87422d…: “Throughout the navy matrix”

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Consider the curve $C$ is traced out by the vector-valued function $\vec {r}(t) = \vector {t^2,2t+1,t+2}, \quad t \in \R$ and $S = \left \{(x,y,z)\in \R ^3 : 9x-2y^2+8z = 13 \right \}.$ Determine whether the curve lies on the surface, intersects the surface at finitely many points, or never intersects the surface. If it intersects the surface at finitely many points, give the $x$ , $y$ , and $z$ -coordinates of the intersection.

The curve lies on the surfaceintersects the surface at finitely many pointsdoes not intersect the surface .

(make sure to use the hint for a detailed walkthrough if you need it)

From the parameterization of $C$ , we find that $\begin{align*} x(t)&= \answer{t^2} \\ y(t)&= \answer{2t+1} \\ z(t)&= \answer{t+2} \\ \end{align*}$

We now must substitute these into the equation that describes $\mathcal {S}$ .

$9x(t)-2y(t)^2+8z(t) = 9 \answer {t^2} -2 \left ( \answer {2t+1} \right )^2 + 8 \left (\answer {t+2} \right )^2 = \answer {t^2+14}.$

Since the equation that describes $S$ requires that $9x-2y^2+8z = 13$ , we must solve

$t^2+14 = 13.$

This holds for $t$ -values, the curve lies on the surfaceintersects the surface at finitely many pointsdoes not intersect the surface .

Suppose now that we redefine our surface $S$ by $S = \left \{(x,y,z)\in \R ^3 : 9x-2y^2+8z = a \right \}.$ Are there any values of $a$ for which the curve will intersect the surface exactly once?

yes no

What is the smallest value $A$ for which the curve will intersect the surface more than once if $a>A$ ? $A = \answer {14}.$ How many times will the curve intersect the surface if $a>A$ ?

once twice three times more than three times, but finitely many infinitely many times It depends on the choice of $a$ .

Now, set $a=18$ and find the intersection points $(x,y,z)$ where the curve intersects the surface.

The intersection points occur when $t=\answer {-2}$ and $t=\answer {2}$ (type the smaller $t$ -value first). The actual intersection poitns are $(x,y,z) = \left (\answer {4},\answer {-3}, 0 \right )$ and $(x,y,z) = \left (\answer {4},\answer {5}, \answer {4} \right )$ .