Arc Length

\begin{matrix} ℒ & = & \sqrt{{(Δ x)}^{2} + {(Δ y)}^{2}} \\ = & (\sqrt{1 + {(\frac{Δ y}{Δ x})}^{2}}) Δ x \\ = & (\sqrt{1 + {(\frac{Δ x}{Δ y})}^{2}}) Δ y \end{matrix}

Let $f (x)$ be a smooth function over the interval $[𝒶, 𝒷]$ . Then the arc length of $f (x)$ from $(𝒶, f (𝒶))$ to $(𝒷, f (𝒷))$ is

ℒ = \int_{𝒶}^{𝒷} (\sqrt{1 + {(\frac{d y}{d x})}^{2}}) d x = \int_{𝒶}^{𝒷} (\sqrt{1 + {(f' (x))}^{2}}) d x

Let $g (y)$ be a smooth function over the interval $[𝒸, 𝒹]$ . Then the arc length of $g (y)$ from $(𝒸, g (𝒸))$ to $(𝒹, g (𝒹))$ is

ℒ = \int_{𝒸}^{𝒹} (\sqrt{1 + {(\frac{d x}{d y})}^{2}}) d y = \int_{𝒸}^{𝒹} (\sqrt{1 + {(g' (y))}^{2}}) d y

f (x) = \frac{4 \sqrt{2}}{3} \sqrt{x^{3}} - 1

between $x = 0$ and $x = 1$ .

f (x) = \frac{1}{12} x^{3} + \frac{1}{x}

between $x = 1$ and $x = 4$ .

f (x) = \frac{1}{2} ℯ^{- x} + \frac{1}{2} ℯ^{x}

between $x = 0$ and $x = 2$ .

y^{3} = x^{2}

between $x = 1$ and $x = 8$ .

Find the arc length function $ℒ (x)$ of the curve

f (t) = t^{2} - \frac{1}{8} ln t

between $t = 1$ and $t = x$ .