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Given x(t) = (2t,t^2,t^3/3), I am asked to "find equations for the osculating planes at time t = 0 and t = 1, and find a parameterization of the line formed by the intersection of these planes."

I have already solved the vector-valued functions for x.

I know this is vague and I'm asking for a lot, but can someone please explain how to solve the problem and also what parameterization is?

Thank you!

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1 Answer

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First find the unit tangent vector

$${\bf T} = \frac{{\bf x'}}{\|{\bf x'}\|}$$

Then the unit normal (or just normal for ease):

$${\bf N} = \frac{{\bf T'}}{\|{\bf T'}\|}$$

And you may need the binormal which is just

$${\bf B} = {\bf T} \times {\bf N} $$

Then your point is ${\bf x}(0)$ and your vector is ${\bf B}(0)$. Then plug into the plane equation:

$$a(x-x_0) + b(y-y_0) + c(z-z_0)=0$$

Example: $x(t) = \langle \cos(t), \sin(t), t \rangle$

I get $${\bf T} = \langle -\frac{1}{\sqrt{2}}\sin(t), \frac{1}{\sqrt{2}}\cos(t), \frac{1}{\sqrt{2}} \rangle$$

Then the unit normal (or just normal for ease):

$${\bf N} = \langle \cos(t), -\sin(t), 0 \rangle$$

And you may need the binormal which is just

$${\bf B}(0) = {\bf T}(0) \times {\bf N}(0) $$

$${\bf B}(0) = \langle -\frac{1}{\sqrt{2}}\sin(0), \frac{1}{\sqrt{2}}\cos(0), \frac{1}{\sqrt{2}} \rangle \times \langle \cos(0), -\sin(0), 0 \rangle $$

$${\bf B}(0) = \langle 0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \rangle $$

I get the point

$x(0) = \langle \cos(0), \sin(0), 0 \rangle = \langle 1, 0, 0 \rangle $

So the plane is

$$0(x-1) + \frac{1}{\sqrt{2}}(y-0) - \frac{1}{\sqrt{2}}(z-0)=0$$

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