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Given a twice-differentiable vector-valued function r(t)and a point t0in its domain, what is the osculating plane at rt0?

Short Answer

Expert verified

The osculating plane atBt0=Tt0Nt0.

Step by step solution

01

Step 1. Given information. 

Consider the given question,

A twice-differentiable vector-valued function is rt0is any point in the domain ofrt.

02

Step 2. Definition of osculating plane at rt0.

Assume rt=xt,yt,ztbe a differentiable vector function on the interval IRsuch that,

T't00where, t0I. Then the osculating plane at rt0is defined as,

Bt0.x-xt0,y-yt0,z-zt0=0

Where, Bt0=Tt0Nt0.

Tt0,Nt0are the orthogonal unit vectors. so their cross product Bt0is another unit vector and is orthogonal to both Tt0,Nt0.

The osculating plane contains the point rt0and has Bt0as its normal vector.

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