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(BIG) HINT: $$\cos(\phi)+\sqrt{\cos^2(\phi)+15}=\frac{2}{\sin(\phi)}\Longleftrightarrow$$ $$\cos(\phi)+\sqrt{\cos^2(\phi)+15}=2\csc(\phi)\Longleftrightarrow$$ $$\sqrt{\cos^2(\phi)+15}=2\csc(\phi)-\cos(\phi)\Longleftrightarrow$$ $$\cos^2(\phi)+15=\left(2\csc(\phi)-\cos(\phi)\right)^2\Longleftrightarrow$$ ...

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You can do linear algebra in terms of $n$-tuples $(x_1,x_2,\ldots, x_n)$ and matrices $\bigl[a_{ik}\bigr]$, or in terms of "abstract" vectors ${\bf x}$ and linear maps $A$. Similarly with derivatives in a multivariate setting: You can consider component functions $f_i$ $(1\leq i\leq m)$ and their partial derivatives $f_{i.k}$, as defined in $(25)$, or you ...

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The function isn't bounded above since $$f(x,0)=2x^4-3x^2+2\rightarrow\infty$$ Since the function is even in both variables we can work only with $\;x,y\ge0\;$ , and $$f(x,y)\xrightarrow[(x,y)\to(0,0)]{} 1+1=2$$ But for any point on the circle $\;x^2+y^2=1\;$ or on the ellipse $\;2x^2+y^2=1\;$ the value of the function is $\;1\;$ : $\;f(x,y)= 1$ Yet ...

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Let $g(x,y)=(x^3,y)$ and $h(x,y)=(\sin x\cosh y,\cos x\sinh y)$, so $f=h\circ g$. Then $g(0,0)=(0,0)$ and $g$ is invertible, and $h$ is locally invertible at $(0,0)$ since $\displaystyle\frac{\partial(h_1,h_2)}{\partial(x,y)}=\cos^2x\cosh^2y+\sin^2x\sinh^2y=\cos^2x(\sinh^2y+1)+\sin^2x\sinh^2y=\sinh^2y+\cos^2x$, so ...

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Along $y = -x$ for example: $$\lim_{(x,y)\rightarrow(0,0)}\frac{ e^{x+y} - x - y}{\sqrt{x^2 + y^2}} = \frac{1}{\sqrt{2}}\cdot{\lim_{x\rightarrow 0}\frac{1}{|x|}} \rightarrow \infty$$

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Consider the function $f(x, y) = x^2 + y^2$. The gradient is $\langle 2x, 2y\rangle$, so if you depart from the origin at any angle and continue straight, you will be parallel to the gradient at any point. So the trajectory isn't unique; you'll have to choose an initial velocity. For any function $f$, I believe we can identify one of these trajectories as a ...

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You are correct until partial difference of $f(\mathbf{x})$. Next, the vector $\mathbf{x} - \nabla f(\mathbf{x})$ is replaced for $\mathbf{y}$. Therefore, \begin{align} \mathbf{y} &=\mathbf{x}-\nabla f(\mathbf{x}) \\ &=\mathbf{x}-(Q\mathbf{x}-\mathbf{b}) \\ &=(E-Q)\mathbf{x}+\mathbf{b} \end{align} Then, $E$ is an identical matrix. So, ...

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As the whole situation has a rotational symmetry w.r.t. to $z$ we can simplify the problem by looking at a plane which goes throu the z-axis. In this image you can see that $R=4$ is the radius of the sphere, $r=2$ is the radius of the cylinder. You can also see that the intersectins are circles. And In this case you can easily find that those circles have ...

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The designation is important. For example, we can write the scalar $I_1$ $$I_1=\int_S (\hat n\cdot \vec F) \,dS$$ the vector $I_2$ $$I_2=\int_S (\hat n\times \vec F) \,dS$$ and the dyadic (tensor, rank 2) $I_3$ $$I_3=\int_S(\hat n\,\vec F) \,dS$$ The notation for $I$, $$I=\int_S \vec F\,$$ is ambiguous without explicit designation. 

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If we write $r:=\vert\vert(x,y)-(x_0,y_0)\vert\vert$, differentiability of $f$ is equivalent to the existence of $(a,b)$ such that $f(x,y)=f(x_0,y_0)+a(x-x_0)+b(y-y_0)+o(r)$. Squaring this gives $f(x,y)^2=f(x_0,y_0)^2+2(a(x-x_0)+b(y-y_0))f(x_0,y_0)+(a(x-x_0)+b(y-y_0))^2+o(r)$ (noting that $(f(x_0,y_0)+a(x-x_0)+b(y-y_0))\times o(r)$ is $o(r)$. By switching ...

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$$3u_x+2u_t=\cos(x)$$ HINT : The equations characteristics are : $$\frac{dx}{3}=\frac{dt}{2}=\frac{du}{\cos(x)}$$ First characteristic : $\frac{dx}{3}=\frac{du}{\cos(x)} \quad\to\quad 3du-\cos(x)dx=0 \quad\to\quad u-\frac{1}{3}\sin(x)=c_1$ Second characteristic : $\frac{dx}{3}-\frac{dt}{2}=0 \quad\to\quad 2x-3t=c_2$ General solution on implicite form, ...

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The best way to view this in my opinion is by looking at the actual ODE you get when you solve the characteristic equation. This ODE is what you get along the path $x=x_0+\frac{3}{2}t$. For each $x_0$ you have $v(t)=u(t,x_0+\frac{3}{2}t)$ with $v(0)=f(x_0)$. It satisfies $\frac{dv}{dt}=\frac{1}{2} \cos(x(t))=\frac{1}{2} \cos(x_0+\frac{3}{2}t)$. The solution ...

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It happens because $f(ax,ay) = af(x,y)$

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It's true that . However, you might note that the same can be said of your ordinary integral on a line: one can easily see if $f$ is a function $[a,b]\rightarrow \mathbb R$, then it's already clear what $$\int_a^bf$$ means, and the $dx$ is just going to be tagging along. However, the advantage of writing $$\int_a^b f(x)\,dx$$ becomes more clear when one has ...

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I would write it as $$\sqrt{\cos^2\phi+15} = \frac{2}{\sin \phi} - \cos\phi$$ Both left and right hand side are positive (where defined), so you can square both sides. You will get to $$\cos^2\phi + 15 = \frac{4}{\sin^2\phi} + \cos^2\phi - 4 \frac{\cos\phi}{\sin\phi}.$$ Cancel what you can, multiply by $\sin^2\phi$, and use double angle formulas to ...

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Here is a simple proof using index notation and BAC-CAB identity. \begin{align} \nabla \times \left( {{\bf{A}} \times {\bf{B}}} \right) &= {{\bf{e}}_i}{\partial _i} \times \left( {{A_j}{{\bf{e}}_j} \times {B_k}{{\bf{e}}_k}} \right)\\ &= {\partial _i}\left( {{A_j}{B_k}} \right){{\bf{e}}_i} \times \left( {{{\bf{e}}_j} \times {{\bf{e}}_k}} \right)\\ ... 1 One way is to recognize that the intersection must satisfy the equation for both planes, and must therefore satisfy their sum: (3x-y+z)+(y+z) = 4+2  3x+2z = 6 $$You can then let x = t, and then 3t+2z = 6, whence we get z = 3-\frac{3}{2}t. You can then rewrite your first equation as$$ y = 3x+z-4 $$to obtain an expression for y in ... 1 The notes that you are following should be corrected by erasing the initial fraction:$$ \frac{dE}{dt}(t) = \int_\Omega (u_tu_{tt} + c(x)^2 \nabla u \cdot \nabla u_t + q(x)uu_t)\, dx. $$In particular, the derivative of (u_t)^2 is obtained exactly as you wrote. As for the middle term, you have instead$$ \frac{\partial}{\partial t}|\nabla u|^2 = ...

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A first integration with respect to $x$ is clearly easier. I suggest you to apply Fubini theorem and reverse the order of integrals. $$\int_{0}^{1}\int_{x}^{1} y^2 \sin(2\pi \frac{x}{y})dydx = \int_{0}^{1} y^2\left( \int_{0}^{y} \sin(2\pi \frac{x}{y})dx \right)dy.$$ And $$\int_{0}^{y} \sin(2\pi \frac{x}{y}) dx = \frac{y}{2\pi} [-\cos(2\pi \frac{x}{y})]_0^y ... 1 Let's think about how this works for a curve in \Bbb R^3. We show that a curve \gamma has torsion identically 0 if and only if \gamma lies in an affine plane. Namely: If there is a constant unit vector \mathbf C so that \mathbf C\cdot(\gamma-\gamma_0) = 0, then we differentiate a few times to find that \mathbf e_1\cdot\mathbf C = \mathbf ... 1 Integrate wrt to x to get that$$\frac{\partial u}{\partial y} = z(y)$$Integrate wrt to x$$u = \int z(y)dy + g(x) = f(y) + g(x)$$for two functions f,g. Now add the boundary conditions to find$$u(x, x^3) = f(x^3 ) + g(x) = \sin x^6$$\frac{\partial u}{\partial x} (x,x^3) = g'(x) = 0 \implies g(x) = B \in \mathbb R hence f(x^3) = \sin x^6 ... 1 I think I know how to make this clear. Consider for example the function of one variable g(x)=\begin{cases}1&x>0\\-1&x<0\end{cases} How would you show that \lim\limits_ {x\to 0}g(x) does not exist? Well, you say that \underbrace{\lim\limits_{x\downarrow 0}g(x)=1}_{\text{the limit from the right}} and ... 1 Either you note that f(z)=e^z and know (or show) that e^z takes all complex values except zero, or you solve explicitly the equations e^x\cos y=a and e^x\sin y=b: These equations give e^x=\sqrt{a^2+b^2} and since the real exponential e^x never vanishes you get a\ne0 or b\ne0. In particular, x=\log\sqrt{a^2+b^2}. If a=0, then you can ... 1 The mistake you seem to be making is that you misunderstood what \Sigma is. It is the cilinder without the two 'caps' at the ends (it is what you would get is you took a piece of A4 paper and folded two sides together). This means that the boundary of \partial \Omega are the two circles at the ends of the cilinder. While checking the rest of your ... 1 HINT: multiplying the first plane by -2 we get the same plane, these two planes are the same 1 I hope I have a possible solution. Le us assume that, for all the components J_i of \boldsymbol{J}, there is a function f_i\in C^4(\mathring{A}) such that \nabla^2 f_i=J_i, with \bar{V}\subset \mathring{A} and \boldsymbol{x}\in \mathring{V} and there exists a \delta such that, for all \epsilon\le \delta, \epsilon >0, the region ... 1 Sketch/Hint: Let me show you one "follow your nose" way to tinker until you reach a conclusion. We want to show that |F(x_k,t_k) - F(x,t)| \to 0 as (x_k,t_k) \to (x,t) in \mathbb R^n \times \mathbb R. Since we know that |f(x_k,s)-f(x,s)| \to 0 as k \to \infty (why?) and f(x,s) is the integrand inside the integral definition F(x,t), we can try ... 1 A permutation matrix is a 0, 1 square matrix with a single 1 in each row and in each column. Clearly, all the row sums and all the column sums of a permutation matrix are 1. Now if P_1,P_2,\dots,P_r are permutation matrices, and c_1,c_2,\dots,c_r are any real numbers, then$$c_1P_1+c_2P_2+\cdots+c_rP_r$$has the property that every row sum and every ... 1 For the first equation, complete the square in z... that is, convert the equation to the form x^2 + y^2 + (z - q)^2 = r^2. That will give you a sphere centered at \{0,0,q\} of radius r. You must express q and r in terms of your a. For the second equation, note that the region is rotationally symmetric about the z axis because the x-y ... 1 A change to polar coordinates will suffice:$$ \lim_{r\to 0}\frac{ e^{r(\sin\theta+\cos\theta)} - r(\sin\theta+\cos\theta)}{r}=\lim_{r\to 0}[\frac{ e^{r(\sin\theta+\cos\theta)}}{r} - (\sin\theta+\cos\theta)]  Which does not exist since $\lim_{r\to 0}\frac{ e^{r(\sin\theta+\cos\theta)}}{r}$ does not exist for any $\theta$.

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