Answer
The function lies on a cone.
Work Step by Step
We have:
\[
x=(1+\cos16t)\cos t, \quad y=(1+\cos16t)\sin t, \quad z=(1+\cos16t)
\]
Let \(a=(1+\cos16t)\):
\[
x=a\cos t, \quad y=a\sin t, \quad z=a \]
Squaring each component:
\[
x^2=(a\cos t)^2, \quad y^2=(a\sin t)^2, \quad z^2=a^2 \\
x^2=a^2\cos^2 t, \quad y^2=a^2\sin^2 t, \quad z^2=a^2 \\
\]
Manipulating \(x^2+y^2\):
\[
x^2+y^2=a^2(\sin^2t+\cos^2t)
\]
Since \((\sin^2t+\cos^2t)=1\), we have:
\[ x^2+y^2=a^2 \]
Additionally, \(z^2=a^2\) so:
\[
x^2+y^2=z^2
\]
This represents a cone oriented along the \(z\)-axis.
