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

r(t)=⟨t,5sin3t,5cos3t⟩,t=π6

Short Answer

Expert verified

Ans: The unit tangent vector to ⟨t,5sin3t,5cos3t⟩at t=π6 is1226,0,-15226

Step by step solution

01

Step 1. Given information:

r(t)=⟨t,5sin3t,5cos3t⟩,t=π6

02

Step 2. Simplifying the Unit tangent vectors :

Consider r(t)=⟨t,5sin3t,5cos3t⟩

First, we compute role="math" localid="1649673535411" r'(t)

r'(t)=ddt⟨t,5sin3t,5cos3t⟩=⟨1,15cos3t,-15sin3t⟩r'(t)=‖⟨1,15cos3t,-15sin3t⟩‖=(1)2+(15cos3t)2+(-15sin3t)2=1+225cos23t+225sin23t=1+225sin23t+cos23t=226

03

Step 3. Finding the Unit tangent vectors: 

The unit tangent vector to r(t)is

T(t)=r'(t)r'(t)=⟨1,15cos3t,-15sin3t⟩226

At t=Ï€6, the unit tangent vector to r(t)is

Tπ6=1,15cos3π6,-15sin3π6226=1,15cosπ2,-15sinπ2226=⟨1,0,-15⟩226=1226,0,-15226

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