Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was looking through my multivariable calculus homework and I saw an example where we needed to find the arclength of a simple space curve. It's very simple:

$$r(t)= cos(t)\hat i + sin(t)\hat j + t\hat k $$

Is the path defined by this curve not the same as the diagonal of the rectangle surface of a cylinder?

The formal length of the curve is given by $\int|r'(t)|dt$ and here for one complete revolution of the curve, the limits of this integral would be from 0 to 2$\pi$.

Doing the math results in a length of $2\pi\sqrt2$.

Using my presumed geometric method, I use the Pythagorean formula and plug in the circumference of the circle and height, but end up with a sum under the radical. In this case, I would plug in $2\pi$ and 1, because at $t=2\pi$, the point is $(1,0,2\pi)$ and at $t=0$, the point is $(1,0,0)$.

In other words, if I drew a diagonal across a rectangle $2\pi$ by $1$, and then curled the surface so it would form a topless & bottomless cylinder, isn't the line defined on the paper the same as the curve defined by $r(t)$?

I'm pretty sure my geometric assumption is wrong, but I'm stumped as to why. Could anyone provide an explanation as to why this is not the case?

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

Are you sure the height is 1? Check that again... (...$+t\hat k$), but your concept is correct =)

Unless you are saying your height is $2\pi$ and your circumference is $1$. In that case check your circumference. The diameter of the circle that makes it is 2.

share|improve this answer
    
Gosh darn it! You got me. I made a blunder. :D The height is $2\pi$ and the circumference is $2\pi$ –  Zchpyvr Nov 1 '12 at 0:02
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.