I've heard that the arc length for certain curves such as \(y=x^2\) and \(y=e^x\) have closed forms. Is it known whether are similar closed forms for the arc lengths of \(x^3\), \(\frac{1}{x}\), or the trigonometric functions?

For \(y = x^{2}\) we have \(S = \displaystyle\int_{a}^{b} \sqrt{1 + 4x^{2}} dx,\) for which we would use trig substitution with \(2x = \tan(\theta).\)

For \(y = e^{x}\) we have \(S = \displaystyle\int_{a}^{b} \sqrt{1 + e^{2x}} dx,\) for which we could substitute \(u = \sqrt{1 + e^{2x}}\) and end up with a (solvable) rational fraction in \(u.\)

However, for the other functions you mention we would end up having to solve some elliptic integrals, and as such we would not be able to get closed form arclength formulas. There aren't actually that many curves that we can get closed forms for; besides the two above, there is also the circle and the curve \(x^{\frac{2}{3}} + y^{\frac{2}{3}} = 1.\)

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TopNewestAs you probably already know, the formula for the arclength \(S\) of a 2D curve \(y = f(x)\) is

\(S = \displaystyle\int_{a}^{b} \sqrt{1 + (f'(x))^{2}} dx.\)

For \(y = x^{2}\) we have \(S = \displaystyle\int_{a}^{b} \sqrt{1 + 4x^{2}} dx,\) for which we would use trig substitution with \(2x = \tan(\theta).\)

For \(y = e^{x}\) we have \(S = \displaystyle\int_{a}^{b} \sqrt{1 + e^{2x}} dx,\) for which we could substitute \(u = \sqrt{1 + e^{2x}}\) and end up with a (solvable) rational fraction in \(u.\)

However, for the other functions you mention we would end up having to solve some elliptic integrals, and as such we would not be able to get closed form arclength formulas. There aren't actually that many curves that we can get closed forms for; besides the two above, there is also the circle and the curve \(x^{\frac{2}{3}} + y^{\frac{2}{3}} = 1.\)

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