How do you find parametric equations for the tangent line to the curve with the...

Step 1
We have parametric equations as functions of t for x, y and z so we can form the derivatives wrt t for each variable:
${\displaystyle x=\cos t \Rightarrow \frac{dx}{dt}=-\sin t}$
${\displaystyle y=4e^{8t} \Rightarrow \frac{dy}{dt}=32e^{8t}}$
${\displaystyle z=4e^{-8t}\Rightarrow \frac{dz}{dt}=-32e^{8t}}$
In order to find the normal at any particular point in vector space we use the Del, or gradient operator:
${\displaystyle \nabla f\left(t\right)=\frac{dx}{dt}\hat{i}+\frac{dy}{dt}\hat{j}+\frac{dz}{dt}\hat{k}}$
${\displaystyle =-\sin t\hat{i}+32e^{8t}\hat{j}-32e^{-8t}\hat{k}}$
So for the particular point ${\displaystyle (1,4,4)}$ we need to establish the corresponding value of t, which by inspection is ${\displaystyle t=0}$ .
So for this particular point, the normal vector to the surface is given by:
${\displaystyle \nabla f\left(0\right)=0\hat{i}+32\hat{j}-32\hat{k}}$
${\displaystyle =32\hat{j}-32\hat{k}}$
${\displaystyle =32(\hat{j}-\hat{k})}$So the tangent line is a line with direction ${\displaystyle \hat{j}-\hat{k}}$ that passes through the point ${\displaystyle (1,4,4)}$ , which therefore has the vector equation:
${\displaystyle \overrightarrow{r}=\left(\begin{array}{c}1\\ 4\\ 4\end{array}\right)+\lambda \left(\begin{array}{c}0\\ 1\\ -1\end{array}\right)}$
Which we can therefore parametrise (using ${\displaystyle \lambda }$ ) as follows
${\displaystyle x=1}$
${\displaystyle y=4+\lambda }$
${\displaystyle z=4-\lambda }$