Find all points of intersection

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Homework Help Overview

The discussion revolves around finding points of intersection between a spheroid and a cone, with a focus on the conditions under which their gradients are orthogonal. The original poster explores the mathematical relationships between these surfaces and expresses confusion regarding the nature of the intersections.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to visualize the problem by defining the surfaces and calculating their gradients. They raise questions about the implications of their findings, particularly regarding the nature of the intersection and the absence of z-coordinates in their derived equations.

Discussion Status

Participants are engaged in clarifying the original poster's reasoning and addressing potential errors in their equations. There is a recognition of the complexity of the problem, with some participants suggesting that the original poster may not be fully utilizing the concept of level curves in their analysis.

Contextual Notes

There is mention of a potential sign error in the equations presented, and a request to move the thread to a different sub-forum, indicating a mix of homework-like inquiry and broader discussion on the topic.

Addez123
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Homework Statement
Find all points where the level surface
$$4x^2+y^2+z^2 = 8$$
and
$$x^2+9y^2=z^2$$
intersects eachother at a 90 degree angle.
Relevant Equations
Surface 1: $$4x^2+y^2+z^2 = 8$$
Surface 2: $$x^2+9y^2 - z^2 = 0$$
First I try to visualize it:
w = Surface 1, is a spheroid
w_2 = Surface 2 is a cone stretching up the z axisThen I calculate their gradients:
$$∇w = (8x, 2y, 2z)$$
$$∇w_2 = (2x, 18y, 2z)$$

The points where they intersect at 90 degrees is when dot product is zero.
$$∇w \cdot ∇w_2 = 0$$
$$16x^2 + 36y^2 - 4z^2 = 0$$
$$z^2 =4x^2 + 8y^2$$
This is a cone stretched differently in x and y axis, but a cone none the less.
Now I need to find where this cone intersects with EITHER the sphere (Surface 1) or the inital cone (Surface 2). Where these intersects, curve 1 and 2 intersect under 90 degree angle. This happens when I set $$z^2 = z_2^2$$.

I use the cone, Surface 2, equation:
$$x^2 + 9y^2 = 4x^2 + 8y^2$$
$$3x^2 - y^2 = 0$$

This is where I get confused. There's no z coordinates so if this was a circle I'd assume it was a cylinder stretching up the whole z-axis.
But the answer is suppose to be either just points, or a curve. It makes no sense that the intersection would create another surface.

What am I doing wrong?
 
Last edited:
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Addez123 said:
Homework Statement:: Find all points where the level curves
...
What am I doing wrong?
My guess is you never make use of this 'level curves' in the problem statement
By the way
Addez123 said:
is a sphere
No. It is a spheroid.
 
BvU said:
My guess is you never make use of this 'level curves' in the problem statement
By the way
No. It is a spheroid.

I do use the level surfaces, especially when I combine surface 2 with the surface representing all points at which the two functions form 90 degree angles:

Addez123 said:
I use the cone, Surface 2, equation:
$$x^2 + 9y^2 = 4x^2 + 8y^2$$
 
From post 1:
##x^2+9y^2=z^2##
You have a sign error in this equation.

Addez123 said:
##∇w_2 = (2x, 18y, 2z)##
 
Mark44 said:
From post 1:
You have a sign error in this equation.
Thanks! I saw that before tho and changed everywhere but apparently forgot to change that line. Eitherway you can see the dot product is still correct.
 
Can someone move this to the calculus sub-forum? This isn't homework.
 
It's enough like a homework question that it should remain here.
 

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