Register to reply

Given a curve, find the equation of the plane the cuve lies in

by b0it0i
Tags: curve, cuve, equation, lies, plane
Share this thread:
b0it0i
#1
Sep18-07, 11:08 PM
P: 36
1. The problem statement, all variables and given/known data

r(t) = ( (1+t)/(1-t), 1/(1-t^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(x-xo) + B(y-yo) + C(z-zo) = 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)(x-1) + (-7)(y-1) + (23/6)(z-1) = 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(x-xo) + B(y-yo) + C(z-zo) = 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
Phys.Org News Partner Science news on Phys.org
Mysterious source of ozone-depleting chemical baffles NASA
Water leads to chemical that gunks up biofuels production
How lizards regenerate their tails: Researchers discover genetic 'recipe'
b0it0i
#2
Sep19-07, 11:52 AM
P: 36
no takers?
StatusX
#3
Sep19-07, 12:25 PM
HW Helper
P: 2,567
Your ideas seem fine. Why not see if they give the same answer?

Guinessrulz
#4
Oct14-07, 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.
b0it0i
#5
Oct15-07, 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(x-xo) + B(y-yo) + C(z-zo) = 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


Register to reply

Related Discussions
Find an equation of the tangent line to the curve at the given point Calculus & Beyond Homework 23
Is there a formula to find the intersection of a plane and a curve at a given point? Calculus & Beyond Homework 2
Prove that a curve lies in a plane Calculus & Beyond Homework 1
How to find out equation of a plane intersecting with other ? Calculus 2
How to find angle of a curve in xy coordinate plane? General Math 5