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- To put it another way, a curve is described as a collection of points that resemble a straight line that passes through two adjacent locations. We are aware that the straight line has zero curvature. Therefore, we can refer to a line as being curved if its curvature is greater than zero.
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May 4, 2014 · $\begingroup$ In two dimensions, curvature can have a sign. In three dimensions, it can become $0$, and of course you can start out with a prescribed curvature function with isolated zeros. The assumption just states that that case is not covered by the theorem.
Problem: Show that if $\kappa(s) = 0$ for all $s$, then the curve $\mathbf{r} = \mathbf{r}(s)$ is a straight line. (Here, $\kappa$ represents curvature.) Attempt at solution: If $\kappa(s) = 0$ for all $s$, then \begin{align*} \frac{d\hat{T}}{ds} = \kappa \hat{N} = 0, \end{align*} where $\hat{N}$ is the unit principal normal. This means ...
How can I show that a given (not necessarily unit-speed) parametrization $\gamma(t)$ of a curve in $\mathbb{R}^3$ which exhibits zero curvature is a line ? What I know is that zero curvature means...
The curvature at a point of a differentiable curve is the curvature of its osculating circle — that is, the circle that best approximates the curve near this point. The curvature of a straight line is zero.
Aug 18, 2023 · Mathematically, the curvature formula (k) of curve r (t) = (x (t), y (t)) at a point (x0, y0) is defined as: k = (x' (t) y” (t) – y' (t) x” (t)) / ( x ′ (t) 2 + y ′ (t) 2) 3 / 2) In this formula, x' (t) and y' (t) represent the first derivatives of x and y with respect to t, and x” (t) and y” (t) represent their second derivatives.
Feb 27, 2022 · In Question 1.2.1.5 of Section 1.2, we found that the spiral \[ \vecs{r} (t) = e^t (\cos t, \sin t) \nonumber \] parametrized in terms of arclength is \[ \textbf{R}(s)=\frac{s}{\sqrt{2}}\left(\cos\Big(\ln\Big(\frac{s}{\sqrt{2}}\Big)\Big)\,,\, \sin\Big(\ln\Big(\frac{s}{\sqrt{2}}\Big)\Big)\right). \nonumber \]
Between these two cases is the case of zero curvature. In this case the surface has a line along which the surface agrees with the tangent plane. For instance, a cylinder has zero curvature, as suggested in Figure 7.1.2(c).
