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6

This is a nice question, and I don't have a comprehensive answer for you, but perhaps this will help a bit. For conservative forces, it's generally convenient to define the scalar potential $V$. The force will satisfy $\vec{F} = \vec{\nabla}V$, and your potential $V = r$ (yay, system is conservative). Since the kinetic + potential energy is constant, and ...


4

No, this is not true. Without getting into great details of construction, here is the idea: We will have a chain of circles, of ever decreasing sizes, such that the circles "converge" to the black point. We will cut small arcs in these circles, and we want them to be small enough so that n-th circle has vertices of n+2-gon. Now we use these cuts to ...


3

Consider an orbit $$\gamma: \quad t\mapsto z(t)=r(t)e^{i\phi(t)}$$ of such a particle. We are interested in the polar representation $$\phi\mapsto r(\phi)\tag{1}$$ of the resulting curve $\hat\gamma\subset {\mathbb C}$. Denoting the differentiation with respect to $t$ by a $\cdot$ and the differentiation with respect to $\phi$ by a $'$ we have $\dot ...


3

In Macaulay2 R=QQ[s,t,x,y,z,MonomialOrder=>Eliminate 2] I=ideal(x-2*s*t*(3*t^4+50*t^2*s^2-33*s^4),y-2*(7*t^6-60*t^4*s^2+15*t^2*s^4+2*s^6),z-(t^2+s^2)^3) gens gb I yields a term free of $s, t$: $625x^6+1875x^4y^2+1875x^2y^4+625y^6-182250x^4yz+364500x^2y^3z-36450y^5z+585816x^4z^2+1171632x^2y^2z^2+585816y^4z^2-41620992x^2z^4-41620992y^2z^4+550731776z^6$ ...


2

The answer is no. For $r, \epsilon_0, \epsilon_1>0$, let $C_{r, \epsilon_0, \epsilon_1}$ be a circle of radius $r$ with two disjoint arcs of lengths $\epsilon_0$ and $\epsilon_1$ deleted. Note that for any $n$, there is a $q_n>0$ such that for all $r, \epsilon_0, \epsilon_1>0$, if ${\epsilon_0\over r}, {\epsilon_1\over r}< q_n$, then $C_{r, ...


1

I started preparing this answer before Narasimham made his edits, wherein the original independent variable name $t$ was replaced by $s$. Since I was nearly done when his edits occurred, I chose to respect the original formulation in my answer; nevertheless, the relationship 'twixt $s$ and $t$ should be clear in what follows. A unit-speed circle in $\Bbb ...


1

Let $A$ denote the area of one of the five triangular "points" and $B$ the area of the central pentagon. Your Green's theorem integral double-counts $B$, so its value is $5A + 2B$, whereas you want to find $5A + B$. Let $t_{0}$ denote the first root of $x(t)$ larger than $\pi$, i.e., the time at which the curve first crosses the negative $y$-axis after ...


1

Using integral of squared second derivatives allows the functional to be minimized to become a quadratic form of the unknowns (i.e., control points of the spline), which eventually will result in a linear equation set, which is easy to solve. This is actually similar to the famous "least square" method which minimizes the sum of the squared errors, instead ...


1

Let $K$ be a compact convex subset of $\mathbb R^2$. If $K$ has no interior points that $K=\partial K$ is just a line segment (or a single point). Assume henceforth that $K$ has an interioir point $a$. Then $\partial K$ is a closed curve, that is there exists a continuous surjective map $\gamma\colon S^1\to \partial K$: By convexity of the intersection of ...



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