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### Section 12.5 : Functions of Several Variables

7. Identify and sketch the level curves (or contours) for the following function.

${y^2} = 2{x^2} + z$

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We know that level curves or contours are given by setting $$z = k$$. Doing this in our equation gives,

${y^2} = 2{x^2} + k$ Show Step 2

For this problem the value of $$k$$ will affect the type of graph of the level curve.

First, if $$k = 0$$ the equation will be,

${y^2} = 2{x^2}\hspace{0.5in} \Rightarrow \hspace{0.5in}y = \pm \sqrt 2 \,\,x$

So, in this case the level curve (actually curves if you think about it) will be two lines through the origin. One is increasing and the other is decreasing.

Next, let’s take a look at what we get if $$k > 0$$. In this case a quick rewrite of the equation from Step 2 gives,

$\frac{{{y^2}}}{k} - \frac{{2{x^2}}}{k} = 1$

Because we know that $$k$$ is positive we see that we have a hyperbola with the $$y$$ term the positive term and the $$x$$ term the negative term. This means that the hyperbola will be symmetric about the $$y$$-axis and opens up and down.

Finally, what do we get if $$k < 0$$. In this case the equation is,

$- \frac{{2{x^2}}}{k} + \frac{{{y^2}}}{k} = 1$

Now, be careful with this equation. In this case we have negative values of $$k$$. This means that the $$x$$ term is in fact positive (the minus sign will cancel against the minus sign in the $$k$$). Likewise, the $$y$$ term will be negative (it’s got a negative $$k$$ in the denominator). Therefore, we’ll have a hyperbola that is symmetric about the $$x$$-axis and opens right and left.

Show Step 3

Below is a sketch of some level curves for some values of $$k$$ for this function. 