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I am in the course of a project, in which I need to solve these two simultaneous equations:

\begin{equation} \sqrt{(1000-y)^2 + x^2} - \sqrt{y^2 + x^2} = 342.371 \end{equation} \begin{equation} \sqrt{(2000-x)^2 + y^2} - \sqrt{y^2 + x^2} = 961.674 \end{equation}

I know that the answers are y = 250, and x = 500, but for some reason, I cannot back it up with any calculations, because they always seem to get faulty. I think it has something to do with the square roots, but I was hoping for a useful hand out there.

Just for the sake of the argument, my methodology consists of: - expanding the formulas - cancel all equal terms in each formula - write one variable in terms of another and... - ... plug in the expression into the second equation.

But my problem already starts with the fact that when I start with the first equation and cancel all equal terms, I am left with an expression containing only one variable. (solving this variable results also in the wrong answer.)

Thanks again for the useful help.

Here my calculations for the first equation, which shall give me an expression to solve the second one: \begin{equation} ((1000 - y)^2 + x^2) - (y^2 + x^2) = 342.371^2 \end{equation} \begin{equation} 1000^2 - 2000y + y^2 + x^2 - y^2 - x^2 = 343.371^2 \end{equation} \begin{equation} 1000^2 - 2000y = 343.371^2 \end{equation} ... and now I am left with an equation without an x in it... and solving this also does not result in the correct solution anyways.

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  • $\begingroup$ What do you mean by cancelling equal terms? $\endgroup$ – GFauxPas Apr 13 '15 at 23:28
  • $\begingroup$ If you expand the expression, and you arrive to the situation e.g. of -x^2 and further on x^2... You can just cancel them out and away. $\endgroup$ – user3604362 Apr 14 '15 at 9:10
  • $\begingroup$ My problem is mainly that e.g. the first expression, as I do that cancelling method, my x variable completely dissapears, and only the y variable is left... and this IS wrong... but what is RiGhT? $\endgroup$ – user3604362 Apr 14 '15 at 9:12
  • $\begingroup$ My first approach is to take the equation, and square it all together... and then I proceed with the expansion and cancelling equal terms...finally giving me an expression... but in this case, something is not quite right... $\endgroup$ – user3604362 Apr 14 '15 at 9:14
  • $\begingroup$ Please show us your method explicitly so we can pin down what goes wrong. I have a hunch you're not squaring correctly. $\endgroup$ – GFauxPas Apr 14 '15 at 10:48
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$\sqrt{(1000-y)^2+x^2}-\sqrt{y^2+x^2}=Q$.

$\sqrt{(1000-y)^2+x^2}=\sqrt{y^2+x^2}+Q$.

$(1000-y)^2+x^2=y^2+x^2+2Q\sqrt{y^2+x^2}+Q^2$.

$(1000-y)^2+x^2-y^2-x^2-Q^2=2Q\sqrt{y^2+x^2}$.

$1000000-2000y-Q^2=2Q\sqrt{y^2+x^2}$.

Now square both sides to get an equation quadratic in $x$ and $y$. Then go through the same steps with the other equation to get a second quadratic in $x$ and $y$. Then solve the system of 2 equations in 2 unknowns.

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  • $\begingroup$ sorry, but I am clearly missing out something... from where does the $2Q\sqrt{y^2 +x^2}$ come from? $\endgroup$ – user3604362 Apr 15 '15 at 13:30
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    $\begingroup$ Binomial theorem: $(a+b)^2=a^2+b^2+2ab$. $\endgroup$ – Lutz Lehmann Apr 15 '15 at 18:12
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    $\begingroup$ Aaaaahhhh! That is my mistake! ahahaha! Thank you thank you thank you! $\endgroup$ – user3604362 Apr 15 '15 at 19:19
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There are two ways to solve this problem.

The first option you can apply to this problem is to convert the x-y coordinates to polar coordinates, such that $x=rcos(\theta)$ and $y=rsin(\theta)$. With that, you will be able to simplify the expression enough to solve this problem the "traditional" way, using substitution. The first expression for example would reduce to: \begin{equation} \sqrt{2000^2 - 4000rcos(\theta) + r^2} - r = 961.674 \end{equation} Nevertheless, you might want to stick to x-y coordinates due to different reasons.

In that case, you will have to use Taylor's theorem to solve this problem. You could go ahead and expand the square root expressions with Taylor's theorem, but then again, Newtons method (which is based on Taylor) can be used to solve this problem numerically, which is much easier.

Here the link to a further explanation of the use of Newtons method: \url{http://recsam.edu.my/cosmed/cosmed09/AbstractsFullPapers2009/Abstract/Mathematics%20Parallel%20PDF/Full%20Paper/M33.pdf}

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