How can I linearize ?

Welcome to PF!

Hi e135193! Welcome to PF!

hmm … (c cosθ, c sinθ) is the point at angle θ on the circle of radius c round the origin.

r1 is the distance from that point, and r2 is the distance from the opposite point, (-c cosθ, -c sinθ).

but I don't understand what you mean by "linearization of the equations around x = θ = 0"

 

Hi e135193!

(btw, anyone who contributes to a thread gets an email whenever someone else replies, so there's no need to use PMs. )

This may help: √(1 + x) = (1 + x/2 + …); sinx = x + … ; cosx = 1 - x²/2 + …

But what I don't understand is that you have three variables, x y and θ, for a 2-dimensional situation.

 

I once walked into a test on differential equations that asked for the "linearization" of a non-linear de and I had no idea what "linearization" meant! Turns out it's easy.

You have
[tex]r_1= \sqrt{(c cos(\theta)- x)^2+ (y- c sin(\theta)^2}[/tex]
[tex]r_2= \sqrt{(c cos(\theta)+ x)^2+ (y+ c sin(\theta)^2}[/tex]
and you say that c and y are constant. So you want linear functions for r₁ and r₂ in terms of x and [itex]\theta[/itex]. You don't really need the full "Taylor's series", you just need the tangent line equations (which are the order one Taylor's Polynomials).

In general, if r= f(x, [itex]\theta[/itex]), then the "tangent plane" approximation, about x= 0, [itex]\theta= 0[/itex], is given by
[tex]r= f(0,0)+ \frac{\partial f}{\partial x}(0,0) x+ \frac{\partial f}{\partial \theta}(0,0)\theta[/tex]
Find the partial derivatives of your functions and plug into that.

 

Hi e135193!

ok, y and c are constants.

So r is a function of x and θ.

What is the derivative ∂r/∂x? It's the rate at which r increases if you increase x (but leave θ fixed).

So if you only go a very small distance (in x), then r depends linearly on x, and the factor is ∂r/∂x. Similarly for ∂r/∂θ.

Work out ∂r/∂x and ∂r/∂θ, and then find their values at (x=0,θ=0).

 

ok, let's start with an example: ∂/∂x of √(x² + px + q) … can you do that?

(if not, just d/dx of √(x² + px + q)? )

 

ok, just try d/dx of x² + px + q.

You can do that, can't you?

Do you then know how to use substitution to get d/dx of √(x² + px + q)?

hmm … gotta go to bed now … :zzz:

 

Hi e135193!

But I don't know your time-zone!

And what part of …

… did you not understand?

Look, if you really can't do d/dx of x² + px + q, then you'd better ask your teacher/TA for special help.

That sort of d/dx is something you should be able to do in your head by now, like seven times eight equals fifty-six.

But if you can do it, show us what it is, and we'll help you through the next steps.

 

If you do not know how to differentiate at all (which is what you seem to be saying) then you should not trying to do a problem like this. This kind of problem is typically given in a calculus course about a year after learning things like how to differentiate x²+ px+ q.

 

Hi e135193!

I'm sorry, but i have no idea how you got that …

are they supposed to be ∂r1/∂x and ∂r2/∂x, or what?

and where did those √(c² + y²) some from?

We'd better start with:

what is d/dx of √(x² + px + q)?

 

oh … now I understand.

Your -cx/√(c² + y²) and your + √(c² + y²) are correct,

but I've no idea how you got the cos(c) and the sin(y) in the middle (presumably from your ∂/∂θ).

Do the ∂/∂θ again, and this time show your working, so we can see where you're going wrong.

 

Hi e135193!

Words almost fail me!

You clearly have no idea how to do elementary calculus.

No-one on this forum will do your work for you.

I've tried to talk you through this problem, but you haven't cooperated, and it turns out you can't even use Mathematica properly, or even recognise when it churns out nonsense.

If you can't do elementary calculus in your head, are you even doing the right course for you?

e135193 , I think you'd better come clean and start a new thread in the Academic & Career Guidance sub-forum, and explain your present academic situation and ask for advice.

There are plenty of people on this forum who have experience of your situation, and who can advise you, if you ask.