Question

Find a vector function that represents the curve of intersection hyperboloid z=x^2-y^2 and the cylinder x^2+y^2=1

Answer 1

I tried isolating the x in the second equation and then plugging that into the first one but it turned out pretty ugly and then I just didn't know where I was going :(

Answer 2

why you just did not sum them ?

Answer 3

No, why should I?

Answer 4

@mathmale

Answer 5

@jackthegreatest

Answer 6

i have no idea : (

Answer 7

I really want to understand how to solve these types of problems.

Answer 8

It is not a plug-and-chug sort of thing. It is a matter of understanding what you are seeing.

z=x^2-y^2 and the cylinder x^2+y^2=1

z = x^2 + y^2 - 2y^2 = 1 - 2y^2 Ever see one of these? z = 1 - 2y^2 These are cross-sections and they are parabolas. No reason the same sort of thing wouldn't work for eliminating the y^2 and seeing what other sorts of cross sections abound.

Answer 9

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

Answer 10

Hey

Answer 11

what did you add and subtract?

Answer 12

z = x^2 - y^2

Add and subtract y^2

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

Simplify Conveniently

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

Given that x^2 + y^2 = 1

z = 1 - 2y^2

Such manipulations are very helpful.

Answer 13

I added and subtracted because I already knew x^2 + y^2 = 1 and I needed that structure. So I built it.

Answer 14

ohhhh makes sense

Answer 15

Now, I can say if y= t, then z=1-2t^2 ?

Answer 16

I suppose. If that has some relevance to the present investigation.

Answer 17

Well, I need a vector function so I have to "parametrize" right?

Answer 18

Thank you a lot, tkhunny.
You made it so clear, for me at least.

Answer 19

Whoah I'm confused, I checked the answer key and it says the vector function is cost i + sin t j + cos2t k how did it end up being trig?

Answer 20

I'm here just to saying "woah" xD

Answer 21

Cylinders have circular cross sections. You will get trigonometric functions.

Answer 22

It makes sense to use cylindrical coordinates
x= r cos t
y = r sin t
z = z

in this case
\[ x^2+y^2 = 1\\ r^2 \cos^2 t + r^2 \sin^2 t = 1\\ r=1\]
and
\[ z= x^2 -y^2 \\= r^2(\cos^2 t - \sin^2 t) \\ = r \cos 2t \\= \cos 2t \]
we get
\[ < \cos t , \sin t , \cos 2t >\]