Question

I have to find if r(t)=(t^2-2t^3,2t-t^2,3t^2-2t^3-4t) is a plane curve. If i take r(0)=(0,0,0) r(1)=(-1,1,-3),r(2)=(-12,0,-12) i get that the plane curve is -x+2y+z=0,
but if i take r(0),r(1),r(-1)=(3,-3,9) i get that the plane curve is 0=0. Is this a point or what is it?

Answer 1

A curve lies in a plane if all the tangent vectors T(t) line in a plane; and that's true if all the normal vectors N(t) lie in a plane. And that's true if all the binormal vectors
B(t) = T(t) x N(t)
are equal up to a constant. The vector B(t) is by construction perpendicular to the plane established by T(t) and N(t).

So calculate B(t) and show it's always in the same direction for all t.

Nice question.

Answer 2

btw, you'll save yourself a lot of work if you don't normalize the vectors. Just take
T(t) = r'(t), etc.

r(t)=(t^2-2t^3,2t-t^2,3t^2-2t^3-4t)
T(t) = r'(t) = (2t-6t^2, 2 - 2t, 6t - 6t^2 - 4)
N(t) = T'(t) = r''(t) = (2 - 12t, -2, 6 - 12 t)

Answer 3

yes, i was tryint to calculate torsion, becaouse if torsion is 0 then curve is a plane curve
but its a lot of work

Answer 4

Given what I just said, it's quite easy now. Just calculate B(t) and show it is always equal, up to a constant. I.e., if B_0 = B(0), then B(t) = f(t)B_0 for some scalar function f(t).

Answer 5

This is what I get:
B(t) = (12t^2 - 24t + 4) (1, -2, -1)

Answer 6

So that means that if i put any number instead of t, i will get the same function---and this means that the curve is plane curve?

Answer 7

It means that for any value of t, the bi-normal points in the same direction; it's just not normalized. Hence using the logic I articulated up top in this thread, that means the curve must lie in a plane, the plane which is perpendicular to B(t)

Answer 8

i get B_0=-4
B_1= 16
B_2=40

can you give me some example for f(t) so B_0=f(t)*B(t)

Answer 9

T(0) = (0,2,-4) and N(0) = (2,-2,6)

B(0) = T(0) x N(0) = (4, -8, -4)

which is consistent with my formula above:
B(t) = (12t^2 - 24t + 4) (1, -2, -1)

If you want to write B(t) = f(t)B(0), then f(t) = 3t^2 - 8t + 1

Answer 10

r(t)= (t^2-2t^3, 2t-t^2, 3t^2-2t^3-4t)
T(t) = r'(t) = (2t-6t^2, 2 - 2t, 6t - 6t^2 - 4)
N(t) = T'(t) = r''(t) = (2 - 12t, -2, 6 - 12 t)