Section 12.5 : Functions of Several Variables
7. Identify and sketch the level curves (or contours) for the following function.
y2=2x2+zShow All Steps Hide All Steps
Start SolutionWe know that level curves or contours are given by setting z=k. Doing this in our equation gives,
y2=2x2+k Show Step 2For this problem the value of k will affect the type of graph of the level curve.
First, if k=0 the equation will be,
y2=2x2⇒y=±√2xSo, 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,
y2k−2x2k=1Because 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,
−2x2k+y2k=1Now, 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 3Below is a sketch of some level curves for some values of k for this function.
