
#1
Sep1807, 11:08 PM

P: 36

1. The problem statement, all variables and given/known data
r(t) = ( (1+t)/(1t), 1/(1t^2), (1/1+t) ) i have proven that the curve is planer now the second part of the problem is to find the equation of the plane that the curve lies in 2. Relevant equations Equation of a plane: A(xxo) + B(yyo) + C(zzo) = 0 3. The attempt at a solution I'm not certain how to go at this problem Idea: I'm thinking you can choose any 3 points, t = 0, t = 2, t = 3 Then: r(0) = (1,1,1) = A r(2) = (3,  1/8, 1/4) = B r(2) = (1/3, 1/3, 1) = C Then Vector BA = (4, 9/8, 3/4) Vector CA = (4/3, 4/3/, 2) then the normal for the plane would be BA x CA BA x CA = (5/4, 7, 23/6) so the equation of the plane would be (5/4)(x1) + (7)(y1) + (23/6)(z1) = 0 thus 5/4 x  7y + 23/6 z = 23/12 or 15x  84y +46z = 23 Idea 2: I'm thinking, since the curve is planer, the tangent at any point of the curve lies in the tangent plane can i choose any point let's say t=4 and since the binormal is normal to the tangent plane would I need to find B(4) which is T(4) x N(4) after I have B(4), could i plug in my "normal" to the tangent plane into the equation A(xxo) + B(yyo) + C(zzo) = 0 this method seems very tedious, trying to find T(t) and N(t) Is this the correct idea(s)? or am i totally off If i am off, are there any hints/ suggestions? thank you 



#2
Sep1907, 11:52 AM

P: 36

no takers?




#3
Sep1907, 12:25 PM

HW Helper
P: 2,566

Your ideas seem fine. Why not see if they give the same answer?




#4
Oct1407, 09:58 PM

P: 1

Given a curve, find the equation of the plane the cuve lies in
I'm curious how you figured it was planer without finding T and N all ready? or for that matter the Binormal as well (using Frenets Formulas). If its planer then your curvature is a constant and your torsion is 0. That being the case you should be able to use your Binormal to find the osculating plane at your choosen point r(t). Ask if you want to hear more or you haven't already completed the problem.




#5
Oct1507, 06:11 PM

P: 36

i completed this awhile ago
i proved it was planer by using the theorem a curve is planar if and only if the torsion is 0 i did the tedious task of computing the torsion, using the invariant formula, and it was indeed 0 thus i concluded it was planar the easy solution was to just "guess" the plane in which the given curve lies in assume the curve is planar Thus the curve is spanned by gamma dot and gamma double dot chose a point t=0 gamma dot (0) gamma double dot (0) took the cross product of the two, which would be the normal of the assumed tangent plane then plug in the normal into the equation A(xxo) + B(yyo) + C(zzo) = 0 once i got that equation, i plugged in the components of gamma x = __ y = __ z = __ into the equation of the plane, and indeed it's 0. thus gamma lies in the plane, and with this process, you also found the equation of the plane it lies in 


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