# Find the equation for a unit speed planar curve alpha

1. Dec 7, 2007

### buzzmath

1. The problem statement, all variables and given/known data
1.Find the equation for a unit speed planar curve alpha such that k(t)=1/t

2.If the earth has radius r and the plane in which a pendulum swings in a city rotates by angle theta in 24 hours what is the distance from the city to the equator?

2. Relevant equations

3. The attempt at a solution

1.Since alpha has unit speed k should equal the second derivative of the norm of alpha. So to find alpha I think I just need to integrate twice from 0 to t and then add a constant. I get a problem because when I integrate 1/s from 0 to t I get ln(s) from 0 to t but ln(0) is not defined. So how would I get around this problem?

2. I don't know where to even start on this problem. Any suggestions?

thanks

Last edited: Dec 7, 2007
2. Dec 7, 2007

### dynamicsolo

For this question, you'll want to look up "Foucault pendulum". (If I'm understanding the description in the question correctly, I know what the answer would be, but I don't know of any city there...)

3. Dec 8, 2007

### buzzmath

I think it is the Foucault pendulum. The question is if you set up a pendulum in a city. The plane on which this pendulum swings rotates by a total angle theta in 24 hours. The earth has radius r. what is the distance from the city, where the pendulum is, to the equator?
So I look at the Foucault pendulum and I came up with that if you take theta to be the angle the plane rotates in 24 hours divide theta by 24 and that is the angle in which the plane rotates in one hour (call this angle phi). then divide 360 by phi which gives you the length of time it takes the plane to rotate in a complete circle. Let t be this value ( the time it takes the plane to rotate in a complete circle). Then we have t=24/sin(latitude) so the latitude is inverse sin(24/t) which gives you the distance from the equator. Does this sound right? I'm a little confused because I think that r, the radius of earth, should be in there someplace.
Any suggestions on problem 1?
thanks

4. Dec 8, 2007

### dynamicsolo

The issue here is that the plane of the pendulum's swing is stationary with respect to "the stars", that is, a reference frame which is not rotating "with respect to the Universe". As seen on the rotating Earth, the plane of the pendulum seems to cycle around once every Earth rotation period (actually a sidereal, not a solar, day). The formula you cite means that this occurs at latitude 90º North or South, which is why I remarked that I knew of no city there. The radius of the Earth enters in only because they appear to be asking for the distance along the Earth's surface from these locations to the Equator.

5. Dec 8, 2007

### dynamicsolo

Am I correct in understanding that k(t) here represents the curvature function? If so, you will need the equation which expresses curvature for a parameterized curve. A unit-speed curve means, I believe, that the parameter is chosen so that ds/dt = 1 , where s represents "arclength" along the curve.

6. Dec 8, 2007

### buzzmath

Yes, I think we're trying to find the distance on the earths surface. Is what I gave correct for the latitude? then could I just use that latitude to find the arclength from the equator to that latitude?
k(t) is the curvature of the function. the equation for the curvature is the second derivative of the norm of the parametization. so ds/dt=1 if s(t) is the parametrization. do the second derivative of s(t) = 1/t which is the curvature. so you would need to integrate right? but then you get ln(t) which isn't defined at 0. so how would you go about doing this?
thanks