# Tag Info

0

I think what I was missing is a dot product, $$\frac{d f(\alpha X)}{d \alpha}=\langle G, X \rangle$$.

1

The chain rule works. Let $\phi(\alpha) = \alpha X$, then $(D (f \circ \phi) (\alpha))(h) = Df(\phi(\alpha)) (D \phi(\alpha)(h))$. Since $\phi$ is linear, $D \phi(\alpha)(h) = \phi(h) = h X$, and so $(D (f \circ \phi) (\alpha))(h) = h Df(\alpha X) (X)$. (Keep in mind that $Df(\alpha X): \mathbb{R}^{m \times n } \to \mathbb{R}$, so $Df(\alpha X) (X)$ is ...

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Why do you need the closest point to $(0,0)$? It shouldn't be difficult to modify the drawing code to handle negative widths. Alternatively, assuming that all of the coordinates are positive, you could do something like if(x1 > x2) std::swap(x1,x2); if(y1 > y2) std::swap(y1,y2); and then call your old code.

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Note that $$r= \sin(\theta) \implies x^2+y^2 = y \implies x^2 + (y-1/2)^2 = 1/4$$ $\theta$ from $\pi/2$ to $\pi$ is the left half of the circle centered at $(0,1/2)$ with radius $1/2$. I trust you can figure out the limits for $x$ and $y$ from here.

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Note that, the cylinder is bounded below by $y=0$ and above by the plane $y=3-x$. The region in the $xz$-plane is an ellipse given by $x^+3z^2=9$ with a major axis on the $x$-axis, then, the volume is given by $$V = \int_{-\sqrt{3}}^{\sqrt{3}}\int_{-\sqrt{9-3z^2}}^{\sqrt{9-3z^2}}\int_{0}^{3-x}dydxdz=4 ... 1 You've forgotten about the chain rule in calculating the Jacobian. For instance:$$\frac{\partial r}{\partial x} = \frac{1}{\cos\theta} + \frac{x\sin\theta}{\cos^2\!\theta}\frac{\partial \theta}{\partial x} = \frac{1}{\cos\theta}\left[ 1 + \frac{\partial}{\partial x}\left(\arccos\left(\frac{x}{\sqrt{x^2+y^2}}\right)\right)\right].$$However, I'd highly ... 4 Your definitions of r at the beginning are getting you into trouble. Bad things happen at \sin \theta = 0 or \cos \theta = 0. Try r = \sqrt{x^2+y^2}. Then$$\frac{\partial{r}}{\partial{x}} = \frac{x}{r}; \frac{\partial{r}}{\partial{y}} = \frac{y}{r}.$$Then, using Fantini's \theta = \tan ^{-1}(\frac{y}{x}), we get$$\frac{\partial ...

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If you want to do this using quadratic form: HINT: $x^TAx = \begin{bmatrix}x & y \\ \end{bmatrix} \begin{bmatrix} 16 & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{3}{16} \\ \end{bmatrix} \begin{bmatrix} x \\ y \\ \end{bmatrix}$ Then diagonalize $A$ and solve for the eigenvalues and eigenvectors.

2

A substitution of the form $u=e^{-at}v$ with $a=c^2/(2D)$ transforms the equation into the telegraph equation $$\frac1{c^2}v_{tt}-v_{xx}=bv$$ with $b=a/(2D)$. The telegraph equation is a much studied equation.

3

The Lagrange multipliers are absolutely perfect for this!

2

Note that $f(0,y,0) = \begin{cases} {1 \over y^2}, & y \neq 0 \\ 0, & \text{otherwise} \end{cases}$.

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The equation of the line can be written as: $$\frac{x-1}{-2}=\frac{y+1}{4}=\frac{z-3}{-2}=t$$ which tells us that one possible point on the desired plan would be $$A:~(-1,+1,-3)$$ and also the leading vector of the line is: $$\vec{w}:=(-2,4,-2)$$ On the other hand, the normal vector of the given plan is $\vec{u}:= (2,2,0)$. Now find the vector $$\vec{n}:= ... 3 And an even simpler method for finding this integral is replacing x with u^2 , y with v^2 and z with w^2. Instead of this :$$\int_{0}^1\int_{0}^{\left(1-\sqrt x\right)^2}\int_{0}^{\left(1-\sqrt x -\sqrt y\right)^2}\,dz\,dy\,dx$$dx = 2u\,du\\dy=2v\,dv\\dz=2z\,dz you will get this : ... 2 Well, by a straightforward computation,$$ \alpha \wedge d\alpha = (\omega - f^{-1}dx^{m+1}) \wedge d (\omega - f^{-1}dx^{m+1})\\ = (\omega - f^{-1}dx^{m+1}) \wedge (d\omega + f^{-2} df \wedge dx^{m+1})\\ = \omega \wedge d \omega + \omega \wedge f^{-2}df \wedge dx^{m+1} - f^{-1}dx^{m+1} \wedge d\omega\\ = \omega \wedge d\omega - f^{-2} (df \wedge \omega ...

1

This is not quite right. The region of integration there is not curved but straight like this: Instead, maybe to Also please look at this question here: Find the volume common to two circular cylinders, each with radius r, if the axes of the cylinders intersect at right angles. (using disk/washer).

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$\langle x, y, 1 \rangle \cdot \langle -2x, -2y, 1 \rangle = -2x^2 - 2y^2 + 1$, and this integrates to $0$ over not $S_1$, but the projection of $S_1$ to the $xy$-plane (which is the unit disk).

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You can check out 1) Vector Calculus, linear algebra and differential forms by Hubbard and Hubbard 2) Advanced Calculus by Gerrald B Folland.

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You want to convey information in higher dimensions on a $2D$ screen. What is generally used for $3D$, is a projection of a two dimensional surface, where $z$ is a function of $x,y$ . For $4D$ color is generally added, so that in addition to height, a color is used to represent the value in the fourth dimension. Other methods used include drawing multiple ...

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(-1,0,1) is the normal vector for the plane x=z (the equation can be written as -1*x + 0*y +1*z =0). You are correct about question 2, you always want the unit normal vector.

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All odd polynomials have at least one real root because of the intermediate value theorem. To prove this just plug in a very large positive number and a very large negative number for $x$ (e.g. $10^{23}$ and $-10^{23})$ and note that corresponding $y$ values will have opposite signs. Then the IVT tells you that there is at least one value of $x$ between the ...

4

If all the roots are real, then it is proven. Otherwise, there is at least one non-real root $z$. But then $\bar{z}$, the conjugate, is also a root (can you show this?). Thus, there are an even number of non-real roots. Since there are three roots in $\mathbb{C}$, it follows that at least one must be real.

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No, $a_i$ is a scalar valued function that takes a vector among its arguments. Just because the input is a vector, the output need not be. But the collection of all $a_i$, $(a_1,\dots,a_n)$, can be considered as a vector-valued function, and often it is convenient to do so. Then the PDE takes the form $$-\operatorname{div} A(x,u(x),\nabla u(x)) + ... 2 No. Defining a(x,u(x),\nabla u(x)) := |\nabla u(x)| takes a vector-valued argument to a real number. I would say$$\frac{\partial}{\partial x_i}\nabla u(x) = (u_{x_1x_i}, u_{x_2x_i}, ..., u_{x_nx_i})$$but the covention used in the book might be different. 2 You really do need to know the definition in order to figure this problem out. That is, the limit definition:$$ \dfrac{\partial f}{\partial x}(0,0) = \lim_{h\to 0} \dfrac{f(h,0) - f(0,0)}{h} = \lim_{h\to 0} \dfrac{\dfrac{h \cdot 0}{h^2+0} - 0}{h} = \lim_{h\to 0}\dfrac{0}{h} = 0 $$and so the partial derivative with respect to x is zero. Similarly,$$ ...

0

If f(x) refers to z, and f'(x) refers to $\frac{\partial z}{\partial x}$, what you did is right. You treat the y's as constants and differentiate with respect to x.

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You answer is correct. When we want to take the partial derivative of a function with respect to one variable we consider all the other variables as a constants. We also avoid to use $f^\prime$ notation because we don't know which variable you are taking derivatives if you use the prime notation. We use either $f_x$ or $\frac{\partial}{\partial x}f$. A ...

1

Yes , He used simple method of derivative and differentiate as usual like df(x)/dx and assuming y as constant k .

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1) Evaluation of $I_{\max}(k,n)$ It is clear that $\max x = x$ so $I_{\max}(1,n)=\int\limits_0^1 x_1^n\,dx_1=\frac{1}{n+1}$. Also it is obvious that $\max\limits_{1\le i\le k}x_i=\max\left(x_k,\max\limits_{1\le i\le k-1}x_i\right)$. Then ...

2

Your surface is defined by a polynomial equation $f(x,y,z) = 0$. You could interpret this situation geometrically in the following way: $f:R^3\to R$ gives a color to every point of the 3D space (imagine you have a color palette indexed by real numbers). Assume white is indexed at $0$, then your surface is the locus of points with white color. The ...

3

Indeed, while $n>3$ and $r\to 0^+$ then $$\frac{r^{3-n}}{3-n}\to\infty$$ And clearly we should keep $n$ not to touch $3$.

0

$$\nabla z=\left\langle\frac{\partial z}{\partial x},\frac{\partial z}{\partial y}\right\rangle^T=\langle12x^2+3y,12y^2+3x\rangle^T$$ At $(-1,1,-3)$, we have $\nabla z=\begin{pmatrix}15\\9\end{pmatrix}$ so we can conclude that the plane is of the form $z=15x+9y+c$. Solving for $c$, we have \begin{align}-3&=15(-1)+9(1)+c\\3&=c\end{align} So the ...

6

Hint: Notice that the numerator factors as follows: $$x^3 - y^3 = (x - y)(x^2 + xy + y^2)$$ Now factor the denominator and cancel the $(x-y)$ factor.

1

Having a non-zero differential is not sufficient to get what you want. If you assume that the differential of $f$ at $0$ has rank $2$, then the local inversion theorem says that the map $f$ is a local diffeomorphism in a neighborhhod of $0$. In particular, there exists a neighborhood $U$ of $(0,0)$ such that if $(x,y) \in U$ with $f(x,y) = f(0,0)$ then ...

0

Let $p=(p_1,\dots,p_n)$ be the "Pole", i.e., center of projection, lying on the unit sphere. The role of the $x_n$-coordinate is now taken by the linear functional $\varphi$ defined by $x\mapsto x\cdot p$. Note that $\varphi(p)=1$. Given a point $x$ on the sphere, rescale the vector $x-p$ so that the $\varphi$ value of the scaled vector is $-1$. Then add ...

1

for your general education, i will instead compute $\frac{d a^x }{dx}$, then you should be able to apple the chain rule to get your result by using the fact $(x^n)' = nx^{n-1}$ : $a \in \mathbb{R}$ $$\frac{d a^x}{dx } = \lim_{\Delta x \to 0} \frac{a^{\Delta x + x} - a^x}{\Delta x} = a^x \bigg( \lim_{\Delta x \to 0} \frac{a^{\Delta x}-1}{\Delta x } \bigg) ... 0 You can also take the logarithm of both sides and compute the derivative. y = 2^\sqrt{t} gives \log(y)= \sqrt{t}\ \log2 Then y' / y = \log{2} / (2 \sqrt{t}) and then y' = 2^\sqrt{t} \log{2} / (2 \sqrt{t}). This only requires to remember what is the derivative of a logarithmic function. 3 Sounds like Chain rule. First, notice that$$ 2^{\sqrt{t}} = \left(e^{\log 2}\right)^{\sqrt{t}} = e^{\log 2\sqrt{t}} $$Then$$ \frac{dy}{dt} = \frac{d}{dt}\left(e^{\log 2\sqrt{t}}\right) = \left(e^{\log 2\sqrt{t}}\right)\frac{d}{dt}\left(\log 2\sqrt{t}\right) = 2^\sqrt{t}\frac{\log 2}{2\sqrt{t}} $$Where \log is the natural logarithm. 0 The differential of f at the given point P is a linear map, respresented by a 1\times 2 matrix, namely f'(P)=\left(\displaystyle\frac{df}{dx}(P),\ \frac{df}{dy}(P)\right). Now, by assumption, df/dx(P)=6, and let b:=df/dy(P) the missing coordinate. The other piece of information tells us that$$\pmatrix{6&b}\cdot\pmatrix{1/\sqrt5 \\ 2/\sqrt5} ...

1

The directional derivative of $f(x,y)$ along some unit vector $\mathbf{v}$ is $\nabla f(x,y) \cdot \mathbf{v}$. Applying this to the given in the problem: $$\frac{1}{\sqrt{5}}\cdot\frac{\partial f}{\partial x} + \frac{2}{\sqrt{5}}\cdot\frac{\partial f}{\partial y}=\frac{16}{\sqrt{5}}$$ Using the value ${\partial f}/{\partial x} = 6$, we get ${\partial ... 1 I assume the function is defined to be$0$at$(0,0)$. Otherwise, it makes not senseto talk about differentiability at this point. I claim the derivative exists and is$\nabla f=\langle 0, 0\rangle$. To prove it, observe that $$\left|\frac{(x^{2}+y^{2})sin\left(\frac{1}{\sqrt{(x^{2}+y^{2})}}\right)-0}{\sqrt{x^{2}+y^{2}}}\right|=\sqrt{(x^{2}+y^{2})}\ ... 1 If k=(0,0,1), then the integral I is actually two-dimensional and over the space \{u_3=0\} and the integrand is \exp(-\alpha(w_1-u_1)^2)\cdot\exp(-\alpha(w_2-u_2)^2). The change of variable$$(u'_1,u'_2)=(u_1-w_1,u_2-w_2) $$shows that I is the square of$$ \int_\mathbb R\mathrm e^{-\alpha v^2}\mathrm dv=\sqrt{\frac{\alpha}\pi}, $$hence$$ ... 1 The Dirac delta distribution is non-zero only when its argument is 0. So integrating over all space is really not precise. Since k is constant, the only contribution to the integral is in the space$\mathbf{k}\cdot \mathbf{u} = 0$which you should be able to prove is a plane orthogonal to k. You should then be able to do a change of variables which lowers ... 0 You cannot solve a second-order differential equation by separating variables as you did. You're supposed to know (or have been shown) that the general solution of$\dfrac{d^2x}{dt^2}+x=0$is$x(t)=c_1 \cos t + c_2\sin t$. An alternative solution fitting multivariable calculus is to note that$G(x,y)=(x,y)\$ is orthogonal to the flow line at the point ...

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