77c327…: “Less a stormy quadrilateral”

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Consider $F(x,y) = 4x^2+2y^2+5$ .

The domain is $\R$ $\R ^2$ $z>5$ $z \geq 5$ .
The range is $\R$ $\R ^2$ $z>5$ $z \geq 5$ .

Give a description of the level curve that passes through $(1,2)$ in terms of $x$ and $y$ .

The level curve that passes through the point $(1,2)$ is depicted in the figure below.

$\begin{align*} F(x,y) &= \answer{17} \\ 4x^2+2y^2 &= \answer{12} \end{align*}$

This level curve is a linea planea parabolaa circlean ellipsea hyperbola .

Does the curve $C$ parameterized by $\vec {r}(t) = \vector {2t,t}$ lie on a level curve of $F(x,y)$ ?

Yes. No.

Give a parameterization of the curve on the surface that lies above $C$ .

$\vec {R}(t) = \vector {\answer {2t},\answer {t},\answer {18t^2+5}}$

If $\vec {r}(t) = \vector {2t,t}$ lies on a level curve of $F(x,y)$ , then $F(x,y)$ should be constant along $C$ . Along $C$ , we have $x=\answer {2t}$ and $y=\answer {t}$ so $F(x(t),y(t)) = \answer {18t^2+5}$ , which isis not constant.

To find the curve on the surface $z=F(x,y)$ above the curve, note that $x(t) = 2t$ and $y(t)=t$ . We can use the function $F(x,y)$ to write $z$ in terms of $t$ .