Finding the curve of graph intersections

[Lucci]

Homework Statement

A circular cylinder of radius 1 and the hyperbolic paraboloid, z=y²-x², intersect. Which vector function r(t)=x(t)i+y(t)j+z(t)k has the curve of intersection as its graph if x(0)=0.

Homework Equations

The Attempt at a Solution

I know that intersection is a circle with a changing z value (the question provided a graph), so my x and y values will either be cos(t), sin(t) or vice versa.

So, if you use sin(t) and cos(t) for the x and y values, you get
z=cos² (t)-sin²(t)
and z = cos(2t)

So we have r(t)=<sin(t),cos(t),cos(2t)>

However, this would mean that when y=0, t=pi/2
So when t=pi/2, we should have point (1,0,1), but on the graph provided, the point at y=0 is clearly(positive x value,0, NEGATIVE z value)

I am confused. :/

 

[LCKurtz]

Hello Lucci. Welcome to PF. But I'm afraid there aren't any mind readers here. Since you didn't give us the equation of the cylinder nor a picture of the graph, how can we be expected to help you? Also, I very much doubt the intersection of a cylinder with a hyperbolic paraboloid is a circle. At least give us, word-for-word, the exact statement of the problem.

 

[Lucci]

Aye.
I apologize. I'm still getting used to this thing.
This is the picture of the graph...

And when I mean a circle, I sort of imagine parameterized curves as being drawn by a moving t value. This t value will go in a circle around the z axis (x²+y² while simultaneously moving up and down as z increases/decreases.

 

[HallsofIvy]

In other words, the cylinder is given by $x^2+ y^2= 1$. If you add that to $z= x^2- y^2$ you get $z+1= 2x^2$ or $z= 2x^2- 1$.

Standard parametric equations for the unit circle are $x= cos(\theta)$ and $y= sin(\theta)$.

 

[LCKurtz]

Homework Statement

A circular cylinder of radius 1 and the hyperbolic paraboloid, z=y²-x², intersect. Which vector function r(t)=x(t)i+y(t)j+z(t)k has the curve of intersection as its graph if x(0)=0.

Homework Equations

The Attempt at a Solution

I know that intersection is a circle with a changing z value (the question provided a graph), so my x and y values will either be cos(t), sin(t) or vice versa.

So, if you use sin(t) and cos(t) for the x and y values, you get
z=cos² (t)-sin²(t)
and z = cos(2t)

So we have r(t)=<sin(t),cos(t),cos(2t)>

However, this would mean that when y=0, t=pi/2
So when t=pi/2, we should have point (1,0,1), but on the graph provided, the point at y=0 is clearly(positive x value,0, NEGATIVE z value)

I am confused. :/

Aye.
I apologize. I'm still getting used to this thing.
This is the picture of the graph...

And when I mean a circle, I sort of imagine parameterized curves as being drawn by a moving t value. This t value will go in a circle around the z axis (x²+y² while simultaneously moving up and down as z increases/decreases.

OK, thanks for the graph Lucci. I have a couple of observations. First, the question should have asked for "a parameterization" not "which parameterization" because there are many that would technically work. And, of course, as I believe you understand, it isn't the graph that is a circle but the xy parameter curve that is. And last, HallsOfIvy's parameterization doesn't satisfy your requirement that x(0) = 0.

Now to answer your question. The parameterization you gave is correct. It correctly parameterizes the unit circle and x(0) = 0 as required. What is confusing you is that your parameterization starts at the y-axis and goes around the other way than what you are used to. Looking down on the xy plane it goes around the circle clockwise. So at t = pi/2 you should get (1,0,-1) as the graph shows.