Complex Equation Solver

[Dejhan_Tulip]

Hi everyone !

Lately in this Statics class I have been running into problems with complex equations. For example, today I have to solve this 2 equations and 2 unknows: (The unknowns are N and x; Note that itan = arctan = tan^-1)

(1) 0.4N - 75cos(itan(12/x)) = 0
(2) N - 180 + 75sin(itan(12/x)) = 0

They seem not "that" hard but they are if I try to solve them by hand

I was wondering if anyone could give me the answer to N and x for those equatios, and also if anyone knows any software that would solve those kind of equations I would really appreciate it if you can give me the name, website, anything Thank you very much.

Have a good one,

--D.T.

[Yan Wang]

The TI89 or HP49G+ calculators have multivariable equation solvers built in. But I don't feel like pulling mine out.

As for software, my copy of Mathematica can easily do it. Along with most anything else I need related to math or physics.

[mtrawls]

Never fear, this isn't a complex equation at all! (Why, you'd have to have the sqrt of -1 for that to be the case ).

To get the answer, it is a simple trig "trick" -- i.e., once you see it it's not that hard. Think about what the inverse tangent of 12/x represents. It equals some angle from a right triangle whose "opposite" side is 12 and whose "adjacent" side is x (because tan theta is opp/adj). Then you want to find the cos (and sin) of this angle. Well, cos is adj/hyp and sin is opp/hyp -- and from the previous argument, you know both adj and opp, so all that is left is to find the hypotenouse. Just form the triangle ...

    |\
12  | \  hyp
    |_(\
     x

But from the pythagorean theorum you know that hyp = sqrt(144+x^2). Well, then you can get the system of equations in terms of x without any nasty trig functions, which should make it pretty simple to solve (I'm assuming you can solve systems of equations and the trig was throwing you off ... if not, just ask for an explanation!)

Oh, and an 89 is good for just about anything Or mathematica, if you have the software (somewhat pricey). But nothing beats pen and paper!

[Dejhan_Tulip]

Originally Posted by Yan Wang

The TI89 or HP49G+ calculators have multivariable equation solvers built in. But I don't feel like pulling mine out.

As for software, my copy of Mathematica can easily do it. Along with most anything else I need related to math or physics.

Can you tell me why did you bother replying a post like that ??

You can do it, but you are not going to do it. You know how to do it, but you don't feel like it. You have the software that would be able to compute it, but you don't use it to get the answer and post it.

I wonder why you are even reading posts... dude, go to bed, watch some TV, and do something you "feel" like doing it.

--D.T.

P.S. I really would like someone to help me out in this one, this HW is kind of important. Thanks a lot in advance

[Yan Wang]

Originally Posted by Dejhan_Tulip

Can you tell me why did you bother replying a post like that ??

You can do it, but you are not going to do it. You know how to do it, but you don't feel like it. You have the software that would be able to compute it, but you don't use it to get the answer and post it.

I wonder why you are even reading posts... dude, go to bed, watch some TV, and do something you "feel" like doing it.

--D.T.

P.S. I really would like someone to help me out in this one, this HW is kind of important. Thanks a lot in advance

Because you asked the following:

and also if anyone knows any software that would solve those kind of equations I would really appreciate it if you can give me the name, website, anything

And though I was too LAZY to do it myself, I did contribute to the thread by answering your question regarding software or "anything" that would solve the problem you proposed. I did not post a completely non-informative response as you seem to be suggesting.

[Dejhan_Tulip]

Originally Posted by mtrawls

Never fear, this isn't a complex equation at all! (Why, you'd have to have the sqrt of -1 for that to be the case ).

To get the answer, it is a simple trig "trick" -- i.e., once you see it it's not that hard. Think about what the inverse tangent of 12/x represents. It equals some angle from a right triangle whose "opposite" side is 12 and whose "adjacent" side is x (because tan theta is opp/adj). Then you want to find the cos (and sin) of this angle. Well, cos is adj/hyp and sin is opp/hyp -- and from the previous argument, you know both adj and opp, so all that is left is to find the hypotenouse. Just form the triangle ...

    |\
12  | \  hyp
    |_(\
     x

But from the pythagorean theorum you know that hyp = sqrt(144+x^2). Well, then you can get the system of equations in terms of x without any nasty trig functions, which should make it pretty simple to solve (I'm assuming you can solve systems of equations and the trig was throwing you off ... if not, just ask for an explanation!)

Oh, and an 89 is good for just about anything Or mathematica, if you have the software (somewhat pricey). But nothing beats pen and paper!

Thanks a lot for your reply !!

However, that is exactly what I did and the equation didn't get any simpler

If you do that and solve for X you have still cosines and sines involved and while it seems to get a little simpler it doesn't

I thought I was kind of good at doing this, I already passed all Calculus classes but this got me.

Any clarification, solution, procedure will be greatly appreciated.

Thanks a lot,

--D.T.

[mtrawls]

Originally Posted by Dejhan_Tulip

Thanks a lot for your reply !!

However, that is exactly what I did and the equation didn't get any simpler

If you do that and solve for X you have still cosines and sines involved and while it seems to get a little simpler it doesn't

Given the system of equations:

(1) 0.4N - 75cos(itan(12/x)) = 0
(2) N - 180 + 75sin(itan(12/x)) = 0

Then, working on them as described above (e.g., cos(itan(12/x)) = x/sqrt(144+x^2) ...), reduces to:

(1) 0.4N - 75x/sqrt(144+x^2) = 0
(2) N - 180 + 900/sqrt(144+x^2) = 0

Which eliminates the sines/cosines in the equations. Solving for x here shouldn't give you any sines/cosines, but rather a number (provided the system is consistent). Now, that might not be "simple" by some standards ... but it should be solvable -- if not by hand, then more quickly by a calculator/"dumb box." I'd recommend starting by multiplying (2) by -4/10 and adding the two equations together, which would then allow you to solve for x. Then plug this value back into one of the equations and solve for N.

[Yan Wang]

I was bored a couple minutes ago and decided to solve this thing...

Answer: x = 10.52, n = 123.60 (to two decimal places)

In the last step, before finding x, I just used the quadratic equation. To find n, I just plugged x back into both the given equations and voila, n was the same in both. So it works.

Solution attached below as jpeg. Note that my handwriting depreciates greatly the farther along I get into any math problem

[Dejhan_Tulip]

Thanks a lot guys !!!! You all ROCK !!!!