Question

Use traces to sketch and identify the surface.

\(y = {z^2} - {x^2}\)

Short answer

The surface equation \(y = {z^2} - {x^2}\) is a hyperbolic paraboloid centred at origin.

The surface equation \(y = {z^2} - {x^2}\) is sketched.

Step-by-step solution

equation of hyperbolic paraboloid

Formula for equation of a hyperbolic paraboloid:

Consider the standard equation of a hyperbolic paraboloid along the negative\(y\)-axis.

\(\frac{y}{b} = \frac{{{z^2}}}{{{c^2}}} - \frac{{{x^2}}}{{{a^2}}}.....(1)\)

Consider the given surface equation.

\(y = {z^2} - {x^2}...........(2)\)

Modify the equation.

\(\frac{y}{1} = \frac{{{z^2}}}{1} - \frac{{{x^2}}}{1}.........{\rm{ (3) }}\)

By comparing equation (3) with (1), the given surface equation satisfies the equation of a hyperbolic paraboloid along the negative \(y\)-axis.

Thus, the surface equation \(y = {z^2} - {x^2}\) is a hyperbolic paraboloid centred at origin.

Find \(a,{\rm{ }}b\) and \(c\) 

Find \(a,{\rm{ }}b\)and \(c\) by comparing equation (3) with equation (1).

\(\begin{array}{l}{a^2} = 1\\a = \sqrt 1 \\a = \pm 1\end{array}\)

The value of \(a\) is \( \pm 1\).

\(b = 1\)

The value of \(b\) is 1.

\(\begin{array}{l}{c^2} = 1\\c = \sqrt 1 \\c = \pm 1\end{array}\)

The value of \(c\) is \( \pm 1\).

Substitute \(k\) for \(x\) in equation (2)

Case i:

Let \(x = k\).

Substitute \(k\) for \(x\) in equation (2),

\(\begin{array}{l}y = {z^2} - {k^2}\\y - {z^2} = - {k^2}\\y = - {k^2} + {z^2}\end{array}\)

This expression represents the parabola along the positive \(y\)-direction for any values of \(k\). So, the surface equation having a trace of parabola for \(x = k\).

Substitute \(k\) for \(y\) in equation (2)

Case ii:

Let \(y = k\).

Substitute \(k\) for \(y\) in equation (2),

\(\begin{array}{l}k = {z^2} - {x^2}\\{z^2} - {x^2} = k......(4)\end{array}\)

Modify equation (4) for \(k = 0\)

\(\begin{array}{l}{z^2} - {x^2} = 0\\(z + x)(z - x) = 0\\z + x = 0,z - x = 0\end{array}\)

\(z = - x,z = x\)

So, the surface equation (4) has two intersecting line equations for \(k = 0\)Similarly, the surface equation (4) has a family of a hyperbola traces for \(k \ne 0\)

Substitute \(k\) for \(z\) in equation (2)

Case iii:

Let \(z = k\).

Substitute \(k\) for \(z\) in equation (2),

\(y = {k^2} - {x^2}\)

observe above cases and plot graph 

Trace of this expression represents a parabola along the negative \(y\)-direction. So, the resultant trace of the surface equation \(y = {z^2} - {x^2}\) represents a hyperbolic paraboloid, which is lie on the origin point \((0,0,0)\).

Draw the surface equation \(y = {z^2} - {x^2}\) as shown in Figure 1.

Figure 1

Thus, the surface equation \(y = {z^2} - {x^2}\) is sketched.