Horizontal tangent to wolfram alpha's heart-shaped graph

[nomadreid]

Thanks for the recommendation, Svein. But when you write

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?

 

[Svein]

valid for what?

For a tangent vector.

As a solution for dy/dx?

No. dy/dx is an expression for the tangent when y is expressed as a function of x.

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.

 

[nomadreid]

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 v_t=[(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.

 

[Svein]

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 $\frac{dy(t)}{dt}=0$.