New answers tagged

1

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


0

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, ...


0

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.


1

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 ...


1

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


1

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 $z$ ...


0

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 ...


1

(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.


1

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 ...


1

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 ...


2

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 $z=0$. $$\int_0^{\pi/2} \int_0^2 (3-r\sin\theta) rdrd\theta$$


1

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.


2

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 $z\ge 0$.


0

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 ...


0

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 ...


0

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 ...


0

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 $\;U(x,y,z)\;$: $$U_x=6xy+4xz\implies U=3x^2y+2x^2z+C(y,z)\;,\;\;C(y,z)\;\text{a constant in}\;x\implies$$ $$3x^2+2yz=U_y=3x^2+C_y(y,z)\implies C(y,z)=y^2z+K(z)=\text{ constant in y}\implies$$ $$2x^2+y^2=U_z=2x^2+y^2+K'(z)\implies K(z)=T=\text{ constant}\implies$$ ...


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) ...


0

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.


2

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 ...


0

$$\frac{\partial f}{\partial x}=2xy+y$$ $$\frac{\partial f^2}{\partial x \partial y}=\frac{\partial }{\partial y}\bigg(\frac{\partial f}{\partial x}\bigg)=\frac{\partial }{\partial y}\bigg(2xy+y\bigg)=2x+1$$ And $$\frac{\partial f}{\partial y}=x^2+2y+x$$ $$\frac{\partial f^2}{\partial y \partial x}=\frac{\partial }{\partial x}\bigg(\frac{\partial f}{\partial ...


1

I assume you want to show: Given such $f=\begin{bmatrix}f_{1}\\f_{2}\\ \cdots\\f_{m}\end{bmatrix}$. There exist local coordinates $$\xi_1:U_{1}\rightarrow V_{1}~with~x^{*}\in U_{1}\subset U,~0\in V_{1}~and~U_{1},V_{1}\subset \mathbb{R}^{n}$$ and $$\xi_{2}:U_{2}\rightarrow V_{2}~with~f(x^{*})\in U_{2},~0\in V_{1}~and~U_{2},V_{2}\subset \mathbb{R}^{m}$$ such ...


1

AT first of all it is not continuous. Consider the sequence of points $(\frac{1}{n},\frac{1}{n})$ the values of the functions are constant i.e. $\frac{1}{2}$ , but to be continuous its limit as $ n \rightarrow \infty $ it has to be zero . So it is not continuous . Hence not differentiable.


1

Let $y = tx$ then $f(x,tx) = {t x^2 \over (1+t^2) x^2} = { t \over 1+t^2}$. Choose $t = 0$ and let $x \to 0$, the limit is $0$. Choose $t = 1$ and let $x \to 0$, the limit is ${1 \over 2}$.


0

First, consider integrating for the volume of a circle along a line. $$2\int_0^r (r^2-x^2)^.5dx$$ This one's somewhat involved, as is every subsequent integral that yields an increase in pi power, because you have to use polar on the square root, but you eventually get: $$A=\pi r^2$$ Now, we want to find the volume of a sphere by adding up the volumes of ...


0

This belongs to an Abel equation of the second kind. Let $x=e^{-t}$ , Then $\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{\dfrac{dy}{dt}}{-e^{-t}}=-e^t\dfrac{dy}{dt}$ $\therefore-e^ty\dfrac{dy}{dt}+e^ty+k=0$ $y\dfrac{dy}{dt}-y=ke^{-t}$ This belongs to an Abel equation of the second kind in the canonical form. Please follow the method in ...


1

Both are fine. We take the absolute value of the determinant.


0

You are correct. Generally for a composition $g = f\circ h$ it holds that $[Jg](x) = [Jf]\big(h(x)\big) \cdot [Jh](x)$, where $Jf$ is the Jacobian of $f$, and $[Jf](x)$ is the Jacobian of $f$ evaluated at $x$. This is simply the chain rule. In your case $g$ is a function of a single variable and thus its Jacobian is simply the usual derivative. We have $g ...


0

Let $F(x,y)=x^2+2y^2$. Then the limits are $F(x,y)-2 \leq z\leq 6-F(x,y)$, such that $2F(x,y)\leq 8$ and therefore, $F(x,y)\leq 4$. Let's notice that $x^2+2y^2 \leq 4$, and therefore, $y^2 \leq \frac{4-x^2}{2}$, and therefore, $-\sqrt{\frac{4-x^2}{2}}\leq y \leq \sqrt{\frac{4-x^2}{2}}$. On the other hand, since, $x^2\leq 4$ as well, then $-2\leq x\leq 2$. ...


0

I assume you are implying $a^tx-b>0$ to ensure convexity (see Michael's comment). For simplicity I also assume that $Q$ is Hermitian but the solution can be easily generalized. If $b \le 0$, then $x=\mathbf{0}$ is trivially a solution. If $b > 0$, you can follow the approach below (forgive some sloppiness in the maths, for brevity). Setting $\nabla ...


1

In this case, simple geometric considerations tell you that you have a minimum: the objective function is the square of the distance from the origin and the constraint is the equation of a plane. More generally, one can examine the bordered Hessian to determine the nature of the stationary points found via Lagrange multipliers. In this case, $Hf = 2I_3$ ...


0

Cylindrical coordinates is polar coordinates but in 3D. First you have to find projection in XY Plane which can be done so by eliminating z from above equations. Doing so you will get $$(x-\frac{a}{3})^2 - (\frac{4y}{\sqrt 3})^2 = (\frac{2a}{3})^2$$ Also your Z limits will be $ \sqrt{\frac{r^2cos^2\theta}{4}) + r^2sin^2\theta}$ and $\frac{a-rcos\theta}{4}$ ...


2

My solution is another way of stating @Jack D'Aurizio's solution. We want to rotate coordinates so that the plane $x+4z=a$ is horizontal. This can be achieved through $$\begin{array}{rl}x^{\prime}&=\frac4{\sqrt{17}}x-\frac1{\sqrt{17}}z\\ z^{\prime}&=\frac1{\sqrt{17}}x+\frac4{\sqrt{17}}z\end{array}$$ The inverse transformation is ...


1

Well you can always compare it to other values. For example, take $x=1, y=1, z=10$, $x+y+z=12$, and $f(x)=1+1+100=102>48$ So 48 has to be a minimum.


3

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} ...


2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} ...


0

Recall that $$B(x,y) = 2\int_0^{\pi/2}(\sin\theta)^{2x-1}(\cos\theta)^{2y-1}\,\mathrm{d}\theta, \qquad \Re(x)>0,\ \Re(y)>0 $$ we have $$ \int_0^{\frac{\pi}{2}} \cos (\theta) ^{2k} \,\mathrm d\theta=\frac{1}{2}B\left(\frac{1}{2},k+\frac{1}{2}\right)=\frac{\sqrt{\pi}\,\Gamma\left(k+\frac{1}{2}\right)}{2\Gamma(k+1)}\qquad\text{for}\;\;k\in\Bbb ...


0

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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