Find parametric equations of the tangent line to the circular helix
$$x=\cos \color{red}{t},\quad y=\sin \color{red}{t},\quad z=\color{red}{t}$$ Where $\color{red}{t}=\color{red}{t_0}$, and use that result to find parametric equations for the tangent line at the point
where $\color{red}{t}=\color{red}{\pi}$.
Solution:
The vector equation of the tangent line at $\color{red}{t}=\color{red}{t_0}$ is
\begin{align}
\textbf{r}=&\cos \color{red}{t_0}\vec{i}+\sin \color{red}{t_0}\vec{j}+\color{red}{t_0}\vec{k}+\color{green}{t}[-\sin \color{red}{t_0}\vec{i}+\cos \color{red}{t_0}\vec{j}+\vec{k}] \quad \text{Using }\textbf{r}=\vec{r}(t_0)+t\vec{r'}(t_0)\\
&=(\cos \color{red}{t_0}-\color{green}{t}\sin \color{red}{t_0})\vec{i}+(\sin \color{red}{t_0}+\color{green}{t}\cos \color{red}{t_0})\vec{j}+(\color{red}{t_0}+\color{green}{t})\vec{k}
\end{align}
Thus, the parametric equations of the tangent line at $\color{red}{t}=\color{red}{\pi}$ are$$x=-1,\quad y=-\color{green}{t}, \quad z=\pi+\color{green}{t}$$
My question:
A parametrization of the line through a point $r(t_0)$ and parallel to the vector $\vec{r'}(t_0)$ is the parametrization of the tangent line at $t_0$ $$\textbf{r}=\vec{r}(t_0)+t\vec{r'}(t_0)$$ Isn't we want
the line $\textbf{r}$ to pass through $\vec{r}(t_0)$ When $t=t_0$ then it should be $$\textbf{r}=\vec{r}(t_0)+(t-t_0)\vec{r'}(t_0)\qquad -(1)$$ But using $(1)$ I get different answer. Actually My confusion began with these following paragraph which I found at mathinsight
A parametrization of the line through a point a and parallel to the vector $v$ is $l(t)=a+tv$. Setting $a=c(t_0)$ and $v=c′(t_0)$, we obtain a parametrization of the tangent line: $$l(t)=c(t_0)+tc′(t_0)\quad (2)$$
However, we typically want the line given by $l(t)$
to pass through $c(t_0)$ when $t=t_0$. So we usually change the parametrization slightly to
$$l(t)=c(t_0)+(t−t_0)c′(t_0)\quad (3)$$
Maybe explaining the difference between $(2)$ and $(3)$ clear my whole confusion.
Best Answer
and let your parametric equations of circular helix is $\mathbf c(t)=⟨\cos t,\sin t,t⟩$. Which give you $\mathbf c(\pi)=⟨-1,0,\pi⟩$
Now from here maybe your confusion start why $\mathbf{r_1(\pi)}$ isn't same as $\mathbf c(\pi)$. Please note that $\mathbf{r_1(t)}$ and $\mathbf{r_2(t)}$ are the different parametrization of same tangent line. To make it convenient mathinsight change the parameter slightly to make sure $\mathbf{r_2(t)}$ match $\mathbf {c(t)}$ at every point $t$ as mentioned. By the way you can see that $\mathbf{r_1(0)}=\mathbf {c(\pi)}$.
Am I correctly caught you$?$