# Is the derivative of a vector perpendicular always?

1. Dec 29, 2013

### arman.raina

I'm learning vectors. I read somewhere that if a vectors magnitude is constant, then its derivative is perpendicular. However, in polar co-ordinates, I learnt something else.
The distance of a particle in orbit from a focus is r. if /r/ varies with t even then dr/dt is v and it is perpendicular to r. Further if /v/ varies, then dv/dt is a and is is perpendicular to v and parallel to r. Must the magnitude of the vector be constant or not for the derivative to be perpendicular?

2. Dec 29, 2013

### Simon Bridge

Welcome to PF;
I'm afraid none of that is making sense - if the radial distance varies with time, then the time derivative will be the radial velocity. The radial velocity will be in the direction that the radius changes... the radial direction.

$\dot{\vec r} = \frac{d}{dt}(r \hat r + \theta \hat \theta) = \dot r \hat r + \dot \theta \hat \theta=v\hat r + \omega \hat \theta$

"perpendicular to a vector" is not very precise anyway - there's an infinite number of vectors perpendicular to any particular vector. The derivative of a vector function gives the gradient of the function - the slope of the tangent.

In circular motion r does not change with time, so it's time-derivative is zero ... but the perpendicular (we'd say "tangential") component of the velocity is still non-zero.

In elliptical motion - the velocity and acceleration will have radial and tangential components that vary. The result is that the velocity and acceleration vectors rarely point perpendicularly to the radius vector:

Notice that the net force points opposite to the radial direction all the time but there are only two places where the velocity vector is perpendicular to that.

Aside: you can use LaTeX markup to represent vectors like this $\vec{v}$ or just bold-face them like this v.

3. Dec 30, 2013

### scoobmx

If the vector magnitude really is constant in time then you can prove it like this:
$| \vec{r} |^{2} = \vec{r} \cdot \vec{r} = c$
$\frac {d}{dt} \vec{r} \cdot \vec{r} = 2 \vec{r} \cdot \frac{d \vec{r}}{dt} = 0$
The last equality means r and its derivative are orthogonal.

4. Dec 30, 2013

### arman.raina

Sorry about not making sense. I think that is a result of my confusion.
In that video from around 3:40 onwards it says that r and a are parallel. I assumed that this is a result of v being perpendicular to both and being limited to the x-y plane. This I thought must result from v being the derivative of r and a being the derivative of v. My problem is that the magnitudes of all the vectors are variable. And the proof above assumes that the magnitudes of the vectors being derived is constant.
This is the definition of v used in the video.
v=dr/dt
There is an error in the video where it says A=2pi*r^2 but I don't think it matters.
I re-read your diagram. Does this mean that r and v are not necessarily perpendicular, but r and a are parallel even in an ellipse?

Last edited: Dec 30, 2013
5. Dec 30, 2013

### Simon Bridge

Yes.
The acceleration always points in the direction of the net force.
In orbits, like shown in the vid, the only force is gravity ... and that will always point at the central mass - so the acceleration always points that way too.
The velocity vector, however, always points in the (instantaneous) direction of travel ... that is part of the definition of "velocity".
Since an object in orbit is constantly changing direction, the angle between velocity and acceleration will vary - except for a circular orbit.

You have to be careful about drawing general conclusions from specific situations.

That only happens for a circular orbit.

That is correct, for the total velocity.

Yes. Provided the only force on the object orbiting comes from the thing it is orbiting.

6. Dec 31, 2013

### arman.raina

Thanks. I am beginning to get it. F, a and r are all parallel because F=ma. v is only perpendicular to all these in a circular orbit or during certain instances of an ellipse.
I am still unsure of why acceleration and radius are parallel mathematically. Under what restrictions is the second derivative of a vector always parallel to the original vector? (Like a and r are parallel) Neither a and r are constant...
I placed the restriction that r=(rcosθ,rsinθ,0)
a=d^2r/dt^2
I was left with the following additional restriction
2(dr/dt)(dθ/dt)-(r)(d^2θ/dt^2)=0
I substituted and got V(r)=a(t)/2ω
Is there any physics or mathematics principle that makes one of the above statement true in the case of an elliptical orbit? Or is that statement incorrect?

7. Dec 31, 2013

### scoobmx

The physical principle is the acceleration is toward the gravitation body, which if you place at the origin means it accelerates everything around it straight toward itself, which is the radius direction.

8. Dec 31, 2013

### Simon Bridge

There is no purely mathematical reason that this should be the case - the reason is a mixture of physics and the way the coordinate system was set up.

If you set the coordinate system for the elliptical orbit so that the origin was in the geometric center of the ellipse, then the position vector and the acceleration vector will no longer point in opposite directions.

That would be a circular orbit, with the center at the origin... unless r is a function of time?
What you want is the parametric equation of an ellipse.

Last edited: Dec 31, 2013
9. Jan 1, 2014

### arman.raina

Thanks so much. I think I'm almost there. R is indeed a function of time, so the statement is plausible. The a(t) is actually tangential acceleration. In the case of circular motion the statement this is true whenever, since this simplifies to 0=0. I think I need to rephrase the last part of my question
1) firstly, is the rxa=0 calculated correctly (I'm teaching myself linear algebra, so I'm not very confident)
2) if I start with a statement of truth in this case (rxa=0) then I should end with a statement of truth in (this case I got v(r)=a(t)/2ω) that I can independently verify...it should be evident why this is true aside from the calculation
3) I.e if I were RTP the above statement, without taking rxa=0, how could I?
I understand now how the acceleration is always towards the origin, and thus along the radius. I just want to ascertain that rxa=0 in this case, for it is not universally true. I.e. If I were to accept that the acceleration of an orbiting body is parallel to the radius, then rxa=0 no matter what. However, there seems to be a restriction that the v(r),a(t),ω must be in ratio. I am thinking that with some manipulation, it should resemble a well known physics principle that is applicable in the case of elliptical, orbital motion.

10. Jan 1, 2014

### Simon Bridge

So when you got:
... you meant that the velocity varies with position by the ratio of the tangential acceleration with the angular velocity?

I think you got confused somewhere.

Notation can do that too ... time is not a vector, neither are the parentheses, so don't bold-face them. It is a(t) or $\vec a_\perp(t)$. So v(r) means something different from v(r).

Physically: $$a_\perp(t) = r(t)\frac{d^2}{dt^2}\theta(t)$$

I don't know because you did not show your working.
I think you got confused someplace though. Perhaps you forgot that ω is also a function of time?

There is a bit of an issue in that r(t) and θ(t) are not specified.
You may want to have a look at Kepler's laws.

What are you trying to discover?

11. Jan 5, 2014

### arman.raina

Thank you. All are functions of time. I have not figured out how to write a subscript of 't' on a, to indicate tangential acceleration. so sr/dt=v(r)which is actually v subscript r of t. a⊥=2ω(dr/dt). Does that make sense?
It folllows from 2(dr/dt)(dθ/dt)-(r)(d^2θ/dt^2)=0
r is the magnitude of r. θ is the angle of r.
if r=(rcosθ,rsinθ,0)
v=dr/dt=(dr/dtcosθ-rsinθ(dθ/dt),(dr/dt)sinθ+rcosθ(dθ/dt),0)
a=dv/dt=((d^2r/dt^2)cosθ-(dr/dt)sinθ(dθ/dt)-(dr/dt)sinθ(dθ/dt)-rcosθ(dθ/dt)^2-r(d^2θ/dt^2)sinθ,(d^2r/dt^2)sinθ+(dr/dt)cosθ(dθ/dt)+(dr/dt)cosθ(dθ/dt)-rsinθ(dθ/dt)^2+r(d^2θ/dt^2)cosθ,0)
rXa=(0,0,(rcosθ)((d^2r/dt^2)cosθ-2(dr/dt)sinθ(dθ/dt)-rcosθ(dθ/dt)^2-r(d^2θ/dt^2)sinθ)-(rsinθ)((d^2r/dt^2)sinθ+2(dr/dt)cosθ(dθ/dt)-rsinθ(dθ/dt)^2+r(d^2θ/dt^2)cosθ))=(0,0,0)
z dimension
0=(d^2r/dt^2)rsinθcosθ-2(dr/dt)sin^2θ(dθ/dt)-(r^2)sinθcosθ(dθ/dt)^2-(r^2)(sin^2θ)(d^2θ/dt^2)-(d^2r/dt^2)rsinθcosθ-2(dr/dt)cos^2θ(dθ/dt)+(r^2)sinθcosθ(dθ/dt)^2-(r^2)(sin^2θ)
2r(dr/dt)(dθ/dt)-(r^2)(d^2θ/dt^2)=0 (sin^2θ+cos^2θ=1)

12. Jan 5, 2014

### Simon Bridge

Either: at or $\vec a_t$ ... click on the "quote" button attached to the bottom of this post to see how I did that.
In the advanced editor, there is a button that looks like X2 that will put the "sub" tags around highlighted text ... makes it easier.

Sure - though it can get eye-watering to follow.
v(r) would normally mean that the velocity depends on the radius though ... vr(r) would mean that the radial velocity depends on the radius.
The initial and final radial velocities would be vri and vrf respectively and after that it gets nasty.

But better to write vr and even better $v_r$ ... or just avoid subscripts whenever possible by defining "v is the radial velocity and ω is the angular velocity - the tangential velocity is rω" or whatever takes your fancy .... no subscripts needed see?

Um... $$2\frac{dr}{dt}\frac{d\theta}{dt}-r\frac{d^2\theta}{dt^2}=0:\;\vec{r}=(x,y,z), r=\sqrt{x^2+y^2+z^2}$$
... that what you mean?
... but where did the "2" out the front come from?

Let's see, the LHS would be the time derivative of the tangential velocity: $$\frac{d}{dt}\left( r\frac{d\theta}{dt} \right)=0$$ ... means that the tangential speed is a constant. That what you want?

You then describe something in motion with the following relation:
Now, if I understand you correctly, you man that the radius r varies with time.
If you also want the tangential velocity to be a constant, then the object is moving under power rather than orbiting a central body. (Either that or the radius isn't changing with time.) Is this what you had in mind?