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

1 Answer
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 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-curvehttps://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!