New answers tagged vector-analysis
0
votes
Accepted
Equality in curvilinear coordinates
I found the explanation. The left-hand side is equal to $||\nabla v||$ whereas the denominator of the right-hand side is $||\frac{\partial \pmb{r}}{\partial v}||$ with $\pmb{r}$ the position vector.
$\...
- 154
3
votes
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Gradient of a function defined with euclidean norm
Here’s an alternative method. Since
$$
f(x)=g(u(x)).
$$
with
$$
g(t)=te^t,
$$
and
$$
u(x)=\|x\|^2=\sum_{i=1}^n x_i^2,
$$
we have
$$
\frac{\partial f}{\partial x_i}(x)=g’(u(x))\frac{\partial u}{\...
- 15.5k
2
votes
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Using integration by parts to show $\int_{\Sigma} |\nabla^N_{\Sigma} X|^2 = - \int_{\Sigma} \langle X, \Delta^N_{\Sigma} X \rangle$
The normal bundle along the submanifold in the Riemannian background is a special case of the following situation.
Let $M$ be a suitably good manifold of dimension $n$ with a Riemannian metric $g$. It ...
- 10.3k
0
votes
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Calculate angle between two vectors on specific side of the vectors
So, $\vec{p}_0$, $\vec{p}_u$, and $\vec{p}_v$ define a plane.
If we use $\hat{e}_u$ as the unit $u$ axis vector, then
$$\hat{e}_u = \frac{\vec{p}_u - \vec{p}_0}{\left\lVert \vec{p}_u - \vec{p}_0 \...
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Understanding the idea of a pseudo-gradient vector field
Looking at your 2 statements:
(a) min{||DE(u)||,1}<||v(u)||<2∗min{||DE(u)||,1}
(b) cosφ=⟨v(u),DE(u)⟩||v(u)||⋅||DE(u)||>1/2.
We can see that a pseudo gradient is field which is somewhat ...
1
vote
Gradient of a function defined with euclidean norm
Your computation is correct so far. Continuing it, we get
$$
\frac{\partial}{\partial x_i} \sum_{k=1}^n x_k^2 = 2x_i
$$
and
$$
\frac{\partial}{\partial x_i} \exp\Bigl( \sum_{k=1}^n x_k^2 \Bigr) = 2x_i ...
- 678
2
votes
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What does the parentheses mean in vector calculus?
The image that the OP shares with us basically contains the expression $$\int [\vec a(\dot{\vec b}\cdot \vec a+\vec b\cdot\dot{\vec a})+\dot{\vec a}(\vec b\cdot\vec a)-2(\dot{\vec a}\cdot\vec a)\vec b-...
- 2,473
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votes
Derivative of negative log-likelihood function for data following multivariate Gaussian distribution
The log-likelihood is:
$$\ell=\log\mathcal L=-\frac12\left(\log\left|2\pi\Sigma\right|+\mathrm{tr}(\Sigma^{-1}(y-\mu)(y-\mu)^T\right)$$
where $|A|$ is the determinant of the matrix $A$ and $\mathrm{tr}...
- 1,347
1
vote
Accepted
Gradient of matrix and Hadamart product
The componentwise gradient of a matrix with respect to its own elements is given by
$$\eqalign{
\def\p{\partial}
\def\LR#1{\left(#1\right)}
\def\BR#1{\Big(#1\Big)}
\def\grad#1#2{\frac{\p #1}{\p #2}}
\...
- 27.8k
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Derivative of negative log-likelihood function for data following multivariate Gaussian distribution
However, I'm unable to carry out the correct multiplication operation between the two derivatives in the first term on the right hand size.
This is because it is not a matrix multiplication. Note ...
- 10.3k
1
vote
Accepted
Can the "gradient is perpendicular to the level set" property only hold for one particular point?
tl; dr: The answer to 1. is emphatically no.
Let $\phi$ is an arbitrary bounded, real-valued function is a neighborhood of the origin (in the plane, say), and define
$$
f(x, y) = x + (x^{2} + y^{2})\...
- 71k
2
votes
Orthonormal basis given by the integral of a vector field along a curve orthogonal to the tangent
I am sure Ted Shifrin's answer is correct, and here are the missing details on orthogonality.
From his answer, we can see that we can find $s_1,s_2,$ and $s_3$ such that some linear combination of $N(...
- 1,361
3
votes
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Orthonormal basis given by the integral of a vector field along a curve orthogonal to the tangent
This is an unusual result, and I'm surprised the authors dismissed it as if it were standard and well-known. Here's the main idea. You can sort out the final details. Having the supports be very small ...
- 102k
0
votes
Tensor notation proof of Divergence of Curl of a vector field
This answer uses the rules of tensor calculus with both upper and lower indices.
Let us define the divergence of a tensor field $V^i$ by using the covariant derivative $\nabla_jV^i$; where the curl is ...
- 341
2
votes
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What is my error in this $\nabla_{\vec{v}} f(x,y,z)$ at $\vec{a} = (-1, -1, 4)$ and $\vec{v} = (\frac{\sqrt 2}{2}, \frac{1}{2}, \frac{1}{2})$ problem
You need to be careful to differentiate the form $\sqrt{ax}$ when $a$ can be negative (in this case $a=yz$)
Especially $\sqrt{ax}=\sqrt{a}\sqrt{x}$ does not hold when $a$ is negative. I guess this is ...
- 1,226
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votes
Does fundamental theorem of calculus apply to closed curves? If yes why the closed line integral of a function is not zero?
In the first integral, $\oint A \cdot \mathrm d l$, you are integrating over a vector field. In the second integral, $\int_a^b g' \mathrm d x$, you are integrating over a scalar field. The ...
- 1,149
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votes
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Finding parametrisation of curve with the gradient given
You can use implicit function theorem. Since $\nabla f(a)\neq \vec 0$ (regularity assumption), some of the partial defivatives must be nonzero. Without loss of genenerality let's assume that $\partial ...
- 1,779
1
vote
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Verify Stokes' theorem for $F=2yzi-(x+3y-2)j+(x^2+z)k$ where $S\subset \mathbb{R}^3$
Please note that at the intersection of both surfaces, $x^2 + y^2 = x^2 + z^2 \implies y = z$
Surface $x^2 + y^2 = a^2$ should be parametrized as,
$ \phi(z, t) = (a \cos t, a \sin t, z)$ with normal ...
- 50.6k
0
votes
Intuition behind curl identity $ \nabla\times (\nabla\times A)=\nabla (\nabla \cdot A)-\nabla ^2A $
How could one explain this result in intuitive terms?
One heuristic way is to treat the operator $\nabla$ formally as a vector and think about the Lagrange formula for the triple cross product of ...
- 2,385
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Rotating polar velocity vector fields
As mentioned in the link solution, in cartesian coordinates one can write the rotated vector field in the form
$$E_{new}(\pmb v) = R_\alpha(E(R_\alpha^{-1}(\pmb v))) $$
where $\pmb v = (x,y,z)$ and $...
- 938
0
votes
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Rotating polar velocity vector fields
It will turn out that the rotation in polar coordinates is nothing else than a shift
of coordinate functions by an angle $\alpha$. Note that we usually shift a function
$g$ on $\mathbb R$ by $\alpha$ ...
- 4,409
0
votes
Derivative of negative log-likelihood function for data following multivariate Gaussian distribution
Using differential,
\begin{eqnarray}
dL
&=&
\frac{\partial L}{\partial \mathrm{vec}(\mathbf{\Sigma})}:d\mathrm{vec}(\mathbf{\Sigma})+
\frac{\partial L}{\partial \pmb{\mu}}:d\pmb{\mu} \\
&...
- 2,168
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