Question

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.

Answer 1

The unit tangent vector will be "$T(4) = <\frac{5}{21} ,\frac{4}{21} ,\frac{20}{21} >$".

Given:

→ $r(t) = , t=4$

By differentiating the components, we get

$r'(t) = <2t-3, 4, \frac{1}{3},3t^2+\frac{1}{2}.2t >$

$r'(t) = <2t-3, 4, t^2+t>$

$r'(4) = <2(4)-3, 4, (4)^2+4>$

$r'(4) = <5, 4, 20>$

On finding the magnitude, we get

→ $|r'(4)| = \sqrt{(5)^2+(4)^2+(20)^2}$

            $= \sqrt{25+16+400}$

            $= \sqrt{441}$

            $= 21$

Hence,

The unit tangent vector is:

→ $T(4) = \frac{r'(4)}{ |r'(4)|}$

By substituting the values,

           $= \frac{<5,4, 20>}{21}$

           $= <\frac{5}{21} , \frac{4}{21} ,\frac{20}{21} >$

Thus the answer above is correct.  

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