How does one parametrize the function z^2=sqrt(x^2+y^2)/3?

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How does one parametrize the function z^2=sqrt(x^2+y^2)/3?

Mathematics
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BTW its a Conic frustum
Swing z^2 on the other side and find derivative
actually I typed it wrong its z=sqrt(x^2+y^2)3

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Other answers:

ugh z=sqrt(x^2+y^2)/3
find derivative, find magnitude of derivative
The derivative with respect to what?
This function need to be a vector, so partial fractions to get

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