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Unit tangent vectors: Find the unit tangent vector for the given function at the specified value of t.

r(t)=⟨3sin2t,3cos2t⟩,t=π3

Short Answer

Expert verified

Ans: Thus the unit tangent vector to ⟨3sin2t,3cos2t⟩at t=π3is-12,-32

Step by step solution

01

Step 1. Given information: 

r(t)=⟨3sin2t,3cos2t⟩,t=π3

02

Step 2. Simplifying the Unit tangent vectors :

Consider r(t)=⟨3sin2t,3cos2t⟩

First, we compute

r'(t)=ddt⟨3sin2t,3cos2t⟩=3ddt(sin2t),ddt(cos2t)=3⟨2cos2t,-2sin2t⟩=6⟨cos2t,-sin2t⟩r'(t)=‖6⟨cos2t,-sin2t⟩‖=6(cos2t)2+(-sin2t)2=6cos22t+sin22t

=6sincesin22t+cos22t=1

03

Step 3. Finding the Unit tangent vectors: 

The unit tangent vector to r(t)is

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

At t=Ï€3, the unit tangent vector is

Tπ3=cos2π3,-sin2π3=cos120°,-sin120°=-12,-32

Thus the unit tangent vector to ⟨3sin2t,3cos2t⟩at t=π3 is

-12,-32

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