Find the unit tangent vector T(t) at the point with the given value of the parameter t. r(t) = t2 − 3t, 1 + 4t, 1 3 t3 + 1 2 t2 , t = 4.

The unit tangent vector will be "[tex]T(4) = <\frac{5}{21} ,\frac{4}{21} ,\frac{20}{21} >[/tex]".Given:→ [tex]r(t) = , t=4[/tex]By differentiating the components, we get[tex]r'(t) = <2t-3, 4, \frac{1}{3},3t^2+\frac{1}{2}.2t >[/tex][tex]r'(t) = <2t-3, 4, t^2+t>[/tex][tex]r'(4) = <2(4)-3, 4, (4)^2+4>[/tex][tex]r'(4) = <5, 4, 20>[/tex]On finding the magnitude, we get→ [tex]|r'(4)| = \sqrt{(5)^2+(4)^2+(20)^2}[/tex]             [tex]= \sqrt{25+16+400}[/tex]             [tex]= \sqrt{441}[/tex]             [tex]= 21[/tex] Hence,The unit tangent vector is:→ [tex]T(4) = \frac{r'(4)}{ |r'(4)|}[/tex]By substituting the values,            [tex]= \frac{<5,4, 20>}{21}[/tex]            [tex]= <\frac{5}{21} , \frac{4}{21} ,\frac{20}{21} >[/tex]Thus the answer above is correct.  Learn more about tangent here: