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推导过程:
$$x=a\cos(t),y=b\sin(t)$$
$$\dot x=-a\sin(t),\dot y=b\cos(t)$$
$$\ddot x=-a\cos(t),\ddot y=-b\sin(t)$$
$$\kappa=\frac{\dot x\ddot y-\ddot x\dot y}{(\dot x^2+\dot y^2)^{3/2}}=\frac{ab}{(a^2\sin^2(t)+b^2\cos^2(t))^{3/2}}$$
$$r=\frac1\kappa=\frac{\left(\left(a\sin t\right)^{2}+\left(b\cos t\right)^{2}\right)^{3/2}}{ab}$$
法向量$=(-\dot y,\dot x)=(-b\cos(t),-a\sin(t))$
单位法向量$=\frac1{(a^2\sin^2(t)+b^2\cos^2(t))^{1/2}}(-b\cos(t),-a\sin(t))$
曲率中心$=(a\cos(t),b\sin(t))+r\cdot\text{单位法向量}=$
$$=(a\cos(t),b\sin(t))+\frac{\left(\left(a\sin t\right)^{2}+\left(b\cos t\right)^{2}\right)^{3/2}}{ab}\frac1{(a^2\sin^2(t)+b^2\cos^2(t))^{1/2}}(-b\cos(t),-a\sin(t))$$
$$=(a\cos(t),b\sin(t))+\frac{\left(a\sin t\right)^{2}+\left(b\cos t\right)^{2}}{ab}(-b\cos(t),-a\sin(t))$$
$$=\left(\frac{\left(a^2-b^2\right) \cos ^3(t)}{a},-\frac{\left(a^2-b^2\right) \sin ^3(t)}{b}\right)$$ |
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