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.

GWE

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.

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

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)

On 7 May 2006 08:04:09 -0700, "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

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.

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)

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.