Question about a Tangent Vector

[smodak]

In the following book, please look at equation 3.16. Why are the components of the tangent vector given by
uⁱ = dxⁱ/dt? I understand the velocity components would be dxⁱ/dt and the velocity vector would be a tangent vector. Is that the same reasoning the author uses? The book is normally crystal clear, but, this specific discussion confused me.

https://books.google.com/books?id=7zFalCF_LiEC&pg=PA48&lpg=PA48&dq=tangent+vector+parabola+dx/dt&source=bl&ots=Q2m0RBTbkI&sig=moW1rwceNUEop-SkG5rLfkeAZhY&hl=en&sa=X&ved=0ahUKEwjrgee4qKbMAhWCw4MKHeE7D3gQ6AEIJTAC#v=onepage&q=tangent vector parabola dx/dt&f=false

Thanks in advance for your help!

 

[Let'sthink]

dvi/dt will be acceleration component!

 

[smodak]

dvi/dt will be acceleration component!

Sorry. I meant dxⁱ/dt. But the question is still the same.

 

[Let'sthink]

That is how we define velocity.

 

[smodak]

That is how we define velocity.

I understand. What I am asking is why the components of a tangent vector to a parametric curve x = f(t) and y = g(t) would be dx/dt and dy/dt?

 

[Chestermiller]

I understand. What I am asking is why the components of a tangent vector to a parametric curve x = f(t) and y = g(t) would be dx/dt and dy/dt?

Yes. The velocity vector is tangent to the trajectory of the particle.

 

[Let'sthink]

For constant unit vectors it is obvious for varying unit vectors as in curvilinear coordinates dxi/dt should include the change in direction part too.

 

[Chestermiller]

For constant unit vectors it is obvious for varying unit vectors as in curvilinear coordinates dxi/dt should include the change in direction part too.

Not for the special case of the expressing the velocities. A differential position vector along the trajectory is $$\vec{ds}=\vec{a_1}dx^1+\vec{a_2}dx^2$$ where the a's are the coordinate basis vectors. So the velocity is given by:$$\vec{v}=\frac{\vec{ds}}{dt}=\vec{a_1}\frac{dx^1}{dt}+\vec{a_2}\frac{dx^2}{dt}$$
If you want to express the accelerations in curvilinear coordinates, however, you need to include the derivatives of the coordinate basis vectors. The difference is that ds is already a differential vector, while v is not.

 

[Let'sthink]

Yes I understand. Thank you Sir Insights Author.

 

[smodak]

Yes. The velocity vector is tangent to the trajectory of the particle.

Thanks. How can this be proved? How can we prove that the components of a tangent vector to a parametric curve x = f(t) and y = g(t) would be dx/dt and dy/dt?

Let's say that that curve is a simple parabola y = -x².
In parametric form (as a function of time t): x = t , y = √(1-t²)
How can we prove that the components of any tangent vector to the above curve would be dx/dt and dy/dt?

 

[Chestermiller]

Thanks. How can this be proved? How can we prove that the components of a tangent vector to a parametric curve x = f(t) and y = g(t) would be dx/dt and dy/dt?

Let's say that that curve is a simple parabola y = -x².
In parametric form (as a function of time t): x = t , y = √(1-t²)

The parametric form you wrote down does not agree with the equation you wrote down for the curve.

How can we prove that the components of any tangent vector to the above curve would be dx/dt and dy/dt?

Those are not the components of any tangent vector to the curve. But, the components of any tangent vector to the curve will be proportional to these. For example, the components of a unit vector tangent to the curve are given by $\frac{\frac{dx}{dt}}{\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}}$ and $\frac{\frac{dy}{dt}}{\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}}$

 

[pixel]

How can we prove that the components of any tangent vector to the above curve would be dx/dt and dy/dt?

Consider a point x(t),y(t) on the curve at time t. After a time Δt, we are now at the point x(t+Δt),y(t+Δt) and the vector connecting the two points, Δr, which is approximately in the tangent direction, has components x(t+Δt)-x(t) and y(t+Δt)-y(t). As we let Δt -> 0, the velocity, v, is given by dr/dt and its components ( x(t+Δt)-x(t) )/Δt and ( y(t+Δt)-y(t) )/Δt as Δt -> 0 are just dx/dt and dy/dt. So dx/dt and dy/dt are the components of the tangent vector.

 

[DrGreg]

How can we prove that the components of a tangent vector to a parametric curve x = f(t) and y = g(t) would be dx/dt and dy/dt?

Well, what is your definition of a tangent vector?

 

[smodak]

The parametric form you wrote down does not agree with the equation you wrote down for the curve.

You are absolutely correct. Those are for a circle. I was not thinking :).

Those are not the components of any tangent vector to the curve. But, the components of any tangent vector to the curve will be proportional to these. For example, the components of a unit vector tangent to the curve are given by $\frac{\frac{dx}{dt}}{\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}}$ and $\frac{\frac{dy}{dt}}{\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}}$

I understand now. Thanks.

I was really confused by what the author wrote in the book in my originally linked page.

 

[smodak]

Well, what is your definition of a tangent vector?

Tangent vector: Anyvector that is tangent to a curve or surface at a given point

 

[DrGreg]

Tangent vector: Anyvector that is tangent to a curve or surface at a given point

That doesn't really help. What's your definition of "tangent to a curve or surface"?

(If your answer in post #14 means that you think that the problem of this thread is solved, there's no need to respond.)

 

[smodak]

That doesn't really help. What's your definition of "tangent to a curve or surface"?
(If your answer in post #14 means that you think that the problem of this thread is solved, there's no need to respond.)

Tangent = "just touches" a curve at a single point. A vector is a tangent vector to a curve if it just touches a curve at a single point. I believe, It is a vector that is in (or exactly opposite to) the direction of motion. Is that not correct?

Also #14 sort of helps but I am still a bit unclear about what the authors Øyvind Grøn, Arne Næss are saying from the link in my OP. It almost seems like they are just saying that the components of the tangent vectors to the curve are dx/dt and dy/dt without offering any reasoning. I must be missing something obvious.

 

[pixel]

It almost seems like they are just saying that the components of the tangent vectors to the curve are dx/dt and dy/dt without offering any reasoning. I must be missing something obvious.

Did you read my post #12 above?

 

[smodak]

Did you read my post #12 above?

Actually I had missed it. I think I understand what you mean.

 

[smodak]

Did you read my post #12 above?

If I understand you correctly, according to your logic, velocity is a tangent vector and the components of the velocity tangent vector are dx/dt and dy/dt. However, this does not automatically mean that any tangent vector would have components dx/dt and dy/dt as Øyvind Grøn, Arne Næss seem to indicate in the book. This is the crux of my confusion.

 

[pixel]

If I understand you correctly, according to your logic, velocity is a tangent vector and the components of the velocity tangent vector are dx/dt and dy/dt. However, this does not automatically mean that any tangent vector would have components dx/dt and dy/dt as Øyvind Grøn, Arne Næss seem to indicate in the book. This is the crux of my confusion.

Okay, I think I see what you're saying. Yes, the velocity is one particular tangent vector that has a particular physical meaning. There are other tangent vectors that have the same direction but differ in their magnitude. They all have to have the same slope, however, which is given by the ratio of the y-component to the x-component. Note that dy/dt divided by dx/dt equals dy/dx, which is the slope. So any vector with components in the same proportion as dx/dt and dy/dt will be a tangent vector, but they will have different magnitudes. Only the one tangent vector with components dx/dt and dy/dt will be the velocity.

 

[smodak]

Okay, I think I see what you're saying. Yes, the velocity is one particular tangent vector that has a particular physical meaning. There are other tangent vectors that have the same direction but differ in their magnitude. They all have to have the same slope, however, which is given by the ratio of the y-component to the x-component. Note that dy/dt divided by dx/dt equals dy/dx, which is the slope. So any vector with components in the same proportion as dx/dt and dy/dt will be a tangent vector, but they will have different magnitudes. Only the one tangent vector with components dx/dt and dy/dt will be the velocity.

Bingo. Unfortunately, this is not very clearly written in that book although the rest of the logic in the book is fabulous (so far).