Horizontal tangent to wolfram alpha's heart-shaped graph

Click For Summary

Discussion Overview

The discussion revolves around finding horizontal and vertical tangents to a heart-shaped curve defined parametrically. Participants explore the mathematical implications of the curve's equations and the behavior of tangents at specific points, while questioning the outputs from Wolfram Alpha.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents the heart-shaped curve's parametric equations and seeks to identify points where horizontal tangents occur by solving for dy/dt = 0.
  • Another participant notes that horizontal tangents occur at t = 0.908 and suggests that the root plot should also show t = 2π - 0.908.
  • Concerns are raised about the absence of certain real solutions and the presence of complex roots in Wolfram Alpha's output.
  • Participants discuss the implications of dx/dt = 0 and its relation to vertical tangents, with one participant expressing confusion about the expected intersections with the x-axis.
  • There is a debate about the differentiability of the curve at points where dx/dt = 0, with differing views on whether the curve can be treated as a function at those points.
  • One participant questions the validity of using arctan(y/x) to identify angles associated with the curve, leading to a request for clarification on the reasoning behind this approach.
  • Another participant suggests that the parametric nature of the curve allows for direct computation of dy/dt and dx/dt without needing to derive from polar coordinates.

Areas of Agreement / Disagreement

Participants express differing views on the behavior of tangents and the outputs from Wolfram Alpha. There is no consensus on the correct interpretation of the tangent points or the differentiability of the curve at specific locations.

Contextual Notes

Participants highlight limitations in the outputs from Wolfram Alpha, particularly regarding the identification of real versus complex solutions. The discussion also touches on the mathematical properties of the curve and the assumptions made in analyzing its tangents.

Who May Find This Useful

This discussion may be of interest to those studying parametric equations, differential geometry, or mathematical analysis, particularly in the context of curves and tangents.

  • #31
Thanks for the recommendation, Svein. But when you write

Svein said:
since a tangent vector in the y-axis direction is just as valid as any other tangent vector

valid for what? As a solution for dy/dx? It sounds as if you are saying that all tangent vectors, no matter which direction, are equivalent to one another. Perhaps you meant that a tangent vector, positioned on the y-axis, is the same as the same tangent vector, positioned anywhere else?
 
Physics news on Phys.org
  • #32
nomadreid said:
valid for what?
For a tangent vector.
nomadreid said:
As a solution for dy/dx?
No. dy/dx is an expression for the tangent when y is expressed as a function of x.
nomadreid said:
Perhaps you meant that a tangent vector, positioned on the y-axis, is the same as the same tangent vector, positioned anywhere else?
No. A tangent vector is a vector, tangent to a curve.
 
  • #33
You are saying "a tangent vector in the y-axis direction is just as valid as a tangent vector as any other tangent vector". I'm not sure I am parsing that right. Are you saying that if we are looking for any old tangent, we might as well take one in the vertical direction, if it exists, as one in some other direction? That would be true, but we are not looking for any tangent vector vt=[(48 sin2t (cos t)), - 13 sin t +10 sin (2t) +6 sin (3t) + 4sin(4t)] , but specifically the ones that will be the direction vectors (0,a) and (b,0) (a,b, non-zero) for vertical and horizontal tangents.
 
  • #34
nomadreid said:
but specifically the ones that will be the direction vectors (0,a) and (b,0) (a,b, non-zero) for vertical and horizontal tangents.
Using the formulas in #26, a tangent vector of the type (0, a) is given by \frac{dx(t)}{dt}=0 and a tangent vector of type (b, 0) is given by \frac{dy(t)}{dt}=0.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
9K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
8
Views
6K
Replies
31
Views
12K
  • · Replies 18 ·
Replies
18
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
9K