Question

\(x = {e^{ - t}}cost,\;\;\;y = {e^{ - t}}sint,\;\;\;z = {e^{ - {t_4}}},\;\;\;(1,0,1)\)

Short answer

The parametric equations of the line are \(x = 1 - t,{\rm{ }}y = t,{\rm{ }}z = 1 - t.\)

Step-by-step solution

Step 1: To Find the tangent vector

First we need to find the tangent vector \({{\bf{r}}^\prime }(t)\)

We have

\(r(t) = {e^{ - t}}costi + {e^{ - t}}sintj + {e^{ - t}}k\)

Differentiate to get the tangent vector

\({r^'}(t) = {e^{ - t}}( - cost - sint)i + {e^{ - t}}( - sint + cost)j - {e^{ - t}}k\)

Note that the point \((1,0,1)\) corresponds to \(t = 0.\) This is because when we put \(t = 0\) in \({\bf{r}}(t),\) we get \((1,0,1)\) as the position.

\({r^'}(0) = {e^{ - 0}}( - cos(0) - sin(0))i + {e^{ - 0}}( - sin(0) + cos(0))j - {e^{ - 0}}k \)

\({r^'}(0) = - i + j - k\)

Step 2: Final proof

Remember that : Equation of a line passing through a point with position vector \(a,\) and parallel to the vector \(b\) is

\(r(t) = a + tb\)

Hence equation of the tangent line at \((1,0,1)\) is

\(r(t) = \langle 1,0,1\rangle + t\langle - 1,1, - 1\rangle \)

\(r(t) = \langle 1 - t,t,1 - t\rangle \)

Hence the parametric equations of the line are

\(x = 1 - t,\;\;\;y = t,\;\;\;z = 1 - t\)