Horizontal tangent to wolfram alpha's heart-shaped graph

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The discussion focuses on finding horizontal and vertical tangents for a heart-shaped curve defined by specific parametric equations. Users explore the values of t that yield horizontal tangents, noting that Wolfram Alpha provides one solution but misses the symmetric counterpart. Concerns are raised about the presence of complex roots and the behavior of vertical tangents, with some participants debating the differentiability of the curve at certain points. The conversation highlights the importance of understanding parametric equations and the conditions under which tangents can be determined. Overall, the thread emphasizes the intricacies of analyzing parametric curves in calculus.
  • #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?
 
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  • #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.
 

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