Cartesian to Polar Vectors

sponsoredwalk

Hi I pretty much can't get past the first chapter in my physics book until I master the vector representation of polar coordinates.

Every explanation I've read thus far has confused me, I keep thinking in Cartesian terms so I think it'd be great to convert a vector equation from cartesian to polar description and then differentiate (with somebodies help, which I need!).

I have a cartesian based vector,

r(t) = (2t^2)i + (3t - 2)j

I don't know how to get this into polar form, do I take the magnitude, how does that work out with the t's? Do I throw sine's and cosines in? How?


v(t) = (4t)i + (3)j
a(t) = 4i

Do these facts help at all?

raah!

Are you looking for full polar coordinate expression or just x=RcosA type thing?

With the unit vectors nothing is more helpful than a diagram. The diagrams are everywhere, but anyway. If A is the angle between r(t) and the positive x axis, then R(unit vector)= cosA i+ SinA j. And O(theta)= -SinAi + CosAj.

I shoulda probably used latex for that. But if you have a diagram, and go over thr trigonometry of it and stuff it you should be able to arrive at those expressions. Well that was a very bad explanation actually.

Oh and if you want to turn them into polar you go. r(t)=rr where r is the magnitude of r(t). And you can see that this is r(t)= rcosA+ rsinA.

Well, someone else will do it properly I'm sure

sponsoredwalk

Ok I've calculated some stuff, what do I do with it



EDIT: Thanks for the links, but they are very integration-ey, I'm just trying to get a grasp of the differentiation, it's kind of a personal quest

raah!

You want the velocity of the polar coordinates? You certainly don't need to find those values at points like that. It will help you understand the geometry alright, but everythign about the trajectory of the particle, vector, whatever is contained in your expression for r there.

It goes v= r'r + O'O

This follows from the earilier definition of the unit vectors r and O. (omeans theta) the ' means differentiated with respect to time.

Those links are second year as well, you don't reall need to know about how to swop between cartesian and polar coordinates to use them. Also, one time a tutorial guy said to us that that unit vector polar coordinate expression was not very useful at all.

And I suppose to find the dependence of angle on time just use O = arctan x(t)/y(t)

I'm not really sure though, I don't think I've ever done this without a jacobian before. Probably have.

sponsoredwalk

I'm not at the level to do Jacobians, I'm vaguely aware of what they do but from Kleppner & Kolenkows Mechanics book & the level of itI'm 100% certain there's no advanced methods to using this formulation.

raah! wrote: »

Oh and if you want to turn them into polar you go. r(t)=rr where r is the magnitude of r(t). And you can see that this is r(t)= rcosA+ rsinA.

Well, you see that I've calculated the magnitude of r(t) at seperate instances, but how are you supposed to do it generally?

Yeah polar coordinates are a pain in the #al## but I'm on a personal quest

At every step it confuses me, I can't even get the thing into a general form polar-ly so that I can take the derivative to get velocity.

Am I supposed to take the first position vector, second?

Why would anybody use these things if they only describe instances & not give a general description like the cartesian flavouring does...?

Well, it has to be possible!

raah!

Ah, the magnitude of a vector is (x(t)^2 = y(t)^2) ^1/2. And that's just a scalar, which depends on time, and so is general. I had kleppner and kolenkow too for my first year mechanics.

On page 31 there is a "method 2" which was what I was used to talk about derivatives and stuff of position vectors expressed in polars. Tbh the book isn't very good. It does things in bad order, and the problems are just irritating too. I'm basing this judgement on a comparision between this book and my full physic's books treatment of mechanics. It was alot slower, but there was never anythign that you would not understand because it explained everything.

Obviously I have too much free time, and I did the r in the first polar equation, from there it is just standard differentiation, once you have a value for theta with respect to t as well.

Anyway r=1/2(4t^4+9t^2-12t+4)^-1/2 x(16t^3+18t-12), and O= 3/2t^-1 -t^-2. So those are two scalars which will be in the polar form of the position vector. The rest is just differentiating with the chain rule. From here you can get an expression for rr, then v in polars, and then a in polars. These are done in kleppner and kolenkow, and also I was still learning about them in chapter 2 as well.

I thinkt he most difficult aspect is the differentiation of the vectors. And when you come to the acceleration, you'll see loads of familar ones that can exist, like the one for circular motion v^2/r, but also strange ones. Like centrifugal acceleration and stuff.

sponsoredwalk

So the magnitude of

r(t) = (2t^2)i + (3t - 2)j

Would be;

|r|=r = √[(2t²)² + (3t - 2)²]
= √/B]4t^4+ 9t[B]²[/B] -12t + 4[B

???

I probably got that wrong!

What physics book are you comparing it to btw?

raah!

Oh yeah that's right. I did the rate of change of the magnitude with respect to time by accident.

I have university physics by young and freedmen.

sponsoredwalk

Alright, we're getting in deep here!

R(unit vector)= cosA i+ SinA j. And O(theta)= -SinAi + CosAj.

Oh and if you want to turn them into polar you go. r(t)=rr where r is the magnitude of r(t). And you can see that this is r(t)= rcosA+ rsinA.

r(t)=rr

r = √/B]4t^4+ 9t[B]²[/B] -12t + 4[B

r(unit vector) = cosA i+ SinA j

So!

r(t)=rr ---> √/B]4t^4+ 9t[B]²[/B] -12t + 4[B [cosA i+ SinA j]

Where A is the angle.

I'm at a loss now how to get the angle A

I'll guess O = [cosA i+ SinA j]


v= r'r + O'O

Before I actually do this ridiculous differentiation, it's just the product rule on r(t)=rr = √/B]4t^4+ 9t[B]²[/B] -12t + 4[B [cosA i+ SinA j] right?


Btw: Thanks for putting up with me,


Edit: I have both of those books too, I would say Kleppner & Kolenkow is more advanced than University Physics. I have to say I'd be lost without Young and Freedman, it is a great book.

raah!

Oh sorry, I only wrote A earlier because I still can't write theta.

So A(t) = arctanj/i. This looks ridiculous, but in all the question sin kleppner and kolenkow that involve polars, the angular velocity is given, that is A(t)'. I've never seen an arctan given. But again, another thing is that they rarely ask you to change from one set of coords to another that much too.

Now the unit vector O(theta) is perpendicular to r(unit vector) and is equal to (-sinAi + CosAJ). The best way to see this is the diagram on page 29.

Yes the product rule on those too things. And when you differentiate CosAi+ sinAj with respect to time, you get -A'sinAi+ A'CosAj which is equal to A'O.

Np, I'm getting good revision too, and obviously I'm not very busy

Edit: Also all this differentiation is done in kleppner and kolenkow, some of those proof methods you should jsut ignore, method two is the best. And once you know what r' and O' are you don't need to refer back to the cartesian expressions anymore. I went along with the book when I was doing it.

And yeah, really kleppner and kolenkow is much better, I've actually never read the mechanics sections of university physics... it's I've found with other things they explain thigns too much if anything.

What college/course are you in by the way?

sponsoredwalk

Yeah, the examples they give in the book already have like "cos" and "sin" there. I'm just trying to build up the whole idea from a basic cartesian standpoint.

I think I'm doing this mainly because after years and years thinking in terms of Descartes coordinates it's just ingrained & tough to stamp out.
If I ever get lost in Polar land I'll be able to swim back to Cartesian shores & do it all again you know

I can't find a single example on the net of what we're doing here, and for good reason lol, look at the size of the expression already!


θ(t) = arctan(j/i) = arctan((3t - 2)/(2t²)

r(t)=rr ---> √/B]4t^4+ 9t[B]²[/B] -12t + 4[B [cosA i+ SinA j]

Will become

√/B]4t^4+ 9t[B]²[/B] -12t + 4[B [cosθ i+ Sinθ j]

but θ is dependent on the original expression, found by arctan(j/i) , and is a function of time. So;

r(t) = √/B]4t^4+ 9t[B]²[/B] -12t + 4[B /COLOR][/B]cos[B][[/B]arctan[B]([/B](3t - 2)[B]/[/B](2t[B]²)][COLOR=DarkRed]i[/COLOR] + [/B]sin[B][[/B]arctan[B]([/B](3t - 2)[B]/[/B](2t[B]²)[COLOR=Red]j[/COLOR][COLOR=Blue

haha!!!

There is no way in hell that what I've written is right, is there?

Then, we would differentiate this using a heck load of chain rule and product rule!!!


EDIT: I'm not in a course, I'm just trying to teach myself all of this crazy stuff & am going to go for Theoretical Physics in trinity in a year for the over 23 thing you know. Most of the work is just maths lol, it's ridiculous sometimes (as we can plainly see with this monstrous calculation!).

How about you, what course are you doing?

raah!

It is correct, but those are cartesian coordinates. The Cosθi+ Sinθj are equal to r(with a hat on it, the unit vector).

It talks about it in the book, that even though r=rr does not explicitly depend on θ, it's there in the defintion of the unit vector in the r direction.

You could do all the differentiation that way, and then sub back in r and θ, but it would be alot more difficult. In polars θ(t) never actually appears, it is subsumed in the definitions of the unit vectors. θ' does appear however. Which would be hte derivative of that artan thing.

But yeah you've done nothing wrong there, you are just not using hte polar coordinates unit vectors. Though they are there , and equal to

[cos[arctan((3t - 2)/(2t²)]i + sin[arctan((3t - 2)/(2t²)j] = r
[-sin[arctan((3t - 2)/(2t²)]i + cos[arctan((3t - 2)/(2t²)]j = θ

Did you get this "θ" from word or is there a sequence of things you press to get it ?

gerardduff

sponsoredwalk wrote: »

There is no way in hell that what I've written is right, is there?

You're right.
So, it looks a bit messy from here on, I mean if you now want to compute the time derivative. It should look like this...

r(t) = A(t)*[cosθ(t)+sinθ(t)]

where A(t) means A is a function of t and likewise for θ. So we are going to have a 'function of a function' or chain rule coming in if we want to differentiate r(t). And that's what happens...

The asterix means multiply ( I just put it there for clarity to show where the product is)

So, applying the product rule,

r'(t)=A'(t)[cosθ +sinθ] + A(t)[-sinθ* dθ/dt+cosθ *dθ/dt]

where again, the Asterix means multiply.

The asterix is not normally used to signify multiplication (except in computing)...I just used it here so I could refer to it, and you hopefully could see better what's going on.

You're doing fine so far, hope I haven't screwed you up with this explanation!
You still have to do the differentiation though...but you know what θ(t) or rather tanθ(t) is so you'll get terms like

sec^2(θ) dθ/dt= d/dt[(3t-2)/2t^2]
which is a fairly straight forward 'quotient' rule on the right hand side.

sponsoredwalk

Haha, I have a list of symbols in notepad file on my desktop.

Here are the most common offenders;

²ƒ ¹ ⅓³Σ∞ °∂ ∫√ ¼ ½ ∀× ÷ ← → ⇒ δ ∇∃ε Δ≈ ≠ ±γθψω
² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁿ ⁽ ⁾ ₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋ ☺ ᵟᵞ/ₔₓ ᵟ ᵋ ᶿ ᶥ ᶲ

I started reading the chapter there & I get confused again.

I don't know why but the 10 pages or so on polar coordinates seem to haunt me.

Seriously though, you're helping to make everything a lot clearer.

Looking at the picture on page 29, by the orthogonality of r & θ and by the geometry of the angle I understand why θ has the -sin & cos thing going on.

Also, yeah I see in the book that the r in rr is the basis vector (with the hat). That is what I mean in what follows, r(t) is position & r is unit vector.

r(t)=rr is shorthand for r(t)=√/B]4t^4+ 9t[B]²[/B] -12t + 4[Br

which is shorthand for

r(t)=√/B]4t^4+ 9t[B]²[/B] -12t + 4[B [cosθ i+ Sinθ j]

which is shorthand for

r(t)=√/B]4t^4+ 9t[B]²[/B] -12t + 4[B /COLOR][/B]cos[B][[/B]arctan[B]([/B](3t - 2)[B]/[/B](2t[B]²)][COLOR=DarkRed]i[/COLOR] + [/B]sin[B][[/B]arctan[B]([/B](3t - 2)[B]/[/B](2t[B]²)[COLOR=Red]j[/COLOR][COLOR=Blue

I think a big part of my misunderstanding was the θ. Look at that dependence, if the calculation you'd have to do to find the position at any particular time would be a good bit of work.
The main task in this whole debacle was to at least see this craziness going on! lol

However, I don't see how to fit
θ=[-sin[arctan((3t - 2)/(2t²)]i + cos[arctan((3t - 2)/(2t²)]j

into the position description? the bold θ,"θ" is a basis vector.

I assume you write this for when it is used in the velocity description, right?

the unbolded θ is held constant for whatever particular time you want to find the position the θ will pop out by evaluating what is inside the brackets of r (the whole arctan stuff).

Am I right about this?


EDIT: Thanks Gerard, yeah I haven't differentiated yet because I want to be absolutely 100% clear on my description of the position.
You see there's still a small source of confusion, but if the above is all right then we'll get cracking with the fluxions!

gerardduff

You use the θ=blah blah in 't' whenever you want to swap between coordinate systems.
Otherwise, eg...when doing differentiation..., just use the simplest expression. I mean, inverse tan is not very user friendly....and that's being nice about it.

You seem to be progressing well, even mentioning basis vectors and recognising intuitively the reason for the minus sign when you differentiate the cosine function.

Methinks you'll do well next year.

<scurries back behind calculator>

raah!

sponsoredwalk wrote: »

EDIT: I'm not in a course, I'm just trying to teach myself all of this crazy stuff & am going to go for Theoretical Physics in trinity in a year for the over 23 thing you know. Most of the work is just maths lol, it's ridiculous sometimes (as we can plainly see with this monstrous calculation!).

How about you, what course are you doing?

Ahhh, I was doing theoretical physics, and was half way through second year, but then I came back to cork for financial reasons and all that. Now I'm going to do second year in ucc when it starts again, So I'm pretty much doing exactly what you are at the moment, teaching myself physics and mechanics and working through all the books I have. Mechanics as taught by kovacs and from that book was widely considered to be the hardest module of first year too, and polar coordinates are one of the hardest parts of mechanics.

sponsoredwalk

Haha, for some reason I read radiator instead of calculator

Assuming the bold theta, shows it's face in the derivative I'll press on with this mess & try to crack the velocity situation.

r(t)=rr

r(t)=√/B]4t^4+ 9t[B]&#178;[/B] -12t + 4[B [cosθ i+ Sinθ j]

r(t)=√/B]4t^4+ 9t[B]&#178;[/B] -12t + 4[B /COLOR][/B]cos[B][[/B]arctan[B]([/B](3t - 2)[B]/[/B](2t[B]&#178;)][COLOR=DarkRed]i[/COLOR] + [/B]sin[B][[/B]arctan[B]([/B](3t - 2)[B]/[/B](2t[B]&#178;)[COLOR=Red]j[/COLOR][COLOR=Blue

I know it's going to be extremely messy but I'm going to differentiate and analyze all three seperate formulations, it' just got to be done in order to be sure! I'll make it as easy on your eyes as possible!

I'll do this one first;
r(t)=√/B]4t^4+ 9t[B]&#178;[/B] -12t + 4[B [cosθ i+ Sinθ j]

= 1/2 /B]4t^4+ 9t[B]&#178;[/B] -12t + 4[B ^-1/2 COLOR=Black]16t^3 + 18t[/COLOR/B]cos&#952; [B]i[/B]+ Sin&#952; [B]j][/B][COLOR=Blue] + [/COLOR][B]&#8730;[[/B]4t^4+ 9t[B]&#178;[/B] -12t + 4[B [-sinθ dθ/dt i + cos θ dθ/dt j ]

Now, if we look at dθ/dt that is indicated in the Kleppner book as θ with a dot over it to signify the time derivative.

Also, we notice that if I factor dθ/dt (or θ dot!)

= /B][/COLOR]1/2[/COLOR][/COLOR]&#8730;[B][[/B]4t^4+ 9t[B]&#178;[/B] -12t + 4[B] COLOR=Black]16t^3 + 18t[/COLOR/B]cos&#952; [B]i[/B]+ Sin&#952; [B]j][/B][COLOR=Blue] + [/COLOR][B]&#8730;[[/B]4t^4+ 9t[B]&#178;[/B] -12t + 4[B [-sinθ i + cos θ j ]dθ/dt

We notice, θ sitting there!

I'll try to rewrite the above expression as cleanly as possible.

v(t) = [1/2r]r'r + rθθ

Which is very similar to the Kleppner & Kolenkow version.

What do you guys think, have I confused or assumed anything stupid?

gerardduff

You've gone a few steps ahead there,

For starters I have,

sec^2 (θ) dθ/dt=(8-6t)/4t^3

Now, in terms of θ and t we should have,

r'=[dA/dt +A(t) θ']cosθ + [dA/dt-A(t) θ']sinθ

where dA/dt is 1/2[4t^4+9t^2-12t+4]^(-1/2)[16t^3+18t-12]

So it is pretty awful looking.

In cartisian we have simply,

r'=4t (i direction)+3 (j direction).

gerardduff

I'm not sure this is the right way of going about this problem. You have to consider how your basis vectors vary with respect to each other. If i and j are the basis vectors in cartesian, what are the basis vectors in polar? They are not the same. Also, what happens to i as you increase [Latex]\theta[/Latex].

Iderown

Sponsored Walk,

The Cartesian unit basis vectors i and j are constant - independent of the problem variable t.

The polar coordinate form consists of two scalars : Magnitude r, and the angle that r makes with the x axis.
Both these quanties are time varying in your problem.

The vector form of r was your starting point. It can be written as (I'm going to have to learn Latex!)

r = IrI x Unit vector r , both factors are time varying.

When you differentiate this (with respect to time) to get the velocity of the particle, the product rule gives a vector sum of components in directions ( Unit vector r ) and (Time rate of change of unit vector r )

A unit vector has length of 1 and you can think of it as the radius of a unit circle. The only way in which a point on the circumference of a unit circle can change is to "travel" to a neighbouring point on the circumference. For a "small" change, this move is at right angles to the original direction of (unit vector r ).

It is when you differentiate the r = Time function of i and j that the transverse direction appears in the result.

You have seen just how messy this sort of differentiation can be. You would need to know all the rules!

Real life mechanical problems mostly come in two forms.

1. The projectile motion type of your example. It is best to stick with the Cartesian vectors here. If necessary convert the resulting velocities and accelerations to polar form after differentiating.

2. Solar system gravitational problems where the force on a body is directed along the direction r. Things are simpler here when the the Cartesian forms are not used. Likewise in problems involving rotating machinery - moments of inertia.


Used to use a wee book "Mechanics" by Smith and Smith although it was considered difficult by some.

gerardduff

Iderown wrote: »

The Cartesian unit basis vectors i and j are constant - independent of the problem variable t.

I agree with this statement.

I'd go on to say that you can, indeed, keep these unit vectors as your basis throughout the calculation. (I had a look at Synge this morning...which i highly recommend and have as an ecopy if anybody wants it..)

Iderown wrote: »

r = IrI x Unit vector r , both factors are time varying.

When you differentiate this (with respect to time) to get the velocity of the particle, the product rule gives a vector sum of components in directions ( Unit vector r ) and (Time rate of change of unit vector r )

What you are saying here is that,

[LATEX]\displaystyle r=|r|\hat{r}[/LATEX]

so, when you differentiate r wrt t, you get,

[LATEX]\displaystyle\frac{dr}{dt}=|r|\frac{d\hat{r}}{dt} + \hat{r}\frac{d|r|}{dt}[/LATEX]

Iderown wrote: »

A unit vector has length of 1 and you can think of it as the radius of a unit circle. The only way in which a point on the circumference of a unit circle can change is to "travel" to a neighbouring point on the circumference. For a "small" change, this move is at right angles to the original direction of (unit vector r ).

It is when you differentiate the r = Time function of i and j that the transverse direction appears in the result.

Again, I agree with this statement, but what does it mean for this problem is the question that has to be asked.
I'll have a think about this, I might just post a solution to the problem later or on the weekend.

raah!

sponsoredwalk wrote: »

v(t) = [1/2r]r'r + rθθ

Which is very similar to the Kleppner & Kolenkow version.

What do you guys think, have I confused or assumed anything stupid?

The 1/2 there is part of r', that's why your expression is different

r'= 1/2 /B]4t^4+ 9t[B]&#178;[/B] -12t + 4[B ^-1/2 COLOR=black]16t^3 + 18t[/COLOR

sponsoredwalk

Hi guys.

I figured out latex so I'll repost a full derivation of the ENTIRE process so that we can all read with ease the glory of polar coordinate derivations.

[latex] \overline{r} (t) = (2t^2)i + (3t - 2)j [/latex]

[latex] convert \ to \ polar \ coordinates \ of \ the \ form \ \overline{r}(t) = \ r \hat{r} \ then \ differentiate![/latex]

1: First we want to calculate |r| = r.

[latex] r \ = \ | \overline{r} | \ = \ \sqrt{(2t^2)^2 + (3t - 2)^2} \ = \sqrt{4t^4 + 9 t^2- 12t + 4} [/latex]

2: We know our unit vector will be;

[latex] \hat{r} \ = \ cos \theta \ + \ \sin \theta [/latex]

3. Calculate θ!

[latex] \theta \ = \ \arctan( \frac{y}{x} ) \ = \ \arctan(\frac{3t - 2}{2t^2} ) [/latex]

4. We see that [latex] \hat{r} [/latex] must be;

[latex] \hat{r} \ = \ cos \theta \ + \ \sin \theta \ = \ \cos [\arctan(\frac{3t - 2}{2t^2} )] \ + \ \sin [\arctan(\frac{3t - 2}{2t^2} )][/latex]

5. Put the position vector all together, (all are equivalent).

[latex] \overline{r}(t) = \ r \hat{r} [/latex]

[latex] \overline{r}(t) \ = \ [ \sqrt{4t^4 + 9 t^2- 12t + 4} ] \hat{r}[/latex]

[latex] \overline{r}(t) \ = \ [ \sqrt{4t^4 + 9 t^2- 12t + 4} ] [\ cos \theta \ + \ \sin \theta ] [/latex]

[latex] \overline{r}(t) \ = \ ( \sqrt{4t^4 + 9 t^2- 12t + 4} ) (cos \theta ) \ + \ ( \sqrt{4t^4 + 9 t^2- 12t + 4} ) (sin \theta ) [/latex]

[latex] \overline{r}(t) \ = \ ( \sqrt{4t^4 + 9 t^2- 12t + 4} ) ( \cos [\arctan(\frac{3t - 2}{2t^2} )] ) \ + \ \sin ( \arctan(\frac{3t - 2}{2t^2} ) ] ) [/latex]

[latex] \overline{r}(t) \ = ( \sqrt{4t^4 + 9 t^2- 12t + 4} ) ( \cos [\arctan(\frac{3t - 2}{2t^2} ) ] ) \ + ( \sqrt{4t^4 + 9 t^2- 12t + 4} ) ( \ \sin [\arctan(\frac{3t - 2}{2t^2} ) ] ) [/latex]

Alright, so I've set up the position vector as best I can. I'm assuming there's nothing wrong so I'll now take the derivative to find the velocity vector. I think I'll do the same thing as last time and watch out for the [latex] \frac{1}{2} [/latex] that I stupidly left in last time, (idk why I did...).

6. Using;

[latex] \overline{r}(t) \ = \ [ \sqrt{4t^4 + 9 t^2- 12t + 4} ] [\ cos \theta \ + \ \sin \theta ] [/latex]

I'll differentiate to obtain an equation of the form;

[latex] \overline{v} (t) \ = \ \frac{d \overline{r} }{dt} \ = \ r \frac{d \hat{r}}{dt} \ + \ \frac{dr}{dt} \hat{r} [/latex]

Here I go!

[latex] \frac{d \overline{r} }{dt} \ = \ ( \sqrt{4t^4 + 9 t^2- 12t + 4} ) (- sin \theta \frac{d \theta} {dt} \ + \ cos \theta \frac{d \theta}{dt} ) \ + \ ( \frac{16t^3 \ + \ 18t \ - \ 12}{2 \sqrt{4t^4 + \ 9t^2 - 12t + 4} } ) (\ cos \theta \ + \ \sin \theta ) [/latex]

[latex] \frac{d \overline{r} }{dt} \ = \ ( \sqrt{4t^4 + 9 t^2- 12t + 4} ) (- sin \theta \ + \ cos \theta ) \frac{d \theta}{dt} \ + \ ( \frac{16t^3 \ + \ 18t \ - \ 12}{2 \sqrt{4t^4 + \ 9t^2 - 12t + 4} } ) (\ cos \theta \ + \ \sin \theta ) [/latex]


I didn't dare try to bring in the arctan and it's craziness into this one, (yet...).

7. Look for patterns!

Well,

[latex] r \ = \ \sqrt{4t^4 + 9 t^2- 12t + 4} [/latex]

[latex] \dot{r} \ = \ ( \frac{16t^3 \ + \ 18t \ - \ 12}{2 \sqrt{4t^4 + \ 9t^2 - 12t + 4} } ) [/latex]

[latex] \hat{r} \ = \ cos \theta \ + \ \sin \theta [/latex]

[latex] \frac{ d \hat{r}}{dt} \ = \ ( - \sin \theta \ + \ \cos \theta ) \frac{ d \theta}{dt}[/latex]

We'll set;

[latex] \hat{ \theta } \ = \ (- sin \theta \ + \ cos \theta )[/latex]

&

[latex] \dot{ \theta} \ = \ \frac{d \theta}{dt} [/latex]

to get

[latex] \frac{ d \hat{r}}{dt} \ = \ \hat{ \theta } \dot{ \theta}[/latex]

8. Write the final, simplified equation for velocity.

[latex] \overline{v} (t) \ = \ r \dot{ \theta} \hat{ \theta } \ + \ \dot{r} \hat{r} [/latex]

Comparing the result of 8. with the result of 6. using 7. to be sure we've labelled everything correctly I'd say we have taken a deceptively simple equation in cartesian form, converted it to it's equivalent polar coordinate description & take the derivative of this to obtain the velocity.

Q.E.D.

Now you guys can tell me if I missed anything, or skipped a step, or skipped a mathematical step, or assumed something that I shouldn't have etc...

I still have a few questions lol but we'll finish this off first

gerardduff

Nice work.

I'd just like to make a clarifying comment about the basis vectors.

So, we started with [latex]\hat{i}[/latex] and [latex]\hat{j}[/latex]. Which are clearly mutually orthogonal.

Then we make the change of coordinate system and introduce the basis vectors [latex]\hat\theta[/latex] and [latex]\hat{r}[/latex].

where,

[latex]\hat\theta=\cos\theta + \sin\theta[/latex]

and,

[latex]\hat{r}=\cos\theta - \sin\theta[/latex].

These are also mutually orthogonal (you can see this graphically, but I have an algebraic proof below).

There's actually a lot you can say about this choice in terms of the underlying mathematics. To understand the motivation for this choice is to understand quite a lot of mathematics. I'm referring here to linear independence. That is, both [latex]\hat\theta[/latex] and [latex]\hat r[/latex] are linear combinations of the cos and sin functions. And they have the added feature of being mutually independent. The term independence is used in mathematics in lots of different contexts, here it means that you cannot write either vector in terms of the other.

The proof (of orthogonality), which contains a useful method, goes like this,

[latex]\hat{r}= \cos\theta + \sin\theta
=\sqrt{2}\cos(\theta + \phi)[/latex]
where,
[latex]\tan\phi=1..........\phi=\frac{\pi}{4}[/latex]

where I've used the formula cos(A + =cosAcosB-sinAsinB

Likewise,

[latex]\hat\theta=\sqrt{2}\cos(\theta-\phi)[/latex]

So, the angle between [latex]\hat {r}[/latex] and [latex]\hat\theta[/latex] is

[latex]\frac{\pi}{4}-(-\frac{\pi}{4})=\frac{\pi}{2}[/latex].

Orthogonality implies independence.

Maybe that preempts your next question.

sponsoredwalk

So, my derivation holds no kinks. That's encouraging!

Yeah, I've only very recently gotten a deeper understanding of linear independence with a proper Linear Algebra book & I am still trying to figure out some of the ins and outs.

Thanks for pre-empting me so thoroughally lol. My knowledge of the proof starts and ends with the following test for linear independence;

[latex] a \overline{u} \ + \ b \overline{v} \ = \ 0 [/latex]

Where u & v form a basis if the above equation is satisfied.

That's all I know so far and it makes sense in that if the above equation wasn't equal to zero you could solve and express one vector as a scalar multiple of the other. The geometric interpretation of this is that vector arrow u points in a different direction to the vector arrow v if they are linearly independent.

However, they have always clearly stated such and such forms a basis & I don't thnk I've ever derived a basis from scratch in terms of components or anything...

I've just assumed that the way to form a basis is to use the above test to check for linear independence. Then, if I am to form the i j basis vectors I'd say, "vectors i & j are linearly independent vectors, or they form an orthonormal basis. This would be the most I can say in terms of defining a basis.

If you understand what I'm talking about it would help so much if you could explain your derivation in terms of what I've told you, as it's all I know so far. It could be that I understand what you've written, but just not as you've written as it looks complicated.

As far as the [latex] \hat{ \theta } [/latex] is concerned, it is not present in the original position vector due to the fact that it is already assumed to be known as part of the angle within [latex] \theta [/latex] to find where [latex] \overline{r} (t)[/latex] even is. I understand that it pops out of the derivative as a mathematical consequence. In terms of position it just seems irrelevant to me, other than I know geometrically it is at a 90 degree angle from the polar vector [latex] \hat{r} [/latex] (which points parallel to the cartesian position vector that goes from the origin to the position).

Coincidentally; In every book there is a picture of the polar coordinates originating from the head of the position vector, [latex] e_r \ and \ e_ \theta [/latex] which I assume are just the position vector [latex] \hat{r} \ and \ \hat{ \theta} [/latex] Yeah?

gerardduff

I know what you are talking about. I'll try to explain.

This matter of linear independence is really a subtle one, especially when you go beyond [latex]R^2[/latex] and [latex]R^3[/latex]. The only condition, necessary and sufficient for linear dependence is the one you have quoted, namely,

sponsoredwalk wrote: »

[latex] a \overline{u} \ + \ b \overline{v} \ = \ 0[/latex]

a and b nonzero.

The little proof I gave there is specific to this problem. I just showed that the unit basis vectors (aka orthonormal vectors) [latex]\hat{r}, \hat\theta[/latex] are in fact orthogonal in the way we understand from everyday life.

The fact that they are linearly independent can only be demonstrated rigorously by putting them through the machinery you alluded to, i.e. that a linear combination of them can only be made to equal zero if the scalars a and b in your formula are both zero, i.e. a=b=0

As for [latex] e_r \ and \ e_ \theta [/latex], they are exactly the same as [latex] \hat{r} \ and \ \hat{ \theta} [/latex].

So yes is the answer to your last Q.

sponsoredwalk

Awesome.

It may have taken three days but we got there in the end!

I've been watching a vector calculus lecture on this topic & I understood almost every part of it.

I'm going to mull some of this stuff over for a few days and I'll come back should there be any problems or questions.

I hope this whole experience was as interesting to you guys as it was to me, . I mean, there was no other way for me to comprehend the situation without doing this drudgery & I'm immensely thankful you took the time to help me out.

:cool: