# Thread: Arc Length and Curvature

1. ## Arc Length and Curvature

This is my first post, so hello forums . Anyway, I am trying to relearn calculus, took a hiatus for a while, and I am currently taking beginners multi-variable calculus and I hit an example that I do not understand in my book (probably cause my diff/integral calculus could use some boning up ).

This involves Arc Length and Curvature:

r(t) = cos(t)i + sin(t)j + tk from point (1,0,0) to (1,0,2 $\pi$)

they show that $\mid r\prime (t)\mid = \sqrt{(-sin (t))^2 + cos^2 (t) + 1} = \sqrt{2}$

and then they say "The arc from (1,0,0) to (1,0,2 $\pi$) is described by the parameter interval $0 \leq t \leq 2\pi$ and so [from formula 3], we have...

$L = \int_0^{2\pi} \mid r\prime (t)\mid dt\ = \int_0^{2\pi} \sqrt{2} dt = 2\sqrt{2\pi}$

I do not understand how they got $2\sqrt{2\pi}$?
shouldn't it be $2\pi\sqrt{2}$?

Thank you for any and all input

2. Originally Posted by Warrenx
This is my first post, so hello forums . Anyway, I am trying to relearn calculus, took a hiatus for a while, and I am currently taking beginners multi-variable calculus and I hit an example that I do not understand in my book (probably cause my diff/integral calculus could use some boning up ).

This involves Arc Length and Curvature:

r(t) = cos(t)i + sin(t)j + tk from point (1,0,0) to (1,0,2 $\pi$)

they show that $\mid r\prime (t)\mid = \sqrt{(-sin (t))^2 + cos^2 (t) + 1} = \sqrt{2}$

and then they say "The arc from (1,0,0) to (1,0,2 $\pi$) is described by the parameter interval $0 \leq t \leq 2\pi$ and so [from formula 3], we have...

$L = \int_0^{2\pi} \mid r\prime (t)\mid dt\ = \int_0^{2\pi} \sqrt{2} dt = 2\sqrt{2\pi}$

I do not understand how they got $2\sqrt{2\pi}$?
shouldn't it be $2\pi\sqrt{2}$?

Thank you for any and all input
Dear Warrenx,

It should be $2\pi\sqrt{2}$. I think in the book they mean, $2\sqrt{2}\pi$ but accidently made the squareroot sign a little longer making it $2\sqrt{2\pi}$

Hope this helps you.

3. Originally Posted by Sudharaka
Dear Warrenx,

It should be $2\pi\sqrt{2}$. I think in the book they mean, $2\sqrt{2}\pi$ but accidently made the squareroot sign a little longer making it $2\sqrt{2\pi}$

Hope this helps you.
Thank you very much Mr. Sudharaka, was bugging me so much I stopped reading the section.

4. That's the reason it is standard (except in this textbook, apparently!) to put numbers or variable before the square root symbol- that is while we typically write "2x", we write " $x\sqrt{2}$".