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We are given a curve in ${\mathbb R}^3$: $$\gamma: \quad s\mapsto{\bf x}(s)\ ,$$ parametrized with respect to arc length $s$. Assume that $\ddot {\bf x}(s)\ne{\bf 0}$. Along this curve the so-called Frenet frame, a moving orthonormal frame, is defined as follows: Begin with $$T=T(s):=\dot{\bf x}(s)\ .$$ This is a unit vector for all $s$. Therefore $\dot ... 2 Since$B = T \times N$, differentiate with respect to$s$, obtaining $$\frac{dB}{ds} = -\tau N = \frac{dT}{ds} \times N + T \times \frac{dN}{ds} = T \times \frac{dN}{ds}$$ since$dT/ds= \kappa N$and$N \times N = 0$. There are some conclusions to be drawn from this:$dN/ds$is not in the direction of$T$, else this cross product would be null.$dN/ds$is ... 2 How about this, a way to make Zorn's Lemma work. Let$\mathcal{K}$be the collection of all nonempty closed subsets$K \subset [a,b]$with the following properties: The first point$a_K$of$K$satisfies$\phi(a)=\phi(a_K)$. The last point$b_K$of$K$satisfies$\phi(b)=\phi(b_K)$. For each interior component$(c,d)$of$[a,b]-K$we have ... 1$\frac{dN}{ds}$is orthogonal to$N$by differentiating the identity$N\cdot N=1$. So,$\frac{dN}{ds}$is a combination of$T$and$B$, i.e.$\frac{dN}{ds}=x\,T+y\,B$. Comparing the "known identity"$T\cdot \frac{dN}{ds}=-\kappa$with $$T\cdot \frac{dN}{ds}=x\,(T\cdot T)+y\,(T\cdot B)=x$$ ($T$is perpendicular to$B$by the definition of$B$) you get ... 1 If you know that the two curves you have resulted from splitting a single initial curve, then you don't need both these curves; knowing either and the cut position is enough to restore the full initial curve. Have a look at this post on Stack Overflow. It discusses how you cut a curve. What you want to do is do the reverse. To find control points$Q_i$for ... 1 You ask: At this point a dog starts running toward the woman from (0,0) they are both running at constant speed, the dogs path is curved and we wish to find the length of the curve until the dog reaches the woman. Path length is given by $$s = \int_0^T ds = \int_0^T \frac{ds}{dt} dt = \int_0^T v(t) dt = \Big[ v(t) t \Big]_0^T - \int_0^T v'(t) t dt = ... 1 Let's make sure we understand splitting, before we talk about joining. To subdivide a Bézier curve into two, you use the deCasteljau algorithm, as illustrated in the figure below. Suppose we are given a curve defined by four control points A, P, Q, G, and a splitting a parameter value u \in [0,1] (which you called t_{\text{cut}}). To split ... 1 The definitions are not equivalent. Yes, if you impose the condition that \gamma'(t)\ne 0 for all t, you'll have a non-zero tangent vector, and hence a tangent line, at each point. For differential geometry, one wants this condition—so that one can reparametrize by arclength, for example. If you consider \gamma(t)=(t^2,t^3), you see that the curve is ... 1 The curves are definitely not ellipses. I'll just discuss the bottom corner, a (which I'll call F). Coordinatize, with A = (-p,0), B = (p, 0 ), C = (c, h ). Let the "lower" trisectors from A and B meet at the "bottom" Morley vertex, F = (x,y). Then$$\tan \angle CAB = \frac{h}{p+c} \qquad \tan\angle CBA = \frac{h}{p-c}\tan \angle FAB ... 1 Hint: the matrix with rows$(x,y,z)$,$(1,1,0)$, and$(1,2,1)$would have determinant$0$, since the three position vectors, from the origin to each of these three points, are coplanar. 1 If you have a matrix$M$composed of vectors$v_1, v_2,... v_n$: $$M = \begin{bmatrix} \uparrow & \uparrow & ... & \uparrow \\ v_1 & v_2 & ... & v_n\\ \downarrow & \downarrow & ... & \downarrow\end{bmatrix}$$ Then one consequence of$M$having a determinant of zero is that the vectors$v_1, v_2,... v_n\$ are not linearly ...