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OpenStudy (anonymous):
Can someone explain level sets and help me figure this question out?
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OpenStudy (anonymous):
Question:
OpenStudy (anonymous):
@wio can you help? been waiting for an hour now for someone to help.
OpenStudy (anonymous):
Let \(z=1\) and mess with the equation to try to find out what sort of equation it is.
OpenStudy (anonymous):
Also \(z=1\) works as well.
OpenStudy (anonymous):
What do you mean by mess with the equation?
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OpenStudy (anonymous):
Try to get it into a recognizable form.
OpenStudy (anonymous):
like \[
ax^2+by^2=c^2
\]This is an ellipse
OpenStudy (anonymous):
\[
y=ax^2+bx+c
\]This is a parabola
OpenStudy (anonymous):
\[
y=mx+b
\]A line
OpenStudy (anonymous):
So the first one: \[
xy+z^2=1\\
\]Well \[
xy=1-z^2
\]And \(1-z^2\) is a constant so:\[
xy=C\implies y=\frac Cx
\]
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OpenStudy (anonymous):
This is not on the list.
OpenStudy (anonymous):
Next one \[
z=\frac{1}{x+y-1}
\] \[
\frac 1z=x+y-1
\] \[
\frac{z+1}{z}=x+y
\]This ends up being: \[
y+x=C\implies y=-x+C
\]This is a line
OpenStudy (anonymous):
Since the slope will always be \(-1\) it's parallel lines.
OpenStudy (anonymous):
ok I see thanks
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