assuming you meant t=π/6
The unit tangent vector
T(t) = r'/|r'| =
cot(t) i - tan(t) j
-------------------------
tan^2(t) + cot^2(t)
The unit normal is T'/|T'| =
-(csc^2(t) i + sec^2(t) j)
----------------------------------
csc^4(t) + sec^4(t)
which at t = π/6 is
-(4i + 4/3 j)/(16 + 16/9)
= -4/3 (3i+j) * 9/156
= 1/13 (3i+j)
as always, check my math
The plane curve is given by the equation R(t)= ( ln sin t)i + ( ln cos t)j.
find the unit normal vector to the curve t= ¤Ð/6.
1 answer