Question

Find the unit tangent vectorat the point with the given value of the parameter

Short answer

The tangent vector is

\[T(0) = \frac{3}{5}j + \frac{4}{5}k\]

Step-by-step solution

Step 1: Solution of the 

Apply derivative on each side

\[\begin{array}{l}{{\bf{r}}^\prime }(t) = \frac{d}{{dt}}(\cos t{\bf{i}} + 3t{\bf{j}} + 2\sin 2t{\bf{k}})\\ = \frac{d}{{dt}}(\cos t){\bf{i}} + \frac{d}{{dt}}(3t){\bf{j}} + 2\frac{d}{{dt}}(\sin 2t){\bf{k}}\\ = - \sin t{\bf{i}} + (3){\bf{j}} + 2(2\cos 2t){\bf{k}}\\{{\bf{r}}^\prime }({\rm{t}}){\rm{ }} = - \sin t{\bf{i}} + 3{\bf{j}} + 4\cos 2t{\bf{k}}\end{array}\]

Step 2: Solution of the 

Putin

\[\begin{array}{l}{{\bf{r}}^\prime }(0) = - \sin (0){\bf{i}} + 3{\bf{j}} + 4\cos 2(0){\bf{k}}\\ = (0){\bf{i}} + 3{\bf{j}} + 4\cos (0){\bf{k}}\\ = 3{\bf{j}} + 4(1){\bf{k}}\\{{\bf{r}}^\prime }(0) = 3{\bf{j}} + 4{\bf{k}}.{\rm{ }}\end{array}\]

Step 3: Solution of the magnitude of 

The magnitude of the

\[\begin{array}{l}\left\| {{{\bf{r}}^\prime }(0)} \right\| = \sqrt {{3^2} + {4^2}} \\ = \sqrt {9 + 16} \\ = \sqrt {25} \\\left\| {{{\bf{r}}^\prime }(0)} \right\| = 5{\rm{ }}\end{array}\]

Step 4: Solution of the tangent vector

The tangent vector is

\[\begin{array}{l}T(0) = \frac{{r'(0)}}{{||r'(0)||}}\\T(0) = \frac{{3j + 4k}}{5}\\T(0) = \frac{3}{5}j + \frac{4}{5}k\end{array}\]