Question

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

Answer 1

The arclength is around `1.41068`.

Explanation:

To calculate the actual arclength, we'll need to get an integral in the form of `intsqrt((dx)^2+(dy)^2)`, based on the Pythagorean theorem:

![https://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/line-integrals-for-scalar-functions-articles/a/arc-length-part-2-parametric-curve]()

(This link does a great job of explaining arc lengths of function graphs.)

To get that integral, first, calculate `dy`:

`color(white)=>y=sin(x+pi/12)`

`=>dy=cos(x+pi/12)*d/dx[x+pi/12]dx`

`color(white)(=>dy)=cos(x+pi/12)*1dx`

`color(white)(=>dy)=cos(x+pi/12)dx`

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

`color(white)=int_0^((3pi)/8) sqrt( (dx)^2 + (dy)^2)`

`=int_0^((3pi)/8) sqrt( (dx)^2 + (cos(x+pi/12)dx)^2)`

`=int_0^((3pi)/8) sqrt( (dx)^2 + cos^2(x+pi/12)(dx)^2)`

`=int_0^((3pi)/8) sqrt( (dx)^2(1+ cos^2(x+pi/12)))`

`=int_0^((3pi)/8) dxsqrt(1+ cos^2(x+pi/12))`

`=int_0^((3pi)/8) sqrt(1+ cos^2(x+pi/12))` `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!