The Arc Length of a Smooth, Planar Curve and Distance Traveled BC only

A short piece of curve has horizontal extent $dx$ and vertical extent $dy$, so $ds = \sqrt{(dx)^{2} + (dy)^{2}}$, and factoring $dx$ out gives $L = \int_a^b \sqrt{1 + (f'(x))^{2}}\,dx$. The $1$ is the factored $(dx)^{2}$, the square comes from the Pythagorean theorem, and the root is the theorem itself, so a formula rebuilt from the triangle keeps all three. For $y = x^{3/2}$ on $[0, 5]$ the true length is $\frac{335}{27} \approx 12.407$, while dropping the $1$ gives $11.180$ and dropping the root gives $33.125$. Substituting $f$ for $f'$ measures nothing: length depends on steepness, not height.

Two checks cost nothing. On the straight line $y = 3x$ over $[0, 4]$ the formula must reproduce the distance formula, and both give $4\sqrt{10}$. And an arc length is at least the horizontal extent and at least the straight-line distance between the endpoints, which rules out $11.180$ immediately, since it falls below the chord length $\sqrt{150}$. For a curve described as $x = g(y)$ the integrand is $\sqrt{1 + (g'(y))^{2}}$ with $y$-limits, and for a parametric path it is $\sqrt{(x')^{2} + (y')^{2}}\,dt$, which has no $1$ because the $1$ was $\left(\frac{dx}{dx}\right)^{2}$ all along. That parametric integral is distance travelled, not displacement: it counts ground covered twice by a path that doubles back.