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#1 2007-07-27 17:57:40

Identity
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Registered: 2007-04-18
Posts: 934

Rotating a Graph

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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#2 2007-07-27 19:13:07

Identity
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Registered: 2007-04-18
Posts: 934

Re: Rotating a Graph

(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.dunno

Last edited by Identity (2007-07-27 19:24:20)

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#3 2007-07-27 19:21:18

John E. Franklin
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Registered: 2005-08-29
Posts: 3,588

Re: Rotating a Graph

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)


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#4 2007-07-27 19:25:46

Identity
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Registered: 2007-04-18
Posts: 934

Re: Rotating a Graph

Thanks John, but what did you use to get that?

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#5 2007-07-27 19:34:49

John E. Franklin
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Registered: 2005-08-29
Posts: 3,588

Re: Rotating a Graph

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)


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#6 2007-07-27 19:45:22

Identity
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Registered: 2007-04-18
Posts: 934

Re: Rotating a Graph

Hmmm, I still don't quite understand... what software did you use for this?

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#7 2007-07-27 19:53:42

John E. Franklin
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Registered: 2005-08-29
Posts: 3,588

Re: Rotating a Graph

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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#8 2007-07-27 20:04:20

Identity
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Registered: 2007-04-18
Posts: 934

Re: Rotating a Graph

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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#9 2007-07-27 20:13:19

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

Re: Rotating a Graph

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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#10 2007-07-27 20:17:49

John E. Franklin
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Registered: 2005-08-29
Posts: 3,588

Re: Rotating a Graph

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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#11 2007-07-27 21:26:27

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

Re: Rotating a Graph

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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#12 2007-07-27 23:04:14

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

Re: Rotating a Graph

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 can’t 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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#13 2007-07-28 15:46:49

Identity
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Registered: 2007-04-18
Posts: 934

Re: Rotating a Graph

Cool matrix, thanks

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