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I have these four equations:

$$\left\{\begin{matrix} A_1 = A_c -D \theta cos(\phi)\\ A_2 = A_c -D \theta cos(\phi-\pi/2)\\ A_3 = A_c -D \theta cos(\phi-\pi)\\ A_4 = A_c -D \theta cos(\phi-3\pi/2) \end{matrix}\right.$$

Variables $A_1$ to $A_4$ are given, D is a constant, and Variables $A_c$, $\theta$, and $\phi$ are unknown.

I need to find $\frac{d \theta}{d A_1}$ to $\frac{d \theta}{d A_4}$, but I have some problems. What I tried was that I used three of these equations and derived $\theta$, $\phi$, and $A_c$, and then differentiated $\theta$ with respect to $A_i$ separately, but I am not sure if this is correct or not.

The thing is that variables $A_i$ are related. This means that if variable $A_1$ changes, at least one of the variables $A_2$ to $A_4$ will also change as there is this relation between the $A_i$ and $A_c$ parameters:

$A_1+A_2+A_3+A_4=4 A_c$

I also get different results when I use three other set of these four equations to derive the parameters. So how can I study the changes of variables $\theta$ and $\phi$ due to changes in variables $A_1$ to $A_4$.

Edit: If I derive variable $\theta$ from the first three equations listed above, I'll have: $$\theta=\frac{\sqrt{2}\, \sqrt{{\mathrm{A1}}^2 - 2\, \mathrm{A1}\, \mathrm{A2} + 2\, {\mathrm{A2}}^2 - 2\, \mathrm{A2}\, \mathrm{A3} + {\mathrm{A3}}^2}}{2\, R}$$

and if I derive variable $\theta$ from the first, second and fourth equations, I'll have:

$$\theta=\frac{\sqrt{2}\, \sqrt{2\, {\mathrm{A1}}^2 - 2\, \mathrm{A1}\, \mathrm{A2} - 2\, \mathrm{A1}\, \mathrm{A4} + {\mathrm{A2}}^2 + {\mathrm{A4}}^2}}{2\, R}$$

,and finally I use $\theta$ derived by Ross Millikan in the answer below which is:

$$\theta=\frac{\sqrt{{\left(\mathrm{A1} - \mathrm{A3}\right)}^2 + {\left(\mathrm{A2} - \mathrm{A4}\right)}^2}}{2\, R}$$

As you notice, the first two solutions do not involve all $A_1$ to $A_4$. All of these solution calculates correct value for $\theta$, but when I differentiate them with respect to $A_1$, you will see different results.

This is how I do this: For $\theta=45 \deg$ and $\phi=0$, I first calculated all $A_1$ to $A_4$ parameters using the four equation in the begining. Now if I use these $A_1$ to $A_4$ parameters in the above solutions for $\theta$, they all would calculate correct $\theta$, but when I differentiate above solutions for $\theta$, and then use $A_1$ to $A_4$ parameters, they determines values of -0.1667, -0.3333, and -0.1667, respectively. And if I repeat this procedure for $\theta=45 \deg$ and $\phi=45 \deg$, then the results would be 0.0, -0.2357, and -0.1179.

My understanding is that the correct one is the third solution that involves all of the $A_1$ to $A_4$ parameters, but then I don't know what is logically wrong with others or how I can relate the two other solutions to the correct one?

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  • $\begingroup$ Is $\theta$ a variable or a function of all the other variables? Please tell us what are the functions (dependent variables) and which are the independent variables. $\endgroup$ – Sergio Parreiras Dec 17 '13 at 15:30
  • $\begingroup$ Variables $A_i$ are dependants and if variable $A_1$ changes, at least one of the variables $A_2$ to $A_4$ will also change. Variables $\theta$, $\phi$, and $A_c$ are independant and can be changed individually. $\endgroup$ – NESHOM Dec 17 '13 at 15:33
  • $\begingroup$ You have missed the point that $A_1+A_3-A_2-A_4=0$ You can use this to eliminate any of the $A$'s that you want, so an expression with only three can be correct. For the same reason, it doesn't make sense to take $\frac {d\theta}{dA_1}$ without recognizing that (at least) one of the $\frac {dA_i}{dA_1}$'s is non-zero. For example you can have $\frac {dA_3}{dA_1}=-1$ and you need to use the chain rule wherever you see an $A_3$ $\endgroup$ – Ross Millikan Dec 17 '13 at 20:24
  • $\begingroup$ So the correct $\frac{d \theta}{d A_1}$ is: $$\frac{d \theta}{d A_1} = \frac{d \theta}{d A_1} +\frac{d \theta}{d A_2}*\frac{d A_2}{d A_1}+\frac{d \theta}{d A_3}*\frac{d A_3}{d A_1}+\frac{d \theta}{d A_4}*\frac{d A_4}{d A_1}$$ where in this case $\frac{d A_2}{d A_1}$=1, $\frac{d A_3}{d A_1}$=-1, and $\frac{d A_4}{d A_1}$=1, am I right? $\endgroup$ – NESHOM Dec 17 '13 at 20:49
  • $\begingroup$ I just realized that $\frac{d A_2}{d A_1}$=0 and also $\frac{d A_4}{d A_1}$=0, and this produces the correct vakue. Now could you please help me on how to find the relation between $A_i$ in case of 5 variables. The only relation between $A_i$ in case of 5 variables is: $$A_1+A_2+A_3+A_4+A_5=5*A_c$$ and I am not sure how to find for example $\frac{d A_2}{d A_1}$ in case of five vars. $\endgroup$ – NESHOM Dec 17 '13 at 21:06
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Note that $\cos(\phi-\pi)=-\cos(\phi), \cos(\phi - \pi/2)=\sin(\phi)$ so $A_3=A_c+D\theta \cos(\phi)$ which gives $A_1+A_3=2A_c=A_2+A_4$ If these are not satisfied your equations are contradictory. You can then write $$A_3-A_1=2D\theta \cos(\phi)\\A_4-A_2=2D\theta \sin(\phi)\\ \phi=\arctan\left(\frac {A_4-A_2}{A_3-A_1}\right)\\ \theta=\frac 1{2D}\sqrt{(A_3-A_1)^2+(A_4-A_2)^2}\\ \theta=\frac 1{2D}\sqrt{(2A_c-2A_1)^2+(2A_c-2A_2)^2}$$ If $A_c$ is fixed, you can now take $\frac {d\theta}{dA_1}$ and $\frac {d\theta}{dA_2}$. The other derivatives are the negatives of these.

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  • $\begingroup$ Thank you so much. This is fantastic. But the thing is that I can produce similar type of equations for 5 variables as: $$\left\{\begin{matrix} A_1 = A_c -D \theta cos(\phi)\\ A_2 = A_c -D \theta cos(\phi-2 \pi/5)\\ A_3 = A_c -D \theta cos(\phi-4 \pi/5)\\ A_4 = A_c -D \theta cos(\phi-6 \pi/5)\\A_5 = A_c -D \theta cos(\phi-8 \pi/5) \end{matrix}\right.$$ Is it possible to derive variables $\theta$ and $\phi$ as functions of $A_1$ to $A_5$? $\endgroup$ – NESHOM Dec 17 '13 at 15:38
  • $\begingroup$ This can be extended to 6, 7, 8, ... variables. So I was looking for kind of generic method to derive the differentiations. $\endgroup$ – NESHOM Dec 17 '13 at 16:13
  • $\begingroup$ When you have constraints like this, you need to specify how they are maintained. When $A_1$ increases, what decreases to maintain the sum? In the four case, we could derive it, but in the other ones you have more freedom. $\endgroup$ – Ross Millikan Dec 17 '13 at 16:22
  • $\begingroup$ So if I use three of these equations and derive $\theta$ and differentiate it with respect to $L_1$. Then try three other equations and derive $\theta$ and its differentiation with respect to $L_1$, will I get same results? The values of $\theta$ derived using any three of the equations are similar, but I am not sure about differentiation of them! $\endgroup$ – NESHOM Dec 17 '13 at 16:34
  • $\begingroup$ Can you join in chat? I think I have given you write access at chat.stackexchange.com/rooms/12030/… $\endgroup$ – Ross Millikan Dec 17 '13 at 16:43

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