In a previous article in the vector series, we cleverly derived the radius of curvature for curves (both plane and space) by combining physics and vectors as: R=\frac{v^2}{a_c}=\frac{|\dot{\vec{r}}|^3}{|\dot{\vec{r}}\times \ddot{\vec{r}}|} \tag{1} Curvature is the reciprocal of the radius of curvature: \rho=\frac{1}{R}. Let us think in reverse: are circles the only plane curves with constant curvature?
The answer seems obvious, but we need to prove it.
Since we are only considering the plane case, we first set \dot{\vec{r}}=(v \cos\theta, v \sin\theta)=z=ve^{i\theta}. Substituting this into (1), we obtain: \frac{\dot{\theta}}{v}=\rho \tag{2}
Note: Here we have used the derivative symbol \dot{\theta}, but we have not yet specified which variable the derivative is with respect to. We find that if \dot{\theta}=\frac{d\theta}{dv}, then (2) is very easy to solve. Thus, we agree that a dot above a function denotes the derivative with respect to the variable v. This is a "post-hoc determination method," because the variable of differentiation here is arbitrary, and an appropriate choice makes it more convenient for us to solve.
Thus, the general solution to (2) is: \theta=1/2 \rho v^2+C_1 \tag{2} Returning to the x, y variables, we have: \frac{dz}{dv}=ve^{i\theta}=ve^{i(1/2 \rho v^2+C_1)} \tag{3} Then: \begin{aligned} z &= \int ve^{i\theta} dv = \int ve^{i(1/2 \rho v^2+C_1)} dv \\ &= \frac{1}{\rho} \int e^{i(1/2 \rho v^2+C_1)} d(1/2 \rho v^2+C_1) = \frac{1}{\rho} e^{i(1/2 \rho v^2+C_1)}+C_2 \end{aligned} \tag{4} By choosing an appropriate translation such that C_2=0, we have: z=\frac{1}{\rho} e^{i(1/2 \rho v^2+C_1)} \tag{5} It is not difficult to verify that |z|=\frac{1}{\rho}. Therefore, this is a circle.
Note that "complex numbers" here serve a "formal" role; they are a tool to combine orthogonal coordinate operations into a single entity for calculation, rather than just simple numbers. In the next article, we will explore similar problems for space curves and planes.
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