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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 ...


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 ...


2

The normal to the inner edge is $$ \frac1{\sqrt{\frac{x^2}{a^4}+\frac{y^2}{b^4}}}\pmatrix{\frac x{a^2}\\\frac y{b^2}}\;. $$ Thus the parametric form of the outer edge is $$ \pmatrix{a\cos\phi\\b\sin\phi}+\frac ...


2

Because turning around to make the trip back corresponds to applying a reflection. A reflection is an orientation inverting operation, so left interchanges with right.


1

Let $$\gamma:(0,1)\to\Bbb R^2,$$ be a smooth parametrized curve such that $\gamma'(t)\neq 0$ for all $t$ (you don't stop during $(0,1)$). Then, for each $t$ there are two unit vectors $N(t)$ orthogonal to $\gamma'(t)$, only one of which makes $\{\gamma'(t),N(t)\}$ an oriented basis for $\Bbb R^2$. This choice of $N(t)$ gives a smooth vector field along ...


1

The solution actually says: From the hypothesis that $k(s) \geq 1$ for all $s$, we know that the simple closed curve $\alpha$ is an oval. I think the mistake is thinking that an oval is a synonym for ellipse. It is actually more general. In the book Modern Differential Geometry of Curves and Surfaces with Mathematica you can see in Definition $6.25$ ...


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

Most of your paper is correct, at the end you would have $$ \int_{-\pi/3}^{\pi/3} \sqrt{\frac{2}{\cos(t)+1}} dt. $$ Now use $$ \cos(2x) = \cos^2(x) - \sin^2(x) = 2 \cos^2(x) - 1, $$ so $$ \cos(2x) + 1 = 2 \cos^2(x), $$ so you can write $$ \int_{-\pi/3}^{\pi/3} \frac{1}{\cos(t/2)} dt, $$ which is easier to solve... $$ \int_{-\pi/3}^{\pi/3} ...



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