Question

Let c be the path c(t) = (2t, t2, log(t)), defined for t > 0. find the arc length of c between the points (2, 1, 0) and (12, 36, log(6)).

Answer 1

The path starts at $t=1$ and ends at $t=6$, so we have an arc length of


$\displaystyle\int_{\mathcal C}\mathrm dS=\int_{t=1}^{t=6}\|\mathbf c'(t)\|\,\mathrm dt$
$=\displaystyle\int_1^6\sqrt{2^2+(2t)^2+\left(\frac1t\right)^2}\,\mathrm dt$
$=\displaystyle\int_1^6\sqrt{4t^2+4+\frac1{t^2}}\,\mathrm dt$
$=\displaystyle\int_1^6\frac1t\sqrt{4t^4+4t^2+1}\,\mathrm dt$
$=\displaystyle\int_1^6\frac1t\sqrt{(2t^2+1)^2}\,\mathrm dt$
$=\displaystyle\int_1^6\frac{2t^2+1}t\,\mathrm dt$
$=\displaystyle\int_1^6\left(2t+\frac1t\right)\,\mathrm dt$
$=t^2+\log t\bigg|_{t=1}^{t=6}$
$=(6^2+\log6)-(1^2+\log1)=35+\log6$