Radius of curvature of plane curves

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Starting from the definition of curvature, find the length of an arc of 1∘1^{\circ } of the Earth's meridian at the pole and at the equator. The meridian may be taken as an ellipse with semi-axes a=6378a=6378 km and b=a−21b=a-21 km (the polar semi-axis).

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Find the radius of curvature of the curve 3ay2=2x33ay^{2}=2x^{3}.

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Find the radius of curvature of the curve y=x3y=x^{3} at the point (1,1)(1,1).

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Find the radius of curvature of the curve x=t2x=t^{2}, y=t3y=t^{3} at the point (1,1)(1,1).

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Find the radius of curvature of the curve y=ach⁡xαy=a\ch \dfrac {x}{\alpha } (the catenary) at the point (0,a)(0,a).

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Find the radius of curvature of the curve y=aln⁡cos⁡xay=a\ln \cos \dfrac {x}{a}.

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Find the radius of curvature of the curve xn+yn=2anx^{n}+y^{n}=2a^{n} at the point (a,a)(a,a).

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Find the radius of curvature of the curve x=acos⁡4tx=a\cos^{4}t, y=asin⁡4ty=a\sin^{4}t at the point (a,0)(a,0).

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Find the radius of curvature of the curve x=a(cos⁡t+tsin⁡t)x=a(\cos t+t\sin t), y=a(sin⁡t−tcos⁡t)y=a(\sin t-t\cos t).

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Find the radius of curvature of the curve x=a(t−sin⁡t)x=a(t-\sin t), y=a(1−cos⁡t)y=a(1-\cos t).

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Find the radius of curvature of the curve r=a(1+cos⁡φ)r=a(1+\cos \varphi ).

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Find the radius of curvature of the curve r2=a2cos⁡2φr^{2}=a^{2}\cos 2\varphi, at φ=0\varphi =0.

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Find the radius of curvature of the curve

{x=a(t+sin⁡t)+bsin⁡t2,y=a(3+cos⁡t)+bcos⁡t2. \begin{cases} x = a(t+\sin t)+b\sin \dfrac {t}{2}, \\ y = a(3+\cos t)+b\cos \dfrac {t}{2}. \end{cases}
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Find the radius of curvature of the curve

{x=ae−t(cos⁡t−sin⁡t),y=ae−t(cos⁡t+sin⁡t). \begin{cases} x = ae^{-t}(\cos t-\sin t), \\ y = ae^{-t}(\cos t+\sin t). \end{cases}
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Find the radius of curvature of the curve

{x=a(3t+sin⁡tcos⁡t(2cos⁡2t+3)),y=2acos⁡4t. \begin{cases} x = a\bigl(3t+\sin t\cos t(2\cos ^{2}t+3)\bigr), \\ y = 2a\cos ^{4}t. \end{cases}
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Prove that for points of the Archimedean spiral r=aφr=a\varphi, as φ→∞\varphi \rightarrow \infty, the difference between the radius vector and the radius of curvature tends to zero.

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Prove that for the curves rn=ancos⁡nφr^{n}=a^{n}\cos n\varphi, the polar normal is n+1n+1 times the radius of curvature.

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Prove that for the curves rn=ansin⁡nφr^{n}=a^{n}\sin n\varphi, the part of the radius vector lying inside the circle of curvature equals 2rn+1\dfrac {2r}{n+1}.

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Prove the equality R=r drdpR=\dfrac {r\, dr}{dp}, where RR is the radius of curvature, rr is the radius vector, and pp is the perpendicular from the pole to the tangent.

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The coordinates of the points of a curve can be expressed in terms of the arc length of the curve: x=x(s)x=x(s), y=y(s)y=y(s). Denoting by Δs\Delta s a small increment of ss, and by Δl\Delta l the corresponding chord, prove the equality

Δl=Δs−124R2Δs3+…, \Delta l = \Delta s - \frac{1}{24R^{2}}\Delta s^{3}+\ldots ,

where RR is the radius of curvature of the curve.

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The point of tangency is taken as the origin, the tangent as the axis OxOx, and the axis OyOy is directed from the point of tangency toward the center of curvature. Prove that in a neighborhood of the point of tangency, i.e.for small values of xx, the following formulas hold:

y1=x22R+ax3+…,y2=x22R+x48R3+…, y_{1} = \frac{x^{2}}{2R}+ax^{3}+\ldots , \qquad y_{2} = \frac{x^{2}}{2R}+\frac{x^{4}}{8R^{3}}+\ldots ,

where RR is the radius of curvature at the point (0,0)(0,0), y1y_{1} is the ordinate of points of the curve, and y2y_{2} is the ordinate of points of the circle of curvature.

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References

[ГюКу03] Гюнтер Н.М., Кузьмин Р.О. Сборник задач по высшей математике. 13-е изд., Издательство «Лань», 2003.