When plotting graphs in polar coordinates, how does one know when to

Click For Summary
SUMMARY

This discussion focuses on determining the characteristics of polar coordinate graphs, specifically when to expect sharp points versus dimples. The examples provided include the polar equations r=1-cos(θ) and r=3/2 + cos(θ). The conversion to Cartesian coordinates is demonstrated using the formulas x=r*cos(θ) and y=r*sin(θ). The analysis of the velocity vector through derivatives reveals that when a=1, the velocity vector at θ=0 is zero, indicating a cusp, while a=3/2 results in a smooth curve.

PREREQUISITES
  • Understanding of polar coordinates and their equations
  • Familiarity with Cartesian coordinate transformations
  • Knowledge of calculus, specifically derivatives
  • Basic graphing skills for polar and Cartesian systems
NEXT STEPS
  • Study the properties of polar curves and their derivatives
  • Learn about cusps and singularities in polar coordinates
  • Explore the implications of different values of 'a' in polar equations
  • Investigate the graphical representation of velocity vectors in polar plots
USEFUL FOR

Mathematicians, physics students, and anyone interested in advanced graphing techniques in polar coordinates.

hale2bopp
Messages
20
Reaction score
0
When plotting graphs in polar coordinates, how does one know when to make the graph sharp (at θ=0) (as in for the graph for r=1-cosθ) as opposed to a dimple (r=3/2 + cos θ) ?
 
Physics news on Phys.org


A graph in polar coordinates is given by r=f(\theta).
Now, we can express such a graph in cartesian coordinates. So, if r=f(\theta), then we can use the formulas

x=r\cos(\theta),~y=r\sin(\theta)

to come up with the following form of the graph in cartesian coordinates:

(f(\theta)\cos(\theta),f(\theta)\sin(\theta)

For example, given r=a-\cos(\theta) (with a constant), we can write this in cartesian coordinates as

((a-\cos(\theta))\cos(\theta),(a-\cos(\theta))\sin(\theta))

Now, the use of this is simpy that we can now investigate our curve using analysis. So, we can find the "velocity vector" at a point by taking derivatives. The derivative of our above function now becomes

(\sin(\theta)(2\cos(\theta)-a),a\cos(\theta)-\cos(2\theta))

Now, if a=3/2, then our derivative in 0 is (0,1/2)

So, we can deduce that in 0, our function is going up with a speed of 1/2.

What if a=1? Then our derivative in 0 is (0,0). This is a weird result. It means that at 0, our velocity vector is zero and thus the curve just stands still. This is the explanation of why you get a sharp point (= a cusp) when a=0, but just a smooth line when a=3/2.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 20 ·
Replies
20
Views
7K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K