Calc 3? Describe the geometry of the set points whose coordinates x,y,z satisfy the equation x^2+y^2+z^2+4x+2y-6z-22=0?
doing quadratic surfaces, are we?
Try completing the square you will get it in this form (x-a)^2 +(y-b)^2 +(z-c)^2 =k so it's a sphere
Any ideas?
Ohh I see I've that before I didn't know if that was what it meant by describing thanks!
"geometry of the set points" is way of asking what shape
What happens if it's only like (x-a)^2 +(y-b)^2=K
then it would be a circle
what does a and b stand for? the transformation it shifts about? and K is the radius right just making sure I understand it right
yes , you are right
okay thanks a lot!
Calc 3 will get a lot more interesting after this stuff
Hey sorry I have one more question, it says show that the intersection of that equation with yz is a circle, does that mean that I make x as 0?
yes,
b/c x=0 on entire y-z plane
so the equation I got was (x+2)^2 + (y+1)^2 + (z+3)^2 = 34 and I'd make it zero so I'd get just (y+1)^2 + (z+3)^2 = 30
ohh I see yeah okay thanks
Okay last thing I swear.... I just want to make sure I'm doing this right b/c I have a test tmr it's just practice
it says describe the intersection of the obj in part a with object in part c..
this is part a : (x+2)^2 + (y+1)^2 + (z+3)^2 = 34 and part c (x+2)^2+(y+1)^2=11.. what does that mean ?
do I combine them? I'm suppose to get 2 circles?
one is sphere while other is circle
Yeah that's what I thought but then it says I'm suppose to obtain two circles. So I assumed that intersection meant somehow they were equal
let me try to plot them
Can't get it to work if I put them together (x+2)^2 and (y+1)^2 cancels
you are supposed to obtain two circle?
It would just be a circle cutting through the sphere
well you know how if a sphere intersects with a plane you get a circle? I'm assuming if a circle intersects a sphere you get two circles??
I don't know it says on the sheet, do you have any ideas how to show that?
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