# What is the arc length of f(t)=(t^2-4t,5-1/t)  over t in [3,4] ?

Nov 16, 2016

Define $x \left(t\right) , y \left(t\right)$ as follows

$\left\{\begin{matrix}x \left(t\right) = {t}^{2} - 4 t \\ y \left(t\right) = 5 - \frac{1}{t}\end{matrix}\right. \implies f \left(t\right) = \left(x \left(t\right) , y \left(t\right)\right)$

Differentiating $x \left(t\right) , y \left(t\right)$ wrt $t$ we get:

$\frac{\mathrm{dx}}{\mathrm{dt}} = 2 t - 4$
$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{1}{t} ^ 2$

Then the arc length of $f \left(t\right)$ over $t \in \left[\alpha , \beta\right]$ is given by;

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

So over $\left[3 , 4\right]$ we have
$L = {\int}_{3}^{4} \sqrt{{\left(2 t - 4\right)}^{2} + {\left(\frac{1}{t} ^ 2\right)}^{2}}$

$\therefore L = {\int}_{3}^{4} \sqrt{4 {t}^{2} - 16 t + 16 + \frac{1}{t} ^ 4}$

This definite integral does not have an intrinsic solution and would need to be solved numerically, using either a computer or estimated using the Trapezium Rule or Simpson's Rule