# What is the arc length of f(x)=sin(x+pi/12)  on x in [0,(3pi)/8]?

Apr 8, 2018

The arclength is around $1.41068$.

#### Explanation:

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 function graphs.)

To get that integral, first, calculate $\mathrm{dy}$:

$\textcolor{w h i t e}{\implies} y = \sin \left(x + \frac{\pi}{12}\right)$

$\implies \mathrm{dy} = \cos \left(x + \frac{\pi}{12}\right) \cdot \frac{d}{\mathrm{dx}} \left[x + \frac{\pi}{12}\right] \mathrm{dx}$

$\textcolor{w h i t e}{\implies \mathrm{dy}} = \cos \left(x + \frac{\pi}{12}\right) \cdot 1 \mathrm{dx}$

$\textcolor{w h i t e}{\implies \mathrm{dy}} = \cos \left(x + \frac{\pi}{12}\right) \mathrm{dx}$

Now, plug this into the aforementioned integral and put the appropriate bounds. You will see that the $\mathrm{dx}$ gets factored out of the radical:

$\textcolor{w h i t e}{=} {\int}_{0}^{\frac{3 \pi}{8}} \sqrt{{\left(\mathrm{dx}\right)}^{2} + {\left(\mathrm{dy}\right)}^{2}}$

$= {\int}_{0}^{\frac{3 \pi}{8}} \sqrt{{\left(\mathrm{dx}\right)}^{2} + {\left(\cos \left(x + \frac{\pi}{12}\right) \mathrm{dx}\right)}^{2}}$

$= {\int}_{0}^{\frac{3 \pi}{8}} \sqrt{{\left(\mathrm{dx}\right)}^{2} + {\cos}^{2} \left(x + \frac{\pi}{12}\right) {\left(\mathrm{dx}\right)}^{2}}$

$= {\int}_{0}^{\frac{3 \pi}{8}} \sqrt{{\left(\mathrm{dx}\right)}^{2} \left(1 + {\cos}^{2} \left(x + \frac{\pi}{12}\right)\right)}$

$= {\int}_{0}^{\frac{3 \pi}{8}} \mathrm{dx} \sqrt{1 + {\cos}^{2} \left(x + \frac{\pi}{12}\right)}$

$= {\int}_{0}^{\frac{3 \pi}{8}} \sqrt{1 + {\cos}^{2} \left(x + \frac{\pi}{12}\right)}$ $\mathrm{dx}$

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 $1.41068$, and that's your answer. Hope this helped!