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Anyone want to give this a try?
Posted by dave-lindell on January 17, 2020 at 2:29 amv?ý+x?ý-458.926?ý-1.412656326vw=0
w?ý+x?ý-552.632?ý-1.045699772wx=0
v?ý+w?ý-630.289?ý+0.468229445vw=0
552.632v+458.926w-630.289x=0
That’s four equations in three unknowns
mathteacher replied 4 years, 3 months ago 10 Members · 49 Replies- 49 Replies
Is that really vw in the first line? It breaks a symmetry that the other lines have.
.You’re right: that should be vx
Dave; do you want a “long hand solution” or “computer solution” ?
JOHN NOLTON
Either or both.
(I have a rough idea what each variable is if you need a “starter” value.)
I tried plugging it into https://www.wolframalpha.com/ , and no solutions exist. Dropping off the last equation, does return a solution though. I did change the vw to vx as well.
Dave, something is strange with your equations. I get the same results that MitchPDT gets.
Can you tell us where you got the equations? Can you check your equations again?
JOHN NOLTON
For some reason the first two lines of the following lyrics came to mind. My mind went back 45+ years to all the fun we had solving such equations by hand in our engineering classes. Rewriting to get things to a neat form for matrix analysis could be mindbending. Liquid refreshments were frequently required, when affordable.
Those were the days my friend
We thought they’d never end
We’d sing and dance forever and a day
We’d live the life we choose
We’d fight and never lose
For we were young and sure to have our way
La la la la la la
La la la la la la
La la la la La la la la la laCould the difficulty be a mis-copied sign?
.Here’s what the graph looks like from the GeoGebra online grapher:
Thanks MathTeacher. I was just going to TRY to graph it but now I have an idea what it looks like.
JOHN NOLTON
The three cylinders intersect in 4 symmetrical points in the real number system. According to Wolfram Alpha, they are:
v = -539.107 and w = 476.296 and x = -125.884
v = -338.967 and w = -457.917 and x = 151.778
v = 338.967 and w = 457.917 and x = -151.778
v = 539.107 and w = -476.296 and x = 125.884
These values don’t exactly satisfy the three equations, perhaps due to rounding, but they are close:
Note that the first one and the last one come closest to satisfying the fourth equation. If something close to either of them was an answer choice on a standardized test, I’d choose it.
Your 2nd and 3rd line above is what I get also using Mathematica V10.0 (I do have a later version but have not put in on the computer yet).
For the 1st line I get -539.074 476.335 -125.827 (don’t know why we get this small difference) but
I will check my entry DATA. Like you say they “don’t exactly satisfy the 3 equ.” I was going to get more digits to see if it was round-off error
but just left it as it is. Close enough for Dave Lindell.
Depending on which Wolfram Alpha URL you go to, the equals sign may be an approximately equals sign. In either case the answers are approximate.
There is a Pro version of Wolfram Alpha that will extend the accuracy. I’ve forgotten the cost, but I think that a year’s worth is roughly equal to the weekly rent for a beach house. To me, that’s not a good trade-off.
This stuff is way over my head, never made it to diffyQ’s. So forgive my naivet’e but what is that grey thing that looks like an overstretched rhombus on a diagonal through the main axis?
If the answer to my question would not be understood by someone not well versed mathematics (ie: me) don’t waste your time. ????
@flga
Yeah, mine, too, save for technology. The funny-looking rhombus is the plane defined by the fourth equation.
Imagine a standpipe fed by two water lines. Those are the three cylinders. Now imagine a bulldozer blade (never mind that it’s curved, this is the ideal math world!) being lowered so that it cuts all three of them. That’s the plane defined by the fourth equation.
The question is, is there a single point or multiple points shared by the three pipes and the bulldozer blade? There are 4 points shared by the pipes; are one or more of these also shared by the blade? Of course, the answer is determined by the slope and direction of the blade. It may or may not find a common point; its equation determines that.
In the real world, it doesn’t matter, ’cause there’s gonna be hell to pay regardless. But it is a nice thought problem.
I have the Pro version. I have a license and have for like 20 years. What the heck you can’t take it with you. Spend as much
as you can cause the wife will get it in the END. BE HAPPY; SPEND!
Here’s a better picture of just the cylinders (pipes) from GeoGebra:
And here’s the one with the plane (blade) added:
And here’s one of just the plane defined by the fourth equation:
OK, here’s where they came from. They are a mixture of Ptolemy’s Theorem and the Law of Cosines while trying to solve for the radius of the inscribed circle on the left. The circumscribed triangle ABC was to be shown by a minimal about of information, but can be solved easily. The line DE is parallel with the base of the triangle. The inscribed circle is tangent to the triangle, the circumscribed circle and the parallel line. I thought they might degenerate into identities. Negative values are not valid.
From the drawing, which I had difficulty plotting because I couldn’t get the inscribed circle to be tangent to the lines I wanted, gives V=275.7?ñ, W=506.0?ñ, and X=610?ñ. I’m betting a better CAD user can plot it perfectly. I try not to use problems that can be solved by AutoCAD.
I may even be on the wrong track to a solution!
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