MHB How can the equation be modified to create a perfect ellipse on the graph?

  • Thread starter Thread starter highmath
  • Start date Start date
Click For Summary
SUMMARY

The equation cos(a) + cos(b) = cos(a + b) does not produce a perfect ellipse on the graph. Instead, it generates an ellipse-like shape due to the nature of the function. To achieve a perfect ellipse, one must ensure that the conditions for a smooth function are met, specifically that the first derivatives at the extremum point (x_0, y_0) are zero and the determinant of the Hessian matrix is positive. Adjusting parameters in tools like Desmos can visually demonstrate how the graph approaches a perfect ellipse as parameter 'a' nears 1.

PREREQUISITES
  • Understanding of trigonometric functions and their graphs
  • Familiarity with calculus concepts, particularly derivatives and Hessian matrices
  • Experience with graphing software, specifically Desmos
  • Knowledge of the properties of ellipses and conic sections
NEXT STEPS
  • Explore the properties of the Hessian matrix in multivariable calculus
  • Learn how to manipulate parameters in Desmos to visualize different graph shapes
  • Study the derivation and properties of conic sections, focusing on ellipses
  • Investigate other equations that yield perfect ellipses and their graphical representations
USEFUL FOR

Mathematicians, educators, students in calculus or geometry, and anyone interested in graphing and visualizing mathematical equations.

highmath
Messages
35
Reaction score
0
The graph of the equation:
cos(a) + cos(b) = cos(a + b)
show when the equation is valid.

The graph show an ellipses-like and not like a "perfect" ellipse. Why?
If I want to change the equation, what can I do to get a perfect ellipse?!
 
Mathematics news on Phys.org
This equation is discussed on Math.StackExchange. The accepted answer says that it is indeed not an ellipse, but the solutions to any equation $f(x,y)=\text{const}$ around $(x_0,y_0)$ looks like an ellipse for a smooth function $f$ if $\frac{\partial f}{\partial x}(x_0,y_0)=\frac{\partial f}{\partial y}(x_0,y_0)=0$ and $$\begin{vmatrix}\frac{\partial^2f}{\partial x^2}f(x_0,y_0)&\frac{\partial^2f}{\partial x\partial y}f(x_0,y_0)\\\frac{\partial^2f}{\partial x\partial y}f(x_0,y_0)&\frac{\partial^2f}{\partial y^2}f(x_0,y_0)\end{vmatrix}>0$$ (i.e., if $(x_0,y_0)$ is an extremum point of $f$).
 
If you move the slider for this Desmos graph, you will see that the graph looks very like an ellipse until the parameter $a$ is quite close to $1$.

[DESMOS]advanced: {"version":5,"graph":{"squareAxes":false,"viewport":{"xmin":-1.3747197653766552,"ymin":-0.9204796366940666,"xmax":6.9623451644014835,"ymax":7.416585293084072}},"expressions":{"list":[{"type":"expression","id":"graph1","color":"#2d70b3","latex":"\\cos x\\ +\\ \\cos y\\ -\\ \\cos\\left(x+y\\right)\\ =\\ a","style":"SOLID"},{"type":"expression","id":"2","color":"#388c46","latex":"a=0","hidden":true,"sliderHardMin":true,"sliderHardMax":true,"sliderMin":"-3","sliderMax":"1","sliderInterval":"0.25","style":"SOLID"}]}}[/DESMOS]
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 45 ·
2
Replies
45
Views
4K