Revolutions of a helix

[KLS2111]

Hello,
The problem is how many revolutions will the circular helix r(t)= cos ti+sintj + 2tk make in a segment of arc length 50 units?

I really have no idea how to approach this problem.
I know that x= cos t
y=sin t
z=2t

What I don't know is if I am supposed to integrate because we are talking about arc length.. and then what would I do with the 50? Any guidance on how I should get started would be appreciated!

 

[galactus]

A helix has the form \(\displaystyle a\cdot cos(t)i+a\cdot sin(t)j+ctk\).

'a' is the radius and it rises \(\displaystyle 2{\pi}c\) units per revolution.


\(\displaystyle \frac{dx}{dt}=-sin(t)\)

\(\displaystyle \frac{dy}{dt}=cos(t)\)

\(\displaystyle \frac{dz}{dt}=2\)

The arc length formula for parametrics:

\(\displaystyle L=\int_{0}^{s}\sqrt{\underbrace{sin^{2}(t)+cos^{2}(t)}_{\text{this is 1}}+4}dt\)

Now, integrate. Then, set equal to 50 and solve for s. s is the total radians revolved. Then, divide that result by 2Pi to find the number of revolutions.