# Reparametrize the curve with respect to arc length measured from the point where t = 0 in the direction of increading t.

$\boldsymbol{ r ( t ) \ = \ e^{ 2t } \ cos( 2t ) \ \hat{ i } \ + \ 2 \ \hat{ j } \ + \ e^{ 2t } sin( 2t ) \ \hat{ k } }$

The aim of this question is to reparametrize the given curve equation.

To solve this question, we will first evaluate the tangent to the above curve by calculating the derivative of the curve. Then we will find the new parameter by fitting the linear curve onto the independent variable. Finally, we will substitute the value of t in terms of the new variable in the above equation to find the reparametrized curve.

Given:

$r ( t ) \ = \ e^{ 2t } \ cos( 2t ) \ \hat{ i } \ + \ 2 \ \hat{ j } \ + \ e^{ 2t } sin( 2t ) \ \hat{ k }$

Taking derivative of the above equation:

$\dfrac{ d }{ dt } \bigg ( r ( t ) \bigg ) \ = \ \dfrac{ d }{ dt } \bigg ( e^{ 2t } \ cos( 2t ) \ \hat{ i } \ + \ 2 \ \hat{ j } \ + \ e^{ 2t } sin( 2t ) \ \hat{ k } \bigg )$

$r’ ( t ) \ = \ \dfrac{ d }{ dt } \bigg ( e^{ 2t } \ cos( 2t ) \bigg ) \ \hat{ i } \ + \ \dfrac{ d }{ dt } \bigg ( 2 \bigg ) \ \hat{ j } \ + \ \dfrac{ d }{ dt } \bigg ( e^{ 2t } sin( 2t ) \bigg ) \ \hat{ k }$

Using the product rule:

$r’ ( t ) \ = \ \left [ \begin{array}{ l } \bigg ( \dfrac{ d }{ dt } ( e^{ 2t } ) \ cos( 2t ) + e^{ 2t } \dfrac{ d }{ dt } (cos (2t ) )\bigg ) \ \hat{ i } \\ + \ \dfrac{ d }{ dt } \bigg ( 2 \bigg ) \ \hat{ j } \\ + \ \bigg ( \dfrac{ d }{ dt } ( e^{ 2t } ) \ sin( 2t ) + e^{ 2t } \dfrac{ d }{ dt } (sin (2t ) )\bigg ) \ \hat{ k } \end{array} \right.$

Evaluating derivatives:

$r’ ( t ) \ = \ \bigg ( 2e^{ 2t } \ cos( 2t ) – e^{ 2t } sin( 2t ) \bigg ) \ \hat{ i } \ + \ ( 0 ) \ \hat{ j } \ + \ \bigg ( 2e^{ 2t } \ sin( 2t ) + e^{ 2t } cos( 2t ) \bigg ) \ \hat{ k }$

$r’ ( t ) \ = \ \bigg ( 2e^{ 2t } \ cos( 2t ) – e^{ 2t } sin( 2t ) \bigg ) \ \hat{ i } \ + \ \bigg ( 2e^{ 2t } \ sin( 2t ) + e^{ 2t } cos( 2t ) \bigg ) \ \hat{ k }$

Now to find the magnitude of the derivative:

$| r’ ( t ) | \ = \ \sqrt{ \bigg ( 2e^{ 2t } \ cos( 2t ) – e^{ 2t } sin( 2t ) \bigg )^2 \ + \ \bigg ( 2e^{ 2t } \ sin( 2t ) + e^{ 2t } cos( 2t ) \bigg )^2 }$

$| r’ ( t ) | \ = \ 2e^{ 2t } \sqrt{ \bigg ( \ cos( 2t ) – sin( 2t ) \bigg )^2 \ + \ \bigg ( \ sin( 2t ) + cos( 2t ) \bigg )^2 }$

$| r’ ( t ) | \ = \ 2e^{ 2t } \sqrt{ cos^2( 2t ) + sin^2( 2t ) – 2 sin( 2t ) cos( 2t ) \ + \ cos^2( 2t ) + sin^2( 2t ) + 2 sin( 2t ) cos( 2t ) }$

$| r’ ( t ) | \ = \ 2e^{ 2t } \sqrt{ 2 \bigg ( cos^2( 2t ) + sin^2( 2t ) \bigg ) }$

$| r’ ( t ) | \ = \ 2e^{ 2t } \sqrt{ 2 }$

Now to reparametrize:

$L \ = \ \int_0^t | r’ ( t ) | \ = \ \int_0^t 2e^{ 2t } \sqrt{ 2 } dt$

$L \ = \ \sqrt{ 2 } \int_0^t 2 e^{ 2t } dt$

$L \ = \ \sqrt{ 2 } \bigg | e^{ 2t } \bigg |_0^t$

$L \ = \ \sqrt{ 2 } \bigg [ e^{ 2t } – e^{ 2(0) } \bigg ]$

$L \ = \ \sqrt{ 2 } ( e^{ 2t } – 1 )$

Also:

$S \ = \ L t$

$S \ = \ \sqrt{ 2 } ( e^{ 2t } – 1 ) t$

$\Rightarrow t \ = \ \dfrac{ 1 }{ \sqrt{ 2 } ( e^{ 2t } – 1 ) } S$

Substituting this value in the given equation:

$r \bigg ( t (s) \bigg ) \ = \left [ \begin{array}{l}\ e^{ 2 \bigg ( \dfrac{ 1 }{ \sqrt{ 2 } ( e^{ 2t } – 1 ) } S \bigg ) } \ cos 2 \bigg ( \dfrac{ 1 }{ \sqrt{ 2 } ( e^{ 2t } – 1 ) } S \bigg ) \ \hat{ i } \\ + \ 2 \ \hat{ j } \\ + \ e^{ 2 \bigg (\dfrac{ 1 }{ \sqrt{ 2 } ( e^{ 2t } – 1 ) } S \bigg ) } sin 2 \bigg (\dfrac{ 1 }{ \sqrt{ 2 } ( e^{ 2t } – 1 ) } S \bigg ) \ \hat{ k } \end{array} \right.$

## Numerical Result

$r \bigg ( t (s) \bigg ) \ = \left [ \begin{array}{l}\ e^{ 2 \bigg ( \dfrac{ 1 }{ \sqrt{ 2 } ( e^{ 2t } – 1 ) } S \bigg ) } \ cos 2 \bigg ( \dfrac{ 1 }{ \sqrt{ 2 } ( e^{ 2t } – 1 ) } S \bigg ) \ \hat{ i } \\ + \ 2 \ \hat{ j } \\ + \ e^{ 2 \bigg (\dfrac{ 1 }{ \sqrt{ 2 } ( e^{ 2t } – 1 ) } S \bigg ) } sin 2 \bigg (\dfrac{ 1 }{ \sqrt{ 2 } ( e^{ 2t } – 1 ) } S \bigg ) \ \hat{ k } \end{array} \right.$

## Example

Evaluate the tangent to the given curve at t = 0.

Recall:

$| r’ ( t ) | \ = \ 2e^{ 2t } \sqrt{ 2 }$

Substituting t = 0:

$| r’ ( 0 ) | \ = \ 2e^{ 2(0) } \sqrt{ 2 }$

$| r’ ( 0 ) | \ = \ 2 \sqrt{ 2 }$