Horizontal tangent to wolfram alpha's heart-shaped graph

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  • #26
Svein
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A short comment: If you have a parametric curve (given in the standard vector form [itex]\vec{r}(t)=(x(t), y(t)) [/itex]), then a tangent vector is given by [itex]\vec{\dot{r}}(t)=(\frac{dx}{dt}(t), \frac{dy}{dt}(t)) [/itex]. The length of the tangent vector is, of course [itex]\lvert \dot{r}(t) \rvert=\sqrt{(\frac{dx}{dt}(t))^{2} +(\frac{dy}{dt}(t))^{2}} [/itex]. As long as the length of the tangent vector is not 0, it is possible to define the unit tangent vector [itex] \vec{T(t)}= \frac{\vec{\dot{r}}(t)}{\lvert \vec{\dot{r}}(t) \rvert} [/itex].
 
  • #27
BvU
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I'm not sure this would be relevant, but using l'Hôpital's rule, limt→0 (dx/dt)/(dy/dt) would be
24[cos(0)+3cos(3*0)]/[-13cos(0)+20cos(2*0)+18cos(3*0)+16cos(4*0)] = 96/41≠0
Please explain. I get something else ...
 
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  • #28
nomadreid
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Oops, BvU. You're right, there is an error there. I will post tomorrow morning (it's late where I am), when I can also look at the new posts by Svein and MrAnchovy. Sorry for the delayed response, but PhysicsForums didn't notify me that there had been new posts since the last time I posted.
 
  • #29
nomadreid
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Last posting, continued.
First, BvU, sorry about the mistake: I miscounted how many times I had taken the derivative. So, correctly done, that limit is zero, right?
Secondly, Svein, the unit tangent vectors for a horizontal or vertical tangent are automatically i and j, resp, no?
Thirdly, MrAnchovy: if I am doing my arithmetic correctly (no guarantee), then at π/2,
dx/dy = [48 sin2 (π/2) cos (π/2)]/[- 13 sin (π/2) +10 sin (π)) +6 sin (3π/2) + 4sin(2π)] =0/(-19) =0,
so that would make it vertical, no?
 
  • #30
Svein
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Secondly, Svein, the unit tangent vectors for a horizontal or vertical tangent are automatically i and j, resp, no?
Yes. When I write (a, b) for a vector, it is the same as [itex]a\cdot \vec{i} + b\cdot \vec{j} [/itex].
Comment: When given a parametric formula for a curve, looking at [itex]\frac{dy}{dx} [/itex] is often meaningless, since a tangent vector in the y-axis direction is just as valid as any other tangent vector. I recommend that you use the formula for the tangent vector.
 
  • #31
nomadreid
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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?
 
  • #32
Svein
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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.
 
  • #33
nomadreid
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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
Svein
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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 [itex]\frac{dx(t)}{dt}=0 [/itex] and a tangent vector of type (b, 0) is given by [itex]\frac{dy(t)}{dt}=0 [/itex].
 

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