Load Gif Files

Step 1 : Start ABG
Pattern Mathematics on Internet.
- Select run at current location.
- Select yes for intall and run.
Step 2 : Select program from menu
- Click start to load subjects in upper box.
- Click subject 05 in upper box.
- Click program 08 in lower box.
- Diagrams for p = 1,3,5,7,9,11
  - 1st Row : Graphs of R = 1 + 1*sin(p*A/8)^1
  - 2nd Row : Graphs of R = 1 + 1*sin(p*A/8)^2
  - 3rd Row : Graphs of R = 1 + 1*sin(p*A/8)^3

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Q2. Example : Find petals of R = sin(p*A/q)

Patterns in software ABG Command Petals

Step 1 : Start ABG
Pattern Mathematics on Internet.
- Select run at current location.
- Select yes for intall and run.
Step 2 : Click Command Petals
- It gives graph of R = sin(A) which is a circle.
- Click Command petals again and it gives graph of R=sin(2*A) which has 4 petals.
- Click Command petals again and it gives graph of R=sin(3*A) which has 3 petals.
- Click Command petals again and it gives graph of R=sin(4*A) which has 8 petals.
- Click Command petals again and it gives graph of R=sin(5*A) which has 5 petals.
- Repeat to click command Petals, we have a conclusion
  - R=sin(n*A) has n petals if n is odd.
  - R=sin(n*A) has 2*n petals if n is even.
We can also using ABG program 01 05 by giving n and M for R=sin(n*A)^M.

Petal Rules for R = sin(p*A/q)^M

If p is odd and q is odd and M is odd, then it has p petals.
If p is odd and q is even and M is odd, then it has 2*p petals.
If p is even and q is odd and M is odd, then it has 2*p petals.

Example : Find number of petals of R = sin(3.1*A)

3.1 = 31/10 and then p = 31 and q = 10.
Hence it has 2*p petals. That is 62 petals.
Diagram is given in program 01 07.

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Q3. Example : Find patterns of R = 1+1*sin(p*A/4)^M

Patterns in software ABG Command Patterns

Step 1 : Start ABG
Pattern Mathematics on Internet.
- Select run at current location.
- Select yes for intall and run.
Step 2 : Click Command Patterns
- Click Command petals again and it gives graph of R=1+1(sin(p*A/4)^M.
- Click Command petals again and it gives graph of R=1+2(sin(p*A/4)^M.
- Click Command petals again and it gives graph of R=1+3(sin(p*A/4)^M.
- Click Command petals again and it gives graph of R=1+4(sin(p*A/4)^M.
- Click Command petals again and it gives graph of R=2+1(sin(p*A/4)^M.
- Click Command petals again and it gives graph of R=2+3(sin(p*A/4)^M.

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Q4. Example : Find patterns of R = 4+2*tan(p*A/q)^M

Patterns in software ABG subject 13

Step 1 : Start ABG
Pattern Mathematics on Internet.
- Select run at current location.
- Select yes for intall and run.
Step 2 : Click start to load subject menu
- Click subject 13 in upper box.
- Click q = 8 in lower box, it gives graph of R=1+1*tan(p*A/4)^M
  - 1st row : R=1+1*tan(p*A/4)^1 for p = 1, 3, 5, 7, 9, 11.
  - 2ed row : R=1+1*tan(p*A/4)^2 for p = 1, 3, 5, 7, 9, 11.
  - 3rd row : R=1+1*tan(p*A/4)^3 for p = 1, 3, 5, 7, 9, 11.
The range windows are only defined for q = 8 in the program.

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Q5. Example : Find domain of R = sin(p*A/q)^M

Patterns in software ABG program 01 06

Step 1 : Start ABG
Pattern Mathematics on Internet.
- Select run at current location.
- Select yes for intall and run.
Step 2 : Click start to load subject menu
- Click subject 1 in upper box.
- Click program 6 in lower box. it gives graph of R=sin(p*A/q)^M
  - Give data for p = 2, q = 1, M = 1.
  - We give 2,1,1 for R = sin(2*A/1)^1.
  - We find A = 0 to 1*pi and graph is not complete on left diagram.
  - We find A = 0 to 2*pi and graph is complete on right diagram.
  - Hence the domain for R=sin(2*A) is 2*pi.
- Repeat and give a conclusion
  - Cycle domain of R=sin(n*A) is pi if n is odd.
  - Cycle domain of R=sin(n*A) is 2*pi if n is even.

Domain Rules for R = sin(p*A/q)^M

If p is odd and q is odd and M is odd, then cycle domain is A = 0 to pi.
If p is odd and q is even and M is odd, then cycle domain is A = 0 to 2*pi.
If p is even and q is odd and M is odd, then cycle domain is A = 0 to 2*pi.

Example : Find number of petals and cycle domain of R = sin(3.1*A)

3.1 = 31/10 and then p = 31 and q = 10.
Hence it has 2*p petals. That is 62 petals.
Hence cycle domain is 2*q*pi = 2*10*pi
Diagram can be obtained in program 01 08.

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Q6. Prove that R = sin(2*A) has 4 petals

The procedures are based on

R = sin(2*n*pi+pi/2)=1 and R = sin(2*n*pi+1.5*pi) = -1.
That is 2*A = 90, 270, 350, ....

The proof

.. Angle 2*A ... Data angle A ... Value of R ... Plot angle between 0 and 360
1. 90 .......... A = 45 ......... R = +1 ....... A = 45
2. 270 ......... A = 135 ........ R = -1 ....... A = 135 + 180 (315)
3. 450 ......... A = 225 ........ R = +1 ....... A = 225
4. 630 ......... A = 315 ........ R = -1 ....... A = 315 + 180 (135)
5. 810 ......... A = 405 ........ R = +1 ....... A = 405 (45)
Conclusion
- Number 1 and number 5 have same |R| and same plot angle.
- Hence it only has 4 petals.
- The cycle domain is A = 405 - 45 = 360 degrees = 2*pi
Note
- If R is negative, the plot angle + 180.
- If plot angle greater than 360 and change it to between 0 and 360.

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Q7. Prove that R = sin(3*A) has 3 petals

The procedures are based on

R = sin(2*n*pi+pi/2)=1 and R = sin(2*n*pi+1.5*pi) = -1.
That is 2*A = 90, 270, 350, ....

The proof

.. Angle 3*A ... Data angle A ... Value of R ... Plot angle between 0 and 360
1. 90 .......... A = 30 ......... R = +1 ....... A = 30
2. 270 ......... A = 90 ......... R = -1 ....... A = 90
3. 450 ......... A = 150 ........ R = +1 ....... A = 150
4. 630 ......... A = 210 ........ R = -1 ....... A = 210 + 180 (30)
5. 810 ......... A = 270 ........ R = +1 ....... A = 270 + 180 (90)
6. 990 ......... A = 330 ........ R = -1 ....... A = 330 + 180 (150)
Conclusion
- Number 1 and number 4 have same |R| and same plot angle.
- Number 2 and number 5 have same |R| and same plot angle.
- Number 3 and number 6 have same |R| and same plot angle.
- Hence it only has 3 petals.
- The cycle domain is 210 - 30 = 180 degrees = pi

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Q8. Prove that R = sin(A) is a circle.

Diagram

Start ABG (See program 01 03)

Proof: The procedures are based

R = sin(2*n*pi+pi/2)=1 and R = sin(2*n*pi+1.5*pi) = -1.
That is 2*A = 90, 270, 350, ....

Prove it has only 1 petal

.. Angle A ..... Data angle A ... Value of R ... Plot angle between 0 and 360
1. 90 .......... A = 90 ......... R = +1 ....... A = 90
2. 270 ......... A = 270 ........ R = -1 ....... A = 270 + 180 (90)
3. 450 ......... A = 450 ........ R = +1 ....... A = 450 (90)
- Number 1 and number 2 have same |R| and same plot angle.
- Number 2 and number 3 have same |R| and same plot angle.
- Hence it only has 1 petals.
- The cycle domain is 270 - 90 = 180 degrees = pi

Prove that it is a circle

In polar coordinates x = r*cos(A) and y = r*sin(A).
Hence R^2 = x^2 + y^2.
R = sin(A) = y/R.
R^2 = y.
x^2 + y^2 = y.
By completing square : x^2 + y^2 - y + 1/4 - 1/4 = 0.
Hence x^2 + (y+1/2)^2 = (1/2)^2
This is a circle : Center at (0,1/2) and radius = 1/2.

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Q9. Prove that R = cos(A) is a circle.

Diagram

Start ABG (See program 01 04)

The procedures are based on

R = sin(2*n*pi)=1 and R = sin(2*pi+pi) = -1.
That is 2*A = 0, 180, 360, ....

Prove it has only 1 petal

.. Angle A ..... Data angle A ... Value of R ... Plot angle between 0 and 360
1. 0 ........... A = 00 ......... R = +1 ....... A = 0
2. 180 ......... A = 180 ........ R = -1 ....... A = 180 + 180 (0)
3. 360 ......... A = 360 ........ R = +1 ....... A = 360 (0)
- Number 1 and number 2 have same |R| and same plot angle.
- Number 2 and number 3 have same |R| and same plot angle.
- Hence it only has 1 petals.
- The cycle domain is 180 - 0 = 180 degrees = pi.

Prove that it is a circle

In polar coordinates x = r*cos(A) and y = r*sin(A).
Hence R^2 = x^2 + y^2.
R = cos(A) = x/R.
R^2 = x.
x^2 + y^2 = x.
By completing square : x^2 - x + 1/4 - 1/4 + y^2 = 0.
Hence (x-1/2)^2 + y^2 = (1/2)^2
This is a circle : Center at (1/2,0) and radius = 1/2.

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Q10. Compare y = sin(x) and R = sin(A)

Diagram

Start ABG See diagrams in ABG 01 01

Function : R = sin(A)

It is a cirlce (See Q8).
Center at (0,1/2).
radius = 1/2.
Equation : (x^2) + ((y - 1/2)^2) = (1/2)^2

Function : y = sin(x)

It is a cine curve.
It has periond 2*pi since sin(2*pi+x) = sin(x).
It is odd function since sin(-x) = -sin(x).
It has amplitude 1. That is y between -1 and 1.

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Q11. Compare y = cos(x) and R = cos(A)

Diagram

Start ABG See diagrams in ABG 01 02

Function R = cos(A)

It is a cirlce (See Q8).
Center at (1/2,0).
radius = 1/2.
Equation : ((x - 1/2)^2) + y^2 = (1/2)^2

Function y = cos(x)

It is a cocine curve.
It has periond 2*pi since cos(2*pi+x) = cos(x).
It is even function since cos(-x) = cos(x).
It has amplitude 1. That is y between -1 and 1.

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Q12. Compare y = sin(x) and y = sinh(x)

Diagram

Start ABG See diagrams in ABG 14 07

Function y = sin(x)

It is a cine curve.
It has periond 2*pi since sin(2*pi+x) = sin(x).
It is even function since sin(-x) = -sin(x).
It has amplitude 1. That is y between -1 and 1.

Function y = sinh(x)

If x less than zero
- It is from -infinite to 0.
- It is concave downward.
If x greater than zero
- It is from 0 to infinite.
- It is concave upward.

Question : Why do we name it as y = sinh(x) ?

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Q13. Compare y = tan(x) and y = sinh(x)

Diagram

Start ABG
See diagrams in ABG 14 13
y = tan(x) between -pi/2 and pi/2
- It is similar as y = sinh(x)
y = sinh(x)
- If x less than zero
  - It is from -infinite to 0.
  - It is concave downward.
- If x greater than zero
  - It is from 0 to infinite.
  - It is concave upward.
Example : x=sin(t) and y = cos(t) is a circle
Example : x=sech(t) and y=tanh(t) is a semi-cicle

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