Question

Anybody know how will be look figure limited next lines:
x^2+y^2+z^2=1;
x^2+y^2=4z^2;
x=0; y=0; (x>=0; y>=0; z>=0)

Answer 1

Notice how both equations have x^2+y^2 in them. We'll take advantage of this


start with equation 1

x^2+y^2+z^2=1

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

-------------------------

move onto equation 2

x^2+y^2=4z^2

1-z^2 = 4z^2 ... replace x^2+y^2 with 1-z^2 (since we've just shown that x^2+y^2=1-z^2 above)


So you now have one equation with one variable and it is


1-z^2 = 4z^2


solve this for z and tell me what you get

Answer 2

yep @jim_thompson5910
got it :)

Answer 3

I received z=1\sqrt5

Answer 4

so

z = 1/sqrt(5)

z^2 = (1/sqrt(5))^2

z^2 = 1/5

--------------------------

now make the proper substitution into the first equation

x^2+y^2+z^2=1

x^2+y^2+1/5=1

and do the same for the second equation

x^2+y^2=4z^2

x^2+y^2=4(1/5)

x^2+y^2=4/5

Answer 5

you now have this system

x^2+y^2+1/5=1
x^2+y^2=4/5

Answer 6

yes, and what it gives to me?

Answer 7

notice that

x^2+y^2+1/5=1

can be simplified like so

x^2+y^2+1/5=1

x^2+y^2=1-1/5

x^2+y^2=5/5-1/5

x^2+y^2=(5-1)/5

x^2+y^2=4/5

Answer 8

so what's really going on is that

x^2+y^2+1/5=1
x^2+y^2=4/5

is the same as

x^2+y^2=4/5
x^2+y^2=4/5

Answer 9

yes, I see it

Answer 10

and those two equations are the same, so you're just solving

x^2+y^2=4/5

for x and y

so this means you'll have an infinite number of solutions because there are infinitely many values of x and y that satisfy x^2+y^2=4/5

Answer 11

the way you can see this is if you solve for y to get

x^2+y^2=4/5

y^2=4/5 - x^2

y = sqrt(4/5 - x^2) or y = -sqrt(4/5 - x^2)

and you can see that you'll get a ton of solutions

Answer 12

yes, but I must make a graph. How it may be look on graph?

Answer 13

Else I have that x=0 and y=0

Answer 14

x^2+y^2=4/5

x^2+y^2=(sqrt(4/5))^2

so this is a circle with radius sqrt(4/5) and is centered at (0,0)

Answer 15

hm, exactly :)

Answer 16

but it must be in 3 dimensional space

Answer 17

well similar to x^2+y^2 = 1 being a circle centered at (0,0) with a radius of 1

x^2+y^2+z^2=1 is a sphere centered at (0,0,0) with a radius of 1

Answer 18

and the graph for the second one is a bit more complicated, but it's shown here

http://www.wolframalpha.com/input/?i=x^2%2By^2%3D4z^2

Answer 19

are you sure, that we can solve this system and except z?

Answer 20

what do you mean

Answer 21

when we made 1-z^2=4z^2 and then solved for z

Answer 22

yes that's perfectly valid

Answer 23

ok, I understand a bit
the first - it is a circle, the second - it is in link
but how it combine

Answer 24

the first x^2+y^2+z^2=1 is a sphere

the second x^2+y^2=4z^2 is shown in the link

Answer 25

combining the two would have you plotting both the sphere and the infinite cone in the same xyz coordinate system

Answer 26

I dont understand what to do then

Answer 27

let me see if I can find a picture of the two combined

Answer 28

ok, I hope that you will find

Answer 29

ok here's one image I was able to come up with

the blue sphere is equation 1

the green surface is the graph of equation 2 and it is actually an infinite cone (that stretches on forever up above and down below)

Answer 30

ok, i will thinking else and then wtite to you if you dont mind

Answer 31

ok what were you thinking?

Answer 32

how it build there