shop math problem

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shop math problem Grant Erwin 05-06-2006
Posted by Dave on May 7, 2006, 1:38 pm
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 mike wrote:
> "Dave" wrote:
> >Grant Erwin wrote:
> >> Dave wrote:
> >> > Grant Erwin wrote:
> >> >
> >> >>This is actually a little worse than shop math. I have solved this; will
confirm
> >> >>the first correct answer.
> >> >>
> >> >>Find an equation of a conic (2nd order function) which goes through the
points
> >> >>(0,0) (2,0.75) and (2.5,5) and is tangent to the X axis at (0,0) and is
nearly
> >> >>vertical at the point (2.5,5). Put it in the form y = F(x); in other
words with
> >> >>y alone on the left hand side of the equation.
> >> >
> >> >
> >> > Got any background to make this interesting? Looks pretty dull.
> >> >
> >> > y= ax^2 + bx + c
> >> > 0= 0 + 0 + c so c = 0
> >> > 0.75 = 4a + 2b
> >> > 5 = 6.25a + 2.5b
> >> >
> >> > Two equations in two unknowns so put a or b in terms of the other and
> >> > plug it in and grind out an answer. Not sure if a "conic" might mean
> >> > you'd really need to start with the form y = ax^2 + bxy + cy^2 + dx + e
> >> > but I guess you'd get the same thing.
> >> >
> >>
> >> Background? Well, I'm musing on ways to make the center piece of a vibratory
> >> polishing bowl.
> >>
> >> It's a little more complicated than your analysis, but here 'tis:
> >>
> >> In my solution, the final equation looks like
> >>
> >> y = Px + Q + SQRT(R*X^2 + S*X + T)
> >>
> >> and for a curve going through (0,0) (2,0.75) and (2.5,5) and tangent to the
x
> >> axis at (0,0) and very nearly vertical at (2.5,5) I got my 5 constants as:
> >>
> >> P = -12.083
> >> Q = 35.208
> >> R = 142.007 ;; there may be slight roundoff or truncation errors in these
> >> S = -850.868
> >> T = 1239.627
> >>
> >> It took me an amazing amount of time flailing at this problem (which I
didn't at
> >> all need to solve; I just like doing math sometimes) before I resorted to
> >> looking up how aircraft designers used to lay out airframes, and found that
back
> >> in the 1950s those guys really knew about 2nd order equations and how to
whip
> >> them into shape.
> >>
> >> GWE
> >
> >Your proposed solution seems to fail the (0,0) test...
> >
> >0 = 0 + Q + SQRT(0 + 0 + T)
>
> Dave, remember Square roots have a + and a - value.
> Mike in BC

Well, that does seem to be a valid explanation for the ambiguity
presented. I'm unclear whether in the problem statement "nearly
vertical" is supposed to imply a vertical asymptote near the point
(2.5,5) ? If not then a simple parabola would seem to be a cleaner and
easier solution.


Posted by nick on May 7, 2006, 2:10 pm
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 Grant Erwin wrote:

> It's a little more complicated than your analysis, but here 'tis:
>
> In my solution, the final equation looks like
>
> y = Px + Q + SQRT(R*X^2 + S*X + T)
>
> and for a curve going through (0,0) (2,0.75) and (2.5,5) and tangent to the x
> axis at (0,0) and very nearly vertical at (2.5,5) I got my 5 constants as:
>
> P = -12.083
> Q = 35.208
> R = 142.007 ;; there may be slight roundoff or truncation errors in these
> S = -850.868
> T = 1239.627
>
> It took me an amazing amount of time flailing at this problem (which I didn't
at
> all need to solve; I just like doing math sometimes) before I resorted to
> looking up how aircraft designers used to lay out airframes, and found that
back
> in the 1950s those guys really knew about 2nd order equations and how to whip
> them into shape.
>
> GWE

I get

48*x^2 + 12*y^2 + 290*x*y -845y =0

or

y=[(845-290x)-SQRT(81796*x^2-490100*x+714025)]/24

as a solution where the tangent is vertical at (2.5,5)

Posted by xray on May 7, 2006, 11:10 am
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>Grant Erwin wrote:
>> This is actually a little worse than shop math. I have solved this; will
confirm
>> the first correct answer.
>>
>> Find an equation of a conic (2nd order function) which goes through the points
>> (0,0) (2,0.75) and (2.5,5) and is tangent to the X axis at (0,0) and is nearly
>> vertical at the point (2.5,5). Put it in the form y = F(x); in other words
with
>> y alone on the left hand side of the equation.
>
>Got any background to make this interesting? Looks pretty dull.
>
>y= ax^2 + bx + c
>0= 0 + 0 + c so c = 0
>0.75 = 4a + 2b
>5 = 6.25a + 2.5b
>
>Two equations in two unknowns so put a or b in terms of the other and
>plug it in and grind out an answer. Not sure if a "conic" might mean
>you'd really need to start with the form y = ax^2 + bxy + cy^2 + dx + e
>but I guess you'd get the same thing.

Not quite so simple, I think.

In your first equation, if b is not = 0 then the curve won't be tangent
to the x axis at 0,0.

I did a bit of looking at some form of the equation for an elipse, which
could have a tangent at 0,0 and nearly vertical at 2.5,5 but so far I
haven't found any major/minor sizes for an ellipse that come close to
working.

I'm not a math wiz but I can't follow what "conic (2nd order function)"
means in this puzzle. The only conic section I can think of that could
have a point with zero slope at 0,0 and a point nearly asymtotic to
vertical (parallel to the y axis) is an ellipse. Such an equation has
terms with both x^2 and y^2 so Grant's required y = f(x) form seems odd.

What kind of slope is "close to vertical" at the 3rd point?

Grant, do you have nonlinear regression software that helped in this
puzzle?


Posted by Dave on May 7, 2006, 2:10 pm
Please log in for more thread options
xray wrote:
> "Dave" wrote:
> >Grant Erwin wrote:
> >> This is actually a little worse than shop math. I have solved this; will
confirm
> >> the first correct answer.
> >>
> >> Find an equation of a conic (2nd order function) which goes through the
points
> >> (0,0) (2,0.75) and (2.5,5) and is tangent to the X axis at (0,0) and is
nearly
> >> vertical at the point (2.5,5). Put it in the form y = F(x); in other words
with
> >> y alone on the left hand side of the equation.
> >
> >Got any background to make this interesting? Looks pretty dull.
> >
> >y= ax^2 + bx + c
> >0= 0 + 0 + c so c = 0
> >0.75 = 4a + 2b
> >5 = 6.25a + 2.5b
> >
> >Two equations in two unknowns so put a or b in terms of the other and
> >plug it in and grind out an answer. Not sure if a "conic" might mean
> >you'd really need to start with the form y = ax^2 + bxy + cy^2 + dx + e
> >but I guess you'd get the same thing.
>
> Not quite so simple, I think.
>
> In your first equation, if b is not = 0 then the curve won't be tangent
> to the x axis at 0,0.
> [...]

Oops. Yeah, you're right. It can't be a parabola.


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