# Math Insight

### Tangent line to parametrized curve examples

#### Example 1

Given the path (parametrized curve) $\dllp(t)=(3t+2,t^2-7,t-t^2)$, find a parametrization of the line tangent to $\dllp(t)$ at the point $\dllp(1)$.

Solution: The line must pass through the point $\dllp(1)=(5,-6,0)$.

The derivative of the path is \begin{align*} \dllp'(t) = 3\vc{i}+2t\vc{j}+(1-2t)\vc{k} \end{align*} When $t=1$, the tangent vector must be \begin{align*} \dllp'(1) = 3\vc{i} + 2\vc{j}-\vc{k}. \end{align*} A parametrization for the tangent line is \begin{align*} \vc{l}(t) &= \dllp(1) + (t-1) \dllp'(t_0)\\ &= (5\vc{i}-6\vc{j}) + (t-1)(3\vc{i} + 2\vc{j}-\vc{k})\\ &= (3t+2)\vc{i} + (2t-8)\vc{j} + (-t+1)\vc{k} \end{align*}

#### Example 2

Suppose a particle is following the path $\dllp(t)=(3t+2,t^2-7,t-t^2)$. At time $t_0=1$, the particle flies off on a tangent. Compute the position of the particle at time $t_1=2$.

Solution: The key observation is that, after the time $t_0$, the position of the particle is given by tangent line $\vc{l}(t)$. From the solution to Example 1, we know that \begin{align*} \vc{l}(t) = (3t+2, 2t-8,-t+1). \end{align*} The particle's position at time $t_1=2$ is \begin{align*} \vc{l}(2) = (8, -4, -1). \end{align*}