Reparametrize the curve with respect to arc length measured from the point where

To reparametrize the curve with respect to arc length measured from the point where $t=0$ in the direction of increasing $t$, we need to find a new parameterization $\alpha (s)$, where $s$ represents the arc length.
The arc length of a curve in three-dimensional space is given by the integral of the magnitude of its velocity vector:
$s=\int \sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2+{\left(\frac{dz}{dt}\right)}^2}dt$
In this case, we have the following parameterization:
$𝐫(t)=e^{2t}\cos (2t)𝐢+2𝐣+e^{2t}\sin (2t)𝐤$
Let's compute the derivatives of $x(t)$, $y(t)$, and $z(t)$:
$\frac{dx}{dt}=\frac{d}{dt}(e^{2t}\cos (2t))=2e^{2t}\cos (2t)-2e^{2t}\sin (2t)$
$\frac{dy}{dt}=\frac{d}{dt}\left(2\right)=0$
$\frac{dz}{dt}=\frac{d}{dt}(e^{2t}\sin (2t))=2e^{2t}\sin (2t)+2e^{2t}\cos (2t)$
Now, let's calculate the magnitude of the velocity vector:
$\Vert \frac{d𝐫}{dt}\Vert =\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2+{\left(\frac{dz}{dt}\right)}^2}$
$=\sqrt{{(2e^{2t}\cos (2t)-2e^{2t}\sin (2t))}^2+0+{(2e^{2t}\sin (2t)+2e^{2t}\cos (2t))}^2}$
Simplifying this expression will give us the integrand for the arc length integral.
$\Vert \frac{d𝐫}{dt}\Vert =\sqrt{{(2e^{2t}\cos (2t)-2e^{2t}\sin (2t))}^2+{(2e^{2t}\sin (2t)+2e^{2t}\cos (2t))}^2}$
Expanding the squares and combining like terms:
$=\sqrt{4e^{4t}{\cos }^2(2t)-8e^{4t}\cos (2t)\sin (2t)+4e^{4t}{\sin }^2(2t)+4e^{4t}{\sin }^2(2t)+8e^{4t}\cos (2t)\sin (2t)+4e^{4t}{\cos }^2(2t)}$
$=\sqrt{8e^{4t}({\cos }^2(2t)+{\sin }^2(2t))}$
$=\sqrt{8e^{4t}}$
Simplifying further:
$=2e^{2t}$
Now we have the integrand for the arc length integral: $2e^{2t}$. We can proceed to integrate this with respect to $t$ to find an expression for $s$ in terms of $t$.
$s=\int 2e^{2t}dt$
Integrating $2e^{2t}$ with respect to $t$ gives:
$s=e^{2t}+C$
where $C$ is the constant of integration.
To determine the value of $C$, we need to evaluate $s$ at the starting point $t=0$. From the original parameterization, we know that $r(0)=e^{2(0)}\cos (0)𝐢+2𝐣+e^{2(0)}\sin (0)𝐤$. Simplifying this expression, we find that $r(0)=𝐢+2𝐣$.
Substituting $t=0$ into the expression for $s$, we get:
$s=e^{2(0)}+C=1+C$
Since $s$ represents the arc length measured from the point where $t=0$, we can conclude that $C$ must be zero. Thus, the expression for $s$ in terms of $t$ is:
$s=e^{2t}$
Now, let's rewrite the original parameterization $𝐫(t)$ in terms of $s$:
$𝐫(t)=e^{2t}\cos (2t)𝐢+2𝐣+e^{2t}\sin (2t)𝐤$
Using the expression for $s$, we substitute $t$ with $\ln (s)$:
$𝐫(s)=e^{2\ln (s)}\cos (2\ln (s))𝐢+2𝐣+e^{2\ln (s)}\sin (2\ln (s))𝐤$
Simplifying further:
$𝐫(s)=s^2\cos (2\ln (s))𝐢+2𝐣+s^2\sin (2\ln (s))𝐤$
Finally, we can write the reparametrized curve in terms of arc length $s$ measured from the point where $t=0$:
$𝐫(s)=s^2\cos (2\ln (s))𝐢+2𝐣+s^2\sin (2\ln (s))𝐤$
This represents the reparametrization of the curve $𝐫(t)$ with respect to arc length measured from the point where $t=0$ in the direction of increasing $t$.