Find equation of plane given curve

[nicknack1016]

Homework Statement

r(t) = (2t)i + (t^2)j + (1 - t^2)k
Find the equation of the plane that this curve lies on.

Homework Equations

The Attempt at a Solution

I've found planes when I had vectors or lines but now with a curve I'm not sure where to even start. Can someone just give me a starting point? I don't want it to be solved for me or anything. I'm just looking for a point to even begin.

 

[LCKurtz]

Try eliminating the parameter from your equations

x = 2t, y = t², z = 1 - t².

 

[gerben]

The plane that r(t) lies on is always perpendicular to some vector, n. This vector, n, is normal to the plane. So find the vector, n, that is always normal to r(t). This vector defines the plane.

 

[nicknack1016]

To find a normal vector wouldn't i need a point as well? Or can i just substitute in some value of T in r(T) to get a point?

 

[LCKurtz]

The plane that r(t) lies on is always perpendicular to some vector, n. This vector, n, is normal to the plane. So find the vector, n, that is always normal to r(t). This vector defines the plane.

r(t) is a position vector from the origin to a point on the curve. The normal n to the plane will not be perpendicular to r(t).

 

[nicknack1016]

solving for T in the equation x = 2t then substituting it into both of those equations gave me...
y = x^2/4, z = 1 - x^2/4

then using the Y equation
z = 1 - y then 1 = z + y
would this be my plane?

 

[LCKurtz]

solving for T in the equation x = 2t then substituting it into both of those equations gave me...
y = x^2/4, z = 1 - x^2/4

then using the Y equation
z = 1 - y then 1 = z + y
would this be my plane?

Yes, it would. The fact that x is missing in the equation just means the plane is parallel to the x axis. x can be any value as long as y + z = 1. Now that you have the equation of the plane, it is a trivial matter to read off the normal.

 

[nicknack1016]

Now I need to find the point where the tangent line at T = 2 intersects the xy-plane.
I found the derivative of r(t)..
r'(t) = 2i + (2t)j - (2t)k
then substituted in 2 for t
r'(2) = <2, 4, -4>
this is the directional vector of the tanget line.
how would i use this to find where it hits the xy-plane?

 

[LCKurtz]

Now I need to find the point where the tangent line at T = 2 intersects the xy-plane.
I found the derivative of r(t)..
r'(t) = 2i + (2t)j - (2t)k
then substituted in 2 for t
r'(2) = <2, 4, -4>
this is the directional vector of the tanget line.
how would i use this to find where it hits the xy-plane?

You can figure out the position r(2) and you have the direction vector, so write the parametric equations of the line. Put the x,y, and z in the plane and figure out what parameter t works and...

 

[nicknack1016]

wow really? haha awesome.
so r(2) = 4i + 4j - 3k
x = 4 + 2t, y = 4 + 4t, z = -3 - 4t
it intersects at z=0 sooo t = -3/4
when t = -3/4...x = 5/2, y = 1, and z = 0
=D thanks for all the help it was very much appreciated!