Find the equation for a unit speed planar curve alpha

In summary: I don't know where to even start on this problem. Any suggestions?In summary,homework equation 1: 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?homework equation 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?
  • #1
buzzmath
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Homework Statement


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?

Homework Equations


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
 
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  • #2
buzzmath said:
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?

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
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
buzzmath said:
Then we have t=24/sin(latitude) so the latitude is inverse sin(24/t) which gives you the distance from the equator.
thanks

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
buzzmath said:
1.Find the equation for a unit speed planar curve alpha such that k(t)=1/t

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
Yes, I think we're trying to find the distance on the Earth's 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
 

1. What is a unit speed planar curve?

A unit speed planar curve is a type of curve in which the speed of the curve is constant at every point. This means that the distance traveled along the curve in a given time interval is always the same, regardless of the specific point on the curve. Unit speed curves are commonly used in mathematics and physics to simplify calculations and to model various phenomena.

2. How do you find the equation for a unit speed planar curve?

To find the equation for a unit speed planar curve, you first need to parametrize the curve using the arc length parameter. This means that the parameter t represents the distance traveled along the curve. Then, you can use the parametric equations x(t) and y(t) to describe the x and y coordinates of the curve as a function of t. Finally, you can eliminate the parameter t to obtain the equation of the curve in terms of x and y.

3. What is the significance of using a unit speed parameter?

Using a unit speed parameter has several advantages. First, it simplifies calculations as the speed of the curve is constant. This allows for easier differentiation and integration when analyzing the curve. Additionally, unit speed curves are invariant under reparametrization, which means that the curve will not change shape if the parameter t is changed. This makes unit speed curves useful in modeling physical systems that are independent of time.

4. Can a unit speed planar curve have multiple equations?

Yes, a unit speed planar curve can have multiple equations. This is because the equation of the curve depends on the chosen parametrization. Different parametrizations can result in different equations that describe the same curve. However, all of these equations will have the same shape and characteristics as they represent the same curve.

5. What kind of curves can be described using a unit speed parameter?

A unit speed parameter can be used to describe a wide range of curves, including lines, circles, ellipses, parabolas, and more complex curves such as spirals and lemniscates. However, not all curves can be described using a unit speed parameter. For example, curves with sharp corners or cusps cannot be described using a unit speed parameter because they do not have a well-defined tangent at every point.

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