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- Thread starter e135193
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tiny-tim

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Is there anyone that can help me about linearization of the equations around x=teta=0 ? It is a bit urgent. I am waiting answer from the calculus experts.

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"

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tiny-tim

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As you see equations are non-linear. I need equations like in the picture below. I heard that it can be made by using taylor series.

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.

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Thanks for your interest to help me.

I'm sorry about the insufficient information. The "y" and "c" terms are constants.

Then, there are only 2 variables, x and θ.

Can you help me further from that because I am not good at mathematics.

I'm sorry about the insufficient information. The "y" and "c" terms are constants.

Then, there are only 2 variables, x and θ.

Can you help me further from that because I am not good at mathematics.

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HallsofIvy

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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 r

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.

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thank you so much for your effort. I'm gonna try to do but its not gonna be easy for me. But the point I smash is that I can not plug that functions into the expression you define above. Can you please a bit help )

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tiny-tim

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thank you so much for your effort. I'm gonna try to do but its not gonna be easy for me. But the point I smash is that I can not plug that functions into the expression you define above. Can you please a bit help )

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

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).

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Thanks for your effort again but I said that I was not good at mathematics.

I can not take the partial derivative :) of that function.

could you please explain a bit further.

kind regards.

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tiny-tim

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Thanks for your effort again but I said that I was not good at mathematics.

I can not take the partial derivative :) of that function.

could you please explain a bit further.

kind regards.

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)? )

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if I said no, would you angry of me ?

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tiny-tim

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You can do

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

hmm … gotta go to bed now … :zzz:

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I'm begging you please. It's very important for me. I have to finish this before the sun rises .

Again I'm saying I'm not good at mathematics.

There is someone who waits for answer.

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tiny-tim

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I have to finish this before the sun rises

But I don't know your time-zone!

And what part of …

… did you not understand?hmm … gotta go to bed now … :zzz:

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

But if you

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HallsofIvy

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tiny-tim

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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)?

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tiny-tim

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The equation I found is that the linearized one of the non-linear equations.

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.

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you say it is wrong. Then show the corrcet solution.

please I have to do this otherwise I will fail and I lose my scholarship.

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tiny-tim

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I used Mathematica for that partial derivative.

Words

You clearly have no idea how to do elementary calculus.

you say it is wrong. Then show the corrcet solution.

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.

please I have to do this otherwise I will fail and I lose my scholarship.

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

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

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