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Solving a basic equation

  1. Sep 27, 2013 #1
    1. The problem statement, all variables and given/known data

    I have found the shortest distance between two points on the conic surface z = 1 - sqrt(x2 + y2) to be sqrt(x2 + y2) = ρ = Asec((θ/√2) + B). What is the actual equation of this path between the points (x,y,z) = (0, -1, 0) and (0, 1, 0)?
    2. Relevant equations



    3. The attempt at a solution

    here we have ρ = 1 for both coordinates. we also have θ = -pi/2 and pi/2. I cannot seem to solve this as i keep going in circles getting 0 = 0 and sometimes even more ridiculous results like pi/2 = -pi/2. I tried using wolfram alpha but they insist A = 0.

    I am taking my expression for ρ to be correct as it was confirmed by the instructor. what the heck is A and B????
     
    Last edited: Sep 27, 2013
  2. jcsd
  3. Sep 27, 2013 #2
  4. Sep 27, 2013 #3
    Here are some thoughts, although this is not something I know much about:

    First, I think your surface is a cone (although I'm not quite sure and maybe you should double check that).

    Your two points (0,1,0) and (0,-1,0) certainly lie on that surface. In fact they lie along a line orthogonal to the x-z axis and going through the point (0,0,0). That is somewhat to the "left" of the origin point of the cone, which I think is (0,0,1).

    Reasonableness check: Does it make sense to you that these two points are on that cone, and equidistant from the x-z plane? Are there any other points on that line which also lie on the cone?

    If this surface is indeed a cone, then the distance between the two points as you travel along the surface must be twice the distance from (0,1,0) to the origin of the cone at (0,0,1). Surely you know how to find the distance between two vectors?

    And again, if it is a cone, I think ##\theta## must be the angle it makes with the x-z axis. Perhaps you know?

    With this information, you can plug your distance into the expression for p, and find out what A is. The B terms with multiples of ##\pi## simply express that the sec is a periodic function with period ##\pi##.

    Can you see now what the equation must be to connect the two points?

    I think when Wolfram said A could be "anything" he may have meant that it is some real number which depends on the two points; given enough points then, A could be anything.
     
  5. Sep 30, 2013 #4
    [itex]
    1=A\sec(\frac{\nicefrac{-\pi}{2}}{\sqrt{2}}+B)[/itex]
    [itex]
    \frac{1}{A}=\sec(\frac{-\pi}{2\sqrt{2}}+B)
    [/itex]
    [itex]
    A=\cos(\frac{-\pi}{2\sqrt{2}}+B)
    [/itex]
    [itex]
    \cos(\frac{\pi}{2\sqrt{2}})\cos(B)+\sin(\frac{\pi}{2\sqrt{2}})\sin(B)=\cos(\frac{-\pi}{2\sqrt{2}})\cos(B)-\sin(\frac{-\pi}{2\sqrt{2}})\sin(B)
    [/itex]
    [itex]
    \cos(\frac{\pi}{2\sqrt{2}})\cos(B)+\sin(\frac{\pi}{2\sqrt{2}})\sin(B)-\cos(\frac{-\pi}{2\sqrt{2}})\cos(B)+\sin(\frac{-\pi}{2\sqrt{2}})\sin(B)=0
    [/itex]
    [itex]
    \cos(B)(\cos(\frac{\pi}{2\sqrt{2}})-\cos(\frac{-\pi}{2\sqrt{2}}))+\sin(B)(\sin(\frac{\pi}{2\sqrt{2}})+\sin(\frac{-\pi}{2\sqrt{2}}))=0
    [/itex]
    [itex]
    \cos(B)(\cos(\frac{\pi}{2\sqrt{2}})-\cos(\frac{-\pi}{2\sqrt{2}}))=-\sin(B)(\sin(\frac{\pi}{2\sqrt{2}})+\sin(\frac{-\pi}{2\sqrt{2}}))
    [/itex]
    [itex]
    \frac{\sin(B)}{\cos(B)}=\frac{\cos(\frac{-\pi}{2\sqrt{2}})-\cos(\frac{\pi}{2\sqrt{2}})}{\sin(\frac{\pi}{2\sqrt{2}})+\sin(\frac{-\pi}{2\sqrt{2}})}
    [/itex]
    [itex]
    \tan(B)=\frac{0}{\sin(\frac{\pi}{2\sqrt{2}})+\sin(\frac{-\pi}{2\sqrt{2}})}=0
    [/itex]
    [itex]
    B=0
    [/itex]

    but why....WHY would my instructor ask us to do this for such an easy question.
     
  6. Sep 30, 2013 #5

    vela

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    What was your thinking in the last step?

    This doesn't work because you actually have 0/0. Regardless of what B is equal to, the first line is always satisfied.
     
  7. Sep 30, 2013 #6
    well in that last step you mentioned, i didn't write out everything. i was a little lazy. basically because theta = -pi/2 and +pi/2, we'll find that A = cos(-pi/2sqrt2 + B) = cos(+pi/2sqrt2 + B). i skipped a couple steps, but basically they're equal. then i used a trig identity cos(A+B) = cos(A)cos(B) - sin(A)sin(B) but i just noticed i put + instead of -. but i don't think it matters because of the negative angle anyways.

    i was given the hint by my instructor to use the identity i mentioned above. I wrote down everything in my post and showed him, and at first he said "you should just find B = 0", suggesting my work was incorrect and complicated. he looked over it again and pointed out that it was actually right, and that in my work, B = 0.

    but yes, we do have 0/0. which further makes me wonder what the heck is even going on.
     
  8. Sep 30, 2013 #7
    Since the cosine is periodic of ##2 \pi ## B has to be a multiple of ##2 \pi ##. 0 works as well as anything else but ## cos( \theta + 2\pi) = cos \theta##. You don't need to go thru all that computation to get B.

    The other question I have is why you took the distance between the 2 points to be 1. If you consider the shortest distance to be just the distance in ##R^2## then the distance is 2. If, as I interpreted it the distance is to be taken on the surface of the conic section, the distance is ##2\sqrt{2}##.
     
  9. Sep 30, 2013 #8

    vela

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    How did you derive the equation? I suspect the constants arose due to integrations. You might find it helpful to use definite integrals to see what A and B geometrically represent.
     
  10. Sep 30, 2013 #9
    hi brmath, rho = sqrt(x^2 + y^2), and we are given two points (0,-1,0) and (0,1,0), so for the first point, we have rho = 1, theta = -pi/2. for the second point, we have rho = 1, theta = pi/2.

    i appreciate your guys' approach(es) to this problem, but at the end of the day i was given hints by my instructor, did it, and found B = 0, which he confirmed. B is just a phase shift. A is a constant of proportionality. i derived the equation by using the equation for arc length basically, but in cylindrical coordinates. i used the euler-lagrange equation to end up with an ODE, solved it by sep. of var., and that's what i got. he told us that was the answer.

    i handed in the assignment on friday so it doesn't really matter, but i sure would love to figure out why i'm getting a few different interpretations.
     
  11. Sep 30, 2013 #10
    Sorry I wasn't able to help, and glad you could work it out with your teacher.

    As to the different approach, quite possibly I didn't understand the problem, which is often the reason people take different approaches -- they are not working on the same problem. Another possibility is that I was working from first priniciples such as what is this surface? and where are these two points on it? so what is the distance between them? I often do that, and sometimes manage to evade long computations; it is even possible you would get the same answer continuing along with what I did.

    He says ##\rho = \sqrt{x^2 + y^2}## is the shortest distance between 2 points on the surface. This is not a distance between two points because there is only one point here (x,y,z). So I didn't understand his ## \rho## and kind of ignored it.

    Next, you are saying ##\theta =\pm \pi /2##, but what is ##\theta ## geometrically? If it is ## \pm \pi/2## then it would be the angle between the vector say (0,1,0) and the xz plane, but what does that have to do with the surface? I would have thought ##\theta## would be the angle between the xz plane, the vertex of the cone and the given point which is ##\pm \pi/4##. It is important to define your terms.

    Actually, my first thought was to compute the arclength along the surface , which looked to me like a line integral; and I wasn't ready to look up or figure out how to parameterize that line. Instead, I thought I could get through it without the heavy machinery. The line integral will of course lead to a diff eq because there are derivatives in the integrand.

    As for the B, you say it is just a "phase shift". Is that any different from saying the cosine is periodic?

    Anyway apparently I misunderstood the problem. Can you enlighten me as to what it really was?
     
  12. Oct 1, 2013 #11

    vela

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    The original post wasn't very clear. Here ##\rho## is simply the radial coordinate in the cylindrical coordinate system ##(\rho,\theta,z)##. The equation ##\rho = A\sec(\theta/\sqrt{2}+B)## is that for a geodesic on the given cone.
     
  13. Oct 1, 2013 #12
    ***
    Well, I went over it again last night, and concluded the following:

    Yes it is a cone centered at (0,0,1).
    The angle between (0,1,0) and (0,0,1) is indeed ##\pi/4##
    The distance between (0,1,0) and (0,-1,0) along the surface is ##2 \sqrt 2## (based on the Pythagoreum theorem)
    If we write ##2 \sqrt 2 = (1/A) cos \pi /4## we get ##2 \sqrt 2 = (1/A) \sqrt 2## and A = 1/2. B can be any multiple of ##2 \pi ##.
    If you want an equation for the curve connecting the two points, it will be 2 straight lines that meet at (0,0,1).

    I hope that is the answer you got.

    The advantage of this method is that it avoids all that computation and solves the problem with common sense.

    The disadvantage is that you don't learn the arclength method, which is no doubt what your prof was trying to teach you.
     
    Last edited: Oct 1, 2013
  14. Oct 1, 2013 #13

    Thank you for the clarification. I really could not make any sense of the ## \rho##.

    I've never worked out a problem like this, and was glad this one was easy enough to kind of power my way thru it. Obviously this method wouldn't work for two random points on the cone, and I too would be forced back into computing the arclength.

    However, once that computation is done, why does one need the formula with the A and B? Is it just to put the answer into a nicer format?
     
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