Finding Arc Lengths of Curves Given by Parametric Equations BC only

Factoring $dt$ rather than $dx$ out of $ds = \sqrt{(dx)^{2}+(dy)^{2}}$ gives $L = \int_{t_1}^{t_2}\sqrt{(dx/dt)^{2} + (dy/dt)^{2}}\,dt$, with $t$-values as the limits and a non-negative integrand that never needs an absolute value. The $1$ of Unit 8's formula has not been dropped: it was $\left(\frac{dx}{dx}\right)^{2}$, and setting $x = t$ recovers Unit 8's integrand exactly, so the two formulas are one. For $x = 3t^{2}$, $y = 2t^{3}$ on $[0,1]$ the length is $2(2\sqrt{2}-1) \approx 3.657$, while dropping the root gives $19.2$, adding the rates gives $5$, and restoring the $1$ gives about $2.42$.

Two bounds catch all three: an arc is at least its chord, here $\sqrt{13} \approx 3.606$, and a monotone arc is at most the sum of its extents, here $5$. The "rates added" error produces exactly that upper bound, which is the statement that a staircase is longer than a diagonal. The integral measures distance travelled rather than curve length, so $x = \cos t$, $y = \sin t$ on $[0, 4\pi]$ returns $4\pi$ for a circle of length $2\pi$: the particle laps twice and both laps are counted. Asking for a curve's length therefore requires an interval on which it is traced once, which is the same caution 9.8 applies to polar area.