Question

How do I find the curve of intersection of this two surfaces z = 1-x^2 and y = 1-x

Answer 1

Curve? Not point?

Answer 2

Well, rearrange them so they look like their more-known forms.
Also, try making agraph of these 2.

Answer 3

yes the intersection of a plane and the other surface hehe.

Answer 4

z=x^2-1 and y=x-1

Answer 5

they look like this: http://assets.openstudy.com/updates/attachments/4f6ea09fe4b0772daa08c3f9-brinethery-1332650156119-tripleintegralproblem.pdf

Answer 6

WAT. I CANT DO THT.

Answer 7

THATS VECTORS>...

Answer 8

Sorry....

Answer 9

i better do my vecotrs work tho

Answer 10

other people, be my savior!

Answer 11

i was thinking systems...but to do it a third equation is needed...is integration applicable?

Answer 12

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.

Answer 13

which math lesson is this? maybe i'll have an idea on how to solve it from there.

Answer 14

Well I was studying triple integrals.

Answer 15

ohh...then this really has something to do with integration...which i am weak at :p sorry

Answer 16

Well the integration part is clear, the problem is to get the cuve of intersection of those surfaces =/

Answer 17

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

Answer 18

Thanks amistre that is one of the results i got. But this one confuse me too:\[z=1-x^2=(1+x)(1-x)=y(1+x)\]

Answer 19

I graphed \[z=y(1+x)\]and I got a weird surface. That confused me.

Answer 20

and the equation \[z=2y-y^2\] it's a surface too. I don't know how does a curve's equation looks like.

Answer 21

the "curve" is the intersection of the plane and the cylindar; but id have to plot them to see it better

Answer 22

Here it is:

Answer 23

yep, thats the setup i had on my paper :)

Answer 24

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.

Answer 25

I can interpret z = 2y - y^2 and x = 0. as a curve. But I'm not sure.

Answer 26

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)

Answer 27

x=1-t
y = t
z=2t-t^2

Answer 28

I did that and I get this:

Answer 29

but with x = t

Answer 30

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

Answer 31

ahh that it's in polar coordinates. How do I do to specify cartesian coordinates?

Answer 32

i dunno yet, wolfram hates me

Answer 33

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---

Answer 34

I got it hehe: http://www.wolframalpha.com/input/?i=parametric+plot+%28t%2C+1-t%2C1-t%5E2%29

Answer 35

Ohh that looks better

Answer 36

that looks like it :)

Answer 37

Well thank you amistre64!

Answer 38

yw