Question

Find parametric equations for the tangent line to the curve\(x = tcost,y = t,z = tsint\) at the point \(( - \pi ,\pi ,0).\) Illustrate by graphing both the curve and the tangent line on a common screen.

Short answer

The parametric equations for the tangent line are \(x = - \pi - t,y = \pi + t{\rm{ }}\) and \(z = - \pi t{\rm{. }}\)

Step-by-step solution

Step 1:To find the vector equation

The vector equation of the curve is

\({\bf{r}}(t) = \langle t\cos t,t,t\sin t\rangle \)

Therefore

\({{\bf{r}}^\prime }(t) = \langle \cos t - t\sin t,\;\;\;1,\sin t + t\cos t\rangle \)

The parameter value corresponding to the point \(( - \pi ,\pi ,0)\) is \(t = \pi \)

Therefore, the tangent vector there is

\({{\bf{r}}^\prime }(\pi ) = \langle \cos (\pi ) - \pi \sin (\pi ),1,\sin (\pi ) + \pi \cos (\pi )\rangle {\rm{ }}\)

\({{\bf{r}}^\prime }(\pi ) = \langle - 1,1, - \pi \rangle \)

Step 2:Final proof 

The tangent line passes through \(( - \pi ,\pi ,0)\) and is parallel to the vector \({{\bf{r}}^\prime }(\pi ).\) Remember that Equation of a line which passes through a point with position vector \(a,\) and is parallel to the vector \(v\) is given by \(r(t) = a + tv\)

Therefore, Equation of the tangent line is

\({\bf{r}}(t) = \langle - \pi ,\pi ,0\rangle + t\langle - 1,1, - \pi \rangle \)

\({\bf{r}}(t) = \langle - \pi - t,\pi + t, - \pi t\rangle \)

Parametric Equations for the tangent line are

\(x = - \pi - t,\;\;\;y = \pi + t,\;\;\;z = - \pi t\)