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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: $1.439$.

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  • $\begingroup$ In this special case, Neile's parabola, see books.google.de/…, it may be done differently. $\endgroup$ Nov 25, 2021 at 10:48
  • $\begingroup$ @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. $\endgroup$
    – bumblebee
    Nov 26, 2021 at 1:10
  • $\begingroup$ It'll be helpful to post your calculations. $\endgroup$ Nov 26, 2021 at 10:01
  • $\begingroup$ Adding it now...Done. $\endgroup$
    – bumblebee
    Nov 26, 2021 at 10:12
  • $\begingroup$ Using the integral you should get 1.439 as well. $\endgroup$ Nov 26, 2021 at 10:23

1 Answer 1

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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.

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  • $\begingroup$ Thanks for the answer, turns out the first time I completely forgot to integrate which is why I got the wrong answer XD $\endgroup$
    – bumblebee
    Jan 11 at 9:58

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