# What is the arclength of (t-3,t+4) on t in [2,4]?

Apr 2, 2018

$A = 2 \sqrt{2}$

#### Explanation:

The formula for parametric arc length is:
$A = {\int}_{a}^{b} \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \setminus \mathrm{dt}$

We begin by finding the two derivatives:

$\frac{\mathrm{dx}}{\mathrm{dt}} = 1$ and $\frac{\mathrm{dy}}{\mathrm{dt}} = 1$

This gives that the arc length is:
$A = {\int}_{2}^{4} \sqrt{{1}^{2} + {1}^{2}} \setminus \mathrm{dt} = {\int}_{2}^{4} \sqrt{2} \setminus \mathrm{dt} = {\left[\sqrt{2} t\right]}_{2}^{4} = 4 \sqrt{2} - 2 \sqrt{2} = 2 \sqrt{2}$

In fact, since the parametric function is so simple (it is a straight line), we don't even need the integral formula. If we plot the function in a graph, we can just use the regular distance formula:

$A = \sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}} = \sqrt{4 + 4} = \sqrt{8} = \sqrt{4 \cdot 2} = 2 \sqrt{2}$

This gives us the same result as the integral, showing that either method works, although in this case, I'd recommend the graphical method because it is simpler.