- #1

Hellken

- 13

- 0

**So what exactly is an osculating plane? r'(t) X r"(t)? Am I right?**

From what I know it is a plane that kisses a point on the curve at point t of curve f(t) and it is a plane normal to r'(t) X r"(t)? Am I right?

So to look for this osculating plane I simply have to find the cross product of r'(t) and r"(t)?

What does it look like on a xyz graph?

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EDIT: By the way per definition of the equation of a plane: a(x-x

_{0}) + b(y-y

_{0}+ c(z-z

_{0}) I need to find

a) a point on the plane (x

_{0},y

_{0},z

_{0})

b) a vector normal to the plane <a,b,c>

Hence if I am looking for the osculating plane I would need:

a) a point on the osculating plane (x

_{0},y

_{0},z

_{0})

b) B(t) which happens to be the vector normal to my osculating plane B(t) = T(t) X N(t), from this I find my desired <a,b,c>

I put it all together and I get my osculating plane of form: a(x-x

_{0}) + b(y-y

_{0}+ c(z-z

_{0}). (Right or wrong?)

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So what exactly is the cross product r'(t) X r"(t) for? Cutting down time so I don't have to go through the tedious process of finding T(t) and N(t)?

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