Revolutions of a helix

KLS2111

New member
Joined
Mar 8, 2007
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22
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!
 
A helix has the form acos(t)i+asin(t)j+ctk\displaystyle a\cdot cos(t)i+a\cdot sin(t)j+ctk.

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


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

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

dzdt=2\displaystyle \frac{dz}{dt}=2

The arc length formula for parametrics:

L=0ssin2(t)+cos2(t)this is 1+4dt\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.
 
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