# What is the arclength of f(t) = (t/sqrt(t^2+2),t/e^(t-2)) on t in [-1,1]?

May 26, 2018

$s = {\int}_{- 1}^{1} \sqrt{\frac{4}{{t}^{2} + 2} ^ 3 + {e}^{4 - 2 t} {\left(1 - t\right)}^{2}} \approx 22.9212$

#### Explanation:

Using the Arc Length Formula, we have

$\textcolor{b l u e}{s = {\int}_{- 1}^{1} \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \mathrm{dt}}$

Since $x \left(t\right) = \frac{t}{\sqrt{{t}^{2} + 2}}$, differentiating and simplifying, we have

$\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{2}{{t}^{2} + 2} ^ \left(\frac{3}{2}\right)$

Similarly,

$\frac{\mathrm{dy}}{\mathrm{dt}} = \left(1 - t\right) {e}^{2 - t}$

Substituting into the formula, we have

$s = {\int}_{- 1}^{1} \sqrt{{\left(\frac{2}{{t}^{2} + 2} ^ \left(\frac{3}{2}\right)\right)}^{2} + {\left(\left(1 - t\right) {e}^{2 - t}\right)}^{2}} \mathrm{dt}$

$= {\int}_{- 1}^{1} \sqrt{\frac{4}{{t}^{2} + 2} ^ 3 + {e}^{4 - 2 t} {\left(1 - t\right)}^{2}}$

This integral unfortunately cannot be expressed in terms of standard functions or constants, so the best you can get is either an infinite series or an approximation.

In this case, the integral evaluates to $s \approx 22.9212$, which is our final answer.

For arclength type questions, please take note that the vast majority of the resultant integrals will, like this one, be unsolvable in terms of standard functions.

### Explanation for Arc Length Formula

To find the arc length $\Delta s$ over a time $\Delta t$, you can try to approximate it with a straight line, using the hypotenuse of the right-angled triangle with sides $\Delta x , \Delta y$, which is the change in the x- and y- coordinates over the time $t$.

$\Delta s \approx \sqrt{\Delta {x}^{2} + \Delta {y}^{2}}$

Of course, this will often be wildly inaccurate since most curves aren't exactly a straight line.

However, for a continuous function, the smaller time interval you choose, the more accurate the above approximation is (you can imagine zooming in on a function's graph: the more you zoom, the closer the function looks like a straight line)

Thus, at the limit when you shorten the time interval to be really (infinitesimally) small, you get

$\mathrm{ds} = \sqrt{{\mathrm{dx}}^{2} + {\mathrm{dy}}^{2}}$

With $\mathrm{ds} , \mathrm{dx} , \mathrm{dy}$ here just representing really small changes in $s , x , y$

To manipulate it to suit the given parametric, we can adjust the equation such that $\mathrm{ds}$ is in terms of $\mathrm{dt}$.

$\mathrm{ds} = \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}} \cdot \mathrm{dt}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}} \cdot \mathrm{dt}\right)}^{2}}$
$= \sqrt{\left({\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}\right) {\mathrm{dt}}^{2}}$
$= \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \mathrm{dt}$

So for the arc length $s$ from $t = a$ to $t = b$, we have, by integrating both sides,

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