Calculus III - Functions of Several Variables

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Section 12.5 : Functions of Several Variables

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8. Identify and sketch the traces for the following function.

$2x - 3y + {z^2} = 1$

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We have two traces. One we get by plugging $x = a$ into the equation and the other we get by plugging $y = b$ into the equation. Here is what we get for each of these.

$\begin{align*}x = a: & \hspace{0.25in}2a - 3y + {z^2} = 1 & \to \hspace{0.25in}y & = \frac{1}{3}{z^2} + \frac{{2a - 1}}{3}\\ y = b: & \hspace{0.25in}2x - 3b + {z^2} = 1 & \to \hspace{0.25in}x & = - \frac{1}{2}{z^2} + \frac{{3b + 1}}{2}\end{align*}$ Show Step 2

Okay, we’re now into a realm that many students have issues with initially. We no longer have equations in terms of $x$ and $y$ . Instead we have one equation in terms of $x$ and $z$ and another in terms of $y$ and $z$ .

Do not get excited about this! They work the same way that equations in terms of $x$ and $y$ work! The only difference is that we need to make a decision on which variable will be the horizontal axis variable and which variable will be the vertical axis variable.

Just because we have an $x$ doesn’t mean that it must be the horizontal axis and just because we have a $y$ doesn’t mean that it must be the vertical axis! We set up the axis variables in a way that will be convenient for us.

In this case since both equation have a $z$ in them and it is squared we’ll let $z$ be the horizontal axis variable for both of the equations.

So, given that convention for the axis variables this means that for the $x = a$ trace we’ll have a parabola that opens upwards with vertex at $\left( {0,\frac{{2a - 1}}{3}} \right)$ and for the $y = b$ trace we’ll have a parabola that opens downwards with vertex at $\left( {0,\frac{{3b + 1}}{2}} \right)$ .

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Below is a sketch for each of the traces.