Question

Use traces to sketch and identify the surface.

\(4{x^2} - 16{y^2} + {z^2} = 16\)

Short answer

The surface \(4{x^2} - 16{y^2} + {z^2} = 16\) is a hyperboloid of one sheet.

Step-by-step solution

given equation

Consider the given surface is \(4{x^2} - 16{y^2} + {z^2} = 16\)

Substitute \(x = k\) in given surface equation

Case (i)

Let \(x = k\).

Substitute \(x = k\) in \(4{x^2} - 16{y^2} + {z^2} = 16\).

\(\begin{array}{l}4{k^2} - 16{y^2} + {z^2} = 16\\ - 16{y^2} + {z^2} = 16 - 4{k^2}\end{array}\)

The surface area represents the trace of hyperbola.

Substitute \(y = k\) in given surface equation

Case (ii):

Let \(y = k\)

Substitute \(y = k\) in \(4{x^2} - 16{y^2} + {z^2} = 16\).

\(\begin{array}{l}4{x^2} - 16{k^2} + {z^2} = 16\\4{x^2} + {z^2} = 16 + 16{k^2}\end{array}\)

The surface area represents the trace of ellipse.

Substitute \(z = k\) in given surface equation

Case (iii):

Let \(z = k\)

Substitute \(z = k\) in \(4{x^2} - 16{y^2} + {z^2} = 16\)

\(\begin{array}{l}4{x^2} - 16{y^2} + {k^2} = 16\\4{x^2} - 16{y^2} = 16 - {k^2}\end{array}\)

The surface area represents the trace of hyperbola.

observe above cases and plot graph

From the above information, it is observed that the graph of \(4{x^2} - 16{y^2} + {z^2} = 16\) is combined behaviour of hyperbola and ellipse.

Since there is a minus sign in the given equation, it indicates one sheet in the hyperboloid, the surface equation \(4{x^2} - 16{y^2} + {z^2} = 16\) is a hyperboloid of one sheet.

Plot the elliptic paraboloid surface \(4{x^2} - 16{y^2} + {z^2} = 16\) as shown below in Figure 1.

From Figure 1, it is observed that the surface is a hyperboloid of one sheet.

From Figure 1, it is observed that the surface is a hyperboloid of one sheet.