What is the arclength of f(t) = (t-2,1/(t-2)) on t in [0,1]?

Oct 16, 2016

Use the equation:

$L = {\int}_{a}^{b} \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \mathrm{dt}$

L ≈ 1.13209

Explanation:

The arc length, L, for parametric equations is:

$L = {\int}_{a}^{b} \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \mathrm{dt}$

$\frac{\mathrm{dx}}{\mathrm{dt}} = 1$
$\frac{\mathrm{dy}}{\mathrm{dt}} = - {\left(t - 2\right)}^{-} 2$
$a = 0$
$b = 1$

$L = {\int}_{0}^{1} \sqrt{{\left(1\right)}^{2} + {\left(- {\left(t - 2\right)}^{-} 2\right)}^{2}} \mathrm{dt}$

$L = {\int}_{0}^{1} \sqrt{1 + {\left(t - 2\right)}^{-} 4} \mathrm{dt}$

L ≈ 1.13209