Are these steps right (intersection curve of 2 cylinders)?

Question

Here are the 2 initial equations:

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

This is where my initial thoughts are going.

First i need to be able to express (x+z=2) in x. so we have this.

x = 2-z

next well want to sub x into (x^2+y^2=1) now we have.

(2-z)^2 + y^2 = 1
4-4z+z^2+y^2 = 1

y^2 + z^2 - 4z = -3

The above equation does not make sence to me.... Please help if you can. Is there differentation to be done somewhere along the line? I am in a coorespondace course and would really appreciate the help. Im finding it hard to understand the textbook

Answer 1

The two equations
x=2-z and
y^2+z^2-4z=-3
together describe the intersection curve. You can write the second as
y^2=4z-z^2-3.
So, for each z, you get two values of y (from thetwo square roots) and one value of x. The curve itself is an ellipse since it is the intersection of the circular cylinder and the plane. Past that, I'm not quite sure what you want to do with this problem. You mention differentiation. Are you trying to find a derivative of one variable with respect to another on the curve?

Answer 2

x^2 + y^2 = 1, x + z = 2
this is the intersection of a cylinder and a PLANE

(2-z)^2 + y^2 = 1
4-4z+z^2+y^2 = 1

y^2 + z^2 - 4z = -3 this is the equation of an ellipse:
y^2 + z^2 - 4z +4= -3 +4
y^2+ (z-2)^2=1
which makes a lot of sense,
it is the intersection of the plane and the cylinder. But actually you need to write the 2 equations:
y^2+ (z-2)^2=1 and x + z = 2
it is like writting y and x as functions of z (the parameter of the curve

Answer 3

You do not have two cylinders in the equations that you have proffered; you have a cylinder and a line in three-space. I will assume that the other equation is x^2 + x^2 = 2, again a cylinder, then simply add the equations to obtain

z^2 + y^2 + 2x^2 = 3

Your line is going to be in 3 space, so solve for z to get

z = +-SQR[3 - 2x^2 - y^2]

where 3 - 2x^2 -y^2 >= 0 or y^2 <= 3 - 2x^2; your answer is z as above with the restriction noted.