MHB Are the Intersections of These 3D Curves Ellipses?

  • Thread starter Thread starter carl123
  • Start date Start date
  • Tags Tags
    Calculus
Click For Summary
The discussion centers on the intersection of two 3D curves defined by z = x² + y² and z = 2x² + 3y² - 1. It explores whether this intersection forms an ellipse and if its projection onto the x-y plane is also an ellipse. Participants suggest that the first curve represents a paraboloid, with cross-sections being circles that increase in size as z increases. The second curve is also analyzed for its geometric properties, with a focus on identifying the type of quadratic surface it represents. Overall, the conversation emphasizes the mathematical relationships and visualizations of these curves in three-dimensional space.
carl123
Messages
55
Reaction score
0
QUESTION 1

Sketch the curves :

z = x2 + y2 and z = 2x2 + 3y2 - 1

a) Is the intersection an ellipse?

b) Is the projection of the intersection onto the x-y plane an ellipse?

QUESTION 2

Sketch the curves:

r1[t] = (t cos [2 pi t], t sin [2 pi t] , t2 ) and r2[t] = (t cos [6 pi t], t sin [6 pi t], t2 ] over the intervals from 0 to 2.

a) There is a quadratic surface that both curves lie on, what sort of surface is it? Give the name?

b) What is an equation for the surface?
 
Physics news on Phys.org
carl123 said:
QUESTION 1

Sketch the curves :

z = x2 + y2 and z = 2x2 + 3y2 - 1

a) Is the intersection an ellipse?

b) Is the projection of the intersection onto the x-y plane an ellipse?

QUESTION 2

Sketch the curves:

r1[t] = (t cos [2 pi t], t sin [2 pi t] , t2 ) and r2[t] = (t cos [6 pi t], t sin [6 pi t], t2 ] over the intervals from 0 to 2.

a) There is a quadratic surface that both curves lie on, what sort of surface is it? Give the name?

b) What is an equation for the surface?

Have you tried anything?

To sketch $\displaystyle \begin{align*} z = x^2 + y^2 \end{align*}$, think about every cross section parallel to the x-y plane. First of all, for obvious reasons, $\displaystyle \begin{align*} z \geq 0 \end{align*}$. Also obviously no matter what z value you pick, you are going to have something of the form $\displaystyle \begin{align*} x^2 + y^2 = r^2 \end{align*}$ (r a constant), therefore each cross section must be a circle. But since as you move up the z axis, you are increasing z, that means r is increasing as well. So each cross section is a circle, getting bigger in radius as you move up the z axis. That seems like a paraboloid to me...

z '=' x'^'2 '+' y'^'2 - Wolfram|Alpha

What do you think the second surface would look like and be classified as?

As for finding their intersection, well you already have both of them as z in terms of x and y, so you can set them equal to each other...
 
Last edited:

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 1 ·
Replies
1
Views
6K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K