Arc lengths leading to elliptic integrals

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Prove that the arc length of the ellipse

x=acos⁡t,y=bsin⁡t x=a\cos t ,\quad y=b\sin t

equals the length of one wave of the sine curve y=csin⁡xby=c\sin \dfrac {x}{b}, where c=a2−b2c=\sqrt{a^{2}-b^{2}}.

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Approximately find the perimeter of an ellipse with semi-axes a=10a=10 and b=6b=6.

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Find the arc length of a single half-wave of the sine curve y=sin⁡xy=\sin x (0≤x≤π0 \leq x \leq \pi), accurate to 0.010.01.

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Express the arc length of the following ellipse as a series expanded in positive integer powers of the eccentricity:

x=acos⁡t,x=bsin⁡t(0⩽t⩽2π). x=a\cos t,\quad x=b\sin t \quad (0 \leqslant t \leqslant 2\pi ).
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Express, in terms of elliptic integrals, the length of the arc of the curve:

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(1)

y=asin⁡ωx,0⩽x⩽x0⩽π/(2ω)y = a \sin \omega x, 0 \leqslant x \leqslant x_{0} \leqslant \pi /(2 \omega );

(2)

x=at−bsin⁡t,y=a−bcos⁡t,a>0,b>0,0⩽t0⩽t⩽πx = a t-b \sin t, y = a-b \cos t, a > 0, b > 0,0 \leqslant t_{0} \leqslant t \leqslant \pi;

(3)

r2=2a2cos⁡2φ,0⩽φ⩽φ0<π/4r^{2} = 2 a^{2} \cos 2 \varphi , 0 \leqslant \varphi \leqslant \varphi_{0} < \pi / 4 (lemniscate);

(4)

x2/a2+y2/b2=1,0⩽x⩽x0⩽a,y⩾0,a>bx^{2} / a^{2}+y^{2} / b^{2} = 1,0 \leqslant x \leqslant x_{0} \leqslant a, y \geqslant 0, a > b;

(5)

x2/a2−y2/b2=1,0⩽y⩽y0,x>0x^{2} / a^{2}-y^{2} / b^{2} = 1,0 \leqslant y \leqslant y_{0}, x > 0.

Find the length of the curve:

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(1)

x2/a2+y2/b2=1x^{2} / a^{2}+y^{2} / b^{2} = 1;

(2)

r2=2a2cos⁡2φr^{2} = 2 a^{2} \cos 2 \varphi;

(3)

r=acos⁡φ+br = a \cos \varphi +b (limaçon);

(4)

r=asin⁡nφ,n∈N,n>1r = a \sin n \varphi , \quad n \in \mathbb {N}, \quad n > 1.

Prove that the length of the ellipse x2/a2+y2/b2=1,a>bx^{2} / a^{2}+y^{2} / b^{2} = 1, a > b, equals the length of the sinusoid y=a2−b2sin⁡(x/b),0⩽x⩽2πby = \sqrt{a^{2}-b^{2}} \sin (x / b), 0 \leqslant x \leqslant 2 \pi b.

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Prove that the length of the arc of the curve

x=at−bsin⁡t,y=a−bcos⁡t,0⩽t⩽2π,a>0,b>0x = a t-b \sin t, \quad y = a-b \cos t, \quad 0 \leqslant t \leqslant 2 \pi , \quad a > 0, \quad b > 0, equals the length of the ellipse with semi-axes a+ba+b and ∣a−b∣\left|a-b\right|.

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Prove that the length ss of an ellipse with semi-axes aa and bb satisfies the inequalities

π(a+b)<s<π2(a2+b2) \pi (a+b) < s < \pi \sqrt{2\left(a^{2}+b^{2}\right)}
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Find the length of Viviani's curve

x=Rsin⁡2t,y=Rsin⁡tcos⁡t,z=Rcos⁡t. x = R \sin ^{2} t, \quad y = R \sin t \cos t, \quad z = R \cos t.
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Using tables, find, with an error no greater than 0.1, the arc length of the curve x2+y2+z2=1,y2+z2=yx^{2}+y^{2}+z^{2} = 1, y^{2}+z^{2} = y.

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Let ll denote the length of the perimeter of an ellipse with unequal semi-axes aa and bb. Give two different proofs (with and without the use of conformal mapping) for the inequality

l>π(a+b). l > \pi (a+b).
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Remark.
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The right-hand side is a convenient approximation to ll when aa and bb are nearly equal.

References

[Dem97] Demidovich B.P. Сборник задач и упражнений по математическому анализу. 13-е изд., Издательство Московского университета / Издательство "ЧеРо", 1997.
[ККЧ+03a] Кудрявцев Л.Д., Кутасов А.Д., Чехлов В.И., Шабунин М.И. Интегралы. Ряды. 2-е изд., ФИЗМАТЛИТ, 2003.
[ККЧ+03b] Кудрявцев Л.Д., Кутасов А.Д., Чехлов В.И., Шабунин М.И. Функции многих переменных. 2-е изд., ФИЗМАТЛИТ, 2003.
[PóSz98] Pólya G., Szegő G. Problems and Theorems in Analysis. Springer, 1998.