Solving Orthogonal Families Homework Problem

  • Thread starter Verdict
  • Start date
  • Tags
    Orthogonal
In summary, the conversation discusses a problem involving finding the product of the slopes of two families of curves, one being a family of hyperbolas and the other being a family of ellipses. The equation for the family of curves is given as x^2/(A^2+lambda^2) - y^2/|B^2+lambda| = 1. The conversation also touches on different approaches and methods for solving the problem, including using implicit differentiation and contour plots. A website is also mentioned that provides a potential solution, but with a typo in the first denominator.
  • #1
Verdict
117
0

Homework Statement


Naajjmloos.jpg

Homework Equations


The product of the slopes has to be equal to -1.

The Attempt at a Solution


Well, as the function, call it u, is equal to a constant, the derivative of u with respect to x is
the partial derivative with respect to x + (the partial derivative of u with respect to y * dy/dx) = 0

So dy/dx = - (∂u/∂x) / (∂u/∂y)

But from here on I don't know how to continue. Could anyone provide a hint?
 
Physics news on Phys.org
  • #2
I apologize, I had not realized that it was in the wrong section before.

To be precise, I found that
dy/dx = [tex]-\frac{x \left(B^2+L\right)}{y \left(A^2+L^2\right)}[/tex]

Before imposing anything on a, lamda and b.

And thus, If I call the lamba of the first family L and the lamba of the second family L2:

[tex]\frac{x^2 \left(B^2+L\right) \left(B^2+\text{L2}\right)}{y^2 \left(A^2+L^2\right) \left(A^2+\text{L2}^2\right)}=-1[/tex]

Is what I have to show, if they are indeed orthogonal.

Should I express Y in terms of X and plug that in?
 
Last edited:
  • #3
Verdict said:
I apologize, I had not realized that it was in the wrong section before.

To be precise, I found that [tex]\frac{dy}{dx} =-\frac{x \left(B^2+L\right)}{y \left(A^2+L^2\right)}[/tex]
Before imposing anything on a, lamda (λ) and b.
I've been looking at this problem since shortly after you posted it. It does look quite difficult.

Here's my take on it. -- nowhere near to getting a solution.

For the case [itex]\displaystyle \ -A^2<\lambda<-B^2\ :[/itex] (Yes, I prefer using the upper case for a & b, but like to stick with lower case Greek for λ.)
λ < -B2 so then B2 + λ < 0 .

So the equation for the family of curves becomes:

[itex]\displaystyle \frac{x^2}{A^2+\lambda^2}-\frac{y^2}{\left|B^2+\lambda\right|}=1\ .[/itex]

Also, A2>B2 and λ2 > 0 so of course, A2 + λ2 > | B2 + λ | .

This equation describes a hyperbola, centered at (0, 0) with semi-major axis of [itex] \displaystyle \ \sqrt{A^2+\lambda^2\ } [/itex] along the x-axis and semi-minor axis of [itex] \displaystyle \ \sqrt{\left|B^2+\lambda\right|\ } [/itex] along the y-axis.

Similarly you can show that for λ > -B2 > -A2, B2 + λ > 0, so that the family of curves described are ellipses.

So, it seems to me that the problem makes most sense, if we consider A and B to be each fixed at some value and that the "families" are to be parametrized by λ .


I realize that this is very far from a solution. -- It's even far from beginning a solution. -- But, hopefully it helps understanding what's being asked for.

SammyS
 
  • #4
Edit 2: Indeed, the first labda is NOT supposed to be squared, that is a typo! In that case I know of a method on how to solve it, it basically just follows the website I posted below.

Dear SammyS,

Thank you for your response. Indeed, I also thought that it would be most natural to think of λ to be the changing factor, while keeping A and B constant. Interpreting what it meant has proven to be a bit problematic for me, so your input is very useful.

The 'book' I use in my course - complex analysis - is Schaum's Outline of Complex Variables, which is basically just a summary type of book. It does have 1 worked out exercise dealing with orthogonal families, and that is the approach I followed. It uses implicit differentiation to find dy/dx, for both of the equations, and then you can see that the product is -1 without too much hassle.
This problem is clearly not in the same ballpark, as it is very nontrivial (if it can even be seen?) that the product is -1 in this case. I'm going to work out a few examples with values of a and b, and see if that gets me anywhere..

Alright, so what this shows me is that the approach does not work, as far as I can tell. I don't get that the product is -1, for specific values of a, b and the lamda's. Guess I'll have to think of something else..

Making a few contour plots in mathematica, for different values of lamda, the first family is indeed a family of hyperbola's and the second one a family of ellipses.

Now, I don't understand the approach, but http://planetmath.org/hyperbolasorthogonaltoellipses treats how to show that a certain family of ellipses, not too much unlike the one I have here, is perpendicular to a family of hyperbola's.

Looking at it, it does show some promise, but I can't reproduce the line after (3). Applying the same method to my system seems to be working, up to that point, where I extract a quantity that is not equal to zero.

Edit: Alright, so I used the method posted on that website. It ALMOST works! The only issue is that, in the second denominator, lamda is not squared, while the first one is. So if the first one is also not supposed to be squared, then it works.


Edit 2:
Indeed, the first labda is NOT supposed to be squared, that is a typo!
 
Last edited:

Related to Solving Orthogonal Families Homework Problem

What is an orthogonal family?

An orthogonal family is a set of vectors in a vector space that are mutually perpendicular to each other. This means that the dot product of any two vectors in the family is equal to 0.

What is the purpose of solving orthogonal families homework problems?

The purpose of solving orthogonal families homework problems is to gain a deeper understanding of vector spaces and their properties. It also helps in developing problem-solving skills and critical thinking abilities.

How do you determine if a set of vectors is orthogonal?

To determine if a set of vectors is orthogonal, you can calculate the dot product of every pair of vectors in the set. If the dot product is equal to 0 for every pair, then the vectors are orthogonal.

What are some applications of orthogonal families in real life?

Orthogonal families have various applications in fields such as physics, engineering, and computer science. For example, in physics, orthogonal vectors are used to represent forces acting on an object, and in computer graphics, they are used to create three-dimensional images.

What are some strategies for solving orthogonal families homework problems?

Some strategies for solving orthogonal families homework problems include using geometric interpretations, applying the properties of orthogonal vectors, and simplifying the calculations by using special cases such as unit vectors or zero vectors.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
357
  • Calculus and Beyond Homework Help
Replies
4
Views
170
Replies
9
Views
750
  • Calculus and Beyond Homework Help
Replies
10
Views
2K
  • Calculus and Beyond Homework Help
Replies
28
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
752
  • Calculus and Beyond Homework Help
Replies
8
Views
510
  • Calculus and Beyond Homework Help
Replies
2
Views
979
  • Calculus and Beyond Homework Help
Replies
3
Views
930
  • Calculus and Beyond Homework Help
Replies
5
Views
3K
Back
Top