find the principal unit normal vector to the curve at the sp

find the principal unit normal vector to the curve at the specified value of the parameter

r(t)=ti+((6)/(t))j t=2

Solution

dr/dt=i-(6/t^2)j is a tangential vector. When t=6, r=6i+j, and dr/dt=i-(1/6)j m=(r)x(dr/dt)=-2k is a normal to the plane spanned by vectors i and j. (dr/dt)x(m)=(1/3)i+2j is the outward normal to the curve in the i-j plane at the point defined by t=6. Finally, the unit normal vector at that location is equal to n={[(1/3)i+2j]/sqrt[(1/3)^2+(2)^2]} =(i+6j)/sqrt(37)