Find the unit tangent vector T(t ) at the point with the given value of the parameter t. r(t)=⟨t2−2t,1+3t,13t3+12t2⟩

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Find the unit tangent vector T(t ) at the point with the given value of the parameter t.  r(t)=⟨t2−2t,1+3t,13t3+12t2⟩

2k1enyvp · Accepted Answer

Let r(t) be a differentiable vector-valued function and v(t) = r‘(t) be the velocity vector. Then we define the unit tangent vector as the unit vector in the direction of the velocity vector, that is,T(t)=v(t)||v(t)||=r′(t)||r′(t)||Here,  the given vector function is,r(t)=⟨t2−2t,1+3t,13t3+12t2⟩Differentiate the above vector function component-wise, we get,r′(t)=⟨2t−2,3,t2+t⟩Substitute t=2r′(2)=⟨2⋅2−2,3,22+2⟩r′(2)=⟨2,3,6⟩Find the magnitude (or the norm), that is,|r′(2)|=22+32+62|r′(2)|=4+9+36|r′(2)|=49=7Unit tangent vector is T(2)=r′(2)|r′(2)|=⟨2,3,6⟩7=⟨27,37,67⟩Result:The unit tangent vector corresponding to given vector function isT(2)=⟨27,37,67⟩

Find the unit tangent vector T(t ) at the point with the given value of the para

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