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0

Let's use these spherical coordinates ($\rho$ is radius, $\phi$ is colatitude, $\theta$ is longitude): $$ \left\{ \begin{align} x &= \rho \sin \phi \cos \theta \\ y &= \rho \sin \phi \sin \theta \\ z &= \rho \cos \phi \end{align} \right. $$ Notice that $x^2 + y^2$ is a simpler expression than either $x$ or $y$. Try to simplify it. (Hover over ...


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If $(x,y) \neq {\bf 0}$, then $f$ is differentiable, because $x,y, \sin$ and $(x^2+y^2)^{-1}$ are differentiable. And in this situation we have: $$\begin{align*}\frac{\partial f}{\partial x}(x,y) &= y\sin\left(\frac{1}{x^2+y^2}\right) - 2x^2y\cos\left(\frac{1}{x^2+y^2}\right)\frac{1}{(x^2+y^2)^2} \\ \frac{\partial f}{\partial y}(x,y) &= x ...


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Eliminate $ z - x^2 $ to get $ x = y -1/y $. Let $ y = t, x = 1/t -t $, simplify to get $ z = 1/t^2 -2.$


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Summary of comments: Set them equal to each other, $x^2-y^2 = x^2+xy-1$ Simplify the resulting equation: $0=y^2+xy-1$ The equation is easier to solve for $x$: $x=(1-y^2)/y$ Make $y=t$ as the parameter, obtaining $x=(1-t^2)/t$ and $y=t$, and also $z=x^2-y^2 = (1-t^2)^2/t^2 - t^2$. Note that the intersection consists of two curves, corresponding to the ...


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$$ x^2 - y^2 = x^2 + xy - 1 \\ 0 = y^2 + xy - 1 $$ Apply the quadratic formula over $y$. $$ y = \frac{x \pm \sqrt{x^2 - 2(1)(-1)}}{2(1)} = \frac{x \pm \sqrt{x^2 + 2}}{2} $$ Alternatively, we can solve for $x$. $$ 0 = y^2 + xy - 1 \\ x = \frac{1 - y^2}{y} $$ From there, you can continue.


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Since $\displaystyle \lim_{(x,y) \to (a,b)} f(x,y) =l$, for every $\epsilon >0$ there exists $\delta > 0$ such that if $|x-a| < \delta$ and $|y-b| < \delta$ then $$|f(x,y) - l| < \epsilon.$$ Let $$l_a(y) = \lim_{x \to a} f(x,y).$$ Then with $|y-b| < \delta$, $$|f(x,y) - l| < \epsilon \implies \lim_{x \to a} |f(x,y) - l| \leqslant ...


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One can use paths, or particular sequences $(x_n,y_n)\to(0,0)$, in order to prove that such a limit does not exist. Producing just two different paths or sequences with different limits of the expression in question closes the case. But in the case at hand you (correctly) conjecture that the limit does exist, and you want to prove this. Under these ...


2

$$\frac{\partial z}{\partial x} = \frac{1}{y}f'(x/y)$$ and $$\frac{\partial z}{\partial y} = -\frac{x}{y^2}f'(x/y)$$ by the chain rule. All together: $$x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial z} = \left(\frac{x}{y} - \frac{x}{y}\right)f'(x/y) = 0$$


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It seems that the differential of a function and the gradient tend to mix with one another in many people's minds. We start with the definition of directional derivative. Let $f:\mathbb{R}^n\to\mathbb{R}$ be a function, and let $x,v\in\mathbb{R}^n$. One can define another function $f_{x,v}:\mathbb{R}\to\mathbb{R}$ by$$f_{x,v}(t)=f(x+tv).$$Intuitively, ...


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How to do it: In this simple example it is actually possible to give an explicit parametrization of the curve of intersection $\gamma$ in terms of elementary functions. But in general you cannot count on such luck. Fortunately it is possible to compute this directional derivative without computing $\gamma$. We are given two surfaces $$S_1:\quad ...


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If you plug $z^2=x^2+y^2$ into $2x^2+2y^2-z^2=25$, you will get the intersection $x^2+y^2=25, z=5$, which is a circle. The parametric equation is then $(5\cos{t},5\sin{t},5), t\in[0,2\pi]$. And yes, you can use that to find directional derivative.


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You can find the tangent planes to both curves at (1,1,1), then take the cross product of the two normal vectors to get the direction vector of the tangent line, since this will give you a vector that lies on both tangent planes.


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If a, b, c are the number of products, then you can set it up as a system of equtions subject to the constraint (147,174 and 238). The number of product A, B, C will be like below It would be that the maximum for A+B+C = 2+10+17 = 29. That will be the maximum number of products you can get Thanks Satish


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Maximizing $A+B+C=18(x+y+z)$ amounts to mazimizing each element, namely $x$ and $y$ and $z$ seperately. So you must use whatever you have in your hands.


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Given that the curve lies inside the intersection of the two surfaces given by $$g(x, y, z) = 2x^2 + 2y^2 - z^2 - 25 = 0$$ and $$ h(x, y, z) = x^2 + y^2 - z^2 = 0, $$ then that curve will have tangent vector normal both to $\nabla g$ and $\nabla h$ at $(3, 4, 5)$. So you can use cross product for example to find that. If you do not want to use ...


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Assuming that tha motion starts in the fixed point A(a,0), we can write the coordinate of the center B of the moving circle as \begin{equation} ((a-b) \cos(t), (a-b) \sin(t)) \end{equation} The coordinate of the moving point P relative to B are: \begin{equation} (b \cos(\phi), -b \sin(\phi)) \end{equation} where the minus sign is due to the fact that $\phi$ ...


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Answer for the second one $$\frac{d}{dt}\bigg(\frac{\partial \mathbf r_i}{\partial q_j}\bigg) = \frac{\partial \mathbf{\dot r}_i}{\partial q_j}$$ $$\frac{\partial \mathbf{\dot r}_i}{\partial q_j} = \frac{\partial \bigg(\sum_{k=1}^m \frac{\partial \mathbf{r}_i}{\partial q_k}\frac{dq_k}{dt}\bigg)}{\partial q_j} = \sum_{k=1}^m \frac{\partial ...


1

In the neighborhood of $(0,0)$ the manifold $S$ has a bijective parametric representation of the form $$S:\quad x\mapsto \bigl(x,h(x)\bigr)$$ with $h(0)=h'(0)=0$. Consider any function $p:\>(x,y)\mapsto p(x,y)\in{\mathbb R}$ defined in a neighborhood of $(0,0)$ and its pullback to $S$ defined by $$\hat p(x):=p\bigl(x, h(x)\bigr)\ .$$ Using the chain rule ...


1

You are integrating over an ellipse, which is hard, relative to integrating over a circle. So first I would change your ellipse into a circle, using $u = \frac{x}{2a}$ and $v = \frac{y}{2b}$. Then your domain of integration is $\mathcal{D}_{1}:= \left\{(u,v): u^2+v^2\le 1\right\}$, and your integral $I$ transforms to \begin{align} I= 4ab ...


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Split the integral into to parts: from 0 to $\pi/2$ and to $\pi/2$ to $\pi$, the second is equal to the first (you can prove it by the following substitution: $t = \pi - \theta$). To calculate the first integral you can divide all by $(cos(\theta))^2$ and then make the substitution $x = tan(x)$. From there use simple fractions and you're done. You could ...


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As a rule of thumb, it's better to use $$\frac12\int_{\partial B}(-y\,dx+x\,dy)$$ whenever trig functions are involved. That will give a simpler expression once you use standard trig identities. Comment: It's more straight forward in this particular case to do the area integral directly in polar coordinates: $$\int_0^{2\pi}\int_0^{1+\cos\theta} ...


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The current approach seems reasonably efficient to me: From this point, if you expand out the dot product, you'll get just three terms, each of them manageable integrals involving sines and cosines. You can handle parts of the integrals quickly using the fact that the interval has length $2\pi$, which is the period of the $\sin$ and $\cos$ functions that ...


1

$$\iiint6xy\,dV=6\int_0^1\int_0^{\sqrt x}\int_0^{1+x+y} xy\,dz\,dy\,dx$$


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In single variable calculus, if $f:\mathbb R \to \mathbb R$, then \begin{equation} f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}. \end{equation} A very useful way to think about $f'(x)$ is this: \begin{equation} \tag{$\spadesuit$} f(x + \Delta x) \approx f(x) + f'(x) \Delta x. \end{equation} One of the advantages of equation ...


3

I suppose $t$ is a fixed number. I recommend you to sketch a $ns$ plane. We know that $t-n<s<t$. So, draw the line $s=t-n$ and the line $s=t$. Since $0<n<\infty$ then you have a portion of the $ns$ of the plane as the region on which you will integrate. That region looks like a triangular region, but it's unbounded of course. If you want to ...


0

There are many functions that have a local minimum on this curve. It would be more common to ask about functions of a given form with one or more parameters. Also your conditions for a local minimum only give a minimum along the $y$ direction. Usually you would require that $\frac {dF}{dx}=0$ and that all the second derivatives are greater than zero.


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That means u is a constant, while v can vary within those bounds. So the image is a segment in uv space.


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The gradient is the vector who's components are the partial derivatives of f(a). Since for every x in B(a) where f is differentiable, f(x) is less then or equal to z= f(a), then z= f(a) is a local maximum for f in B(a). But this means the value of every partial derivative with respect to a unit vector y in B(a) is going to be zero at z=f(a) and this means in ...


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Hint: If $f'(a;y)$ is the directional derivative of $f$ at $a$ in the direction $y$, you have $f'(a;-y) = -f'(a;y)$


2

It means that if you consider $f(t,y_1,\dots,y_n)$, the partial derivatives $\partial f/\partial y_i$ exist and are continuous for all $i=1,\dots,n$. The first line of your question is really wrong. You are considering a composite function $F(t) = f(t,y(t))$; this is a function of $t$ only.


1

If $$\frac{y^2-x^4}{y^2+x^4}=k$$ solve for $y^2$ first; this gives $$y^2=x^4\space\frac{1+k }{1-k}$$ So, $$y=\pm x^2\sqrt{\frac{1+k }{1-k}}$$ So, two parabolas which exist only if $\cdots$


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$$ \frac{y^2-x^4}{y^2+x^4}=k\implies y^2=\frac{1+k}{1-k}\,x^4,\quad k\ne1. $$ Do you recognize those curves?


0

Try approaching the limit by the line $y=k\cdot x$. Applying the standard limit $$\lim_{t\rightarrow 0} \frac{\sin(t)}{t}$$ You can prove that the limit depends on the constant K, hence does not exist


1

let $$u = e^{6t}.$$ then we have $$(u^2 + 1)T = (\sqrt 2 u, u^2, 1) \tag 1$$ differentiating $(1)$ with respect to $u,$ gives $2uT + (u^2+1)\frac{dT}{du} = (\sqrt 2, 2u, 0)$ which can be simplifies to give $$(u^2 + 1)^2\frac{dT}{du} = (u^2 + 1)(\sqrt 2, 2u, 0) - 2u(\sqrt 2 u, u^2, 1) =(\sqrt 2(1-u^2),2u,-2u) $$ therefore $$N = ...


1

You simply substitute in the point $(x,y)=(0,0)$. $$ \lim\limits_{(x, y)\to (0, 0)}\frac{x^3y-xy^3-x}{1-xy} = \frac{(0^3\cdot0) - (0\cdot0^3) -0}{1- (0\cdot0)} = \frac{0}{1}=0 $$ This comes from the definition $$\lim \limits_{(x,y)\to (a,b)}f(x,y)=f(a,b) = L$$ given than $L<\infty$, that is, $L$ is finite


0

There has been a typo in the definition of the line segment: It should be that the line is $$f(t)=at+(1-t)b$$ where $a$ and $b$ are arbitrary points in $\mathbb{R}^n$. This is a classic way to represent the line between two points. The boundary is important because when we evaluate $f$ at $t=0$, we just get the point $a$, and when we evaluate it at $t=1$, ...


1

You did a mistake when computing your expression. The variable $x$ multiplies $f_{u}(v,u)$, so that the correct expression is \begin{equation} f_u(v,u) x (2 a -1)+f_v (v,u)= y \end{equation} Now expressing x and y as function of v, u: \begin{equation} f_u(v,u) v (2 a -1)+f_v (v,u)= u - a v^2 \end{equation} if you set $a=\frac{1}{2}$ you get: \begin{equation} ...


1

I would do this: \begin{equation} 0\le\frac{\left|x\right|^\alpha |y|}{x^2+y^2}=\frac{\left|x\right|^{2}}{x^2+y^2} \lvert y\rvert \lvert x \rvert^{\alpha-2} \le \lvert y\rvert \lvert x \rvert^{\alpha-2} \xrightarrow{(x,y)\to(0,0)}0 \end{equation} because $\displaystyle\frac{\left|x\right|^{2}}{x^2+y^2}\le 1$. Where am I mistaken?


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}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} ...


1

Hint: A useful strategy when we have $x^2+y^2$ in the denominator is to go to polar coordinates, letting $x=r\cos\theta$ and $y=r\sin\theta$.


3

We have $$0\le\frac{\left|x\right|^\alpha |y|}{x^2+y^2}=\left|x\right|^{\alpha-1}\frac{\left|x\right| |y|}{x^2+y^2}\le\frac12 \left|x\right|^{\alpha-1}\xrightarrow{(x,y)\to(0,0)}0$$ and conclude by the squeeze theorem.


2

Wait, $F(t)$ is divergent. By a change of variable, $F(t) = t^2 F(1)$, and $F(1)$ is divergent since $$\int_0^1 \int_0^1 e^{\frac{x}{y^2}}\, dx\, dy = \int_0^1 y^2(e^{\frac{1}{y^2}} - 1)\, dy,$$ which is divergent. Indeed, for all $x\ge 0$, $e^x - 1 = \int_0^x e^t\, dt \ge \int_0^x 1\, dt = x$. Hence, for all $x\ge 0$, $e^x - 1 \ge \int_0^x t\, dt = ...


0

What if $f(x)=x^2$ and $x\in (0,\infty)$. Would the following two be equivalent: $$(1)\;\;\;|x|\rightarrow0,\;\;\; \frac{x^2}{x}=x\rightarrow0$$ $$(2)\;\;\;\;\;\;\;\lim_{|x|\rightarrow0} \frac{x^2}{x}=x=0$$


2

$x^2+y^2 = y$ describes an infinite cylinder $$x^2 + \left(y - \frac{1}{2}\right)^2 = \left(\frac{1}{2}\right)^2$$ The interior of intersection of two regions is thus given by $$ \mathcal{R} = \left\{ x^2 + y^2 + z^2 \leqslant 1, x^2 + \left(y - \frac{1}{2}\right)^2 \leqslant \left(\frac{1}{2}\right)^2, x \geqslant 0, y \geqslant 0, z \geqslant 0 ...


0

Compute the plane by the three points $(x,y,z)$, $(1,0,2)$, $(0,2,2)$ and check that it is tangent to the surface at $(x,y,z)$. The normal vector to the plane is given by $N_p=(x-1,y-0,z-2)\times(1-0,0-2,2-2)$. The normal vector to the surface is given by $N_s=(2x,2y,1)$. These two vectors are parallel, $N_p\times N_s=0$. $$-x^2+4xy+y^2-4y-1=0,\\ ...


0

tangent plane at $(a, b, 1 - a^2 - b^2)$ is $$2a(x-a)+2b(y-b) + (z +a^2 + b^2 - 1) = 0 $$ since the point $(1,0,2)$ is to be on the plane we get $2a(1-a) + 2b(-b) + (a^2 + b^2 + 1) = 0$ which can be simplified to yield $a^2 + b^2 -2a - 1 = 0.$ in the same way making the point $(0,2,2)$ on the plane requires $2a(-a) + 2b(2-b) + (1+a62 + b^2) = 0.$ on ...


1

Sure. For example, supposing the intersection of the domain with every vertical line is an interval containing zero, you could just take $$ \mathbf F(x,y,z) = (0, 0, \int_0^z f(x,y,t)\,dt) $$ For a domain of a more complex shape it could be more tedious to patch a solution together from smaller ones, but the existence of an $\mathbf F$ isn't really at ...


1

Let a rectangle be defined by two points $[a,b]$ in $\mathbb{R}^{n}$ as the the set of $x$ such that $a\le x\le b$. (The inequalities must hold for every component.) If for some component $a_i>b_i$, the rectangle is void. Now the intersection of the rectangles $[a,b]$ and $[c,d]$ is the set of $x$ such that $a\le x\le b\land c\le x\le d$. In other ...



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