Given the vector r(t)=cos⁡T,sin⁡T,ln⁡(cos⁡T) and point (1, 0, 0) find vectors T, N and B at that point. Vector T is the unit tangent vector, so the derivative r(t) is needed. Vector N is the normal unit vector, and the equation for it uses the derivative of T(t). The B vector is the binormal vector, which is a crossproduct of T and N.

Question

Given the vector r(t)=cos⁡T,sin⁡T,ln⁡(cos⁡T) and point (1, 0, 0) find vectors T, N and B at that point.  Vector T is the unit tangent vector, so the derivative r(t) is needed.   Vector N is the normal unit vector, and the equation for it uses the derivative of T(t).   The B vector is the binormal vector, which is a crossproduct of T and N.

Ezra Herbert · Accepted Answer

The derivative of the vector is,r′(t)={−sin⁡5,cos⁡t,−tan⁡t}The unit tangent vector, T at the given point is obtained as follows.T(t)=r′(t)||r′(t)||T(t)=⟨−sin⁡t,cos⁡t,−tan⁡t⟩(−sin⁡t)2+(cos⁡t)2+(−tan⁡t2T(t)=⟨−sin⁡tsec⁡t,costsect,−tantsec⁡t⟩T(t)|1,0,0=⟨−0.45,1,0⟩The derivative of the unit tangent vector, T is,T′(t)=⟨−cos⁡t−tan⁡tsin⁡tsec⁡t,−sin⁡2t,−cos⁡t⟩The normal unit vector, N at the given point is obtained as follows.N(t)=T′(t)||T′(t)||N(t)=⟨−cos⁡t−tan⁡tsin⁡tsec⁡t,−sin⁡2t,−cos⁡t⟩(−cos⁡t+tan⁡tsin⁡t)2+sin2(2t)sec2(t)+1sec⁡(t)N(t)|1,0,0=⟨0.366,0,−0.70⟩The binormal vector, B at the given point is obtained as follows.B(t)=T(t)xxN(t)B(t)|1,0,0=[ijk−0.45100.3660−0.70]=(−0.70)i−(0.315)j−(0.366)k

Given the vector r(t) = { cosT, sinT, ln (CosT) } and point (1, 0, 0) find vectors T, N and B at that point.Vector T is the unit tangent vector, so the derivative r(t) is needed.

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