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4

Building on Grumpy's comment, changing the order of integration in $\int_0^1 \int_0^s t^t (1/s) \ dt\ ds$ gives $$\int_0^1 \int_t^1 t^t (1/s)\ ds\ dt = \int_0^1 t^t (-\ln t)\ dt.$$ The claim in the question thus amounts to showing that $\int_0^1 t^t (1+\ln t)\ dt = 0$, which follows readily from the substitution $u = t \ln t$.

4

A coordinate mapping (such as polar coordinates in the plane) is only unique if the implicit function theorem is valid on the region. The implicit function theorem in n-dimensional Euclidean space states that if the Jacobian matrix, i.e. the matrix of partial derivatives, of the coordinatized mappings defining the graph of the function on an open subset of ...

3

There are topological constraints, even in the plane. For instance, an annulus cannot be continuously parametrized by the unit square or unit disk because an annulus is not simply connected. On the other hand, the Riemann mapping theorem says that every non-empty simply connected open subset of the plane can be nicely parametrized by the open disk. Here ...

3

Hint: border of $R$ has 4 parts, $[0,1]\times \{0\}$, $[0,1]\times \{2\}$, $\{0\}\times [0,2]$, $\{1\}\times [0,2]$. Moreover you can solve this without gradient: note that $f$ is decreasing in $x$ and increasing in $y$ in your domain...

2

The boundary is: $\{(x,0): 0\le x\le 1\}$ and $\{(x,1): 0\le x\le 1\}$; $\{(0,y): 0\le y\le 2\}$ and $\{(1,y): 0\le y\le 2\}$

2

Note that the original region of integration $R$ is the trapezoid with vertices $(0,0), (1,0), (1,1), (1/2,1)$. (a) is incorrect because the region of integration is a strict superset of $R$, containing all points in $R$ yet also including the triangle $2 > 2x > y > 1$. (b) is notationally incorrect because the inner integral is evaluated with ...

2

No, you cannot. Let $f(x,y)=xy$ and $g(x,y)=x^2y^2$. Then $f(1,1)=g(1,1)$, but $\frac{\partial f}{\partial x}(1,1)=1$ and $\frac{\partial g}{\partial x}(1,1)=2$ Addendum: Letting $f(x,y)=x(y-1)$, $g(x,y)=x^2(y-1)^2$, and $(a,b)=(1,2)$ gives a counterexample where $a\neq b$. In fact, the property you want is usually false; the values $f(a,b)$ and $g(a,b)$ ...

2

Firstly, $\int_0^{\beta}dx \int_0^{\beta}dy \, f\left( |x-y| \right)$ is the integral of $g(x,y) = f(|x-y|)$ over $[0,\beta]^2$. Since $g(x,y) = g(y,x)$, we have $$\int_0^{\beta}dx \int_0^{\beta}dy \, f\left( |x-y| \right) = 2 \int_0^{\beta}dx \int_0^{x}dy \, f\left( x-y \right)$$ i.e. the integral of $g(x,y)$ over $[0,\beta]^2$ is equal to $2$ times is ...

2

Your bounds are correct. Your integral in spherical coordinates becomes... $$= \int_0^{2\pi} \int_0^{\pi/2} \int_0^{\cos(\phi)} \rho \cdot \rho^2 \sin(\phi) \,d\rho \,d\phi \,d\theta$$ Since nothing involves $\theta$, that integral can be factored out. $$= \int_0^{2\pi}\,d\theta \int_0^{\pi/2} \int_0^{\cos(\phi)} \rho^3 \sin(\phi) \,d\rho \,d\phi = ... 1 This function g(x,y) is not one-to-one, and so has no inverse. Each of the points (1,1) and (a,b) map to the same ordered pair (c,d) where (approximately) a=2.61229...,\ b=0.039772... and the output ordered pair (c,d) is about c=14.77811, \ d=2.71828... So this means one cannot find x,y uniquely for this particular pair (c,d) so there ... 1 In the multivariable setting, f' is a linear transformation, i.e. a matrix. They mean that this matrix is invertible, i.e. has nonzero determinant. 1 I'm going to suggest that you restrict your attention to the partial derivatives, i.e., how \ln(f(x_1, \ldots, x_n)) varies when you vary x_i. For this, you can think of this as a one-variable calculus problem, and the result is$$ \frac{d\ln(f(\mathbf x))}{dx_i} = \frac{1}{f(\mathbf x)} \frac{df(\mathbf x)}{dx_i}. $$By long tradition, this derivative ... 1 Hint: Let a be a point on the plane, for convenience let's say a=\begin{pmatrix}24\\0\\0\end{pmatrix}. Now, b and c must be vectors that are normal to \begin{pmatrix}2\\3\\4\end{pmatrix}, that is for which \overrightarrow{b}\cdot\begin{pmatrix}2\\3\\4\end{pmatrix}=0 and \overrightarrow{c}\cdot\begin{pmatrix}2\\3\\4\end{pmatrix}=0, where ... 1 Hint In dimension two we have (a,b)\perp (-b,a). This idea can be translated to 3D as follows:$$(a,b,c)\perp (-b,a,0); (a,b,c)\perp (-c,0,a); (a,b,c)\perp (0,-c,b).$$Can you get from this two vectors perpendicular to the normal vector of the plane (2,3,4) satisfying the required conditions? 1 the region of integration is simply$$ \{(x,y,z) : 0\le x\le z, 0\le y\le z, 0\le z\le 1 \} $$Now you can rewrite the integrals in the following ways:$$ \int_{x}\int_y\int_z\\ \int_{x}\int_z\int_y\\ \int_{y}\int_x\int_z\\ \int_{y}\int_z\int_x\\ \int_{z}\int_y\int_x\\ \int_{z}\int_x\int_y\\ $$the step is everytime the same: for the most outer integral, ... 1 You can apply the derivative on identities only. For example, we know that$$(x-1)^2 = x^2 - 2x + 1 $$right? Taking derivatives, we get$$2(x-1) = 2x - 2,$$which is true. Now, this serves as an explanation for why, in general, we can't use derivatives to solve equations. Suppose I want to solve the equation$$\frac{x^3}{3} - \frac{5x^2}{2} + 6x = 0 $$... 1$$ f(x,y,z)=xyz-1. $$Then$$ (f_x,f_y,f_z)=(yz,zx,xy)=\left(\frac{1}{x},\frac{1}{y},\frac{1}{z}\right), $$is perpendicular to the surface xyz=1. 1 Let us try the change of variables$$(x, \delta) = (x,x-y)$$so that$$ \int g(x,y) dxdy = 2 \int g(x, x-\delta) dx d\delta $$Application:$$ \frac1\beta\int_{[0,\beta]^2} f(|x-y| ) dxdy = \frac1\beta\int_{[0,\beta]} dx \int_{x-\beta}^x d\delta f(|\delta| )\\ = \frac1\beta\int_{[-\beta,\beta]} d\delta \int_{\max(0,\delta)}^ {\min(\beta + \delta, ...

1

For a function on 3-space, it's a level-surface rather than a level curve, but you're right, it's at level 4. The other fact that you need to know is that hte gradient of your function is perpendicular to the level surface. Since $$\nabla f (x, y, z) = ( (1+4y)/(5z^2 + 3), 2/(5z^2+3), -(x+2y+4xy)/(5z^2+3)^2)$$ we get $$\nabla f (6, -1, 1) = ( (-3)/8, ... 1 As you showed, the vector field is conservative, so it doesn't matter which path you take, the only thing you need are the starting and end point. First, as \mathrm{F} is conservative, you have to calculate a function f such that \nabla f=\mathrm{F}. An easy way to do this is using this formula:$$\displaystyle f(x,y) = ...

1

Existence of partial derivative does not guarantee the existence of derivative. But if each partial derivative is continuous at point $(a,b)$ then it has derivative at $(a,b)$. Unfortunately, partial derivative of $f$ respect to $x$ is not continuous, since $$f_x(x,y)=\begin{cases} \frac{4}{3}x^{1/3}\sin(y/x)-x^{-2/3}y\cos(y/x)&\text{if }x\neq 0\\ ... 1 Let's connect your notation to geography: \theta is latitude with 0 being North Pole, \pi/2 the equator, \pi is South Pole \phi is the longitude, perhaps 0\le \phi\le 2\pi with 0 being the Greenwich meridian. (This does not really matter, all meridians look the same on the globe). Unlike meridians, the circles of constant latitude do not ... 1 First, one computes the "coordinate basis vectors." The coordinate basis vectors are not necessarily unit vectors. Rather, given a function \mathbf r(r, \theta, \phi) that maps the coordinates to positions, one obtains the basis vectors as follows:$$\mathbf e_r = \frac{\partial \mathbf r}{\partial r}, \quad \mathbf e_\theta = \frac{\partial \mathbf ...

1

Your result from directly differentiating is incorrect. $y$ is a function of $t$, which is equal to $x$. So $y$ is a function of $x$. Therefore, the derivative of $F=x+y$ with respect ot $x$ is the result from the chain rule. In this case, you could write $F$ as $F=x + \cos(x)$, which may make things more clear.

1

The claim is true. Assume $F$ is not continuous, so without loss of generality, it is discontinuous at $0$, and $f(0)=0$. Thus there is some $\epsilon>0$ such that for each $\delta>0$ there is some $x\in B(0,\delta)\subset\mathbb{R}^2$ with $|F(x)|>\epsilon.$ For any natural $n$ let $x_n\in B(0,1/2^n)$ such that $|F(x)|>\epsilon$. Let $c$ be the ...

1

Here's another way. Define the following variables: \begin{align} \tilde x&=x\\ \tilde y&=y \cos\alpha -z \sin\alpha\\ \tilde z&=-y\sin\alpha +z \cos\alpha \end{align} with $\alpha=\tan^{-1} (c)$. With these variables, the curve is of the form $$(\tilde x, \tilde y, \tilde z)= \left(\cos t,\frac{\sin t}{\cos\alpha},0\right)= \left(\cos ... 1 As Paul said, the boundary consists of four line segments:$$ L_1: x=0,2\ge y\ge 0  L_2: x=1,2\ge y\ge 0  L_3: y=0,1\ge x\ge 0  L_4: y=2,1\ge x\ge 0 $$Let's look at L_1, the left border. To find the maxima and minima of f on L_1, you use the equation x=0 to think of f(x,y) as a single variable function, f(0,y). To find the ... 1 Since the line is perpendicular to the plane, the normal vector \langle 1,3,1\rangle for the plane is parallel to the line, so the line has vector equation \langle x,y,z \rangle=\langle 1,0,6\rangle + t\langle 1,3,1\rangle and parametric equations x=1+t, \;\;y=3t, \;\;z=6+t. 1 x^4+6y^2-4xy^3=x^4+2y^2(3-2xy) is positive for 0<|xy|<1.5, so f(x,y) has the minimum -1 for x=y=0 1 Hint: Assume that x=\pm\epsilon and y=\pm \delta, where \epsilon, \delta close to 0 and assume without loss of generality that \delta < \epsilon. Then$$x^4+6y^2-4xy^3 \ge \epsilon^4+6\delta^2-4\epsilon\delta^3 Now for appropriate choice of $\epsilon$ and $\delta$ it is not difficult to see that the term ...

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