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Binormal vectors and osculating planes: Find the binormal vector and equation of the osculating plane for the given function at the specified value of t.

r(t)=sint,cost,t=0

Short Answer

Expert verified

The binomial vector is 0,0,-1.

The equation of the osculating plane is z=0.

Step by step solution

01

Step 1. Given Information  

We are given,

r(t)=sint,cost,t=0

02

Step 2. Finding the binormal vector. 

Finding the binormal vector,

r(t)=sint,costr'(t)=cost,-sintr'(t)=(cost)2+(-sint)2=T(t)=r'(t)r'(t)=cost,-sint=cost,-sint

At t=0,

role="math" localid="1649754634630" T(t)=T(0)=cos0,-sin0=1,0T'(t)=-sint,-costT'(t)=(-sin)2+(-cost)2=N(t)=T'(t)T'(t)=-sint,-costN(t)=-sint,-cost

03

Step 3. Finding the binormal vector. 

At t=0,

N(t)=N(0)=-sin0,-cos0=0,-1

Now,

The binormal vector is given by,

B(0)=T(0)N(0)=1,00,-1=ijk1000-10=k(-1)=-k=0,0,-1

Hence, the binormal vector is 0,0,-1.

04

Step 4. Finding the equation of the osculating plane  

The equation of the osculating at r(t=0)is defined by,

B(0)x-x(0),y-y(0),z-z(0)=00,0,-1x-0,y-,z=0-z=0z=0

Hence, the equation is z=0.

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