Question

Hello,

Could someone give me a hint or a tool (not the answer) on how to find the parametric form of the intersection curve of these two equations z = x2 −y2 and z = 2+(x−y)2?

Answer 1

The intersection is an hyperbola right?

Answer 2

Does the answer lie single variable calculus or do I need to use what we learnt about parametric equations in multivariable calculus?

Answer 3

Please help

Answer 4

what does x2 and y2 mean? Is it x^2 and y^2?

I'd start with eliminating z. That gives us the values of x,y for which the two surfaces intersect. Find a parametrization of this curve on the x,y plane and finally use one of the equations to express z with this parameter.

That would be my plan anyway, haven't actually solved it yet.

Answer 5

Well the thing is, that's what I did but it yields this equation: xy-y^2=1 which is an hyperbola but I don't know how to deal with the xy to find the parametrization of this very hyperbola. Yeah I meant y^2 and x^2. By the way thanks and if you're motivated take a look at the second exercice of problem set 4 ex. 2 of multivariable calculus 2010. thx mate

Answer 6

You can express x as a function of y and use t=y as a parameter. I.e xy=y^2+1 so x=y+1/y if y is not equal to 0. Then you can give the vector form of the function as (x(t),y(t),z(t)).

Answer 7

How come you can do this substitution y=t ?

Answer 8

Think back to what it means to parametrize a curve. If any y determines exactly one point on the curve than we can just chose that to be our parameter.

Answer 9

Thx it worked. But I don't understand what you say about any y determining exactly one point on the curve then you can use it as a parameter? The parametrizations I had done so far were really different, I had to build a position vector.

Answer 10

Could you tell me what I should read to help me understand what you meant with "if y determines exactly a point on the curve then it's your parameter" ?

Answer 11

For any parametrization, your goal is always to have everything it terms of a single variable. The substitution \[y = t\] wasn't necessary, you could've written everything in terms of \[y\] as well. The only difference is that most parametrizations are done in the form \[f(t)\]