinder, it is easier to work with cylindrical coordinate. $y=r\sin \theta.$ Also, notice that the hyperplane intersect the cylinder above $z=0$. $$\int_0^{\pi/2} \int_0^2 (3-r\sin\theta) rdrd\theta$$

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Since you reference $f'$, $f$ must be differentiable. The inverse function theorem can actually be relaxed from $C^1$ functions to differentiable functions (see here), so the argument you had in mind still applies.

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I have trouble to understand what surface is meant here: (Large version) Your first equation gives the cone (red), the second the plane and the third the half-space $z\ge 0$.

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I don't know about vector calculus, but here's a good way of doing it: Definition 0. Let $V$ denote a vector space over the reals. Then $A \subseteq V$ is a plane iff there exists a two-dimensional linear subspace $X$ of $V$, together with an element $v \in V$, such that $A = X+v$. Definition 1. Let $V$ denote a vector space over the reals. Consider ...

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The critical points are points $a \in \mathbb R^2$ such that $\mathrm {grad}\, f(a) = 0$. Now $\mathrm{grad}\, f(x,y) = (2xy + y , x^2 + 2y + x)$. You need to solve $$\begin{cases}2xy + y = 0 \\x^2 + 2y + x = 0\end{cases} \iff \begin{cases}y(2x+ 1) = 0\\x(x + 1) + 2y = 0\end{cases} \iff \begin{cases}y= 0 \,\,\text{or}\,\, x = -\frac{1}{2}\\x(x+1) + 2y = ... 0 You missed a critical point. You have:$$f_x = 0 \iff 2xy+y = 0 \iff y(2x+1) = 0 \iff y = 0 \,\vee x = -\tfrac{1}{2}$$From y=0 you already found the critical points (0,0) and (-1,0), substitution of x = -\tfrac{1}{2} into f_y = 0 gives a third critical point: \left(-\tfrac{1}{2},\tfrac{1}{8}\right). For (0,0), \det H = -1 < 0 so that is ... 1 I don't know what a "rotational field" is. In the first place your F=(-y,x,0) is a plane field: It's z-component is \equiv0, and the two other components do not depend on z. Therefore we understand everything about F if we understand what happens in the (x,y)-plane. It is a basic fact of plane analytic geometry that$$j:\quad {\mathbb ...

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Your cost function can be simply $$f(x, y) = x^{1/4} y^{3/4}$$ since the profit, $$p(x, y) = 20 - 8 f(x, y)$$ (that is 10% of sales less than the advert cost of 20). Formulate the problem as $$\arg \underset{x, y}{\max} \, x^{1/4} y^{3/4}$$ Subject to: $x + y = 20$. Therefore, $$L(x, y, \lambda) = x^{1/4} y^{3/4} - \lambda (x + y - 20)$$ Please ...

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You need to formulate the problem first. the first thing is to write out objective function $$\min_{square \in C} Area_{square}$$ where implicit constraint C is a set of all inscribed squares. For inscribed square, its edge length should satisfy: $$l\cos \theta + l \sin\theta=L$$ Thus, the problem becomes $$\min_{\theta \in (0,\pi/2)} ... 0 Hint. You may use polar coodinates, x=r \cos \theta, y=r \sin \theta, then you initial function writes$$ \frac{\sin (r^2)}{r^2} $$and consider r \to 0. 0 Hint: Change to polar coordinates and use a well-known limit. One does not even need to go to polar coordinates, but that is a useful move in general when the denominator is x^2+y^2. (Your solution is correct.) 0 As mentioned in other answer, curl\,\vec F=0\; , so we can try to find its potential \;U(x,y,z)\;:$$U_x=6xy+4xz\implies U=3x^2y+2x^2z+C(y,z)\;,\;\;C(y,z)\;\text{a constant in}\;x\implies3x^2+2yz=U_y=3x^2+C_y(y,z)\implies C(y,z)=y^2z+K(z)=\text{ constant in y}\implies2x^2+y^2=U_z=2x^2+y^2+K'(z)\implies K(z)=T=\text{ constant}\implies$$... 1 Indeed \text{curl}\, \vec F=0, so \vec F is conservative. Then you have$$ \int_\Gamma \vec F\cdot d\vec r=g(1,1,2)-g(0,0,0), $$where g is a potential function for \vec F. One such function is g(x,y,z)=3x^2y+y^2z+2x^2z. Then$$ \int_\Gamma \vec F\cdot d\vec r=g(1,1,2)-g(0,0,0)=3+2+4=9. $$6 By setting x=au and y=bv the problem boils down to computing$$ I(a,b) = ab\iint_{\mathbb{R}^2}\sqrt{u^2+v^2} e^{-(u^2+v^2)}\,du\,dv = 2\pi ab \int_{0}^{+\infty} \rho^2 e^{-\rho^2}\,d\rho = \pi a b\cdot\Gamma\left(\frac{1}{2}\right).$$1 Since \Bbb R^3 is simply connected, such \phi will exist if and only if \nabla \times F = 0 , so it is always good to check if this is the case, so you don't waste time. You want to solve$$ \begin{cases} \frac{\partial \phi}{\partial x}(x,y,z) = 2xy + 4xz \\ \frac{\partial \phi}{\partial y}(x,y,z) = x^2 + 6yz \\ \frac{\partial \phi}{\partial z}(x,y,z) ...

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The task poses a system of partial derivatives: $$\partial_x \phi = 2xy + 4xz \\ \partial_y \phi = x^2 + 6yz \\ \partial_z \phi = 2x^2 + 3y^2$$ Integration gives $$\phi = x^2y + 2x^2z + f(y,z) \\ \phi = x^2y + g(x,z) + 3y^2z \\ \phi = h(x,y) + 2x^2z + 3y^2z$$ Comparison gives $$\phi = x^2 y + 2 x^2 z + 3 y^2 z + C$$ where $C$ is some constant.

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Just follow the definition: $\nabla \phi = (\frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z})$ Hence, you have to solve: $$\begin{cases} 2xy + 4xz =\frac{\partial \phi}{\partial x} \\ x^2 + 6yz = \frac{\partial \phi}{\partial y} \\ 2x^2 + 3y^2 = \frac{\partial \phi}{\partial z} \end{cases}$$ The first ...

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If you want to do this rigorously, you need to show that $$|f(x_0+h,y_0+k) - f(x_0,y_0) - \partial_xf(x_0,y_0)h - \partial_yf(x_0,y_0)k|= o(||(h,k)||).$$ Using the triangle inequality, $$|f(x_0+h,y_0+k) - f(x_0,y_0) - \partial_xf(x_0,y_0)h - \partial_yf(x_0,y_0)k| \\ \leqslant |f(x_0+h,y_0+k) - f(x_0,y_0+k) - \partial_xf(x_0,y_0)h| + | f(x_0,y_0+k)- ... 1 The objective function is non-negative. If you take \{x_n\} and \{y_n\} to be orthogonal, it is clear that the minimum is equal to 0. Take a_n>0 such that \sum_{n=2}^\infty a_n^2=s<\infty and let$$ x_n=a_n,\quad y_n=-a_n,\quad n\ge2, $$and x_1, y_1 such that x_1y_1=s. Then \sum_{n=1}^\infty x_ny_n=s-\sum_{n=2}^\infty a_n^2=0. 1 Assuming your * is matrix multiplication$$J = \pmatrix{4cx+1 & 0\cr 0 & 4cy+1\cr}$$so this will be a contraction with the Euclidean norm in any convex region where |4cx+1|<1 and |4cy+1|<1. For this to be true in your rectangle (x,y) \in [0.93, 1.52] \times [0.41, 1] you need -25/76 < c < 0. 0 Of course both o(dx) and o(dy) are o(||(dx, dy)||). Then you have only to prove that  \epsilon(dx)dy is o(||(dx, dy)||). Now$$ \frac{|\epsilon(dx)dy|}{||(dx, dy)||}\leq\frac{|\epsilon(dx)||dy|}{|dy|}=|\epsilon(dx)|\rightarrow 0 $$1 If we parameterize x and y as both equal to t, we are not guaranteed that the derivative with respect to t equals the derivative with respect to x or y. Using the rule for total derivative:$$ \frac{df}{dt}=\frac{\partial f}{\partial x}\frac{d x}{dt}+ \frac{\partial f}{\partial y}\frac{d y}{dt}. $$2 The direction vector v is a unit vector:$$ v = (1/\sqrt{2}) \, (1,1)^t $$and as f has a total derivative we get the directional derivative$$ \partial f / \partial v = \DeclareMathOperator{grad}{grad} \grad f \cdot v = (1,1)^t \cdot (1/\sqrt{2}) \, (1,1)^t = (1 / \sqrt{2}) \, 2 = \sqrt{2}  Checking the change of $f$ for the unit step in ...

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