How to find a point on a curve at which the tangent vector is parallel to the radius vector?

I am confused as to how to find a point on any given curve at which the tangent vector is parallel to the radius vector. For example, the curve is represented as r(t) = (1-8t)i+t^2j+2e^(2t) k. A detailed and comprehensive solution would be great. Thank you in advance for your time and efforts.

filling – How to fill a lot of circles in a circle with a diameter of 60mm, and all the inner circles are tangent

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dg.differential geometry – Orthogonality in Wasserstein tangent space for discrete measures with equal mass

Let say I have $N$ discrete probability measures $(mu_1,…,mu_N)$ where each of them has $n$ points in $mathbb{R}^2$ of equal mass.

Let $P(mathcal{X})$ be the space of these probability measures on a Hilbert space $mathcal{X}$ (which is $mathbb{R}^2$ in our case if I’m not mistaken).

We can define the Wasserstein barycenter:

$overline{mu} = argmin_{nu in P(mathcal{X})} sum_i^N alpha_i W_2^2(mu_i, nu)$.

Under the 2-Wasserstein metric $W_2$ and with $alpha_i = frac{1}{N},_{i=1..N}$ (in our case).

I’m interested to grap the intuition with the tangent space at $overline{mu}$. Especially in the notion of orthogonality in this tangent space.

Let say I have a vector (or velocity) field $v_1 in L^2(overline{mu}, mathcal{X})$ (the tangent space at $overline{mu}$), containing $n$ vectors, that move all the $n$ masses of $overline{mu}$ in a direction $v_1^i,_{i=1..n}$ respectively.

What would it mean to have a velocity field $v_2 in L^2(overline{mu}, mathcal{X})$ orthogonal to $v_1$ in the $L^2(overline{mu}, mathcal{X})$ sense? It is just that we need to have $v_2^i perp v_1^i,_{i=1..n}$ in the $mathbb{R}^2$ sense?

I’m interested in any reference that could help (especially because I could need to cite something) and any explications.

Thank you very much.

Find tangent line going through a point outside the curve in multivariable calculus

Determine the equations for all lines tangent to the ellipse

$$ f(x,y)=x^2+xy+y^2=1 $$

and passing through the point (0,2)

I’m a bit stuck on this question.
I was thinking if I have an unknown point on the curve called (a.b). Then I can plug it in the curve.
And if I also find the derivative of it, which is:
$$(2a+b,a+2b) $$
But from here I really don’t know how to proceed. Maybe if I can somehow set it up as simultaneous equations and find the unknown point, from there I can easily find the tangent line.

at.algebraic topology – $pi_{2n-1}(operatorname{SO}(2n))$ element represents the tangent bundle $TS^{2n}$, not torsion and indivisible?

Question: Is the element $alpha$ in $pi_{2n-1}(operatorname{SO}(2n))$ representing the tangent bundle $TS^{2n}$ of the sphere $S^{2n}$ indivisible and not torsion?

My understanding so far —

An $operatorname{SO}(2n)$ bundle over $S^{2n}$ corresponds to an element in $pi_{2n}operatorname{BSO}(2n) =pi_{2n-1}operatorname{SO}(2n)$.

Not torsion: There does not exist any integer $m > 0$ such that $malpha$ is a trivial element.

Indivisible: There does not exist any integer $k > 1$ and any element $beta$ in $pi_{2n-1}operatorname{SO}(2n)$ such that $alpha=kbeta$.

Ref: Mimura, Toda: Topology of Lie groups. Chapter IV Corollary 6.14.

manifolds – An intuitive explanation of Tangent space

Given a manifold M, I imagine the tangent space at p, $T_p (M)$ as follows:

  1. Consider a higher dimensional vector H space where M is embedded.
  2. Look at all smooth curves in M that pass through p.
  3. Compute the velocity vector of the curve at p. These vectors will lie in higher dimensional vector space H.
  4. Together these vectors form the tangent space.

The formal definition however is as follows:

For any given smooth curve $lambda$ in M passing through p, define $v_{p, lambda}$ as a linear map that maps smooth functionals on M to a real number as follows:
$$v_{p, lambda}(f) = (fcirc lambda)'(0)$$
where $lambda(0) = p$.
The set of all such linear maps form the tangent space.

How do I reconcile the intuitive picture with the actual definition?

multivariable calculus – Understanding the isomorphism between tangent spaces and derivations

I have a problem with the last line of the following proof that shows that tangent space $T_p(mathbb{R}^n)$ is isomorphic to the space of derivations at point $p$, $mathcal{D}_p(mathbb{R}^n)$:
enter image description here
enter image description here

I fail to understand the last line which shows surjectivity. How does it follow from $D = D_v$, with $ v=langle Dx^1, Dx^2, ldots, Dx^nrangle$

differential geometry – Concrete example of a tangent (space) vector velocity on a sphere (S^2)

The text I am using states without proof that the tangent vector representing the velocity field due to a rigid rotation about the x-axis is:
V= -sin{phi}, partial_theta -cot{theta} cos{phi} , partial_phi

where $theta$ is the standard polar angle (measured from the z-axis) and $phi$ is the standard azimuthal angle.

I tried to get to get this from what I think is the representation of $V$ in Cartesian coordinates
V= V^y partial_y+ V^zpartial_z = -z , partial_y + y ,partial_z

based on the rotation matrix for infinitesimal rotations about the x-axis.
I thought I would then calculate the components of $V$ in spherical polar coordinates via
V^theta = left(frac{partialtheta}{partial y} right)V^y +left(frac{partial theta}{partial z} right)V^z

V^phi = left(frac{partialphi}{partial y} right)V^y +left(frac{partial phi}{partial z} right)V^z

since that is how coordinates are supposed to transform. However, I don’t get the above result.

I think that I have a fundamental misunderstanding here and need a clue as where to start.

dg.differential geometry – Dirac operator on a 5 dimensional tangent manifold with a $Spin(3)$-bundle

In p.3 of Witten paper from this Physics Letters B, Volume 117, Issue 5, 18 November 1982, Pages 324-328 Physics Letters B, 117(5), 324–328, he says that about the Dirac equation on a 5-dimensional
tangent space manifold locally with $O(5)$ tangent bundle and with an extra $Spin(3)=SU(2)$-bundle.

enter image description here

My concerns

  • He said the gamma matrices $gamma^i$ to be real, symmetric 8 X 8 matrices

question 1: why these $gamma^i$ not 4 X 4 matrices for Dirac operator on the 5 dimension space? Note that we only need 4 X 4 matrices, not 8 X 8 matrices?

question 2: why $gamma^i$ are real? why symmetric?

  • He said the anti-hermitian generators $T^a$ of $SU(2)=Spin(3)$ are real, anti-symmetric matrices

question 3: I agree $T^a$ are anti-hermitian. But why $T^a$ are real? thus anti-symmetric? (Of course, if $T^a$ is real and anti-hermitian, then $T^a$ has to be anti-symmetric.)

Naivel, the anti-hermitian generators 2 X 2 matrices of SU(2) are pseudoreal as:
i sigma_1 =i sigma_mathrm{x} =
i begin{pmatrix}
end{pmatrix}, quad
i sigma_2 =i sigma_mathrm{y} =
i begin{pmatrix}
0& -i \
end{pmatrix}, quad
i sigma_3 = isigma_mathrm{z} =
i begin{pmatrix}

calculus – The triangle formed by the $x$-axis, the tangent to $y=1/x$ at a 1st-quadrant point, and the line from the origin to that point is isosceles

Here is Prob. 45, Sec. 3.2, in the book Calculus With Analytic Geometry by George F. Simmons, 2nd edition:

Let $P$ be a point on the first-quadrant part of the curve $y = 1/x$. Show that the triangle determined by the $x$-axis, the tangent at $P$, and the line from $P$ to the origin is isosceles, and find its area.

My Attempt:

Let the point $P$ be given by $P = left( a, frac{1}{a} right)$, where $a > 0$.

Then the line from $P$ to the origin has equation
$$ y = frac{ frac{1}{a} }{a} x = frac{1}{a^2} x. tag{1} $$

The slope of the tangent line to the curve $y = 1/x$ at the point $P$ is
left( frac{d y}{d x} right)_{x = a} = -frac{1}{a^2},

and thus the equation of the tangent line is
y = -frac{1}{a^2} (x-a) + frac{1}{a} = -frac{1}{a^2} x + frac{2}{a}. tag{2}

And, this line intersects the $x$-axis at the point $(x, 0)$, where
-frac{1}{a^2} x + frac{2}{a} = 0,

that is,
x = frac{ frac{2}{a} }{ frac{1}{a^2} } = 2a.

Thus the tangent line to the curve at the point $(a, 1/a)$ intersects the $x$-axis at the point $A$ given by
$$A = (2a, 0). $$

And, the point of intersection of the tangent line to the curve at point $P$ and the line from $P$ to the origin is of course the point $P = (a, 1/a)$ itself.

Thus the three vertices of our triangle are $O = (0, 0)$, $A = (2a, 0)$, and $P = (a, 1/a)$, where $a > 0$. So the sides of our triangle have lengths
left| overline{OA} right| = 2a,

left| overline{AP} right| = sqrt{ (2a-a)^2 + (0-1/a)^2 } = sqrt{ a^2 + 1/a^2},

left| overline{PO} right| = sqrt{ (a-0)^2 + (1/a-0)^2 } = sqrt{a^2 + 1/a^2}.

Thus the sides $AP$ and $PO$ are congruent, showing that our triangle is indeed isosceles.

Finally, since the base of our triangle is the side $OA$ and the altitude is the vertical line segment from the $x$-axis to the point $P$, therefore the area of our triangle is
frac{1}{2} times (2a) times frac{1}{a} = 1.

Is what I have done correct and clear in each and every detail? Or, are there any issues of accuracy or clarity?