Question

If r ( t ) = ( e^2t, e^-2t , te^2t )find T ( 0 ), r'' ( 0 ),and r' ( t ) x r'' ( t )

Short answer

The solution is

\(\begin{array}{l}r'(t) = ({e^t},(1 + t){e^t},(1 + 2{t^2}){e^{{t^2}}})\\T(1) = \left( {\frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}} \right)\\r''(t) = {e^t},(1 + {t^3}){e^t} + 4t{e^{{t^2}}}\\r'(t) \times r''(t) = ({e^t},(1 + t){e^t},(1 + 2{t^2}){e^{{t^2}}})({e^t},(1 + {t^3}){e^t} + 4t{e^{{t^2}}})\end{array}\)

Step-by-step solution

Step 1: Solution of the r' ( t )

Apply derivative on each side

\(\begin{array}{l}{{\bf{r}}^\prime }(t) = \frac{d}{{dt}}({e^{2t}},{e^{ - 2t}},t{e^{2t}})\\r'(t) = ({e^t},(1 + t){e^t},(1 + 2{t^2}){e^{{t^2}}})\end{array}\)

Step 2: Solution of the r' ( 0 )

Put t = 0 in r' ( 0 )

\(\begin{array}{l}r'(0) = ({e^0},(1 + 0){e^0},(1 + 2{(0)^2}){e^{{{(0)}^2}}})\\r'(0) = (1,1,1)\end{array}\)

Step 3: Solution of the magnitude of | r' ( 0 ) |

The magnitude of the r' ( 0 )

\(\begin{array}{l}\left\| {{{\bf{r}}^\prime }(0)} \right\| = \sqrt {1 + 1 + 1} \\\left\| {{{\bf{r}}^\prime }(0)} \right\| = \sqrt 3 {\rm{ }}\end{array}\)

Solution of the tangent vector

The tangent vector is T ( 1 )

\(\begin{array}{l}T(0) = \frac{{r'(0)}}{{||r'(0)||}}\\T(1) = \left( {\frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}} \right)\end{array}\)

Step 5: Solution of the r'' ( t )

Derivative of r' ( t )

\(\begin{array}{l}r'(t) = ({e^t},(1 + t){e^t},(1 + 2{t^2}){e^{{t^2}}})\\r''(t) = \frac{d}{{dt}}({e^t},{e^t} + t{e^{{t^2}}} + 2{t^2}{e^{{t^2}}})\\ = {e^t},{e^t} + {t^3}{e^t} + 4t{e^{{t^2}}}\\r''(t) = {e^t},(1 + {t^3}){e^t} + 4t{e^{{t^2}}}\end{array}\)

Step 6: Solution of the r' ( t ) x r'' ( t )

Multiplication of r' ( t ) x r'' ( t )

\(r'(t) \times r''(t) = ({e^t},(1 + t){e^t},(1 + 2{t^2}){e^{{t^2}}})({e^t},(1 + {t^3}){e^t} + 4t{e^{{t^2}}})\)