# Thread: length of a helix

1. ## length of a helix

hey guys, i've been asked to consider the helix:

r(t) = 4cos(pi x t)i + 3sin(pi x t)j + tk

where i, j and k are unit vectors in the x, y and z planes respectively and t is a parameter. i need to sketch it and calculate its length from (-4,0,1) to (0,3,5/2).

any help would be greatly appreciated

2. Originally Posted by ben08
hey guys, i've been asked to consider the helix:

r(t) = 4cos(pi x t)i + 3sin(pi x t)j + tk

where i, j and k are unit vectors in the x, y and z planes respectively and t is a parameter. i need to sketch it and calculate its length from (-4,0,1) to (0,3,5/2).

any help would be greatly appreciated
the integral for arclength is

$S= \int_{t_0}^{t_1}\sqrt{ \left( \frac{\partial{x}}{\partial{t}} \right)^2+ \left( \frac{\partial{y}}{\partial{t}} \right)^2+ \left( \frac{\partial{z}}{\partial{t}} \right)^2}$

where x,y,z are the components of the vector function r(t)

Note that $t_0=1 \iff r(1)=(-4,0,1)$
and
$t_1=\frac{5}{2} \iff r(5/2)=(0,3,\frac{5}{2})$

3. Originally Posted by ben08

the formula for the length of a curve is

Length = integral of sqrt(r'(t).r'(t))dt from "t0" to "t end"

i.e. you take the dot product of r'(t) and r'(t), then take the square root of that. You then integrate w/ respect to t between the endpoints given in the question. the main thing i'm confused with is the integration step, you end up having to integrate a sin^2(x) and i'm not sure what you get.
if you need help with an integral please post the integral.

p.s this trig formula may help

$\int \sin^2(t)dt=\int \frac{1}{2}\left(1-\cos(2t) \right)dt$

4. $
L= \int_{1}^{5/2}\sqrt{A}
$

where A = (25(pi^2)/2) - (7(pi^2)/2)xcos(pi x 2t) + 1)

that's how far i've got it down to, now im just not sure how to integrate $\sqrt{cos(pi2t)}$

obviously i don't completely understand how to write integrals, so that's the best i can give you... i'm also pretty unsure of this answer, because when i integrate it and sub in the values i get a two fractions where i divide by zero, so i think i must be wrong...

5. Originally Posted by ben08
$
L= \int_{1}^{5/2}\sqrt{A}
$

where A = (25(pi^2)/2) - (7(pi^2)/2)xcos(pi x 2t) + 1)

that's how far i've got it down to, now im just not sure how to integrate $\sqrt{cos(pi2t)}$

obviously i don't completely understand how to write integrals, so that's the best i can give you... i'm also pretty unsure of this answer, because when i integrate it and sub in the values i get a two fractions where i divide by zero, so i think i must be wrong...
As far as I can see, you need to calculate $\int_{1}^{5/2} \sqrt{16 \pi^2 \sin^2 (\pi t) + 9 \pi^2 \cos^2 (\pi t) + 1} \, dt$ ........

You will not have much luck getting an exact answer, I'm afraid.

Does this question require an exact answer? I doubt it very very much as it will involve the elliptic integral of the second kind. I suggest you ask your Monash lecturer or tutor what answer is required .....