Answer
Tangent line: \[x=-\pi -s, \quad y=\pi+s, \quad z= 0-\pi s\]
Work Step by Step
We are given:
\[
x=t\cos{t}, \quad y=t, \quad z=t\sin{t}; \quad (-\pi, \pi, 0) \\
or \\
\vec{r}(t)= \langle t\cos{t}, t, t\sin{t}\rangle
\]
We need to determine the value of \(t\) corresponding to the point \((-\pi, \pi, 0)\).
Let's start with \(y\) as it is the simplest to calculate:
\[
y=t \implies t= \pi
\]
Now we verify that \(t=\pi\) satisfies \(x\) and \(z\):
\[
x=\pi\cos{\pi} = -\pi \\
z=\pi\sin{\pi}= 0
\]
We conclude that the point \((-\pi, \pi, 0)\) occurs when \(t=\pi\).
Taking the derivative of \(\vec{r}(t)\), we have:
\[
\vec{r}'(t)=\langle \cos{t}-t\sin{t}, 1, t\cos{t}+\sin{t} \rangle
\]
Now we find \(\vec{r}'(\pi)\), that is the tangent vector at \(t=\pi\).
\[
\vec{r}'(\pi)=\langle \cos{\pi}-\pi \sin{\pi}, 1, \pi \cos{\pi}+\sin{\pi} \rangle \\
\vec{r}'(\pi)=\langle -1-0, 1,-\pi+0\rangle \\
\vec{r}'(\pi)=\langle -1,1,-\pi\rangle
\]
The tangent line passes through the point \((-\pi, \pi, 0)\) with direction vector \(\vec{r}'(\pi)=\langle -1,1,-\pi\rangle\)
\[
x=-\pi -s, \quad y=\pi+s, \quad z= 0-\pi s
\]
