A parameterization of $y = 6 x ^ { 2 }$ is

A) $x ( u , v ) = u$ $y ( u , v ) = 6 v ^ { 2 }$ $z ( u , v ) = - v$
B) $x ( u , v ) = u$ $y ( u , v ) = - 6 v ^ { 2 }$ $z ( u , v ) = v$
C) $x ( u , v ) = u$ $y ( u , v ) = 6 v ^ { 2 }$ $z ( u , v ) = v$
D) $x ( u , v ) = u$ $y ( u , v ) = 6 u ^ { 2 }$ $z ( u , v ) = v$
E) $x ( u , v ) = u$ $y ( u , v ) = - 6 u ^ { 2 }$ $z ( u , v ) = v$

Question

A parameterization of $y = 6 x ^ { 2 }$ is

A) $x ( u , v ) = u$ $y ( u , v ) = 6 v ^ { 2 }$ $z ( u , v ) = - v$
B) $x ( u , v ) = u$ $y ( u , v ) = - 6 v ^ { 2 }$ $z ( u , v ) = v$
C) $x ( u , v ) = u$ $y ( u , v ) = 6 v ^ { 2 }$ $z ( u , v ) = v$
D) $x ( u , v ) = u$ $y ( u , v ) = 6 u ^ { 2 }$ $z ( u , v ) = v$
E) $x ( u , v ) = u$ $y ( u , v ) = - 6 u ^ { 2 }$ $z ( u , v ) = v$

Quizplus · Accepted Answer

The correct parameterization of $y = 6x^2$ involves setting $x = u$, which directly leads to $y = 6u^2$, and $z = v$ can be any function since $z$ is not specified by the equation $y = 6x^2$. This is exactly what option D represents.

A Parameterization Of y=6x2y = 6 x ^ { 2 }y=6x2

A Parameterization Of $y = 6 x ^ { 2 }$