# What is the arclength of f(t) = (cos2t-sin2t,tan^2t) on t in [pi/12,(5pi)/12]?

May 22, 2018

14.69 (no closed form)

#### Explanation:

$f \left(t\right) = \left(\cos 2 t - \sin 2 t , {\tan}^{2} t\right)$

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 $\Delta x \left(t\right) , \Delta y \left(t\right)$, 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, you get

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

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

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

Since $\frac{\mathrm{dx}}{\mathrm{dt}} = - 2 \sin 2 t - 2 \cos 2 t , \frac{\mathrm{dy}}{\mathrm{dt}} = 2 \tan \left(t\right) {\sec}^{2} \left(t\right)$,

$\mathrm{ds} = \sqrt{{\left(- 2 \sin 2 t - 2 \cos 2 t\right)}^{2} + {\left(2 \tan \left(t\right) {\sec}^{2} \left(t\right)\right)}^{2}} \mathrm{dt}$

Factorising a 2 out of the entire expression and simplifying, we have

$\mathrm{ds} = 2 \sqrt{{\left(\cos 2 t + \sin 2 t\right)}^{2} + {\left(\tan \left(t\right) {\sec}^{2} \left(t\right)\right)}^{2}} \mathrm{dt}$

To find the total arc length from $\frac{\pi}{12}$ to $\frac{5}{12} \pi$, we can add up tiny segments of $\mathrm{ds}$ from $t = \frac{\pi}{12}$ to $t = \frac{5}{12} \pi$

Mathematically, this is equivalent to integrating from $t = \frac{\pi}{12}$ to $t = \frac{5}{12} \pi$

$s = {\int}_{\frac{\pi}{12}}^{\frac{5}{12} \pi} 2 \sqrt{{\left(\cos 2 t + \sin 2 t\right)}^{2} + {\left(\tan \left(t\right) {\sec}^{2} \left(t\right)\right)}^{2}} \mathrm{dt}$

Now this expression has no closed form and is approximately 14.69