# I Question about a Tangent Vector

1. Apr 23, 2016

### 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.

2. Apr 23, 2016

### Let'sthink

dvi/dt will be acceleration component!

3. Apr 23, 2016

### smodak

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

4. Apr 23, 2016

### Let'sthink

That is how we define velocity.

5. Apr 23, 2016

### smodak

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?

6. Apr 23, 2016

### Staff: Mentor

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

7. Apr 23, 2016

### 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.

8. Apr 23, 2016

### Staff: Mentor

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.

9. Apr 23, 2016

### Let'sthink

Yes I understand. Thank you Sir Insights Author.

10. Apr 24, 2016

### smodak

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?

11. Apr 24, 2016

### Staff: Mentor

The parametric form you wrote down does not agree with the equation you wrote down for the curve.
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 $\frac{\frac{dy}{dt}}{\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}}$

12. Apr 24, 2016

### pixel

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.

13. Apr 24, 2016

### DrGreg

Well, what is your definition of a tangent vector?

14. Apr 24, 2016

### smodak

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

I understand now. Thanks.

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

15. Apr 24, 2016

### smodak

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

16. Apr 24, 2016

### DrGreg

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.)

17. Apr 24, 2016

### smodak

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.

18. Apr 24, 2016

### pixel

Did you read my post #12 above?

19. Apr 24, 2016

### smodak

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

20. Apr 25, 2016

### smodak

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.

21. Apr 25, 2016

### pixel

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.

Last edited: Apr 25, 2016
22. Apr 25, 2016

### smodak

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