Items tagged with geometry

I have to plot 10 squares each one inscribed into each other. After the first one, which is a square with a side of 10m, each following square has got each edges wich are distant from the previous edges (side of the previous square)/4 metres, on the

 

sides of the previous square

How can i draw it?


 

Dear all,

I have the following question, this code:

restart:
with(DifferentialGeometry):
DGsetup([w1,w2],N):
eq1 := ExteriorDerivative(w1);  
eq2 := ExteriorDerivative(w1) &wedge ExteriorDerivative(w2);
eq1 &wedge eq2;

Gives the error:
Error, (in DifferentialGeometry:-Tools:-DGzero)  given degree, 3, exceeds that of frame dimension, 2

Unfortunately, I am not so familiar with differential geometry but as far as I know dw1 \wedge  (dw1 \wedge  dw2) = 0 should be correct.

Thank you for your help
best
baustamm1

Hello,

i need to represent a circle using the command [geometry] circle because i have a problem and i have to calculate an 

intersection between two circles whit command intersection() which is ruled by Package geometry.

For example, with the first circle  i did like the help page says

 But it's Error:

centre of circle(3300;2200)

radius=110

I don't understand the eror. Can you help me?
Thanks in advance.

I want to calculate the ratio of the length of day and night for every latitude on earth ?
but i confused on using Maple in a wise way for finding the formula !
this is my demonstration :

shekofte000.mw
 

Equations

 

the grat circle that divides the earth's surface into two dark and bright sides

[sin(t)*cos(tilt), cos(t), sin(t)*sin(tilt)]

[sin(t)*cos(tilt), cos(t), sin(t)*sin(tilt)]

(1.1)

circle of revolving of a point on earth in 24 hours

[sin(t)*cos(Latitude), cos(t)*cos(Latitude), sin(Latitude)]

[sin(t)*cos(Latitude), cos(t)*cos(Latitude), sin(Latitude)]

(1.2)

Visualization of dark and bright side the of earth

 

Explore(plots[display](plots[spacecurve]({[sin(t)*cos(tilt), cos(t), sin(t)*sin(tilt), color = red], [sin(t)*cos(Latitude), cos(t)*cos(Latitude), sin(Latitude), color = blue]}, t = 0 .. 2*Pi, scaling = constrained, thickness = 4, labels = [x, y, Latitudez], labeldirections = [horizontal, horizontal, vertical], axes = frame), plottools[rotate](plottools[hemisphere]([0, 0, 0], 1, capped = false, color = green, grid = [10, 10], style = surface), 0, tilt, 0), plottools[rotate](plottools[hemisphere]([0, 0, 0], 1, capped = false, color = black, grid = [10, 10], style = surface), 0, Pi+tilt, 0)), parameters = [tilt = 0 .. Pi, Latitude = -(1/2)*Pi .. (1/2)*Pi], initialvalues = [tilt = (1/2)*Pi+.409, Latitude = 1.16])

``


 

Download shekofte000.mw

 

Hello :)

When points of triangle, A, B, C (R^2) are given, how to find area of triangle; equation and radius of circle which passes traingle(A,B,C) and height(AH).

or where can i find information about how to do this?

thank you so much:)

Hi.

This might not be a maple question. But I know maple has very smart and many plotting tools as well as this comunity is full of very skillfull people. So I concluded that I might get an answer here.

Lets say I got some sphere that is cut off by any abirutary plane, how would I go about plotting this?

For the purpose of this example the sphere is centred at the origin with radius of 4 and are cut of by the plane z=2-y.

 

Hi, I was wondering, as stated in the title, if it is possible to plot/draw a triangle knowing only the sides and angles, and not the coordinates for each point making up its corners. If not, what would be the easiest way, to calculate the coordinates using the sides and angles (assuming I know the value for each side and corner) and then plot/draw it?

I'm rather green when it comes to using Maple, so if you could explain it in a simple way that would be appreciated. 

how to Prove that the circumference of a circle of radius r is 2πr
on maple ?????

n := 5:
z1 := exp(2*3.14*I*k1/n)*cosh(z)^(2/n);
z2 := exp(2*3.14*I*k2/n)*sinh(z)^(2/n);
xx := Re(z1);
yy := Re(z2);
uu := cos(alpha)*Im(z1) + sin(alpha)*Im(z2);

i find that the 3d graph has many intersection points to itself

how to find these intersection points of calabi yau ?

 

 

Pick any point P and let Q = (1, 1, 1). [Your choice for P can be anything other than the origin or

Find an equation for the line l1 that passes through P and the origin. Plot the line segment formed by P and the origin in Maple 

How would I plot this?

I try to define an ellipse using the geometry package

with(geometry):

ellipse(e1,['foci'=[[0,1],[4,1]], 'MajorAxis' = 8],[x,y]);

 

I get the message:  Error, (in geometry:-ellipse) wrong type of arguments

but the documentation tells me that I can define an ellipse this way...

 

Please help me solve this question using maple (need the steps in solving).

  The geometry of the triangle
  Romanova Elena,  8 class,  school 57, Kazan, Russia

       Construction of triangle and calculation its angles

       Construction of  bisectors
      
       Construction of medians
      
       Construction of altitudes


> restart:with(geometry):      

The setting of the height of the triandle and let's call it "Т"
> triangle(T,[point(A,4,6),point(B,-3,-5),point(C,-4,8)]);

                                  T

        Construction of the triangle
> draw(T,axes=normal,view=[-8..8,-8..8]);

Construction of the triangle АВС

> draw({T(color=gold,thickness=3)},printtext=true,axes=NONE);     
Calculation of the distance between heights А and В - the length of a side АВ

> d1:=distance(A,B);

                           d1 := sqrt(170)

        
        Calculation of the distance between heights В and С - the length of a side ВС
> d2:=distance(B,C);

                           d2 := sqrt(170)

       The setting of line which passes through two points А and В
> line(l1,[A,B]);

                                  l1

       Display the equation of line l1
> Equation(l1);
> x;
> y;

                         -2 + 11 x - 7 y = 0

        The setting of line which passes through two points А and С
> line(l2,[A,C]);

                                  l2

       Display the equation of line l2
> Equation(l2);
> x;
> y;

                          56 - 2 x - 8 y = 0

         The setting of line which passes through two points В and С
> line(l3,[B,C]);

                                  l3

        Display the equation of line l3
> Equation(l3);
> x;
> y;

                          -44 - 13 x - y = 0

        Check the point А lies on line l1
> IsOnLine(A,l1);

                                 true

        Check the point А lies on line l1
> IsOnLine(B,l1);

                                 true

        Calculation of the andle between lines l1 and l2
> FindAngle(l1,l2);

                              arctan(3)

        The conversion of result to degrees
> b1:=convert(arctan(97/14),degrees);

                                      97
                               arctan(--) degrees
                                      14
                     b1 := 180 ------------------
                                       Pi

        Calculation of decimal value of this angle
> b2:=evalf(b1);

                      b2 := 81.78721981 degrees

        Calculation of the andle between lines l1 and l3
> FindAngle(l1,l3);

                             arctan(3/4)

       The conversion of result to degrees
> b3:=convert(arctan(97/99),degrees);

                                      97
                               arctan(--) degrees
                                      99
                     b3 := 180 ------------------
                                       Pi

        Calculation of decimal value of this angle
> b4:=evalf(b3);

                      b4 := 44.41536947 degrees

       Calculation of the angle between lines l2 and l3
> FindAngle(l2,l3);

                              arctan(3)

       The conversion of  result to degrees
> b5:=convert(arctan(97/71),degrees);

                                      97
                               arctan(--) degrees
                                      71
                     b5 := 180 ------------------
                                       Pi

        Calculation of decimal value of  this angle
> b6:=evalf(b5);

                      b6 := 53.79741070 degrees

        Check the sum of all the angles of the triangle
> b2+b4+b6;

                         180.0000000 degrees

        Analytical information about the point А
> detail(A);
   name of the object: A
   form of the object: point2d
   coordinates of the point: [4, 6]
          Analytical information about the point В
> detail(B);
   name of the object: B
   form of the object: point2d
   coordinates of the point: [-3, -5]
          Analytical information about the point С
> detail(C);
   name of the object: C
   form of the object: point2d
   coordinates of the point: [-4, 8]

   The setting of heights of the triangle points A,B,C and let's call it "Т"

   with(geometry):
> triangle(ABC, [point(A,7,8), point(B,6,-7), point(C,-6,7)]):
        The setting of the bisector of angle А in triandle АВС
> bisector(bA, A, ABC);

                                  bA

        Analytical information about the bisector of angle А in the triandle
> detail(bA);
   name of the object: bA
   form of the object: line2d
   assume that the name of the horizonal and vertical                    axis are _x and _y
   equation of the line: (15*170^(1/2)+226^(1/2))*_x+(-13*226^(1/2)-170^(1/2))*_y+97*226^(1/2)-97*170^(1/2) = 0

        Construction of the triangle
> draw(ABC,axes=normal,view=[-8..8,-8..8]);

 Construction of the triangle ABC

> draw({ABC(color=gold,thickness=3)},printtext=true,axes=NONE);     

 Construction of the bisector of angle А

> draw({ABC(color=gold,thickness=3),bA(color=green,thickness=3)},printtext=true,axes=NONE);    

The setting of the bisector of angle В in the triangle АВС

> bisector(bB, B, ABC);

                                  bB

       Analytical information about the bisector of angle B in the triandle
> detail(bB);
   name of the object: bB
   form of the object: line2d
   assume that the name of the horizonal and vertical                    axis are _x and _y
   equation of the line: (-15*340^(1/2)-14*226^(1/2))*_x+(-12*226^(1/2)+340^(1/2))*_y+97*340^(1/2) = 0

         Construction of the bisector of angle В
>draw({ABC(color=gold,thickness=3),bA(color=green,thickness=3),bB(color=red,thickness=3)},printtext=true,axes=NONE);    



    The setting of the bisector of angle С in the triangle АВС

> bisector(bC, C, ABC);

                                  bC

        Analytical information about the bisector of angle С in the triangle
> detail(bC);
   name of the object: bC
   form of the object: line2d
   assume that the name of the horizonal and vertical                    axis are _x and _y
   equation of the line: (14*170^(1/2)-340^(1/2))*_x+(13*340^(1/2)+12*170^(1/2))*_y-97*340^(1/2) = 0

        Construction of the bisector of angle С
>draw({ABC(color=gold,thickness=3),bA(color=green,thickness=3),bB(color=red,thickness=3),bC(color=blue,thickness=3)},printtext=true,axes=NONE);  

 Calculation of the point of intersection of the bisectors and let's call it "О"

> intersection(O,bA,bB,bC);coordinates(O);

                                  O


     7 sqrt(85) - 3 sqrt(2) sqrt(113) + 3 sqrt(85) sqrt(2)
  [2 -----------------------------------------------------,
       sqrt(85) sqrt(2) + sqrt(2) sqrt(113) + 2 sqrt(85)

          -16 sqrt(85) - 7 sqrt(2) sqrt(113) + 7 sqrt(85) sqrt(2)
        - -------------------------------------------------------]
             sqrt(85) sqrt(2) + sqrt(2) sqrt(113) + 2 sqrt(85)

       Construction of the bisectors and  marking of the point of intersection  "О" in the triandle
>draw({ABC(color=gold,thickness=3),bA(color=green,thickness=3),bB(color=red,thickness=3),bC(color=blue,thickness=3),O},printtext=true,axes=NONE);
> restart:
> with(geometry):
       The setting of the heights of the triangle points A,B,C and let's call it "Т"
> point(A,7,8),point(B,6,-7),point(C,-6,7);

                               A, B, C

        Let's call "Т1"
> triangle(T1,[A,B,C]);

                                  T1

        Construction of "Т1"
> draw(T1(color=gold,thickness=3),axes=NONE,printtext=true);
  The setting of the median from the point В in the trianglemedian(mB,B,T1,B1);
> median(mb,B,T1);

                                  mB


                                  mb

        Construction of the median from the point В
> draw({T1(color=gold,thickness=3),mB(color=green,thickness=3),mb},printtext=true,axes=NONE);

The setting of the median from the point А in the trianglemedian(mA,A,T1,A1);
> median(ma,A,T1);

                                  mA


                                  ma

        Construction of the median from the point А
>draw({T1(color=gold,thickness=3),mB(color=green,thickness=3),mA(color=magenta,thickness=3),ma},printtext=true,axes=NONE);
The setting of the median from the point С in the trianglemedian(mC,C,T1,C1);
> median(mc,C,T1);

                                  mC


                                  mc

        Costruction of the median from the point С
>draw({T1(color=gold,thickness=3),mB(color=green,thickness=3),mA(color=magenta,thickness=3),mA,mC(color=maroon,thickness=3)},printtext=true,axes=NONE);




Calculation of the point of  intersection of the median and let's call it "О"

>intersection(O,ma,mb,mC);coordinates(O);

                                  O


                              [7/3, 8/3]

        Construction of medians and marking of the point of  intersection "О" in the triangle
>draw({T1(color=gold,thickness=3),mB(color=green,thickness=3),mA(color=magenta,thickness=3),mA,mC(color=violet,thickness=3),O},printtext=true,axes=NONE);
> restart:with(geometry):
> _EnvHorizontalName:=x:_EnvVerticalName=y:       The setting of the heights of the triangle points A, B, C  and let's call it "Т"
> triangle(T,[point(A,7,8),point(B,6,-7),point(C,-6,7)]);

                                  T

       Construction of the triangle
> draw(T,axes=normal,view=[-8..8,-8..8]);


The setting of the altitude in the triangle from the point Сaltitude(hC1,C,T,C1);
> altitude(hC,C,T);

                                 hC1


                                  hC

        Analytical information about the altitude hC from the point С in the triangle
> detail(hC);
   name of the object: hC
   form of the object: line2d
   assume that the name of the horizonal and vertical                    axis are _x and _y
   equation of the line: -99+_x+15*_y = 0

        Construction of the altitude from the point С
> draw({T(color=gold,thickness=3),hC1(color=green,thickness=3),hC},printtext=true,axes=NONE);     

  The setting of the altitude in the triangle from the point Аaltitude(hA1,A,T,A1);
> altitude(hA,A,T);

                                 hA1


                                  hA

        Analytical information about the altitude hA from the point А in the triangle
> detail(hA);
   name of the object: hA
   form of the object: line2d
   assume that the name of the horizonal and vertical                    axis are _x and _y
   equation of the line: -28-12*_x+14*_y = 0

        Construction of the altitude from the point А
>draw({T(color=gold,thickness=3),hC1(color=green,thickness=3),hA1(color=red,thickness=3),hA1},printtext=true,axes=NONE);       The setting of the altitude from the point В

> altitude(hB1,B,T,B1);
> altitude(hB,B,T);

                                 hB1


                                  hB

        Analytical information about the altitude hB from the point В in the triangle
> detail(hB);
   name of the object: hB
   form of the object: line2d
   assume that the name of the horizonal and vertical                    axis are _x and _y
   equation of the line: -71+13*_x+_y = 0

        Consruction of the altitude from the point В
>draw({T(color=gold,thickness=3),hC1(color=green,thickness=3),hA1(color=red,thickness=3),hB1(color=blue,thickness=3),hB1},printtext=true,axes=NONE);     
 Calculation of the point of intersection of altitudes and let's call it "О"

>intersection(O,hB,hA,hC);coordinates(O);

                                  O


                               483  608
                              [---, ---]
                               97   97

        Construction of altitudes and marking of the point of intersection "О" in the triangle
>draw({T(color=gold,thickness=3),hC1(color=green,thickness=3),hA1(color=red,thickness=3),hB1(color=blue,thickness=3),hB1,O},printtext=true,axes=NONE);




 

 

 

 

 

 

 

 

 

 

 

 

 

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