Length of Arch of Sine Function

Theorem

The length of one arch of the sine function:

$y = \sin x$

is given by:

$L = 2 \sqrt 2 \map E {\dfrac {\sqrt 2} 2}$

where $E$ denotes the incomplete elliptic integral of the second kind.

Proof

Let $L$ be the length of one arch of $y = \sin x$.

Then:

\(\ds L\)

\(=\)

\(\ds 2 \int_0^{\pi/2} \sqrt {1 + \paren {\map {\frac \d {\d x} } {\sin x} }^2} \rd x\)

Definition of Length of Curve

\(\ds \)

\(=\)

\(\ds 2 \int_0^{\pi/2} \sqrt {1 + \cos^2 x} \rd x\)

Derivative of Sine Function

\(\ds \)

\(=\)

\(\ds 2 \int_0^{\pi/2} \sqrt {2 - \sin^2 x} \rd x\)

Sum of Squares of Sine and Cosine

\(\ds \)

\(=\)

\(\ds 2 \sqrt 2 \int_0^{\pi/2} \sqrt {1 - \paren {\frac {\sqrt 2} 2} \sin^2 x} \rd x\)

extracting $\sqrt 2$

\(\ds \)

\(=\)

\(\ds 2 \sqrt 2 \map E {\dfrac {\sqrt 2} 2}\)

Definition of Complete Elliptic Integral of the Second Kind

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $1$: The Nature of Differential Equations: $\S 5$: Falling Bodies and Other Rate Problems: Problem $6$