Determine the arc length of the parametric curve given by the following parametric equation: $φ(t)= (\sqrt{t}, t+1, t)$ $t\in[10,20]$

In order to do this I simply tried it to solve it by the formula of arc lenght.

Given the formula, $ L=\int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 +\left(\frac{dy}{dt}\right)^2 +\left(\frac{dz}{dt}\right)^2 } \;dt$

I get that $L=\int_{10}^{20} \sqrt {2+\frac{1}{4t}}; dt=\int_{10}^{20}\sqrt {\frac{8t+1}{4t}} \; dt$

Then, I tried to get prettier the integral function by multiplying in the argument of the square root by $4t/4t$ (I eliminated square root in the denominator) getting this not very satisfying result:

$L=\int_{10}^{20}\sqrt {\frac{(4t)^2+t}{2t}} \; dt$

I really don't know how to solve that integral, so I wonder if someone comes out with some idea about how to do it.

In the other hand, I tought that maybe there is a nicer parametrization of that particular parametrization, e.g. a reparametrization, where we could solve the integral more easily, but in that case I'm not sure about how to find it. Any ideas?

  • $\begingroup$ Take $T^2=2+1/4t$ then $4t=\dfrac{1}{T^2-2}$ and ... $\endgroup$ – Bumblebee May 10 '18 at 16:30
  • $\begingroup$ No, it doesn't work. There's an error I just solved in the final expression of the integral $\endgroup$ – Neisy Sofía Vadori May 10 '18 at 16:33
  • $\begingroup$ This variable transformation can solve your integral. Trust me and try one more time. $\endgroup$ – Bumblebee May 10 '18 at 16:37
  • $\begingroup$ You were right. Thanks! $\endgroup$ – Neisy Sofía Vadori May 10 '18 at 17:03

Hint. Given that $\varphi (t) \in \mathbb{R}^3$, you rather have $$ L=\int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 +\left(\frac{dy}{dt}\right)^2 +\left(\frac{dz}{dt}\right)^2 } \;dt $$ that is $$ L=\int_{10}^{20} \sqrt{\frac{1}{4t} +2} \,dt. $$Then make the substitution $$ u=\sqrt{\frac{1}{4t} +2} $$ to obtain a rational fraction as the integrand which is standard.

  • 1
    $\begingroup$ Thanks a lot, I just solved the integral. $\endgroup$ – Neisy Sofía Vadori May 10 '18 at 17:03
  • $\begingroup$ @NeisySofíaVadori You are very welcome. $\endgroup$ – Olivier Oloa May 10 '18 at 18:33

The correct computation is

$$\int\sqrt{\frac 1{4t}+1+1}\,dt=\int\sqrt{8u^2+1}\,du$$ where $t=u^2$.

Then, with $\sqrt8u=\sinh v$,

$$\frac1{\sqrt8}\int\cosh^2v\,dv=\frac1{\sqrt8}\int\frac{\cosh2v+1}2\,dv=\frac1{4\sqrt8}(\sinh2v+2v)=\frac1{2\sqrt8}(\sinh v\cosh v+v)$$


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