# Different way to compute arc length of vector function

Compute the length of the arc given by $$\textbf{w}(t) = \langle t^2,0,t^3 \rangle$$ for $$0\leq t \leq 1$$.

I know this could easily be done via the formula: $$\displaystyle \int_{a}^{b} \sqrt{{\big(\frac{dx}{dt}}\big)^{2} + \big(\frac{dy}{dt}\big)^{2} + \big(\frac{dz}{dt}\big)^{2}}\,dt$$.

I want to know if there is another method of computing the arc length without using this formula?

Using Neile's parabola

Using: $$\big(\frac{1}{27}\big) \times (4 + 9t^2)^{3/2} - \frac{8}{27}$$ I got $$\big(\frac{1}{27}\big) \times (4 + 9(1)^2)^{3/2} - \frac{8}{27}$$. Plugging this into a calculator yields: $$1.439$$.

• In this special case, Neile's parabola, see books.google.de/…, it may be done differently. Nov 25, 2021 at 10:48
• @MichaelHoppe: When I used the method given by you I got $1.439$ but when I used the formula I got $3.605$. Have I used the method wrong? Could you please explain. Nov 26, 2021 at 1:10
• It'll be helpful to post your calculations. Nov 26, 2021 at 10:01
• Adding it now...Done. Nov 26, 2021 at 10:12
• Using the integral you should get 1.439 as well. Nov 26, 2021 at 10:23

I know this could easily be done via the formula: $$\displaystyle \int_{a}^{b} \sqrt{{\big(\frac{dx}{dt}}\big)^{2} + \big(\frac{dy}{dt}\big)^{2} + \big(\frac{dz}{dt}\big)^{2}}\,dt$$.
when I used the formula, I got $$3.605.$$
\begin{align} &\int_{a}^{b} \sqrt{{\big(\frac{dx}{dt}}\big)^{2} + \big(\frac{dy}{dt}\big)^{2} + \big(\frac{dz}{dt}\big)^{2}}\,dt \\=&\int_0^1\sqrt{(2t)^2+0+(3t^2)^2}\,\mathrm dt \\=&\int_0^1\sqrt{9t^4+4t^2}\,\mathrm dt \end{align} Substituting in $$u=t^2:$$ \begin{align} =&\frac12\int_0^1\sqrt{9u+4}\,\mathrm du \\=&\frac1{27}\left[(9u+4)^\frac32\right]_0^1 \\=&1.440,\end{align} as required.