# Arc length of vector function - the integral seems impossible

• overpen57mm
In summary: So in summary, you have an integral over ##\|(e^x \cdot F(x))'\|=\|e^x(F'(x)+F(x))\|=2e^x \sqrt{(\cos 2x -\sin 2x)^2+(\sin 2x+\cos 2x)^2+1^2}## where ##F(x)=(\cos 2x,\sin 2x,1).##

#### overpen57mm

Homework Statement
Find the arc length from 0-3pi for v(x)=(e^x cos(2x), e^x sin(2x), e^x)
Relevant Equations
Arc length formula for vector equations
The vector equation is ## v(x)=(e^x cos(2x), e^x sin(2x), e^x) ##

I know the arc-length formula is ## S=\int_a^b \|v(x)\| \,dx ##

I found the derivative from a previous question dealing with this same function, but the when I plug it into the arc-length function I get an integral that I've tried and tried but just can't get anywhere with. The complexity of the problem also makes me think that I might be approaching it from the wrong direction. Here is the integral as I understand it: $$\int{\sqrt{ (e^{2x}) ( (-2\sin(2x) + \cos(2x))^2 + (2\cos(2x)+\sin(2x))^2 + 1 ) }} \, dx$$

I would appreciate any tips on the integral or the problem as a whole, if there's another way to solve it that I haven't seen.

You have an integral over ##\|(e^x \cdot F(x))'\|=\|e^x(F'(x)+F(x))\|=2e^x \sqrt{(\cos 2x -\sin 2x)^2+(\sin 2x+\cos 2x)^2+1^2}## where ##F(x)=(\cos 2x,\sin 2x,1).##

Last edited:
overpen57mm
fresh_42 said:
You have an integral over ##\|(e^x \cdot F(x))'\|=\|e^x(F'(x)+F(x))\|=2e^x \sqrt{(\cos 2x -\sin 2x)^2+(\sin 2x+\cos 2x)^2+1^2}## where ##F(x)=(\cos 2x,\sin 2x,1).##
Thanks, I guess I was blinded by the derivative I got from the previous question.