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Plane curves

  1. Sep 17, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the curve whose curvature is 2, passes through the point (1,0) and whose tangent vector at (1,0) is [1/2 , (√3)/2 ].

    3. The attempt at a solution

    I know I must use the Fundamental theorem of plane curves but I don't know how to apply it correctly here.
     
  2. jcsd
  3. Sep 17, 2013 #2
    Some hints:

    Let us call that curve f(x(t),y(t)). Start by looking up the definition of curvature. [ If you've been confused by whatever textbook you are using try http://www.millersville.edu/~rumble/Math.457/chapter1.pdf [Broken], -- it's very nicely laid out.] What do you think it means that the curvature is constant.

    Next, look up how the tangent vector is computed.

    Now do some thinking: why is knowing the curvature, the value and tangent vector at (1,0) sufficient to construct the curve?

    A way to start this thinking is to look at the problem in 1 dimension. Suppose you have a function g(x). In analogy to your problem, lets say we have the value of f(0). What is the tangent line at x = 0? Suppose the curvature is constant. How would you reconstruct this single variable function from that information?

    Once you get that far, maybe you can see what to do in 2 dimensions. I'm not sure you need to hit this one with a heavy theorem.
     
    Last edited by a moderator: May 6, 2017
  4. Sep 17, 2013 #3
    Thanks for that link.

    "What do you think it means that the curvature is constant?" - If the curvature is constant then it has the same values at all points.

    "Look up how the tangent vector is computed." - Yup, I know how to compute tangent vector.

    "Now do some thinking: why is knowing the curvature and tangent vector at (1,0) sufficient to construct the curve?" - Since I know what the tangent vector is at (1,0) then I know where the curve is moving at (1,0). Since I know the curvature I know how fast it moves?
     
  5. Sep 17, 2013 #4
    Yes, the curvature is the same at all points. What kind of a curve is that? Could it be w shaped? elliptic?

    You know, I missed something -- they gave you the value of the function at (1,0). I have edited my original response - please read it again. I strongly recommend that you try this in 1 dimension. Assume you know the value of the function at x = 0, the tangent line at x = 0 and that the curvature is a constant say 2. What function is this? You can figure it out completely from this information.

    The advantage of figuring that out is that it will guide you in reconstructing your function in 2 variables.

    Now your use of the term "moving" -- a very physics oriented approach to understanding curves and fine when you have a physics problem about motion. But you are in a math world right now, so we express velocity and acceleration in terms of derivatives. You have to think that way to solve this problem.
     
  6. Sep 17, 2013 #5
    "Yes, the curvature is the same at all points. What kind of a curve is that?"

    - A circle or a sphere?

    "Suppose you have a function g(x). In analogy to your problem, lets say we have the value of f(0). What is the tangent line at x = 0?"

    - The tangent line will just be the derivative of the function f(0) thus f'(0) will be our tangent line at x = 0.

    "Suppose the curvature is constant. How would you reconstruct this single variable function from that information?"

    - I am not sure?

    "But you are in a math world right now, so we express velocity and acceleration in terms of derivatives. " "Why is knowing the curvature and tangent vector at (1,0) sufficient to construct the curve?"

    - Hmm, I am not sure how else to express it without using moving? What will knowing the curvature and tangent vector at (1,0) sufficient to construct the curve?

    "Assume you know the value of the function at x = 0, the tangent line at x = 0 and that the curvature is a constant say 2. What function is this?"

    - I am not sure? Can you elaborate further please. I really want to understand this.
     
  7. Sep 17, 2013 #6
    First, you are right -- in a 2 dim situation this is a circle. So you can almost get this without too much extra effort. You need to know the center and radius of your circle. The curvature will tell you the radius (check back on the link I sent you if you are not sure). The other information will allow you to locate the center. Do you know the formula for a circle of radius r centered at the point (a,b)?

    Now back to the 1-dim case because understanding what you are doing is important, and this will help a lot. The tangent line at 0 is not really f'(0), which is a number, not a line. What does that number represent? How can you construct the tangent line from it?

    Continuing in 1-dimension, there is a property of f which allows you to compute the curvature. Do you know what that is? (I'm sure you studied it in 1st year calc, but you may have forgotten -- if so look it up -- go to the index in your calc book and use "curvature" -- or look it up online ).

    Re "moving" let's do some physics. Suppose h(t) is a function that describes the distance traveled during the time from 0 to t. What is the velocity in terms of h(t)? What is the acceleration in terms of h(t)? In other words, what are the mathematical constructs that describe the motion?
     
  8. Sep 17, 2013 #7
    "The curvature will tell you the radius"

    - I know the radius will equal R=1/k, where k is the curvature?

    Do you know the formula for a circle of radius r centered at the point (a,b)?

    - Is it (a-x)^2 + (b-y)^2 = r^2?


    "What does that number represent? How can you construct the tangent line from it?"

    - I am not sure?

    "Continuing in 1-dimension, there is a property of f which allows you to compute the curvature."

    - Is it arclength?

    "What is the velocity in terms of h(t)? What is the acceleration in terms of h(t)? In other words, what are the mathematical constructs that describe the motion?"

    - I am not sure where this one is leading to? The velocity will just be the derivative of that function.
     
    Last edited: Sep 17, 2013
  9. Sep 17, 2013 #8
    In general f'(x) is the slope of the tangent line at point x. The curvature at point x is given by the second derivative f''. If f'' is constant that determines what kind of function f must be. The arclength only tells you how long the line would be if you stretched your arc out into a line.

    That's right, the velocity at time t is h'(t) and the acceleration is h''(t). So if you like to think in terms of motion, that is fine, but stay aware that from a mathematical point of view it translates into derivatives.

    Your formula for a circle is correct, altho we usually write x -a rather than a - x; and you are right about the radius. So if you can just figure out where this circle is centered, you are done.
     
  10. Sep 17, 2013 #9
    May I recommend that you go back to your calc 1 book and straighten out this stuff about 1st and 2nd derivatives. It's awfully hard to deal with multi-variate calculus when you have uncertainties about the single variable calc.
     
  11. Sep 17, 2013 #10
    "So if you can just figure out where this circle is centered, you are done. "

    - So if I apply my equation

    (x-a)^2 + (y-b)^2 = r^2, I get
    (1-a)^2 + (0-b)^2 = (1/2)^2
    (1-a)^2 + (b)^2 = 1/4

    From this how can I continue?
     
  12. Sep 17, 2013 #11
    You've got two unknowns, a and b, which are the coordinates of the center of the circle. To find two unknowns you need two conditions, and you have so far used just one -- the fact that the function passes through (1,0). That allows you solve for a in terms of b (or vice versa) but isn't enough to pin down the second value.

    However, you've got some information about this curve that you haven't used. What is it? That information is key to nailing down your a and b.
     
  13. Sep 17, 2013 #12
    I am not getting it. If you show me how to construct it then I will have a concrete example to follow on which then I can use for future use for examples like these.
     
  14. Sep 17, 2013 #13
    You know 3 things about this curve. Your problem was find the curve
    1. whose curvature is 2,
    2. passes through the point (1,0)
    3. and whose tangent vector at (1,0) is [1/2 , (√3)/2 ].

    Have you used #1, the curvature? How?
    Have you used #2, passing thru (1,0)? How?
    Have you used #3? How?

    There is one of these you have not used yet,and you need it to figure out what (a,b) is.

    Sorry not to just hand you the answer, but I'm hoping you'll learn more or it will stick better if you think about things
     
  15. Sep 17, 2013 #14
    I just never had a concrete example of how to do this. We do a lot of proofs instead of examples like these, but I am going to start over again.

    Let us call our curve f(t). The curvature for f(t) is given by k=|f''(t)|=2. Since our curvature is a constant 2, we can think of this curve f(t) as a circle with radius r = 1/2. The equation for a circle is:

    (x-a)^2 + (y-b)^2 = r^2, since the circle passes through (1,0) I can plug it in for (x,y) thus I get
    (1-a)^2 + (0-b)^2 = (1/2)^2
    (1-a)^2 + (b)^2 = 1/4

    So we are given the points on the circle but we don't know what the center is but we are given the tangent vector.

    So at (1,0) the tangent line is (1/2, (√3)/2). I can get the slope of this tangent line by rise over run (simple enough). Thus slope equals m = √3.

    But I get stuck now.
     
  16. Sep 18, 2013 #15
    Hi Lee7, I'm going to give you one further hint, and I hope you can get it together from there. What do you know about the connection between the tangent line to a circle and the radius which connects the origin to the point of tangency? There is a strong connection.

    It is always helpful to look at specific examples, so consider the possibility that your circle is centered at (1/2,0). What is the tangent line at (1,0)? Why doesn't (1/2, 0) work with your hypotheses?

    Best wishes,
    brmath
     
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