# What is the arclength of (t-e^(2t),lnt) on t in [1,12]?

Apr 8, 2018

The arclength is around 2.64891×10^10.

#### Explanation:

First, let's get a sense of how long the arclength might actually be. It will go from

$\left(1 - {e}^{2 \cdot 1} , \ln 1\right) \quad \approx \quad \left(- 6.389 , 0\right)$

to

(12-e^(2*12),ln12)quad~~quad(-2.649×10^10,2.485)

We can tell that this distance will be very big, so that means the arclength is also very big.

To calculate the actual arclength, we'll need to get an integral in the form of $\int \sqrt{{\left(\mathrm{dx}\right)}^{2} + {\left(\mathrm{dy}\right)}^{2}}$, based on the Pythagorean theorem:

(This link does a great job of explaining arc lengths of parametric curves.)

First, split the parametric function into a function of $t$ for $x$ and $y$:

x(t)=t−e^(2t)

$y \left(t\right) = \ln t$

Now, differentiate each one and get $\mathrm{dx}$ and $\mathrm{dy}$ in terms of $\mathrm{dt}$:

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

$\mathrm{dy} = \frac{1}{t} \mathrm{dt}$

Now, plug these into the aforementioned integral with the appropriate bounds. You'll see that the $\mathrm{dt}$ can be factored out of the radical:

$\textcolor{w h i t e}{=} {\int}_{1}^{12} \sqrt{{\left(\mathrm{dx}\right)}^{2} + {\left(\mathrm{dy}\right)}^{2}}$

$= {\int}_{1}^{12} \sqrt{{\left(\left(1 - 2 {e}^{2} t\right) \mathrm{dt}\right)}^{2} + {\left(\frac{1}{t} \mathrm{dt}\right)}^{2}}$

$= {\int}_{1}^{12} \sqrt{{\left(1 - 2 {e}^{2} t\right)}^{2} {\mathrm{dt}}^{2} + \frac{1}{t} ^ 2 {\mathrm{dt}}^{2}}$

$= {\int}_{1}^{12} \sqrt{{\mathrm{dt}}^{2} \left({\left(1 - 2 {e}^{2} t\right)}^{2} + \frac{1}{t} ^ 2\right)}$

$= {\int}_{1}^{12} \sqrt{{\left(1 - 2 {e}^{2} t\right)}^{2} + \frac{1}{t} ^ 2}$ $\mathrm{dt}$

At this point, you should probably plug this integral into a calculator because the function probably doesn't have an antiderivative, and even if it does it would be a pain to calculate.

A calculator should spit out something around 2.64891×10^10, and that's your answer. Hope this helped!