To reparametrize the curve with respect to arc length \( s \), we need to follow these steps:
- Compute the derivative of the position vector \(\mathbf{r}(t)\) with respect to \( t \).
- Find the magnitude of this derivative to get the speed \( \|\mathbf{r}'(t)\| \).
- Integrate the speed from 0 to \( t \) to express the arc length \( s \) as a function of \( t \).
- Solve for \( t \) as a function of \( s \).
- Substitute \( t(s) \) back into the original position vector \(\mathbf{r}(t)\).
Given the curve:
\[
\mathbf{r}(t) = e^{6t} \cos(6t) \mathbf{i} + 6 \mathbf{j} + e^{6t} \sin(6t) \mathbf{k}
\]
First, we need to find the derivative \(\mathbf{r}'(t)\):
\[
\mathbf{r}'(t) = \frac{d}{dt} \left( e^{6t} \cos(6t) \mathbf{i} + 6 \mathbf{j} + e^{6t} \sin(6t) \mathbf{k} \right)
\]
Using the product rule and chain rule:
\[
\mathbf{r}'(t) = \left( \frac{d}{dt} e^{6t} \cos(6t) \right) \mathbf{i} + \left( \frac{d}{dt} 6 \right) \mathbf{j} + \left( \frac{d}{dt} e^{6t} \sin(6t) \right) \mathbf{k}
\]
\[
\mathbf{r}'(t) = \left( 6e^{6t} \cos(6t) - 6e^{6t} \sin(6t) \right) \mathbf{i} + 0 \mathbf{j} + \left( 6e^{6t} \sin(6t) + 6e^{6t} \cos(6t) \right) \mathbf{k}
\]
\[
\mathbf{r}'(t) = 6e^{6t} (\cos(6t) - \sin(6t)) \mathbf{i} + 6e^{6t} (\sin(6t) + \cos(6t)) \mathbf{k}
\]
Next, we find the magnitude of \(\mathbf{r}'(t)\):
\[
\|\mathbf{r}'(t)\| = \sqrt{ \left( 6e^{6t} (\cos(6t) - \sin(6t)) \right)^2 + \left( 6e^{6t} (\sin(6t) + \cos(6t)) \right)^2 }
\]
\[
\|\mathbf{r}'(t)\| = 6e^{6t} \sqrt{ (\cos(6t) - \sin(6t))^2 + (\sin(6t) + \cos(6t))^2 }
\]
Simplify the expression inside the square root:
\[
(\cos(6t) - \sin(6t))^2 + (\sin(6t) + \cos(6t))^2 = \cos^2(6t) - 2\cos(6t)\sin(6t) + \sin^2(6t) + \sin^2(6t) + 2\cos(6t)\sin(6t) + \cos^2(6t)
\]
\[
= 2\cos^2(6t) + 2\sin^2(6t) = 2(\cos^2(6t) + \sin^2(6t)) = 2
\]
Thus:
\[
\|\mathbf{r}'(t)\| = 6e^{6t} \sqrt{2} = 6\sqrt{2} e^{6t}
\]
The arc length \(s\) from \(t = 0\) to \(t\) is given by:
\[
s = \int_0^t \|\mathbf{r}'(u)\| \, du = \int_0^t 6\sqrt{2} e^{6u} \, du
\]
Evaluate the integral:
\[
s = 6\sqrt{2} \int_0^t e^{6u} \, du = 6\sqrt{2} \left[ \frac{e^{6u}}{6} \right]_0^t = \sqrt{2} \left( e^{6t} - 1 \right)
\]
We need to express \(t\) in terms of \(s\):
\[
s = \sqrt{2} \left( e^{6t} - 1 \right)
\]
\[
\frac{s}{\sqrt{2}} = e^{6t} - 1
\]
\[
e^{6t} = \frac{s}{\sqrt{2}} + 1
\]
\[
6t = \ln \left( \frac{s}{\sqrt{2}} + 1 \right)
\]
\[
t = \frac{1}{6} \ln \left( \frac{s}{\sqrt{2}} + 1 \right)
\]
Substitute \(t(s)\) back into \(\mathbf{r}(t)\):
\[
\mathbf{r}(t(s)) = e^{6 \left( \frac{1}{6} \ln \left( \frac{s}{\sqrt{2}} + 1 \right) \right)} \cos \left( 6 \left( \frac{1}{6} \ln \left( \frac{s}{\sqrt{2}} + 1 \right) \right) \right) \mathbf{i} + 6 \mathbf{j} + e^{6 \left( \frac{1}{6} \ln \left( \frac{s}{\sqrt{2}} + 1 \right) \right)} \sin \left( 6 \left( \frac{1}{6} \ln \left( \frac{s}{\sqrt{2}} + 1 \right) \right) \right) \mathbf{k}
\]
Simplify:
\[
\mathbf{r}(t(s)) = \left( \frac{s}{\sqrt{2}} + 1 \right) \cos \left( \ln \left( \frac{s}{\sqrt{2}} + 1 \right) \right) \mathbf{i} + 6 \mathbf{j} + \left( \frac{s}{\sqrt{2}} + 1 \right) \sin \left( \ln \left( \frac{s}{\sqrt{2}} + 1 \right) \right) \mathbf{k}
\]
\[
\boxed{\mathbf{r}(t(s)) = \left( \frac{s}{\sqrt{2}} + 1 \right) \cos \left( \ln \left( \frac{s}{\sqrt{2}} + 1 \right) \right) \mathbf{i} + 6 \mathbf{j} + \left( \frac{s}{\sqrt{2}} + 1 \right) \sin \left( \ln \left( \frac{s}{\sqrt{2}} + 1 \right) \right) \mathbf{k}}
\]
Answered
