anonymous
  • anonymous
How do I find the curve of intersection of this two surfaces z = 1-x^2 and y = 1-x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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NotTim
  • NotTim
Curve? Not point?
NotTim
  • NotTim
Well, rearrange them so they look like their more-known forms. Also, try making agraph of these 2.
anonymous
  • anonymous
yes the intersection of a plane and the other surface hehe.

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NotTim
  • NotTim
z=x^2-1 and y=x-1
anonymous
  • anonymous
they look like this: http://assets.openstudy.com/updates/attachments/4f6ea09fe4b0772daa08c3f9-brinethery-1332650156119-tripleintegralproblem.pdf
NotTim
  • NotTim
WAT. I CANT DO THT.
NotTim
  • NotTim
THATS VECTORS>...
NotTim
  • NotTim
Sorry....
NotTim
  • NotTim
i better do my vecotrs work tho
NotTim
  • NotTim
other people, be my savior!
lgbasallote
  • lgbasallote
i was thinking systems...but to do it a third equation is needed...is integration applicable?
anonymous
  • anonymous
Well I don't want to find a point. I want to find a curve. so I just need a pair of equations. But i can figure out a way of getting the equation of the curve.
lgbasallote
  • lgbasallote
which math lesson is this? maybe i'll have an idea on how to solve it from there.
anonymous
  • anonymous
Well I was studying triple integrals.
lgbasallote
  • lgbasallote
ohh...then this really has something to do with integration...which i am weak at :p sorry
anonymous
  • anonymous
Well the integration part is clear, the problem is to get the cuve of intersection of those surfaces =/
amistre64
  • amistre64
z = 1-x^2 y = 1-x -> x = 1-y z = 1-(1-y)^2 z = 1 - 1+2y-y^2 z = 2y-y^2
anonymous
  • anonymous
Thanks amistre that is one of the results i got. But this one confuse me too:\[z=1-x^2=(1+x)(1-x)=y(1+x)\]
anonymous
  • anonymous
I graphed \[z=y(1+x)\]and I got a weird surface. That confused me.
anonymous
  • anonymous
and the equation \[z=2y-y^2\] it's a surface too. I don't know how does a curve's equation looks like.
amistre64
  • amistre64
the "curve" is the intersection of the plane and the cylindar; but id have to plot them to see it better
anonymous
  • anonymous
Here it is:
1 Attachment
amistre64
  • amistre64
yep, thats the setup i had on my paper :)
anonymous
  • anonymous
So the idea I have on my mind is that the intersection between those two surfaces is a space curve. but none of the equations I have got is a space curve.
anonymous
  • anonymous
I can interpret z = 2y - y^2 and x = 0. as a curve. But I'm not sure.
amistre64
  • amistre64
the projection onto the zy plane is 2y-y^2 i wonder if instead of trying to solve for xyz we introduce a x(t), y(t) and z(t)
amistre64
  • amistre64
x=1-t y = t z=2t-t^2
anonymous
  • anonymous
I did that and I get this:
1 Attachment
anonymous
  • anonymous
but with x = t
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=polar+r%3D%3C1-t%2Ct%2C2t-t%5E2%3E this aint it but its a cool mistake nonetheless
anonymous
  • anonymous
ahh that it's in polar coordinates. How do I do to specify cartesian coordinates?
amistre64
  • amistre64
i dunno yet, wolfram hates me
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=parametric+plot&a=*C.parametric+plot-_*Calculator.dflt-&f2=1-t&f=ParametricPlotCalculator.xfunction_1-t&f3=t&f=ParametricPlotCalculator.yfunction_t&f4=2t-t%5E2&f=ParametricPlotCalculator.zfunction_2t-t%5E2&f5=pi&f=ParametricPlotCalculator.upperrange1_pi&f6=0&f=ParametricPlotCalculator.lowerrange1_0&f7=t&x=7&y=9&f=ParametricPlotCalculator.variable1_t&a=*FVarOpt.1-_**-.***ParametricPlotCalculator.variable2-.*ParametricPlotCalculator.lowerrange2-.*ParametricPlotCalculator.upperrange2---.**ParametricPlotCalculator.zfunction---
anonymous
  • anonymous
I got it hehe: http://www.wolframalpha.com/input/?i=parametric+plot+%28t%2C+1-t%2C1-t%5E2%29
anonymous
  • anonymous
Ohh that looks better
amistre64
  • amistre64
that looks like it :)
anonymous
  • anonymous
Well thank you amistre64!
amistre64
  • amistre64
yw

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