Tangent vectors and parametric curves

Question

Consider the curve $C$ defined by $(x,y,z) = \bar{r}(t)$, where $$\bar{r}(t)=\langle t\sin t, t\cos t, t^2 \rangle~~; t \in \mathbb{R}^3$$

1. Show that $C$ lies on the paraboloid $z= x^2 + y^2$

2. Find a vector tangent to $C$ at the point $\bar{r}(1)$.

3. Find the parametric equation for the line tangent to the curve $\bar{r}$ at the point $(0,-\pi, \pi^2)$.

My attempt:

1. Consider the paraboloid $z = x^2 + y^2$.

   Let \begin{align}x=t\sin t \\ y = t\cos t\end{align}

   Then notice that \begin{align}z &= t^2\sin^2 t + t^2\cos^2t \\ &=t^2(\sin^2t + \cos^2t) \\ &= t^2\end{align}

   Thus $C$ lies on the paraboloid $z = x^2 + y^2$.

2. $\bar{r}'(t) = \langle\sin t + t\cos t, \cos t - t\sin t, 2t \rangle$

   Thus, when $t=1$, the tangent vector must be $$\bar{r}'(1) = \langle\sin (1) + \cos(1), \cos(1) - \sin (1), 2 \rangle$$

3. Notice that at the point $(0,-\pi, \pi^2)$ we have that $t = \pi$.

   Thus the parametric equation of the line tangent to $\bar{r}$ a the point $(0,-\pi, \pi^2)$ is given by \begin{align}\bar{l}(t) &= r(\pi) + (t - \pi)\bar{r}'(\pi) \\ &= \langle -\pi t + \pi^2, -t, 2\pi t - \pi^2 \rangle\end{align}

Are these correct? I am trying to work through some old papers in order to prepare for a test.

Answer 1

1. Correct, you have shown that it was consistent by noticing the trig identity.

2. Correct, since you are just looking for the tangent vector, which is a constant, and simply the direction of the line at that point.

3. Correct, you identified the parametric form of the line perfectly. Good work using $(x-\pi)$ to make the equations simpler.