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Principal unit normal vectors: Find the principal unit normal vector for the given function at the specified value of t.

r(t)=⟨3sin2t,3cos2t⟩,t=π3

Short Answer

Expert verified

Ans: The principal unit normal vector of r(t)=⟨3sin2t,3cos2t⟩at t=π3is-32,12

Step by step solution

01

Step 1. Given information:

r(t)=⟨3sin2t,3cos2t⟩,t=π3

02

Step 2. Simplifying the principal unit normal vector:

The principal unit normal vector of r(t)denoted by N(t)is given by

N(t)=T'(t)T'(t)where T(t)=r'(t)r'(t)

T(t)should be calculated first and then N(t)is to be evaluated. We start by comparing r'(t)first:

r'(t)=⟨6cos2t,-6sin2t⟩r'(t)=(6cos2t)2+(-6sin2t)2=36cos22t+sin22t=6

The unit tangent vector of r(t)is

T(t)=r'(t)r'(t)=⟨6cos2t,-6sin2t⟩6=⟨cos2t,-sin2t⟩

03

Step 3. Finding the principal unit normal vector:

To find N(t)we divide T'(t)by its magnitude.

T'(t)=(-2sin2t)2+(-2cos2t)2=4sin22t2+(-2cos2t)2=2

The principal unit normal vector

N(t)=T'(t)T'(t)=⟨-2sin2t,-2cos2t⟩2=⟨-2sin2t,-cos2t⟩Att=π3,N(t)=Nπ3=-sin2π3,-cos2π3=-32,12

Thus the principle unit normal vector of r(t)at t=Ï€3is

-32,12

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