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Helix Lying on a Cylinder


Date: 11/26/97 at 14:38:11
From: Celina Alvarez
Subject: Equation of a helix lying on a cylinder

I am trying to find a parametric equation for a helix (in 3 space) 
lying on the cylinder, x^2+z^2 = 4, and winding three times around 
the y-axis. The helix starts at the point (-2,3,0) and ends at the 
point (-2,9,0).

I have the general equation:

x(T) = {2Cos[T],2Sin[T],T}  and T goes from 0 to 6Pi (so it goes 
3 times around the y-axis)

However, I can not figure out how to get the helix to start and end at 
the given points or how to get the helix to lie on the given cylinder.


Date: 11/26/97 at 21:57:44
From: Doctor Pete
Subject: Re: Equation of a helix lying on a cylinder

Hi,

Since (Cos[t])^2+(Sin[t])^2 = 1, we see that 
(2 Cos[t])^2 + (2 Sin[t])^2 = 4, so if we let 
x[t] = 2 Cos[t], z[t] = 2 Sin[t], then any point of the form 
(x[t],y,z[t]) will lie on the given cylinder. So the helix will have 
some parameterization

     h[t] = (x[t],y[t],z[t]) = (2 Cos[t], y[t], 2 Sin[t]),

for t in some interval [a,b]. 

Now, we require at the start point t = a,

     h[a] = (2 Cos[a], y[a], 2 Sin[a]) = (-2, 3, 0),

and one possible value of a which satisfies this is a = Pi. 
But since we also require the helix to wind around three times, 
b = 6 Pi + a = 7 Pi. Indeed,

     h[b] = (2 Cos[7 Pi], y[7 Pi], 2 Sin[7 Pi]) = (-2, y[7 Pi], 0).

So we conclude that y[Pi] = 3, and y[7 Pi] = 9.  But we know that 
y[t] must be a linear function in order to get a helix; that is, 
y[t] = mt + n for some values m, n. So

       y[Pi] =  m(Pi) + n = 3
     y[7 Pi] = 7m(Pi) + n = 9,

whereupon subtracting the first from the second, we find 6m(Pi) = 6, 
or m = 1/Pi, and substituting this into the first equation, 1 + n = 3, 
so n = 2. Therefore,

     y[t] = t/Pi + 2,

for t between Pi and 7 Pi. Hence

     h[t] = (2 Cos[t], t/Pi + 2, 2 Sin[t]), t = [Pi, 7 Pi].

This gives one parameterization of the helix. There is another, 
namely the one which goes counterclockwise with respect to the 
positive y-axis,

     g[t] = (2 Cos[t], t/Pi + 2, -2 Sin[t]), t = [Pi, 7 Pi].

Note we can substitute u = t - Pi in either of these to give a 
different parameterization

     H[u] = (2 Cos[u+Pi], (u+Pi)/Pi + 2, 2 Sin[u+Pi])
          = (-2 Cos[u], u/Pi + 3, -2 Sin[u]), u = [0, 6 Pi],

     G[u] = (-2 Cos[u], u/Pi + 3, 2 Sin[u]), u = [0, 6 Pi].

-Doctor Pete,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
College Calculus
High School Calculus

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