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How do I find the axis intercepts of
rotated 60° clockwise about it's point of inflexion?Last edited by Identity (2007-07-27 18:08:34)
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(a,b) rotated through θ° about (c,d) will arrive at (x,y)
Distance from (a,b) to (c,d):
By analysing rotation in a circle, distance from (a,b) to (x,y):
Hence,
And so
This is absurd because you need to know more than the information you can have at any one time.
Last edited by Identity (2007-07-27 19:24:20)
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At x = -0.1322 or -0.1321, the cubic equation rotated 60 degrees about (2,3) should cross the X-axis. Actually more accurate is x = -0.132158 or -0.132159 for inputted x before rotation.
y = (x-2)^3 + 3
xCenter = 2
yCenter = 3
deltaY = yCenter - y
deltaX = xCenter - x
newY = 3 + 0.866025404 * deltaX - 0.5 * deltaY
newX = 2 - 0.5*deltaX - 0.866025404 * deltaY
Last edited by John E. Franklin (2007-07-27 20:10:43)
igloo myrtilles fourmis
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Thanks John, but what did you use to get that?
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X-intercept:
OldPoint(-0.132159, -6.6930123) deltaX=2.132159 deltaY=9.6930123 (delta values are distance from point(2,3), point of rotation, obtained from second derivative to zero I guess.)
NewPoint( -7.46047439 , -0.22891074e-5 )
OldPoint(-0.132158, -6.69299866) deltaX=2.132158 deltaY=9.69299866
NewPoint( -7.46046208 , 0.3664017e-5 )
Y-intercept:
OldPoint(0.82314,1.37004954) deltaX=1.17686 deltaY=1.62995046
NewPoint( -0.85091071e-5 , 3.20421542 )
OldPoint(0.82315,1.37009109) deltaX=1.17685 deltaY=1.62990891
NewPoint( 0.32473929e-4 , 3.20422754 )
Last edited by John E. Franklin (2007-07-27 20:09:04)
igloo myrtilles fourmis
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Hmmm, I still don't quite understand... what software did you use for this?
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I programmed it in BASIC language, using Just Basic, free off internet.
The program shown below has some extraneous stuff in it, commented out with the single quote at left of line. But this is the program I am running.
Just Basic runs in Windows XP fine, it is interpretted as it runs.
'draw a graph program by John, January 1st, 2006
'updated 2007.
'REV AAAQ graph of y = (.22/6)x^3 + .11x + 4 and curvatures kappas
'REV AAAR in progress...
dim color$(7)
color$(1) = "darkred"
color$(2) = "brown"
color$(3) = "darkcyan"
color$(4) = "darkgreen"
color$(5) = "darkblue"
color$(6) = "darkgray"
color$(7) = "black"
open "Maximize This Window To See Graphing" for graphics as #1
print #1, "down"
print #1, "size 1"
print #1, "color "; color$(7)
xenlarge = 20
yenlarge = 20
''''''PRINT THE GRAPH HERE''''''''
'' y = (22/6)x^3 + 11x + 4
'' dy/dx = 11x^2 + 11
'' d2y/dx2 = 22x
''curvature kappa = 22x / (1 + (11x^2 + 11)^2 )^(3/2)
pi = 3.1415926535897932384626433832795028841971
for x = 0.8231 to 0.8342 step .00001
'''''y = cos( (1444*theta^0.05) / (45*4/pi) )
'''x = theta
y = (x-2)^3 + 3
'''''print theta;" ";y
xCenter = 2
yCenter = 3
deltaY = yCenter - y
deltaX = xCenter - x
newY = 3 + 0.866025404 * deltaX - 0.5 * deltaY
newX = 2 - 0.5*deltaX - 0.866025404 * deltaY
print x;"%";y;"&";deltaX,deltaY;"( ";newX;" , ";newY;" )"
yPrintMe = newY
xPrintMe = newX
A$ = "set " + Str$(640 + xPrintMe*xenlarge) + " " + Str$(400-yPrintMe*yenlarge)
print #1, "color black"
print #1, A$
'print #1, "color red"
'print #1, A$
'print #1, "color green"
'print #1, A$
next x
''''''''''''''''''''''''''''''''''''
print #1, "color "; color$(6)
'horizontal x-axis drawn
for xx = -600 to 600
yPixel = 0
xPixel = xx
A$ = "set " + Str$(640 + xPixel) + " " + Str$(400-yPixel)
print #1, A$
next xx
print #1, "color "; color$(6)
'vertical y-axis drawn
for yPixel = -390 to 390
xPixel = 0
A$ = "set " + Str$(640 + xPixel) + " " + Str$(400-yPixel)
print #1, A$
next yPixel
'''''''''print the hash lines on the x-axis''''''''''
ddd = int(635 / xenlarge)
for iterbee = -1 * ddd to ddd step 1
xhash$ = Str$(640 + xenlarge * iterbee)
A$ = "set " + xhash$ + " 400"
print #1, A$
A$ = "set " + xhash$ + " 399"
print #1, A$
A$ = "set " + xhash$ + " 401"
print #1, A$
next iterbee
'''''''''print the hash lines on the y-axis''''''''''
ddd = int(395 / yenlarge)
for iterbee = -1 * ddd to ddd step 1
yhash$ = Str$(400 + yenlarge * iterbee)
A$ = "set " + "640 " + yhash$
print #1, A$
A$ = "set " + "639 " + yhash$
print #1, A$
A$ = "set " + "641 " + yhash$
print #1, A$
next iterbee
for iteriter = 1 to 10
Print "All Done"
next iteriter
''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''
print #1, "flush"
''''''''''print #1, "redraw"
wait
close #1
'''end of file
Last edited by John E. Franklin (2007-07-28 02:50:35)
igloo myrtilles fourmis
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Thanks George, I'll take a look at it. Although, my teacher probably didn't expect us to have a program like that, so I think I should try to find an exact answer too.
edit: I copied and pasted the code into JustBasic. The axes appeared but nothing else happened.
Last edited by Identity (2007-07-27 20:09:06)
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If you take this part of program and translate it into an equation, you might have something.
The .866 is
y = (x-2)^3 + 3
xCenter = 2
yCenter = 3
deltaY = yCenter - y
deltaX = xCenter - x
newY = 3 + 0.866025404 * deltaX - 0.5 * deltaY
newX = 2 - 0.5*deltaX - 0.866025404 * deltaY
I obtained these equations by drawing some sketches and using a little trig.
I'll make a picture of the sketch and post it...
Notice the red arrows go up and down, that's the plus .866 and minus 0.5 ratios.
Notice the green arrows go left and left again. That's the minus 0.5 and minus 0.866 ratios.
Last edited by John E. Franklin (2007-07-27 20:44:46)
igloo myrtilles fourmis
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The reason nothing graphed is because of the line just below where pi=3.14159 is.
That "for-loop" is currently narrowing down the Y-intercept I think when I copied it.
Change that line to go from -3 to +2 or something and step by 0.01 instead.
So change it to this:
pi = 3.1415926535897932384626433832795028841971
for x = -3 to 2 step .01
The pi line isn't being used at this time.
Last edited by John E. Franklin (2007-07-27 20:19:48)
igloo myrtilles fourmis
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From post #9 (two back), we can write an equation for the newX.
newX = 2 - 0.5(2-x) - half_sqrt3(3-y)
Sub in equation for y in terms of x...
newX = 2 - 0.5(2-x) - half_sqrt_of3(3 - 3 - (x-2)^3)
Simplify and set newX equal to zero to find y-intercept.
When I set to zero, I doubled both sides.
0 = 4 - (2 - x) - sqrt3(2-x)^3
To solve, this I don't know how, but I plugged in the 0.82314 value found in program and got this with google calculator.
4 - (2 - .82314) - ((3^0.5) * ((2 - .82314)^3)) = -1.70175115 × 10-5
So that's really close to zero.
So the program seems correct, but precise means to solve cubics I don't know yet.
igloo myrtilles fourmis
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The matrix
rotates all co-ordinate points on the Cartesian plane (expressed as column vectors) through angle θ counterclockwise. So for a 60° clockwise i.e. −60° counterclockwise the transformation matrix is
So
Hence
In the first equation, express x in terms of x′, then substitute for x in the second equation to get y′ in terms of x′; this will then be the equation of your new curve. (Or, if you cant express x in terms of x′, then leave the equation of the new curve in parametric form.)
Last edited by JaneFairfax (2007-07-28 00:19:26)
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Cool matrix, thanks
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