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.

# Tag: tangent

## 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.

$$f(a.b)=a^2+ab+b^2=1$$

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:

- Consider a higher dimensional vector H space where M is embedded.
- Look at all smooth curves in M that pass through p.
- Compute the velocity vector of the curve at p. These vectors will lie in higher dimensional vector space H.
- 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 $p$, $mathcal{D}_p(mathbb{R}^n)$:

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

$$

and

$$

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.

## 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}

0&1\

1&0

end{pmatrix}, quad

i sigma_2 =i sigma_mathrm{y} =

i begin{pmatrix}

0& -i \

i&0

end{pmatrix}, quad

i sigma_3 = isigma_mathrm{z} =

i begin{pmatrix}

1&0\

0&-1

end{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},

$$

and

$$

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?