Mathematics Dictionary
Dr. K. G. Shih
Parametric equations
Subjects
Symbol Defintion
Example : x^2 = square of x
AN 15 00 |
- Outlines
AN 15 01 |
- Parametric equations including trigonometric functions
AN 15 02 |
- Properties of the graph of x = tan(t) and y = sec(t)
AN 15 03 |
- Properties of the graph of x = sec(t) and y = tan(t)
AN 15 04 |
- Ellipse : x = h + a*cos(t) and y = k + b*sin(t)
AN 15 05 |
- x = sin(t) and y = sin(t)
AN 15 06 |
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AN 15 07 |
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AN 15 08 |
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AN 15 09 |
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AN 15 10 |
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Answers
AN 15 01. Parametric equations including trigonometric functions
Diagram of equations
Study Subjects |
diagram of parametric equation
Equations : There are 36 equations, e.g.
1. x = cos(t) and y = sin(t)
2. x = tan(t) and y = sec(t)
3. x = sec(t) and y = tan(t)
Go to Begin
AN 15 02. Properties of the graph of x = tan(t) and y = sec(t)
Properties from rectangular form
Change it to rectangular form using tan(t)^2 + 1 = sec(t)^2
x^2 - y^2 = tan(t)^2 - sec(t)^2
Hence x^2 - y^2 = -1
This is a hyperbola
The principal axis is y = 0
The semi-axese : a = 1 and b = 1
The focal length is Sqr(a^2 + b^2) = Sqr(2)
The center is (0, 0)
The Foci are (0, -Sqr(2)) and (0, Sqr(2))
The vertices are (0, -1) and (0, 1)
The asymptotes are y = x and y = -x
Properties from polar form
The polar form : R = (D*e)/(1 - e*sin(A))
The origin is at top of focus
e = f/a = Sqr(2)/1 = Sqr(2)
D is the distance from focus to the directrix
The equation of directrix is y = -D or y = +D
Find D for R = (D*e)/(1 - e*sin(A))
If A = 270 degrees R = f - a and sin(270) = -1
Then R = -(Sqr(2) - 1) = (D*Sqr(2)/(1 + e)
Then D = -(Sqr(2) - 1)*(1 + e)/Sqr(2)
Then D = -(Sqr(2) - 1)*(1 + Sqr(2))/Sqr(2)
Then D = -(2 - 1)/Sqr(2)
Then D = -1/Sqr(2)
Hence equation of diretrix is y = -1/Sqr(2)
The polar form : R = (D*e)/(1 + e*sin(A))
The origin is at one of the foci
e = f/a = Sqr(2)/1 = Sqr(2)
D is the distance from focus to the directrix
The equation of directrix is y = -D or y = +D
Find D for R = (D*e)/(1 + e*sin(A))
If A = 90 degrees R = f - a and sin(90) = 1
Then R = Sqr(2) - 1 = (D*Sqr(2)/(1 + e)
Then D = (Sqr(2) - 1)*(1 + e)/Sqr(2)
Then D = (Sqr(2) - 1)*(1 + Sqr(2))/Sqr(2)
Then D = (2 + 1)/Sqr(2)
Then D = 1/Sqr(2)
Hence equation of diretrix is y = 1/Sqr(2)
Diagrams
Study Subjects |
Program 08 02 and 08 03
Study Subjects |
Program 15 04
Go to Begin
AN 15 03. Properties of the graph of x = sec(t) and y = tan(t)
Properties from rectangular form
Change it to rectangular form using tan(t)^2 + 1 = sec(t)^2
x^2 - y^2 = sec(t)^2 - tan(t)^2
Hence x^2 - y^2 = +1
This is a hyperbola
The principal axis is x = 0
The semi-axese : a = 1 and b = 1
The focal length is Sqr(a^2 + b^2) = Sqr(2)
The center is (0, 0)
The Foci are (-Sqr(2), 0) and (Sqr(2), 0)
The vertices are (-1, 0) and (1, 0)
The asymptotes are y = x and y = -x
Properties from polar form
The polar form R = (D*E)/(1 - e*cos(A))
The origin is at right side focus
e = f/a = Sqr(2)/1 = Sqr(2)
D is the distance from focus to the directrix
The equation of directrix is x = -D or x = +D
Find D for R = (D*e)/(1 - e*cos(A))
If A = 180 degrees R = f - a and cos(180) = -1
Then R = -(Sqr(2) - 1) = (D*Sqr(2)/(1 + e)
Then D = -(Sqr(2) - 1)*(1 + e)/Sqr(2)
Then D = -(Sqr(2) - 1)*(1 + Sqr(2))/Sqr(2)
Then D = -(2 - 1)/Sqr(2)
Then D = -1/Sqr(2)
Hence equation of diretrix is x = -1/Sqr(2)
The polar form R = (D*E)/(1 + e*cos(A))
The origin is at left side focus
e = f/a = Sqr(2)/1 = Sqr(2)
D is the distance from focus to the directrix
The equation of directrix is x = -D or x = +D
Find D for R = (D*e)/(1 + e*cos(A))
If A = 0 degrees R = f - a and cos(0) = 1
Then R = Sqr(2) - 1 = (D*Sqr(2)/(1 + e)
Then D = (Sqr(2) - 1)*(1 + e)/Sqr(2)
Then D = (Sqr(2) - 1)*(1 + Sqr(2))/Sqr(2)
Then D = (2 + 1)/Sqr(2)
Then D = 1/Sqr(2)
Hence equation of diretrix is x = 1/Sqr(2)
Diagrams
Study Subjects |
Program 08 01 and 08 05
Study Subjects |
Program 15 03
Go to Begin
AN 15 04. The graph of x = h + a*cos(t) and y = h + b*sin(t)
Properties from rectangular form
Change it to rectangular form using cos(t)^2 + sin(t)^2 = 1
((x-h)/a)^2 - ((y-k)/b)^2 = cos(t)^2 + sin(t)^2
Hence ((x-h)/a)^2 - ((y-k)/b)^2 = +1
This is an ellipse
The principal axis is y = 0 if a > b
The semi-axese : a and b
The focal length is Sqr(f^2 - a^2)
The center is (h, k)
The Foci are (h - f, 0) and (h + f, 0)
The vertices are (h - a, 0) and (h + a, 0)
Properties from polar form
The polar form R = (D*E)/(1 - e*cos(A))
The origin is at left side focus
e = f/a and e is less than 1
D is the distance from focus to the directrix
The equation of directrix is x = -D
Find D for R = (D*e)/(1 - e*cos(A))
If A = 180 degrees R = a - f and cos(180) = -1
Then R = -(a - f) = (D*e)/(1 + e)
Then D = -(a - f)*(1 + e)/e
Hence equation of diretrix is x = -D
The polar form R = (D*E)/(1 + e*cos(A))
The origin is at right side focus
e = f/a and e is less than 1
D is the distance from focus to the directrix
The equation of directrix is x = +D
Find D for R = (D*e)/(1 + e*cos(A))
If A = 0 degrees R = a - f and cos(0) = 1
Then R = (a - f) = (D*Sqr(2)/(1 + e)
Then D = (a - f)*(1 + e)/e
Hence equation of diretrix is x = D
Diagrams
Study Subjects |
Program 07 01 and 07 05
Go to Begin
AN 15 05. x = sin(t) and y=sin(t)
Properties of sketch
t = 000 to 090 : Plot from (0,0) to (1,1)
t = 090 to 180 : Plot from (1,1) to (0,0)
t = 180 to 270 : Plot from (0,0) to (-1,-1)
t = 270 to 360 : Plot from (-1,-1) to (0,0)
Properties of graph in oxy coordinate
Domain : x = -1 to 1
Range : y = -1 to 1
The line section is [-1, -1] to [1,1]
The slope is 1
Diagrams
Study Subjects |
Program 15 01
Note : If t from 0 to n*pi, the slow motion will be in S.H.M.
Go to Begin
AN 15 06. x=cos(t) and y=sin(t)
Properties of sketch
t = 000 to 090 : Plot arc of circle from (1,0) to (0,1)
t = 090 to 180 : Plot arc of circle from (0,1) to (-1,0)
t = 180 to 270 : Plot arc of circle from (-1,0) to (0,-1)
t = 270 to 360 : Plot arc of circle from (0,-1) to (1,0)
Properties of graph in oxy coordinate
It is a unit circle
Equation : x^2 + y^2 = 1
Center (0,0) and radius = 1
Diagrams
Study Subjects |
Program 15 02
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AN 15 07. Answer
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AN 15 08. Answer
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AN 15 09. Answer
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AN 15 10. Answer
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AN 15 00. Answer
Pythagorean relation
x = cos(t) and y = sin(t) is unit circle
x = sec(t) and y = tan(t) is unit hyperbola
x = tan(t) and y = sec(t) is unit hyperbola
Conic sections
Circle : x = h + r*cos(t) and y = k + r*sin(t)
Ellipse : x = h + a*cos(t) and y = k + b*sin(t)
Hyperbola : x = h + a*sec(t) and y = k + b*tan(t)
Hyperbola : x = h + a*tan(t) and y = k + b*sec(t)
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