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

Question

Emilia Aguilar · Accepted Answer

The arclength is around #1.41068#.

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:

(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!

Charles Anctil · Answer

undefined To find the arc length of $ f(x) = \sin(x + \frac{\pi}{12}) $ on the interval $ [0, \frac{3\pi}{8}] $, we use the formula for arc length:

$$ L = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} \, dx $$

1. Find the derivative of $ f(x) $:
$$ f'(x) = \cos(x + \frac{\pi}{12}) $$

2. Square the derivative:
$$ [f'(x)]^2 = \cos^2(x + \frac{\pi}{12}) $$

3. Integrate the square root of $ 1 + [f'(x)]^2 $ from $ x = 0 $ to $ x = \frac{3\pi}{8} $:
$$ L = \int_{0}^{\frac{3\pi}{8}} \sqrt{1 + \cos^2(x + \frac{\pi}{12})} \, dx $$

This integral can be difficult to evaluate analytically, so numerical methods or computational tools may be necessary to find the exact value of the arc length.