Hey friend, I want to share my experience, which is to be tutor to my friend. She is Anna Susanti (07305141027), my partner in this moment who learning my explanation. On Friday, 9th January 2009 at 2 pm, in Anna’s boarding home, I try to explain the mathematics topic about Homogeneous Equations. As reference, I use the book from Shepley L. Ross with title “Differential Equations”.
First, I explain about the definition of Homogeneous Equations. Then, I explain an example, theorem and another example to find the solution of Homogeneous Equations.
Definition
The first-order differential equation M (x, y) dx + N (x, y) dx = 0 is said to be homogeneous if, when written in the derivative form (dy / dx) = f (x, y), there exists a function g such that f (x, y) can be expressed in the form g (y / x).
Example 1
The differential equation (x^2 – 3y^2) dx + 2xy dy = 0 is homogeneous. To see this, we first write this equation in the derivative form
dy / dx = (3y^2 – x^2) / 2xy
now observing that
(3y^2 – x^2) / 2xy = (3y / 2x) – (x / 2y) = (3/2) (y / x) – (1/2) (1 / (y / x) )
we see that the differential equation under consideration may be written as
dy / dx = (3/2) (y / x) – (1/2) (1/ (y / x) )
in which the right member is of the form g (y / x) for a certain function g.
Theorem 2.3
If
M (x, y) dx + N (x, y) dy = 0 …………….(2.24)
Is a homogeneous equation, then the change of variables y = vx transforms (2.24) into a separable equatio0n in the variables v and x.
Proof. Since M(x, y) dx + N (x, y) dy = 0 is homogeneous, it may be written in the form
dy / dx = g (y / x)
Let y = vx. Then
dy / dx = v + x dv / dx
and (2.24) becomes
v + x dv / dx = g (v)
or
[v – g (v)] dx + x dv = 0
This equation is separable. Separating the variables we obtain
dv / (v – g (v)) + dx /x = 0 ………………(2.25)
Thus to solve a homogeneous differential equation of the form (2.24), we let y = vx and transform the homogeneous equation into a separable equation of the form (2.25).
From this, we have
int dv / (v – g (v)) + int dx / x = c,
where c is an arbitrary constant. Letting F(v) denote
int dv / (v – g (v))
and returning to the original dependent variable y, the solution takes the form
F (y / x) + ln (x) = c
Example 2
Solve the equation
(x^2 – 3y^2) dx + 2xy dy = 0
We have already observed that this equation is homogeneous. Writing it in the form
dy / dx = - (x / 2y) + (3y / 2x)
and letting y = vx, we obtain
v + x dv / dx = - (1/ 2v) + (3v / 2),
or
x dv / dx = - (1/ 2v) + (v / 2)
and finally,
x dv / dx = (v^2 - 1) / 2v
This equation is separable. Separating the variable, we obtain
2v dv / (v^2 - 1) = dx / x
Integrating, we find
ln (v^2 - 1) = ln x + ln c
and hence
(v^2 - 1) = cx
where c is an arbitrary constant. The reader should observe that no solutions were lost in the separation process. Now, replacing v by y / x we obtain the solutions in the form
(y^2 / x^2) – 1 = cx
or
(y^2 – x^2) = (cx) x^2
If y > x > 0, then this may be expressed somewhat more simply as
y^2 – x^2 = cx^3
And the last, I give Anna an exercise. She can solves that exercise successfully. But when I explained to Anna, I find some difficult, it’s because I explained it in English, so she had a few difficult to understand my explanation. But because this activity, I think now I and Anna become more understands about Homogeneous Equations.
For Anna, Thank you for your attention . . .
Kamis, 15 Januari 2009
Senin, 12 Januari 2009
The definition, explanation and the example of the terms questioning by Hamdi
Assignment 6
1. Istilah matematika : mathematical term
Example : This mathematical term was often used by Greek.
2. Bentuk-bentuk bangun datar dan bangun ruang
The form of the plane figure and the three dimentional figure
Example : We will discuss the form of the plane figure and the three dimentional figure.
3. Sulit menghafal kosa-kata
It is difficult to memorize the vocabulary
Example : I had difficulty to memorize the vocabulary in the English.
4. Sulit memahami suatu bacaan
It’s difficult to understand the reading material
Example : He realized still the difficulty to understand the reading material.
5. Cepat bosan : fast bored
Example : If not carrying out the activity anything, I will feel fast bored.
6. Sulit konsentrasi : it’s difficult to concentration
Example : She had difficulty the concentration if she was in the crowded place.
7. Sangat kurangnya pemahaman mengenai Bahasa Inggris dari sejak SMP dan MA
It’s really the shortage of the understanding about English since the Junior High School and MA.
Example : I really had the shortage of the understanding about English since the Junior High School and MA.
8. Malu mengucapkannya jika salah
It’s embarrassed to said it if being wrong.
Example : Every people will be embarrassed to said it if being wrong.
9. Sangat kurangnya kosa-kata
It’s really the shortage of the vocabulary
Example : I realized still the shortage of the vocabulary that was owned by me.
10. Sulit membagi waktu : It’s difficult to divide time
Example : I experienced the difficulty in dividing time.
11. Cepat lupa : fast forgot
Example : My bad habit is fast forgot.
12. Kurangnya fasilitas yang menunjang
The shortage of supportive facilities
Example : The main reason is because of the shortage of the supportive facilities.
13. Sulit menyusun kalimat yang benar
It’s difficult to compile the true sentence
Example : I had difficulty to compile the true sentence in English.
1. Istilah matematika : mathematical term
Example : This mathematical term was often used by Greek.
2. Bentuk-bentuk bangun datar dan bangun ruang
The form of the plane figure and the three dimentional figure
Example : We will discuss the form of the plane figure and the three dimentional figure.
3. Sulit menghafal kosa-kata
It is difficult to memorize the vocabulary
Example : I had difficulty to memorize the vocabulary in the English.
4. Sulit memahami suatu bacaan
It’s difficult to understand the reading material
Example : He realized still the difficulty to understand the reading material.
5. Cepat bosan : fast bored
Example : If not carrying out the activity anything, I will feel fast bored.
6. Sulit konsentrasi : it’s difficult to concentration
Example : She had difficulty the concentration if she was in the crowded place.
7. Sangat kurangnya pemahaman mengenai Bahasa Inggris dari sejak SMP dan MA
It’s really the shortage of the understanding about English since the Junior High School and MA.
Example : I really had the shortage of the understanding about English since the Junior High School and MA.
8. Malu mengucapkannya jika salah
It’s embarrassed to said it if being wrong.
Example : Every people will be embarrassed to said it if being wrong.
9. Sangat kurangnya kosa-kata
It’s really the shortage of the vocabulary
Example : I realized still the shortage of the vocabulary that was owned by me.
10. Sulit membagi waktu : It’s difficult to divide time
Example : I experienced the difficulty in dividing time.
11. Cepat lupa : fast forgot
Example : My bad habit is fast forgot.
12. Kurangnya fasilitas yang menunjang
The shortage of supportive facilities
Example : The main reason is because of the shortage of the supportive facilities.
13. Sulit menyusun kalimat yang benar
It’s difficult to compile the true sentence
Example : I had difficulty to compile the true sentence in English.
Explain the 8 video
Assignment 4
A. Video 1
TATA BAHASA (GRAMMAR)
Tata bahasa adalah bagaimana menghubungkan bagian-bagian dari bahasa untuk membentuk sebuah kalimat. Kalimat sederhana adalah kalimat dimana semua elemen yang terkandung di dalam kalimat tersebut hanya terdiri dari subjek dan predikat. Subjek menunjukkan kegiatan dari kata kerja utama. Subjek utama adalah kata benda khusus yang menunjukkan sebuah kegiatan.
Contoh:
The happy little child kicked the gnome over the fence.
“The happy little child” merupakan subjek.
“happy little” menghubungkan subjek sederhana child.
“child” menunjukkan kata kerja “kick”
Predikat dari sebuah kalimat terdiri dari:
Main verb (kata kerja utama) dan apapun yang mengikutinya. Gabungan keduanya disebut predikat yang lengkap.
Contoh :
The happy little child kicked the gnome over the fence.
“kicked” merupakan kata kerja (predikat sederhana). Sedangkan “kicked the gnome over the fence” merupakan predikat yang lengkap, karena “the gnome” dan “over the fence” menunjukkan lebih jelas lagi tentang apa yang ditendang dan bagaimana tendangan itu.
Kalimat sederhana dapat diperoleh tanpa subjek dan predikat. Kalimat perintah merupakan kalimat yang ditunjukkan langsung pada orang kedua yang adalah “kamu” atau memerintahkan seseorang untuk melakukan sesuatu.
Contoh:
Kick the gnome over the fence
Kalimat diatas merupakan pedikat yang lengkap dengan “kick” sebagai predikat sederhana dan tidak memiliki subjek.
Lalu siapa sebenarnya yang melakukan kegiatan tersebut?
Jawabannya adalah “kamu”, sehingga kalimat tersebut dapat ditulis menjadi
“Hey you, kick the gnome over the fence!”
Dalam kalimat tersebut kata “Hey you” tidak perlu ditulis karena sudah tersirat.
B. Video 2
KATA KERJA (VERB)
Kata kerja menunjukkan sebuah kegiatan atau untuk menjelaskan sebuah kejadian atau menunjukkan apa yang sedang dilakukan oleh suatu benda atau seseorang. Kata kerja sangat penting ada dalam suatu kalimat.
Contoh dalam sebuah kalimat sederhana:
Dave runs
“Run” merupakan kata kerja yang menunjukkan apa yang sedang dilakukan Dave.
Dalam bahasa Inggris, perubahan bentuk kata kerja untuk menunjukkan siapa yang sedang melakukan kegiatan tersebut.
Misal:
I do, you do, he does, she does, we do, they do
Contoh:
Dave runs
Jika subjeknya diganti menjadi “I”, maka kalimatnya akan berubah menjadi “I run”. Kata kerjanya juga berubah karena subjeknya berubah. Dalam hal ini, subjek “I, you, we, they” menggunakan kata kerja “run”, sedangkan untuk subjek “He, she, it” menggunakan kata keja “runs”.
Kata kerja dalam bentuk “to be”, diantaranya:
“I am, you are, he is, she is, it is” merupakan singular subjek atau kata ganti tunggal, yang diikuti dengan kata kerja tunggal. Sedangkan “we are, they are” merupakan plural subjek atau kata ganti jamak, yang diikuti dengan kata kerja jamak.
Contoh:
Mrs. Midori yodels
“Mrs. Midori” merupakan kata ganti tunggal, maka menggunakan kata kerja tunggal, yaitu “yodels”.
Saudara-saudara Midori seperti Else, Gretel, Heidi merupakan kata ganti jamak, maka menggunakan kata ganti jamak, sehingga dapat ditulis:
Midori’s sister yodel.
C. Video 3
KATA KETERANGAN (ADVERB)
Kata keterangan adalah kata yang menjelaskan kata kerja(verb), kata sifat(adjective), dan kata-kata keterangan lainnya. Adverb digunakan untuk menjawab pertanyaan dengan kata tanya “how?, how often?, when?, to what extend?”.
Sangat mudah untuk membentuk sebuah adverb. Caranya adalah adjective (kata sifat) dan diakhiri dengan akhiran –ly.
Contoh:
Simon might be ‘slow”, but he talks ‘slowly”.
Kata “slow” menjelaskan Simon. “Slow” = adjective.
Kata “slowly” menjawab pertanyaan “how does Simon talk? (bagaimana Simon berbicara)”.“Slowly” = adverb.
The color of these mushrooms is slightly different.
Slightly = slight + (-ly)
Sekarang mari kita pelajari tentang adverb yang diikuti oleh adverb-adverb lainnya.
Contoh:
This mushroom is very definitely poisonous.
“Poisonous” (adjective) dimodifikasikan oleh kata “definitely” (adverb).
Sedangkan “definitely” dijelaskan oleh adverb lain, yaitu kata “very”.
Pada umumnya kata “very” adalah adverb yang digunakan untuk menjelaskan adverb lainnya. “very” digunakan sebagai adverb dan tidak diakhiri dengan akhiran (-ly).
Sekarang kita akan mempelajari tenteng pengecualian dalam penggunaan akhiran “-ly” dalam adverb. Misal, untuk kata sifat (adjective) “ good” dan “fast” bila diubah dalam bentuk adverb, maka kita tidak boleh sembarang mengubahnya menjadi “goodly” dan “fastly” karena kedua kata itu salah.
Untuk kata “good”, bila kita akan menjawab pertanyaan “how?” maka kita menggunakan kata “well” sebagai adverb.
Contoh:
Candace can play the accordion very well.
Kata “well” adalah adverb yang dibentuk dari “good” (adjective).
Dan untuk pertanyaan “how Candace can play?” maka jawabannya adalah “candace’s playing is good”. Sedangkan good adalah kata sifat (adjective) yang menjelaskan gerund “playing”.
Ingat, gerund dapat dibentuk dari “kata kerja(verb) + (-ing)”, dan digunakan sebagai kata benda (noun).
D. Video 4
BASIC TRIGONOMETRY
Trigonometry comes from Greek (trigon and metron). Trigonometry is really study of rectangle and the relationship between the side and the angle of rectangle. There is a right-triangle, with the length of opposite is 3, the length of adjacent is 4, and the length of hypotenuse is 5. And there is angle-x in front of the right-angle, and there is angle-m above the right-angle. What is the value of sinus of x, cosines of x, and tangent of x?
To solve them, use the simple trigonometry, they are SOH CAH TOA.
SOH is defined by “Sinus is Opposite over Hypotenuse”, CAH is defined by “Cosine is Adjacent over Hypotenuse”, and TOA is defined by “Tangent is Opposite over Adjacent”.
So we get,
Sin x = Opp/Hyp = 4/5
Cos x = Adj/Hyp = 3/5
Tan x = Opp/Adj = 4/3
If the angle is m, so we get tan m = ¾, the inverse of tangent x.
E. Video 5
KALIMAT MAJEMUK (COMPOUND SENTENCE)
Contoh kalimat “It is the end of the world as we know it and I feel fine”. Dalam kalimat tersebut terdapat 2 klausa yang dihubungkan dengan 1 kata penghubung, yaitu “and”. Ketika suatu kalimat digunakan sebagai bagian-bagian dalam kalimat yang lebih besar, maka kalimat yang lebih kecil disebut klausa. Ketika sebuah klausa dapat berdiri sendiri dalam sebuah kalimat, maka klausa tersebut disebut “ndependent clause”. Dan jika kita memiliki 2 klausa dalam sebuah kalimat, maka kalimat tersebut disebut kalimat majemuk.
Untuk menggabungkan 2 independent clause, kita dapat menggunakan:
Tanda titik dua (:), ketika klausa ke-2 menjelaskan klausa ke-1.
Contoh:
I love my two sisters, they bake me pie
Untuk menggabungkan kalimat menjadi kalimat majemuk kita gunakan titik dua (:).
Kalimat di atas menjadi: “I love my two sisters: they bake me pie”
Titik koma (;), untuk menggantikan konjungsi.
Contoh:
“It is the end of the world as we know it and I feel fine”
Kalimat di atas dapat dipersingkat menjadi:
“It is the end of the world; I feel fine”
Garis penghubung (-)
Menggunakan garis penghubung karena klausa ke-2 dihubungkan dengan klausa ke-1.
Dengan demikian, dapat disimpulkan bahwa untuk menggabungkan kalimat ada 4 cara, yaitu:
Kata penghubung
Tanda titik dua (:)
Tanda titik koma (;)
Garis penghubung (-)
Kalimat fragmen dapat didefinisikan jika kita mendapat porsi dalam suatu kalimat yang tidak dapat berdiri sendiri sebagai kalimat lengkap.
Contoh:
“My pet komodo dragon is as gentle as a lamb” merupakan kalimat lengkap.
“because he has no teeth” merupakan kalimat fragmen.
Dependent clause adalah klausa yang tidak dapat berdiri sendiri, terdapat pada klausa independent, dan bukan merupakan kalimat lengkap.
Kalimat kompleks terdiri dari:
Klausa dependent + klausa independent
Contoh:
Although Tom sleeps regulery, he is constantly tired.
“Although Tom sleeps regulery” merupakan klausa dependen.
“He is constantly tired” merupakan klausa independent.
“Although Tom sleeps regulary, he is constantly tired” merupakan kalimat kompleks.
F. Video 6
LIMIT BY INSPECTION
There are two conditions, they are:
x goes to positive or negative infinity
Limit involves a polynomial divided by a polynomial
For example:
Limit of the function x cube plus four all over x square plus x plus one as x approaches infinity. This problem is caused of two conditions, they are:
Polynomial over polynomial
x approaches infinity
The key to determine the limits by inspection is in looking at powers of x in the numerator and the denominator.
Remember!
To apply this rules, the things which must we do are:
We must be dividing by polynomials
x has to be approaching infinity
The first shortcut rule
If the highest power of x is greater in numerator, then limit is positive or negative infinity. x cube plus four all over x square plus x plus one as x approaches infinity equals infinity equals positive or negative infinity. From that function we get three at x cube is the highest power of x in numerator. Two at x square is the highest power of x in denominator. Limit of function x cube plus four all over x square plus x plus one as x approaches infinity equals infinity equals positive or negative infinity. It can be happened Since all the number are positive and x is going to positive infinity.
If you cannot tell it, the answer is positive or negative. And the rule is:
Substitute a large number for x
See if you end up with a positive or negative number
Whatever sign you get is the sign of infinitive for the limit
The second shortcut rule
If the highest power of x is in the denominator, then the limit is zero let limit of x square plus three all over x cube plus one as x approaches infinitive equals zero. To at x square is highest power of x in numerator, and three x cube is the highest power of x in denominator
The third shortcut rule
This rule is used when the highest power of x in numerator is same as highest power of x in denominator. Limit of a function as x approach positive or negative infinity has the quotient on the coefficients at the two highest powers.
Remember!
Coefficient is the number that goes with a variable.
Example : two is the coefficient of two times x square, seventy five is the coefficient of seventy five times x fourth
The last shortcut rule
Limit of four time x cube plus x square plus one divided by three times x cube plus four as x approaches infinity has three as highest power at numerator and denominator
According to this rule, with conclude that the coefficient of x cube is over each other for example :
Limit of four x cube plus x square plus one all over three x cube plus four as x approaches infinity equals four over three.
G. Video 7
GRAPH OF A RATIONAL FUNCTION
Graph of a rational function which can have discontinuities because has polynomial in the denominator.
Is possible value x divide by 0
Example :
f(x) equals (x+2) over (x-1)
when x=1,so f(1)= (1+2) over (1-1) equals 3 over zero. That is bad idea.
f(1)=(1+2) over zero is break in function graph.
f(x) = ( x+ 2) over (x-1)
for x=0, f(0)=(0+2) over (0-1) equals -2
insert x=1,so f(1)=1+2 over (1-1) equals 3 over 0, it is impossible.
Rational functions don’t always work in this way! Take graph f(x) = 1 over ( x square plus 1 ). Not all rational functions will give zero in denominator because of the (+1) is never zero.
Rational functions denominator can be zero!
Polynomial have smooth and unbroken curve and for rational function x : zero in the denominator that impossible situation. There is no value for the function.
A break can show up in 2 ways. A simply type break is missing point on the graph.
Example :
y= (x square minus x minus 6) over (x minus 3)
The graph loose like this if x = 3, so (3 square minus 3 minus 6) over (3 minus 3) equals 0 over 0. That is not possible, not feasible, and not allowed.
So that is no way if x = 3. This is a typical example to the missing point syndrome.
y = (3 square minus 3 minus 6) over (3 minus 3) equals 0 over 0
When you see result of 0 over 0 and also tell you direction be possible factor top and bottom of rational function and simplify.
For example:
y = (x square minus x minus 6) over (x minus 3) equals (x minus 3) times (x plus 2) all over (x minus 3) equals (x+2).
H. Video 8
TRIGONOMETRY FUNCTION
Now we will describe aright-triangle. Take x as the angle in front of the right-angle. The length of opposite is four, the length of adjacent is three and the length of hypotenuse is five.
To remember it you can see:
SOH is defined by sine equals opposite over hypotenuse
CAH is defined by cosine equals adjacent over hypotenuse
TOA is defined by tangent equals opposite over adjacent
Trigonometry function is only needed to know the values of side to find measure of an angle, figure out figure of all part of triangle. Things is included in trigonometry function are sine, cosine, tangent, cosecant , secant, cotangent. There are six basic of trigonometry function :
Sides of a triangle
Angel being measured
So we can conclude that opposite equals side opposite to x, adjacent equals side adjacent to x.
Sin x equals opposite over hypotenuse, cosine x equals over hypotenuse, tangent x equals opposite over adjacent, cosecant x equal hypotenuse over opposite, secant x equals hypotenuse over adjacent, cotangent adjacent over opposite.
A. Video 1
TATA BAHASA (GRAMMAR)
Tata bahasa adalah bagaimana menghubungkan bagian-bagian dari bahasa untuk membentuk sebuah kalimat. Kalimat sederhana adalah kalimat dimana semua elemen yang terkandung di dalam kalimat tersebut hanya terdiri dari subjek dan predikat. Subjek menunjukkan kegiatan dari kata kerja utama. Subjek utama adalah kata benda khusus yang menunjukkan sebuah kegiatan.
Contoh:
The happy little child kicked the gnome over the fence.
“The happy little child” merupakan subjek.
“happy little” menghubungkan subjek sederhana child.
“child” menunjukkan kata kerja “kick”
Predikat dari sebuah kalimat terdiri dari:
Main verb (kata kerja utama) dan apapun yang mengikutinya. Gabungan keduanya disebut predikat yang lengkap.
Contoh :
The happy little child kicked the gnome over the fence.
“kicked” merupakan kata kerja (predikat sederhana). Sedangkan “kicked the gnome over the fence” merupakan predikat yang lengkap, karena “the gnome” dan “over the fence” menunjukkan lebih jelas lagi tentang apa yang ditendang dan bagaimana tendangan itu.
Kalimat sederhana dapat diperoleh tanpa subjek dan predikat. Kalimat perintah merupakan kalimat yang ditunjukkan langsung pada orang kedua yang adalah “kamu” atau memerintahkan seseorang untuk melakukan sesuatu.
Contoh:
Kick the gnome over the fence
Kalimat diatas merupakan pedikat yang lengkap dengan “kick” sebagai predikat sederhana dan tidak memiliki subjek.
Lalu siapa sebenarnya yang melakukan kegiatan tersebut?
Jawabannya adalah “kamu”, sehingga kalimat tersebut dapat ditulis menjadi
“Hey you, kick the gnome over the fence!”
Dalam kalimat tersebut kata “Hey you” tidak perlu ditulis karena sudah tersirat.
B. Video 2
KATA KERJA (VERB)
Kata kerja menunjukkan sebuah kegiatan atau untuk menjelaskan sebuah kejadian atau menunjukkan apa yang sedang dilakukan oleh suatu benda atau seseorang. Kata kerja sangat penting ada dalam suatu kalimat.
Contoh dalam sebuah kalimat sederhana:
Dave runs
“Run” merupakan kata kerja yang menunjukkan apa yang sedang dilakukan Dave.
Dalam bahasa Inggris, perubahan bentuk kata kerja untuk menunjukkan siapa yang sedang melakukan kegiatan tersebut.
Misal:
I do, you do, he does, she does, we do, they do
Contoh:
Dave runs
Jika subjeknya diganti menjadi “I”, maka kalimatnya akan berubah menjadi “I run”. Kata kerjanya juga berubah karena subjeknya berubah. Dalam hal ini, subjek “I, you, we, they” menggunakan kata kerja “run”, sedangkan untuk subjek “He, she, it” menggunakan kata keja “runs”.
Kata kerja dalam bentuk “to be”, diantaranya:
“I am, you are, he is, she is, it is” merupakan singular subjek atau kata ganti tunggal, yang diikuti dengan kata kerja tunggal. Sedangkan “we are, they are” merupakan plural subjek atau kata ganti jamak, yang diikuti dengan kata kerja jamak.
Contoh:
Mrs. Midori yodels
“Mrs. Midori” merupakan kata ganti tunggal, maka menggunakan kata kerja tunggal, yaitu “yodels”.
Saudara-saudara Midori seperti Else, Gretel, Heidi merupakan kata ganti jamak, maka menggunakan kata ganti jamak, sehingga dapat ditulis:
Midori’s sister yodel.
C. Video 3
KATA KETERANGAN (ADVERB)
Kata keterangan adalah kata yang menjelaskan kata kerja(verb), kata sifat(adjective), dan kata-kata keterangan lainnya. Adverb digunakan untuk menjawab pertanyaan dengan kata tanya “how?, how often?, when?, to what extend?”.
Sangat mudah untuk membentuk sebuah adverb. Caranya adalah adjective (kata sifat) dan diakhiri dengan akhiran –ly.
Contoh:
Simon might be ‘slow”, but he talks ‘slowly”.
Kata “slow” menjelaskan Simon. “Slow” = adjective.
Kata “slowly” menjawab pertanyaan “how does Simon talk? (bagaimana Simon berbicara)”.“Slowly” = adverb.
The color of these mushrooms is slightly different.
Slightly = slight + (-ly)
Sekarang mari kita pelajari tentang adverb yang diikuti oleh adverb-adverb lainnya.
Contoh:
This mushroom is very definitely poisonous.
“Poisonous” (adjective) dimodifikasikan oleh kata “definitely” (adverb).
Sedangkan “definitely” dijelaskan oleh adverb lain, yaitu kata “very”.
Pada umumnya kata “very” adalah adverb yang digunakan untuk menjelaskan adverb lainnya. “very” digunakan sebagai adverb dan tidak diakhiri dengan akhiran (-ly).
Sekarang kita akan mempelajari tenteng pengecualian dalam penggunaan akhiran “-ly” dalam adverb. Misal, untuk kata sifat (adjective) “ good” dan “fast” bila diubah dalam bentuk adverb, maka kita tidak boleh sembarang mengubahnya menjadi “goodly” dan “fastly” karena kedua kata itu salah.
Untuk kata “good”, bila kita akan menjawab pertanyaan “how?” maka kita menggunakan kata “well” sebagai adverb.
Contoh:
Candace can play the accordion very well.
Kata “well” adalah adverb yang dibentuk dari “good” (adjective).
Dan untuk pertanyaan “how Candace can play?” maka jawabannya adalah “candace’s playing is good”. Sedangkan good adalah kata sifat (adjective) yang menjelaskan gerund “playing”.
Ingat, gerund dapat dibentuk dari “kata kerja(verb) + (-ing)”, dan digunakan sebagai kata benda (noun).
D. Video 4
BASIC TRIGONOMETRY
Trigonometry comes from Greek (trigon and metron). Trigonometry is really study of rectangle and the relationship between the side and the angle of rectangle. There is a right-triangle, with the length of opposite is 3, the length of adjacent is 4, and the length of hypotenuse is 5. And there is angle-x in front of the right-angle, and there is angle-m above the right-angle. What is the value of sinus of x, cosines of x, and tangent of x?
To solve them, use the simple trigonometry, they are SOH CAH TOA.
SOH is defined by “Sinus is Opposite over Hypotenuse”, CAH is defined by “Cosine is Adjacent over Hypotenuse”, and TOA is defined by “Tangent is Opposite over Adjacent”.
So we get,
Sin x = Opp/Hyp = 4/5
Cos x = Adj/Hyp = 3/5
Tan x = Opp/Adj = 4/3
If the angle is m, so we get tan m = ¾, the inverse of tangent x.
E. Video 5
KALIMAT MAJEMUK (COMPOUND SENTENCE)
Contoh kalimat “It is the end of the world as we know it and I feel fine”. Dalam kalimat tersebut terdapat 2 klausa yang dihubungkan dengan 1 kata penghubung, yaitu “and”. Ketika suatu kalimat digunakan sebagai bagian-bagian dalam kalimat yang lebih besar, maka kalimat yang lebih kecil disebut klausa. Ketika sebuah klausa dapat berdiri sendiri dalam sebuah kalimat, maka klausa tersebut disebut “ndependent clause”. Dan jika kita memiliki 2 klausa dalam sebuah kalimat, maka kalimat tersebut disebut kalimat majemuk.
Untuk menggabungkan 2 independent clause, kita dapat menggunakan:
Tanda titik dua (:), ketika klausa ke-2 menjelaskan klausa ke-1.
Contoh:
I love my two sisters, they bake me pie
Untuk menggabungkan kalimat menjadi kalimat majemuk kita gunakan titik dua (:).
Kalimat di atas menjadi: “I love my two sisters: they bake me pie”
Titik koma (;), untuk menggantikan konjungsi.
Contoh:
“It is the end of the world as we know it and I feel fine”
Kalimat di atas dapat dipersingkat menjadi:
“It is the end of the world; I feel fine”
Garis penghubung (-)
Menggunakan garis penghubung karena klausa ke-2 dihubungkan dengan klausa ke-1.
Dengan demikian, dapat disimpulkan bahwa untuk menggabungkan kalimat ada 4 cara, yaitu:
Kata penghubung
Tanda titik dua (:)
Tanda titik koma (;)
Garis penghubung (-)
Kalimat fragmen dapat didefinisikan jika kita mendapat porsi dalam suatu kalimat yang tidak dapat berdiri sendiri sebagai kalimat lengkap.
Contoh:
“My pet komodo dragon is as gentle as a lamb” merupakan kalimat lengkap.
“because he has no teeth” merupakan kalimat fragmen.
Dependent clause adalah klausa yang tidak dapat berdiri sendiri, terdapat pada klausa independent, dan bukan merupakan kalimat lengkap.
Kalimat kompleks terdiri dari:
Klausa dependent + klausa independent
Contoh:
Although Tom sleeps regulery, he is constantly tired.
“Although Tom sleeps regulery” merupakan klausa dependen.
“He is constantly tired” merupakan klausa independent.
“Although Tom sleeps regulary, he is constantly tired” merupakan kalimat kompleks.
F. Video 6
LIMIT BY INSPECTION
There are two conditions, they are:
x goes to positive or negative infinity
Limit involves a polynomial divided by a polynomial
For example:
Limit of the function x cube plus four all over x square plus x plus one as x approaches infinity. This problem is caused of two conditions, they are:
Polynomial over polynomial
x approaches infinity
The key to determine the limits by inspection is in looking at powers of x in the numerator and the denominator.
Remember!
To apply this rules, the things which must we do are:
We must be dividing by polynomials
x has to be approaching infinity
The first shortcut rule
If the highest power of x is greater in numerator, then limit is positive or negative infinity. x cube plus four all over x square plus x plus one as x approaches infinity equals infinity equals positive or negative infinity. From that function we get three at x cube is the highest power of x in numerator. Two at x square is the highest power of x in denominator. Limit of function x cube plus four all over x square plus x plus one as x approaches infinity equals infinity equals positive or negative infinity. It can be happened Since all the number are positive and x is going to positive infinity.
If you cannot tell it, the answer is positive or negative. And the rule is:
Substitute a large number for x
See if you end up with a positive or negative number
Whatever sign you get is the sign of infinitive for the limit
The second shortcut rule
If the highest power of x is in the denominator, then the limit is zero let limit of x square plus three all over x cube plus one as x approaches infinitive equals zero. To at x square is highest power of x in numerator, and three x cube is the highest power of x in denominator
The third shortcut rule
This rule is used when the highest power of x in numerator is same as highest power of x in denominator. Limit of a function as x approach positive or negative infinity has the quotient on the coefficients at the two highest powers.
Remember!
Coefficient is the number that goes with a variable.
Example : two is the coefficient of two times x square, seventy five is the coefficient of seventy five times x fourth
The last shortcut rule
Limit of four time x cube plus x square plus one divided by three times x cube plus four as x approaches infinity has three as highest power at numerator and denominator
According to this rule, with conclude that the coefficient of x cube is over each other for example :
Limit of four x cube plus x square plus one all over three x cube plus four as x approaches infinity equals four over three.
G. Video 7
GRAPH OF A RATIONAL FUNCTION
Graph of a rational function which can have discontinuities because has polynomial in the denominator.
Is possible value x divide by 0
Example :
f(x) equals (x+2) over (x-1)
when x=1,so f(1)= (1+2) over (1-1) equals 3 over zero. That is bad idea.
f(1)=(1+2) over zero is break in function graph.
f(x) = ( x+ 2) over (x-1)
for x=0, f(0)=(0+2) over (0-1) equals -2
insert x=1,so f(1)=1+2 over (1-1) equals 3 over 0, it is impossible.
Rational functions don’t always work in this way! Take graph f(x) = 1 over ( x square plus 1 ). Not all rational functions will give zero in denominator because of the (+1) is never zero.
Rational functions denominator can be zero!
Polynomial have smooth and unbroken curve and for rational function x : zero in the denominator that impossible situation. There is no value for the function.
A break can show up in 2 ways. A simply type break is missing point on the graph.
Example :
y= (x square minus x minus 6) over (x minus 3)
The graph loose like this if x = 3, so (3 square minus 3 minus 6) over (3 minus 3) equals 0 over 0. That is not possible, not feasible, and not allowed.
So that is no way if x = 3. This is a typical example to the missing point syndrome.
y = (3 square minus 3 minus 6) over (3 minus 3) equals 0 over 0
When you see result of 0 over 0 and also tell you direction be possible factor top and bottom of rational function and simplify.
For example:
y = (x square minus x minus 6) over (x minus 3) equals (x minus 3) times (x plus 2) all over (x minus 3) equals (x+2).
H. Video 8
TRIGONOMETRY FUNCTION
Now we will describe aright-triangle. Take x as the angle in front of the right-angle. The length of opposite is four, the length of adjacent is three and the length of hypotenuse is five.
To remember it you can see:
SOH is defined by sine equals opposite over hypotenuse
CAH is defined by cosine equals adjacent over hypotenuse
TOA is defined by tangent equals opposite over adjacent
Trigonometry function is only needed to know the values of side to find measure of an angle, figure out figure of all part of triangle. Things is included in trigonometry function are sine, cosine, tangent, cosecant , secant, cotangent. There are six basic of trigonometry function :
Sides of a triangle
Angel being measured
So we can conclude that opposite equals side opposite to x, adjacent equals side adjacent to x.
Sin x equals opposite over hypotenuse, cosine x equals over hypotenuse, tangent x equals opposite over adjacent, cosecant x equal hypotenuse over opposite, secant x equals hypotenuse over adjacent, cotangent adjacent over opposite.
Explain video Do You Believe in Me?
Assignment 2
Do You Believe In Me?
This video describe about the oration by a child. The auditorium is very crowded. Then the moderator calling one of the candidate. And then applause from the audience is heard so noisy. He speak very loudly. He said “Do you believe in me?”, and then the audience answered “Yes!!” loudly too.
He persuade them to believe on what he said. He is a great boy.
Perbaikan:
Do You Believe In Me?
The video describe a boy who has named Dalton Sherman, he came from Charles Rice Learning Center. He believe in him. He ask “Do you believe in me?” He can stand up in there, fearless and talk to all of the audience. He can do anything, create anything, dream anything, become anything, because they believe in him.
He give a question to Dallas ISD, “Do you believe that us can graduate ready for college or the workplace?” Next week, he and Charles Rice Learning Center will come in their school and what he need from them is to believe that he can reach their highest potential.
He hope that they can believe in their colleagues. They came because they wanted to make a difference. Whether they are a counselor or a librarian or whatever, he persuade that he need them.
He persuade that they must believe in their self. He persuade that what they are doing is not just his generation but that of his children and his children’s children.
He persuade all of the students in Dallas, that we need you. We need them now more than ever. We need they, ladies and gentlemen, we need them to know that what they are doing is very important job in the city today.
He persuade again and again to the audience that he believe in him so they must believe in their self too.
Do You Believe In Me?
This video describe about the oration by a child. The auditorium is very crowded. Then the moderator calling one of the candidate. And then applause from the audience is heard so noisy. He speak very loudly. He said “Do you believe in me?”, and then the audience answered “Yes!!” loudly too.
He persuade them to believe on what he said. He is a great boy.
Perbaikan:
Do You Believe In Me?
The video describe a boy who has named Dalton Sherman, he came from Charles Rice Learning Center. He believe in him. He ask “Do you believe in me?” He can stand up in there, fearless and talk to all of the audience. He can do anything, create anything, dream anything, become anything, because they believe in him.
He give a question to Dallas ISD, “Do you believe that us can graduate ready for college or the workplace?” Next week, he and Charles Rice Learning Center will come in their school and what he need from them is to believe that he can reach their highest potential.
He hope that they can believe in their colleagues. They came because they wanted to make a difference. Whether they are a counselor or a librarian or whatever, he persuade that he need them.
He persuade that they must believe in their self. He persuade that what they are doing is not just his generation but that of his children and his children’s children.
He persuade all of the students in Dallas, that we need you. We need them now more than ever. We need they, ladies and gentlemen, we need them to know that what they are doing is very important job in the city today.
He persuade again and again to the audience that he believe in him so they must believe in their self too.
Translate into English
Assignment 5
BAB 6
TRANSFORMASI GEOMETRI
Materi dalam bab ini:
A. Pengertian Transformasi
B. Translasi (Pergeseran)
C. Rotasi (Perputaran)
D. Refleksi (Pencerminan)
E. Dilatasi (Perkalian)
F. Komposisi dari beberapa Transformasi Geometri beserta Matriks Transformasinya
Kompetensi Dasar:
1. Menggunakan translasi dan transformasi geometri yang mempunyai matriks dalam pemecahan masalah.
2. Menentukan komposisi dari transfomasi geometri beserta matriks transformasinya.
Geometri Transformasi meliputi : Translasi, Rotasi, Refleksi, Dilatasi untuk menentukan Komposisi.
A. Pengertian Transformasi
Untuk memindahkan suatu titik atau bangun pada bidang dapat dilakukan dengan menggunakan transformasi. Transformasi geometri adalah bagian dari geometri yang membicarakan perubahan, baik perubahan letak maupun bentuk dari penyajiannya didasarkan dengan gambar dan matriks. Transformasi pada bidang ada empat macam, yaitu :
1. Translasi (Pergeseran)
2. Refleksi (Pencerminan)
3. Rotasi (Perputaran)
4. Dilatasi (Perkalian)
Translasi, refleksi dan rotasi disebut transformasi isometri artinya, bayangan(bangun hasil) dan transformasi itu kongruen dengan bangun semula. Pada dilatasi, bayangan(bangun hasil) dari transformasi sebangun dengan bangun semula, yaitu diperkecil atau diperbesar.
B. Translasi (Pergeseran)
Translasi atau pergeseran adalah perpindahan titik-titik pada bidang dengan jarak dan arah tertentu. Jarak dan arah tertentu itu diwakili oleh ruas garis berarah(vektor), misalnya AB atau dengan suatu pasangan bilangan, misalnya (a,b) yang berarti absis titik ditambah dengan a, sedangkan ordinatnya ditambah dengan b.
Suatu translasi dengan komponen T=(a,b) akan memetakan titik A(x1,y1) ke titik A’(x1+a,y1+b) yang dinotasikan dengan :
T=(a,b)=A(x1,y1)
A’(x1+a,y1+b)
Catatan:
Apabila suatu fungsi atau kurva mengalami pergeseran, fungsi persamaan atau kurva tersebut akan berubah.
Contoh 1:
Carilah bayangan (peta) titik A(4,3) dan B(5,-1) oleh translasi T=(3,2)!
Jawab:
T=(3,2) : A(4,3), A’(4+3,3+2)=A’(7,5)
T=(3,2) : B(5,-1), B’(5+3,-1+2)=B’(8,1)
Contoh 2:
Oleh translasi T=(a,b), titik A(1,-2) dipetakan menjadi titik A’(4,3).
Tentukan T!
Jawab:
T=(a,b) : A(1,-2), A’(1+A,-2+B)=A’(4,3)
1+a = 4 -2+b = 3
a=3 b=5
Jadi , T=(3,5)
Latihan 1
1. Diketahui segitiga PQR dengan titik-titik sudut P(1,2), Q(5,0) dan R(3,6).
a. Tentukan bayangan segitiga PQR oleh translasi T=(1,-3)!
b. Gambarlah segitiga PQR dan bayangannya pada sistem koordinat cartesius!
c. Tentukan persamaan garis yang melalui titik P dan R!
d. Tentukan persamaan bayangan garis yang melalui titik P dan R!
2. Suatu translasi (a,b) mengakibatkan titik P(5,-2) menjadi titik P’(2,2b)
a. Tentukan nilai a dan b!
b. Dengan translasi (a,b), tentukan bayangan titik A(2,1) dan B(-4,2)!
3. Tentukan persamaan bayangan dari:
a. garis y=2x+3 , bila ditranslasikan oleh (3,-4)
b. garis 3x+4y=6, bila ditranslasikan oleh (3,-4)
4. Tentukan persamaan peta bila lingkaran:
a. x2+y2=16 mengalami pergeseran (-1,1)
b. x2+y2-4x+8y-5=0 mengalami pergeseran (2,-1)
5. Tentukan persamaan bayangan dari kurva di bawah ini!
a. Parabola y2=4x, jika titik puncaknya ditranslasikan oleh (2,1)
b. Elips x2/9+y2/4 =1, jika titik pusatnya ditranslasikan oleh (1,-2)
C. Rotasi (Perputaran)
Rotasi atau perputaran ditentukan oleh:
1. Pusat perputaran
2. besar sudut putar, dan
3. arah sudut putar
Suatu perputaran mempunyai arah positif bila arah perputaran itu berlawanan arah dengan arah putar jarum jam. Arah negatif terjadi bila arah perputaran itu searah dengan arah putaran jarum jam.
Titik P’(x’,y’) adalah peta dari titik P(x,y) oleh rotasi terhadap O sebesar α radian. Misalkan koordinat cartesius P(x,y) ditulis dengan koordinat kutub menjadi titik P(r,θ) maka terdapat hubungan:
x = r cos θ
y = r sin θ
Titik P(x,y) = P(r,θ) diputar sebesar α radian menghasilkan P’(x’,y’) = P’(r,θ+α) sehingga:
x’ = r cos (θ+α)
= r (cos θ cos α – sin θ sin α)
= r cos θ cos α – r sin θ sin α
= x cos α – y sin c………......(1)
y’ = r sin (θ + α)
= r ( sin θ cos α + cos θ sin α)
= r sin θ cos α + r cos θ sin α
= y cos α + x sin α
= x sin α + y cos α………….(2)
Ingat!
Dari (1) dan (2) diperoleh:
cos (θ ± α ) = cos θ = sin θ sin α
sin (θ ± α ) = sin θ cos α ± cos θ sin α
Dari (1) dan (2) diperoleh:
x’ = x cos α – y sin α
y’ = x sin α + y cos α
Jadi, diperoleh rumus:
Bila titik P(x,y) diputar dengan pusat O sebesar α radian akan menjadi titik P’(x’,y’) dimana:
(x’,y’) = (x cos α – y sin α, x sin α + y cos α)
Dengan cara yang sama akan diperoleh, jika titik P(x’,y’) diputar dengan pusat A(a,b) sebesar α radian akan menjadi titik P’(x’,y’) dengan:
( x’-a , y’-b ) = ((x-a)( cos α - sin α ) , (y-b)( sin α + cos α ))
Latihan 2:
1. Tulislah matriks-matriks yang berkaitan dengan rotasi pusat O dan sudut putar sebesar:
a.+ 30˚ d.+120˚
b.+ 45˚ e.+180˚
c.+60˚ f .-150˚
2. Tentukan bayangan titik-titik sudut segitiga ABC dengan A(1,2), B(5,-2) dan C(5,6), jika dirotasikan dengan pusat O sebesar:
a. 2/3 π radian
b.3/2 π radian
3. Tentukan bayangan titik-titik berikut, jika dirotasikan dengan pusat (-1,4) dan sudut putar sebesar 90˚!
a. (2,5) c. (6,-8)
b. (-3,4) d. (-4,-6)
4. Tentukan persamaan bayangan dari hal berikut!
a. Garis 2x+y+3=0 , jika dirotasikan dengan pusat O sebesar 90˚
b. Parabola y2=4x+3, jika dirotasikan dengan pusat O sebesar 180˚
c. Lingkaran x2+y2+2x-4y=20 , jika dirotasikan dengan pusat O sebesar -90˚
5. Tentukan persamaan peta dari persamaan-persamaan berikut jika diputar dengan pusat P (-1,2) dan sudut putar 90˚!
a. Garis y =3x-1
b. Elips x2+2y=8
c. Hiperbola 4x2-9y2=1
D. Refleksi (Pencerminan)
Suatu refleksi ditentukan oleh suatu garis tertentu sabagai sumbu pencerminan yang perlu diperhatikan pada pencerminan adalah jarak bangun mula-mula ke sumbu pencerminan sama dengan jarak bangun bayangannya ke sumbu pencerminan. Sehingga pada pencerminan akan diperoleh sebagai berikut:
1. Bangun bayangan sebangun dengan bangun mula-mula.
2. Keliling bangun bayangannya sama dengan keliling bangun mula-mula.
3. Luas bangun bayangannya sama dengan luas bangun mula-mula.
Segitiga ABC dicerminkan terhadap garis α menghasilkan segitiga A’B’C’.
Maka:
AP = A’P
BQ = B’Q
CR = C’R
Pencerminan terhadap Sumbu Koordinat, Titik dan Garis Linear
Misalkan titik P(x,y) dicerminkan terhadap sumbu X menghasilkan titik P’(x’,y’)
Maka:
x’=x=(1)(x)+(0)(y)
y’=-y=(0)(y)+(-1)(y)
(x’,y’)=(x,-y)
Jadi, titik P(x,y) : P’(x’,y’) ditentukan oleh persamaan matriks:
(x’,y’)=(x,-y)
Dengan cara yang sama seperti contoh diatas, dapat diperoleh matriks-matriks pencerminan sebagai berikut:
1. (x,-y) : Pencerminan terhadap sumbu X
2. (-x,y) : Pencerminan terhadap sumbu Y
3. (-x,-y) : Pencerminan terhadap titik asal O
4. (y,x) : Pencerminan terhadap garis y=x
5. (-y,-x) : Pencerminan terhadap garis y=-x
6. ((x cos 2α + y sin 2α),(x sin 2α – y cos 2α)) : Pencerminan terhadap garis y = x tan α
Ingat!
Pencerminan suatu titik terhadap garis y=k, x=(k=0) dan garis linear selain y=x dan y-x, tidak dapat ditulis dalam bentuk matriks.
Catatan:
Cara menghafalkan matriks transformasi diatas, cukup kita ambil 2 titik pada sumbu X dan sumbu Y, dua titik yang dimaksud adalah titik A(1,0) dan B(0,1). Selanjutnya tinggal memperhatikan pernyataan soalnya.
Contoh:
Matriks yang bersesuaian dengan pencerminan terhadap sumbu X
Titik A(1,0) : A’(-1,0)
B(0,1) : B’(0,1)
In English
Chapter 6
THE GEOMETRY TRANSFORMATION
The material in the chapter is:
G. The Definition of The Transformation
H. The Translation
I. The Rotation
J. The Reflection
K. The Dilatation
L. The Composition from the several transformation of geometry and the matrix transformation
The Basic Competence :
3. Used the translation and the transformation of geometry that had the matrix in the solution.
4. Determined the composition from the transformation of geometry and the matrix transformation.
The Geometry Transformation involve of : The Translation, The Rotation, The Reflection, The Dilatation which to determine The Composition.
C. The Definition of The Transformation
To move a point or figure on the plane could be carried out with used the transformation. The transformation of geometry was part of the geometry that discussed the transform, both the transform in the location and the form and this presentation was based with the picture and matrix. The transformation on the plane had four sorts, that is :
1. The translation
2. The reflection
3. The rotation
4. The dilatation
The translation, the reflection and the rotation were acknowledged as the isometry transformation, meaning that the shadow (figure results) of the transformation was congruent. In the dilatation, the shadow (the figure results) of the transformation was congruent by the figure originally, that is reduced or dilated.
D. The Translation
The translation or the shift was the move of the points to the plane with the distance and the certain direction. The distance and the certain direction were represented by the part of the directed line (the vector),for example AB or with a pair number, for example (a,b) that was significant absis the point was increased with a ,where as ordinat him was increased with b.
A translation with the component of T=(a,b) will map the point of A(x1,y1) to the point of A’(x1+a,y1+b) that
T=(a,b)=A(x1,y1)
A’(x1+a,y1+b)
Note:
If a function or the curve experienced the shift, the regulation function or this curve will change.
Example 1:
Look for the shadow (the map) the point of A(4,3) and B(5,-1) by the translation of T=(3,2)!
Solution:
T=(3,2) : A(4,3), A’(4+3,3+2)=A’(7,5)
T=(3,2) : B(5,-1), B’(5+3,-1+2)=B’(8,1)
Example 2:
By the translation of T=(a,b)the point of A(1,-2) was mapped to the point of A’(4,3).
Determine T!
Solution:
T=(a,b) : A(1,-2), A’(1+A,-2+B)=A’(4,3)
1+a = 4 -2+b = 3
a=3 b=5
So , T=(3,5)
Exercises 1
1. Given that triangle PQR with the vertex of P(1,2), Q(5,0) and R(3,6).
a. Determine the triangle PQR shadow by the translation of T=(1,-3)!
b. Drawing triangle PQR and his shadow in the co-ordinate system cartesius!
c. Determine the line equation that through the point P and R!
d. Determine the line shadow equation that through the point P and R!
2. A translation (a,b) resulted in the point P(5,-2) become to point P’(2,2b)
a. Determine the value of a and b!
b. With the translation (a,b) , determine the point shadow of A(2,1) and B(-4,2)!
3. Determine the shadow equation of:
a. The line y=2x+3 , if it is translation by (3,-4)
b. The line 3x+4y=6 , if it is translation by (3,-4)
4. Determine the map equation when the circle:
a. x2+y2=16 experienced the shift (-1,1)
b. x2+y2-4x+8y-5=0 experienced the shift (2,-1)
5. Determine the shadow equation from the curve below this!
a. The parabola y2=4x, if this peak point is translated by (2,1)
b. The ellipse x2/9+y2/4 =1 , if this central point is translated by (1,-2)
C. The Rotation
The rotation was determined by:
1. The relation centre
2. Measure of the angle turned , and
3. The angle direction turned
A rotation had the positive direction when the rotation direction was against whit the direction turned o’clock. The negative direction happened when the rotation direction a direction with the direction turned o’clock.
The point P’(x’,y’) was the map from the point P(x,y) by the rotation toward O as big as α radian. For example the co-ordinate cartesius the point P(x,y) was written with the pole co-ordinate to the point P(r,θ) so that was gotten by relations:
x = r cos θ
y = r sin θ
The point P(x,y) = P(r,θ) was turned as big as α radian produced P’(x’,y’) = P’(r,θ+α) so that:
x’ = r cos (θ+α)
= r (cos θ cos α – sin θ sin α)
= r cos θ cos α – r sin θ sin α
= x cos α – y sin c………......(1)
y’ = r sin (θ + α)
= r ( sin θ cos α + cos θ sin α)
= r sin θ cos α + r cos θ sin α
= y cos α + x sin α
= x sin α + y cos α………….(2)
Remember!
From (1) and (2) were received:
cos (θ ± α ) = cos θ = sin θ sin α
sin (θ ± α ) = sin θ cos α ± cos θ sin α
From (1) and (2) were received:
x’ = x cos α – y sin α
y’ = x sin α + y cos α
So, was received by the formula:
When the point P(x,y) was turned with the centre O as big as α radian will become the point P’(x’,y’) where:
(x’,y’) = (x cos α – y sin α, x sin α + y cos α)
With the same way will receive , if the point P(x’,y’) was turned with the centre A(a,b) as big as α radian will become the point P’(x’,y’) with
( x’-a , y’-b ) = ((x-a)( cos α - sin α ) , (y-b)( sin α + cos α ))
Exercises 2:
1. Write the matrix which correlated with the centre rotation O and the angle turned as big as:
a.+ 30˚ d.+120˚
b.+ 45˚ e.+180˚
c.+60˚ f .-150˚
2. Determine the angle points shadow of triangle ABC with A(1,2), B(5,-2) and C(5,6), if it is rotate with the center O as big as:
a. 2/3 π radian
b.3/2 π radian
3. Determine the point shadow below this, if it is rotated with the centre (-1,4) and the angle turned as big as 90˚!
a.(2,5) c.(6,-8)
b.(-3,4) d.(-4,-6)
4. Determine the shadow equation from this below case!
a. The line 2x+y+3=0 , if it is rotated with the centre O as big as 90˚
b. The parabola y2=4x+3 , if it is rotated with the centre O as big as 90˚
c. The circle x2+y2+2x-4y=20 , if it is rotated with the centre O as big as 90˚
5. Determine the map equation from the equation below, if it is turned with the P (-1,2) and the angle turned 90˚!
a. The line y =3x-1
b. The ellips x2+2y=8
c. The hyperbola 4x2-9y2=1
D. The Reflection
A reflection was determined by a certain line as the reflection axis which need to be paid attention on the reflection was the model distance at first to the reflection axis was the same as with the model distance of his shadow to the reflection axis so on the reflection will be received as below:
1. The shadow model is congruent with the model at first
2. The circumference his shadow model is the same as with the circumference of the model at first
3. The area of his shadow model is the same as with the area of the model at first.
The triangle ABC was reflected toward the line α produced A’B’C’ triangle.
So:
AP = A’P
BQ = B’Q
CR = C’R
The Reflection toward the co-ordinate axis , the point and the linear line .
Let the point P(x,y) is reflected toward the axis x produced the paint P’(x’,y’)
So:
x’=x=(1)(x)+(0)(y)
y’=-y=(0)(y)+(-1)(y)
(x’,y’)=(x,-y)
So, the point P(x,y) : P’(x’,y’) is determined by the matrix equation:
(x’,y’)=(x,-y)
With the same way like the example above, can be received that the reflection matrix as below:
7. (x,-y) : as the reflection toward the axis x
8. (-x,y) : as the reflection toward the axis y
9. (-x,-y) : as the reflection toward the first point O
10. (y,x) : as the reflection toward the line y=x
11. (-y,-x) : as the reflection toward the line y=-x
12. ((x cos 2α + y sin 2α),(x sin 2α – y cos 2α)) : as the reflection toward the line y = x tan α
Remember!
The reflection a point toward the line y=k, x=(k=0) and the linear line aside from y=x and y-x, can not be written in the matrix form.
Note:
The procedure to memorize the transformation matrix above, only with put two points on the axis x and axis y, two points which mean the point A(1,0) and B(0,1). Furthermore remain pay attention to the problem statement.
Example:
The matrix which according to the reflection toward the axis x.
The point A(1,0) : A’(-1,0)
B(0,1) : B’(0,1)
BAB 6
TRANSFORMASI GEOMETRI
Materi dalam bab ini:
A. Pengertian Transformasi
B. Translasi (Pergeseran)
C. Rotasi (Perputaran)
D. Refleksi (Pencerminan)
E. Dilatasi (Perkalian)
F. Komposisi dari beberapa Transformasi Geometri beserta Matriks Transformasinya
Kompetensi Dasar:
1. Menggunakan translasi dan transformasi geometri yang mempunyai matriks dalam pemecahan masalah.
2. Menentukan komposisi dari transfomasi geometri beserta matriks transformasinya.
Geometri Transformasi meliputi : Translasi, Rotasi, Refleksi, Dilatasi untuk menentukan Komposisi.
A. Pengertian Transformasi
Untuk memindahkan suatu titik atau bangun pada bidang dapat dilakukan dengan menggunakan transformasi. Transformasi geometri adalah bagian dari geometri yang membicarakan perubahan, baik perubahan letak maupun bentuk dari penyajiannya didasarkan dengan gambar dan matriks. Transformasi pada bidang ada empat macam, yaitu :
1. Translasi (Pergeseran)
2. Refleksi (Pencerminan)
3. Rotasi (Perputaran)
4. Dilatasi (Perkalian)
Translasi, refleksi dan rotasi disebut transformasi isometri artinya, bayangan(bangun hasil) dan transformasi itu kongruen dengan bangun semula. Pada dilatasi, bayangan(bangun hasil) dari transformasi sebangun dengan bangun semula, yaitu diperkecil atau diperbesar.
B. Translasi (Pergeseran)
Translasi atau pergeseran adalah perpindahan titik-titik pada bidang dengan jarak dan arah tertentu. Jarak dan arah tertentu itu diwakili oleh ruas garis berarah(vektor), misalnya AB atau dengan suatu pasangan bilangan, misalnya (a,b) yang berarti absis titik ditambah dengan a, sedangkan ordinatnya ditambah dengan b.
Suatu translasi dengan komponen T=(a,b) akan memetakan titik A(x1,y1) ke titik A’(x1+a,y1+b) yang dinotasikan dengan :
T=(a,b)=A(x1,y1)
A’(x1+a,y1+b)
Catatan:
Apabila suatu fungsi atau kurva mengalami pergeseran, fungsi persamaan atau kurva tersebut akan berubah.
Contoh 1:
Carilah bayangan (peta) titik A(4,3) dan B(5,-1) oleh translasi T=(3,2)!
Jawab:
T=(3,2) : A(4,3), A’(4+3,3+2)=A’(7,5)
T=(3,2) : B(5,-1), B’(5+3,-1+2)=B’(8,1)
Contoh 2:
Oleh translasi T=(a,b), titik A(1,-2) dipetakan menjadi titik A’(4,3).
Tentukan T!
Jawab:
T=(a,b) : A(1,-2), A’(1+A,-2+B)=A’(4,3)
1+a = 4 -2+b = 3
a=3 b=5
Jadi , T=(3,5)
Latihan 1
1. Diketahui segitiga PQR dengan titik-titik sudut P(1,2), Q(5,0) dan R(3,6).
a. Tentukan bayangan segitiga PQR oleh translasi T=(1,-3)!
b. Gambarlah segitiga PQR dan bayangannya pada sistem koordinat cartesius!
c. Tentukan persamaan garis yang melalui titik P dan R!
d. Tentukan persamaan bayangan garis yang melalui titik P dan R!
2. Suatu translasi (a,b) mengakibatkan titik P(5,-2) menjadi titik P’(2,2b)
a. Tentukan nilai a dan b!
b. Dengan translasi (a,b), tentukan bayangan titik A(2,1) dan B(-4,2)!
3. Tentukan persamaan bayangan dari:
a. garis y=2x+3 , bila ditranslasikan oleh (3,-4)
b. garis 3x+4y=6, bila ditranslasikan oleh (3,-4)
4. Tentukan persamaan peta bila lingkaran:
a. x2+y2=16 mengalami pergeseran (-1,1)
b. x2+y2-4x+8y-5=0 mengalami pergeseran (2,-1)
5. Tentukan persamaan bayangan dari kurva di bawah ini!
a. Parabola y2=4x, jika titik puncaknya ditranslasikan oleh (2,1)
b. Elips x2/9+y2/4 =1, jika titik pusatnya ditranslasikan oleh (1,-2)
C. Rotasi (Perputaran)
Rotasi atau perputaran ditentukan oleh:
1. Pusat perputaran
2. besar sudut putar, dan
3. arah sudut putar
Suatu perputaran mempunyai arah positif bila arah perputaran itu berlawanan arah dengan arah putar jarum jam. Arah negatif terjadi bila arah perputaran itu searah dengan arah putaran jarum jam.
Titik P’(x’,y’) adalah peta dari titik P(x,y) oleh rotasi terhadap O sebesar α radian. Misalkan koordinat cartesius P(x,y) ditulis dengan koordinat kutub menjadi titik P(r,θ) maka terdapat hubungan:
x = r cos θ
y = r sin θ
Titik P(x,y) = P(r,θ) diputar sebesar α radian menghasilkan P’(x’,y’) = P’(r,θ+α) sehingga:
x’ = r cos (θ+α)
= r (cos θ cos α – sin θ sin α)
= r cos θ cos α – r sin θ sin α
= x cos α – y sin c………......(1)
y’ = r sin (θ + α)
= r ( sin θ cos α + cos θ sin α)
= r sin θ cos α + r cos θ sin α
= y cos α + x sin α
= x sin α + y cos α………….(2)
Ingat!
Dari (1) dan (2) diperoleh:
cos (θ ± α ) = cos θ = sin θ sin α
sin (θ ± α ) = sin θ cos α ± cos θ sin α
Dari (1) dan (2) diperoleh:
x’ = x cos α – y sin α
y’ = x sin α + y cos α
Jadi, diperoleh rumus:
Bila titik P(x,y) diputar dengan pusat O sebesar α radian akan menjadi titik P’(x’,y’) dimana:
(x’,y’) = (x cos α – y sin α, x sin α + y cos α)
Dengan cara yang sama akan diperoleh, jika titik P(x’,y’) diputar dengan pusat A(a,b) sebesar α radian akan menjadi titik P’(x’,y’) dengan:
( x’-a , y’-b ) = ((x-a)( cos α - sin α ) , (y-b)( sin α + cos α ))
Latihan 2:
1. Tulislah matriks-matriks yang berkaitan dengan rotasi pusat O dan sudut putar sebesar:
a.+ 30˚ d.+120˚
b.+ 45˚ e.+180˚
c.+60˚ f .-150˚
2. Tentukan bayangan titik-titik sudut segitiga ABC dengan A(1,2), B(5,-2) dan C(5,6), jika dirotasikan dengan pusat O sebesar:
a. 2/3 π radian
b.3/2 π radian
3. Tentukan bayangan titik-titik berikut, jika dirotasikan dengan pusat (-1,4) dan sudut putar sebesar 90˚!
a. (2,5) c. (6,-8)
b. (-3,4) d. (-4,-6)
4. Tentukan persamaan bayangan dari hal berikut!
a. Garis 2x+y+3=0 , jika dirotasikan dengan pusat O sebesar 90˚
b. Parabola y2=4x+3, jika dirotasikan dengan pusat O sebesar 180˚
c. Lingkaran x2+y2+2x-4y=20 , jika dirotasikan dengan pusat O sebesar -90˚
5. Tentukan persamaan peta dari persamaan-persamaan berikut jika diputar dengan pusat P (-1,2) dan sudut putar 90˚!
a. Garis y =3x-1
b. Elips x2+2y=8
c. Hiperbola 4x2-9y2=1
D. Refleksi (Pencerminan)
Suatu refleksi ditentukan oleh suatu garis tertentu sabagai sumbu pencerminan yang perlu diperhatikan pada pencerminan adalah jarak bangun mula-mula ke sumbu pencerminan sama dengan jarak bangun bayangannya ke sumbu pencerminan. Sehingga pada pencerminan akan diperoleh sebagai berikut:
1. Bangun bayangan sebangun dengan bangun mula-mula.
2. Keliling bangun bayangannya sama dengan keliling bangun mula-mula.
3. Luas bangun bayangannya sama dengan luas bangun mula-mula.
Segitiga ABC dicerminkan terhadap garis α menghasilkan segitiga A’B’C’.
Maka:
AP = A’P
BQ = B’Q
CR = C’R
Pencerminan terhadap Sumbu Koordinat, Titik dan Garis Linear
Misalkan titik P(x,y) dicerminkan terhadap sumbu X menghasilkan titik P’(x’,y’)
Maka:
x’=x=(1)(x)+(0)(y)
y’=-y=(0)(y)+(-1)(y)
(x’,y’)=(x,-y)
Jadi, titik P(x,y) : P’(x’,y’) ditentukan oleh persamaan matriks:
(x’,y’)=(x,-y)
Dengan cara yang sama seperti contoh diatas, dapat diperoleh matriks-matriks pencerminan sebagai berikut:
1. (x,-y) : Pencerminan terhadap sumbu X
2. (-x,y) : Pencerminan terhadap sumbu Y
3. (-x,-y) : Pencerminan terhadap titik asal O
4. (y,x) : Pencerminan terhadap garis y=x
5. (-y,-x) : Pencerminan terhadap garis y=-x
6. ((x cos 2α + y sin 2α),(x sin 2α – y cos 2α)) : Pencerminan terhadap garis y = x tan α
Ingat!
Pencerminan suatu titik terhadap garis y=k, x=(k=0) dan garis linear selain y=x dan y-x, tidak dapat ditulis dalam bentuk matriks.
Catatan:
Cara menghafalkan matriks transformasi diatas, cukup kita ambil 2 titik pada sumbu X dan sumbu Y, dua titik yang dimaksud adalah titik A(1,0) dan B(0,1). Selanjutnya tinggal memperhatikan pernyataan soalnya.
Contoh:
Matriks yang bersesuaian dengan pencerminan terhadap sumbu X
Titik A(1,0) : A’(-1,0)
B(0,1) : B’(0,1)
In English
Chapter 6
THE GEOMETRY TRANSFORMATION
The material in the chapter is:
G. The Definition of The Transformation
H. The Translation
I. The Rotation
J. The Reflection
K. The Dilatation
L. The Composition from the several transformation of geometry and the matrix transformation
The Basic Competence :
3. Used the translation and the transformation of geometry that had the matrix in the solution.
4. Determined the composition from the transformation of geometry and the matrix transformation.
The Geometry Transformation involve of : The Translation, The Rotation, The Reflection, The Dilatation which to determine The Composition.
C. The Definition of The Transformation
To move a point or figure on the plane could be carried out with used the transformation. The transformation of geometry was part of the geometry that discussed the transform, both the transform in the location and the form and this presentation was based with the picture and matrix. The transformation on the plane had four sorts, that is :
1. The translation
2. The reflection
3. The rotation
4. The dilatation
The translation, the reflection and the rotation were acknowledged as the isometry transformation, meaning that the shadow (figure results) of the transformation was congruent. In the dilatation, the shadow (the figure results) of the transformation was congruent by the figure originally, that is reduced or dilated.
D. The Translation
The translation or the shift was the move of the points to the plane with the distance and the certain direction. The distance and the certain direction were represented by the part of the directed line (the vector),for example AB or with a pair number, for example (a,b) that was significant absis the point was increased with a ,where as ordinat him was increased with b.
A translation with the component of T=(a,b) will map the point of A(x1,y1) to the point of A’(x1+a,y1+b) that
T=(a,b)=A(x1,y1)
A’(x1+a,y1+b)
Note:
If a function or the curve experienced the shift, the regulation function or this curve will change.
Example 1:
Look for the shadow (the map) the point of A(4,3) and B(5,-1) by the translation of T=(3,2)!
Solution:
T=(3,2) : A(4,3), A’(4+3,3+2)=A’(7,5)
T=(3,2) : B(5,-1), B’(5+3,-1+2)=B’(8,1)
Example 2:
By the translation of T=(a,b)the point of A(1,-2) was mapped to the point of A’(4,3).
Determine T!
Solution:
T=(a,b) : A(1,-2), A’(1+A,-2+B)=A’(4,3)
1+a = 4 -2+b = 3
a=3 b=5
So , T=(3,5)
Exercises 1
1. Given that triangle PQR with the vertex of P(1,2), Q(5,0) and R(3,6).
a. Determine the triangle PQR shadow by the translation of T=(1,-3)!
b. Drawing triangle PQR and his shadow in the co-ordinate system cartesius!
c. Determine the line equation that through the point P and R!
d. Determine the line shadow equation that through the point P and R!
2. A translation (a,b) resulted in the point P(5,-2) become to point P’(2,2b)
a. Determine the value of a and b!
b. With the translation (a,b) , determine the point shadow of A(2,1) and B(-4,2)!
3. Determine the shadow equation of:
a. The line y=2x+3 , if it is translation by (3,-4)
b. The line 3x+4y=6 , if it is translation by (3,-4)
4. Determine the map equation when the circle:
a. x2+y2=16 experienced the shift (-1,1)
b. x2+y2-4x+8y-5=0 experienced the shift (2,-1)
5. Determine the shadow equation from the curve below this!
a. The parabola y2=4x, if this peak point is translated by (2,1)
b. The ellipse x2/9+y2/4 =1 , if this central point is translated by (1,-2)
C. The Rotation
The rotation was determined by:
1. The relation centre
2. Measure of the angle turned , and
3. The angle direction turned
A rotation had the positive direction when the rotation direction was against whit the direction turned o’clock. The negative direction happened when the rotation direction a direction with the direction turned o’clock.
The point P’(x’,y’) was the map from the point P(x,y) by the rotation toward O as big as α radian. For example the co-ordinate cartesius the point P(x,y) was written with the pole co-ordinate to the point P(r,θ) so that was gotten by relations:
x = r cos θ
y = r sin θ
The point P(x,y) = P(r,θ) was turned as big as α radian produced P’(x’,y’) = P’(r,θ+α) so that:
x’ = r cos (θ+α)
= r (cos θ cos α – sin θ sin α)
= r cos θ cos α – r sin θ sin α
= x cos α – y sin c………......(1)
y’ = r sin (θ + α)
= r ( sin θ cos α + cos θ sin α)
= r sin θ cos α + r cos θ sin α
= y cos α + x sin α
= x sin α + y cos α………….(2)
Remember!
From (1) and (2) were received:
cos (θ ± α ) = cos θ = sin θ sin α
sin (θ ± α ) = sin θ cos α ± cos θ sin α
From (1) and (2) were received:
x’ = x cos α – y sin α
y’ = x sin α + y cos α
So, was received by the formula:
When the point P(x,y) was turned with the centre O as big as α radian will become the point P’(x’,y’) where:
(x’,y’) = (x cos α – y sin α, x sin α + y cos α)
With the same way will receive , if the point P(x’,y’) was turned with the centre A(a,b) as big as α radian will become the point P’(x’,y’) with
( x’-a , y’-b ) = ((x-a)( cos α - sin α ) , (y-b)( sin α + cos α ))
Exercises 2:
1. Write the matrix which correlated with the centre rotation O and the angle turned as big as:
a.+ 30˚ d.+120˚
b.+ 45˚ e.+180˚
c.+60˚ f .-150˚
2. Determine the angle points shadow of triangle ABC with A(1,2), B(5,-2) and C(5,6), if it is rotate with the center O as big as:
a. 2/3 π radian
b.3/2 π radian
3. Determine the point shadow below this, if it is rotated with the centre (-1,4) and the angle turned as big as 90˚!
a.(2,5) c.(6,-8)
b.(-3,4) d.(-4,-6)
4. Determine the shadow equation from this below case!
a. The line 2x+y+3=0 , if it is rotated with the centre O as big as 90˚
b. The parabola y2=4x+3 , if it is rotated with the centre O as big as 90˚
c. The circle x2+y2+2x-4y=20 , if it is rotated with the centre O as big as 90˚
5. Determine the map equation from the equation below, if it is turned with the P (-1,2) and the angle turned 90˚!
a. The line y =3x-1
b. The ellips x2+2y=8
c. The hyperbola 4x2-9y2=1
D. The Reflection
A reflection was determined by a certain line as the reflection axis which need to be paid attention on the reflection was the model distance at first to the reflection axis was the same as with the model distance of his shadow to the reflection axis so on the reflection will be received as below:
1. The shadow model is congruent with the model at first
2. The circumference his shadow model is the same as with the circumference of the model at first
3. The area of his shadow model is the same as with the area of the model at first.
The triangle ABC was reflected toward the line α produced A’B’C’ triangle.
So:
AP = A’P
BQ = B’Q
CR = C’R
The Reflection toward the co-ordinate axis , the point and the linear line .
Let the point P(x,y) is reflected toward the axis x produced the paint P’(x’,y’)
So:
x’=x=(1)(x)+(0)(y)
y’=-y=(0)(y)+(-1)(y)
(x’,y’)=(x,-y)
So, the point P(x,y) : P’(x’,y’) is determined by the matrix equation:
(x’,y’)=(x,-y)
With the same way like the example above, can be received that the reflection matrix as below:
7. (x,-y) : as the reflection toward the axis x
8. (-x,y) : as the reflection toward the axis y
9. (-x,-y) : as the reflection toward the first point O
10. (y,x) : as the reflection toward the line y=x
11. (-y,-x) : as the reflection toward the line y=-x
12. ((x cos 2α + y sin 2α),(x sin 2α – y cos 2α)) : as the reflection toward the line y = x tan α
Remember!
The reflection a point toward the line y=k, x=(k=0) and the linear line aside from y=x and y-x, can not be written in the matrix form.
Note:
The procedure to memorize the transformation matrix above, only with put two points on the axis x and axis y, two points which mean the point A(1,0) and B(0,1). Furthermore remain pay attention to the problem statement.
Example:
The matrix which according to the reflection toward the axis x.
The point A(1,0) : A’(-1,0)
B(0,1) : B’(0,1)
Minggu, 21 Desember 2008
Representing The Video of Learning Mathematics
Video 1
Solve The Problem Graph Math
Given that the next problem is question 13 on page 411.
13. The figure above shows the graph of y=g(x). If the function h is defined by h(x)=g(2x)+2. What is the value of h(1)?
Solution:
We looking for h(1)…
The first information is we can show the graph of y=g(x) in the co-ordinate system.
The next information is h(x)=g(2x)+2. We looking for h(1), so we can substitute x=1 on the equation, so h(1)=g(2)+2. Now we have g(2), we can look on the graph, if x=2 so y=1, so we can get g(2)=1.
So h(x)=g(2x)+2
h(1)=g(2)+2
h(1)=1+2=3
The another function with no function problem, this is question 13, the another question 13 on page 534.
13. Let the function f be defined by f(x)=x+1. If 2f(p)=20, what is the value of f(3p)?
Solution :
We looking for f(3p)… What is f when x = 3p?
Now first information is f(x)=x+1, and then the next piece information is 2f(p)=20, because the question is what is f when x = 3p, so we can start from this equation down here: 2f(p)=20, we can divide by 2, so f(p)=10, and then f(p) is just what is function f(x) of f(p),so we can speak f(p)=p+1=10, then p=9. But this is not the right answer, we looking for f when x = 3p, so if we take this and answer from x = 3p,with p=9, so x=27. Then we can substitute x in the equation f(x)=x+1, so f(27)=27+8=28. And the last answer of looking for is 28.
The next question is function with no function problem, we just take it step by step, this is question 17 on page 412.
17. In the xy–coordinate plane, the graph of x equals y square minus 4 intersect line l at (0,p) and (5,t). What is the greatest possible value of the slope of l?
Solution:
We will be looking for greatest m…
The graph x equals y square minus 4 intersect line l at (0,p) and (5,t), when x=0, y=p and when x = 5, y=t. And the question is what is the greatest slope. So what do we know about the slope? We know that m equals (y-y1)/(x2-x1). The slope is going to be m=(t-p)/5. Numerator is t-p. We can play again the value into this equation intersect at (0,p) and (5,t).
Video 2
Factoring Polynomials
One way to find factors of polynomials is to formed the algebraic long division. For example lets try and see (x-3) is a factor of (x cube minus seven x minus six)? When dividing (x-3) into (x cube minus seven x minus six). First step of the problem make a long division problem for elementary school. There is you dividing (x – 3) into (x cube plus zero x square minus seven x minus six). Zero in there because is no second degree term. Now you must ask yourself what times x give you x cube? Of, course it is x square, and then you multiply (x-3) by x square, which give you (x cube minus 3 x square),you subtract (x cube plus zero x square) to get 3 x square. Bring it down to next term negative seven x, you have (3 x square minus 7 x). Now we begin again, dividing (x-3) into (3 x square minus 7 x). Just looking at the first term 3 x square dividing x is 3x. Multiply (x-3) by 3x. We can get (3 x square minus 9 x). Subtracting you have (2x minus 6). Dividing (x-3) into (2 x minus 6) which equals 2 and without a remainder. So the solution for a long division problem (x cube plus zero x square minus seven x minus six) dividing by (x-3) is (x square plus 3 x plus 2). Since (x-3) divide into (x cube minus 7 x minus 6) with no remainder. Then (x-3) is a factor of (x cube minus 7 x minus 6). The conclusion is (x square plus 3 x plus 2) is also a factor of (x cube minus 7 x minus 6). We know now that (x cube minus 7 x minus 6) equals (x-3) times (x square plus 3 x plus 2). The quadratic expression (x square plus 3 plus 2) can be factored into (x+1) times (x+2). So, (x cube minus 7 x minus 6) equals (x-3) times (x+1) times (x+2). Substitution (x cube minus 7x minus 6) to zero,so we get 0=(x-3)(x+1)(x+2) Thus either (x-3)=0 or (x+1)=0 or (x +2)=0. Solving all this equation of x we get x=3, x=-1, x=-2. The roots of (x cube minus 7 x minus 6) are (3, -1, -2).
Remember!!
¨ 3rd degree equation have 3 roots
¨ Quadratic (2nd degree) equations always have at most 2 roots
¨ A 4th degree equation would have 4 or fewer roots and so on.
¨ The degree of polynomials equation always limits the number of roots.
Long division process for 3rd order polynomial:
1. Find a partial quotient of x square, by dividing x into x cube to get x square.
2. Multiply x square by the divisor and subtract the product from the dividend.
3. Repeat the process until you either “clear it out“ or reach a remainder.
Video 3
Graph of A Rational Function
Graph of a rational function which can have discontinuities because has polynomial in the denominator.
Is possible value x divide by 0
Example :
f(x) equals (x+2) over (x-1)
when x=1,so f(1)= (1+2) over (1-1) equals 3 over zero. That is bad idea.
f(1)=(1+2) over zero is break in function graph.
f(x) = ( x+ 2) over (x-1)
for x=0, f(0)=(0+2) over (0-1) equals -2
insert x=1,so f(1)=1+2 over (1-1) equals 3 over 0, it is impossible.
Rational functions don’t always work in this way! Take graph f(x) = 1 over ( x square plus 1 ). Not all rational functions will give zero in denominator because of the (+1) is never zero.
Rational functions denominator can be zero!
Polynomial have smooth and unbroken curve and for rational function x : zero in the denominator that impossible situation. There is no value for the function.
A break can show up in 2 ways. A simply type break is missing point on the graph.
Example :
y= (x square minus x minus 6) over (x minus 3)
The graph loose like this if x = 3, so (3 square minus 3 minus 6) over (3 minus 3) equals 0 over 0. That is not possible, not feasible, and not allowed.
So that is no way if x = 3. This is a typical example to the missing point syndrome.
y = (3 square minus 3 minus 6) over (3 minus 3) equals 0 over 0
When you see result of 0 over 0 and also tell you direction be possible factor top and bottom of rational function and simplify.
For example:
y = (x square minus x minus 6) over (x minus 3) equals (x minus 3) times (x plus 2) all over (x minus 3) equals (x plus 2).
Video 4
Invers Function
F(x,y) = 0
Function y = f(x) : VLT
1.1 function x=g(y) : HLT : invertible
Look at the graph of function y equals x square
y=2x-1
Look at the graph function!
y=2x-1
1+x=2x
x=1
2x-1=y
2x=y+1
x=1/2(y+1)
x=(1/2)y+1/2
This is a equation, so:
y=(1/2)x+1/2
We have function
f(x)=2x-1
g(x)=(1/2)x+1/2
What is the value of f(g(x))?
So, f(g(x))=2g(x)-1
=2((1/2)x+1/2)-1
=x+1-1 = x
What is the value of g(f(x))?
So, g(f(x))=(1/2)f(x)+1/2
=(1/2)(2x-1)+1/2
=x-(1/2)+(1/2) = x
g=f-1
f(g(x))=f(f-1(x))=x
g(f(x))=f-1(f(x))=x
Example:
y=(x-1)/(x+2)
Solution:
y(x+2)=x-1
yx+2y=x-1
yx-x=-1-2y
(y-1)x=-1-2y
x=(-1-2y)/(y-1)
y=(-1-2x)/(x-1)
@ x=0, y=-1
@ y=0,
-1-2x=0
-2x=1
x=-(1/2)
V Asym x=1
H Asym @ y=-2
Look at the graph function!
Translate The Mathematics Articles
1. English into Indonesian
JOHANNES KEPLER (1571-1630)
Johannes Kepler was born in Weil der Stadt in Swabia, in southwest Germany. His paternal grandfather, Sebald Kepler, was a respected craftsman who served as mayor of the city; his maternal grandfather, Melchior Guldenmann, was an innkeeper and mayor of the nearby village of Eltingen. His father, Heinrich Kepler, was "an immoral, rough and quarrelsome soldier," according to Kepler, and he described his mother in similar unflattering terms. From 1574 to 1576 Johannes lived with his grandparents; in 1576 his parents moved to nearby Leonberg, where Johannes entered the Latin school. In 1584 he entered the Protestant seminary at Adelberg, and in 1589 he began his university education at the Protestant university of Tübingen. Here he studied theology and read widely. He passed the M.A. examination in 1591.
Kepler's teacher in the mathematical subjects was Michael Maestlin (1550-1635). Maestlin was one of the earliest astronomers to subscribe to Copernicus's heliocentric theory, although in his university lectures he taught only the Ptolemaic system. Only in what we might call graduate seminars did he acquaint his students, among whom was Kepler, with the technical details of the Copernican system.
In 1594 Kepler accepted an appointment as professor of mathematics at the Protestant seminary in Graz (in the Austrian province of Styria). He was also appointed district mathematician and calendar maker. Kepler remained in Graz until 1600, when all Protestants were forced to convert to Catholicism or leave the province, as part of Counter Reformation measures. For six years, Kepler taught arithmetic, geometry (when there were interested students), Virgil, and rhetoric. In his spare time he pursued his private studies in astronomy and astrology. In 1597 Kepler married Barbara Müller. In that same year he published his first important work, The Cosmographic Mystery, in which he argued that the distances of the planets from the Sun in the Copernican system were determined by the five regular solids, if one supposed that a planet's orbit was circumscribed about one solid and inscribed in another.
Except for Mercury, Kepler's construction produced remarkably accurate results. Because of his talent as a mathematician, displayed in this volume, Kepler was invited by Tycho Brahe to Prague to become his assistant and calculate new orbits for the planets from Tycho's observations. Kepler moved to Prague in 1600.
Kepler served as Tycho Brahe's assistant until the latter's death in 1601 and was then appointed Tycho's successor as Imperial Mathematician, the most prestigious appointment in mathematics in Europe. He occupied this post until, in 1612, Emperor Rudolph II was deposed. In Prague Kepler published a number of important books. In 1604 Astronomia pars Optica ("The Optical Part of Astronomy") appeared, in which he treated atmospheric refraction but also treated lenses and gave the modern explanation of the workings of the eye; in 1606 he published De Stella Nova ("Concerning the New Star") on the new star that had appeared in 1604; and in 1609 his Astronomia Nova ("New Astronomy") appeared, which contained his first two laws (planets move in elliptical orbits with the sun as one of the foci, and a planet sweeps out equal areas in equal times). Whereas other astronomers still followed the ancient precept that the study of the planets is a problem only in kinematics, Kepler took an openly dynamic approach.
In 1610 Kepler heard and read about Galileo's discoveries with the spyglass. He quickly composed a long letter of support which he published as Dissertatio cum Nuncio Sidereo ("Conversation with the Sidereal Messenger"), and when, later that year, he obtained the use of a suitable telescope, he published his observations of Jupiter's satellites under the title Narratio de Observatis Quatuor Jovis Satellitibus ("Narration about Four Satellites of Jupiter observed"). These tracts were an enormous support to Galileo, whose discoveries were doubted or denied by many. Both of Kepler's tracts were quickly reprinted in Florence. Kepler went on to provide the beginning of a theory of the telescope in his Dioptrice, published in 1611.
During this period the Keplers had three children (two had been born in Graz but died within months), Susanna (1602), who married Kepler's assistant Jakob Bartsch in 1630, Friedrich (1604-1611), and Ludwig (1607-1663). Kepler's wife, Barbara, died in 1612. In that year Kepler accepted the position of district mathematician in the city of Linz, a position he occupied until 1626. In Linz Kepler married Susanna Reuttinger. The couple had six children, of whom three died very early.
In Linz Kepler published first a work on chronology and the year of Jesus's birth, In German in 1613 and more amply in Latin in 1614: De Vero Anno quo Aeternus Dei Filius Humanam Naturam in Utero Benedictae Virginis Mariae Assumpsit (Concerning the True Year in which the Son of God assumed a Human Nature in the Uterus of the Blessed Virgin Mary"). In this work Kepler demonstrated that the Christian calendar was in error by five years, and that Jesus had been born in 4 BC, a conclusion that is now universally accepted. Between 1617 and 1621 Kepler published Epitome Astronomiae Copernicanae ("Epitome of Copernican Astronomy"), which became the most influential introduction to heliocentric astronomy; in 1619 he published Harmonice Mundi ("Harmony of the World"), in which he derived the heliocentric distances of the planets and their periods from considerations of musical harmony. In this work we find his third law, relating the periods of the planets to their mean orbital radii.
In 1615-16 there was a witch hunt in Kepler's native region, and his own mother was accused of being a witch. It was not until late in 1620 that the proceedings against her ended with her being set free. At her trial, her defense was conducted by her son Johannes.
1618 marked the beginning of the Thirty Years War, a war that devastated the German and Austrian region. Kepler's position in Linz now became progressively worse, as Counter Reformation measures put pressure on Protestants in the Upper Austria province of which Linz was the capital. Because he was a court official, Kepler was exempted from a decree that banished all Protestants from the province, but he nevertheless suffered persecution. During this time Kepler was having his Tabulae Rudolphinae ("Rudolphine Tables") printed, the new tables, based on Tycho Brahe's accurate observations, calculated according to Kepler's elliptical astronomy. When a peasant rebellion broke out and Linz was besieged, a fire destroyed the printer's house and shop, and with it much of the printed edition. Soldiers were garrisoned in Kepler's house. He and his family left Linz in 1626. The Tabulae Rudolphinae were published in Ulm in 1627.
Kepler now had no position and no salary. He tried to obtain appointments from various courts and returned to Prague in an effort to pry salary that was owed him from his years as Imperial Mathematician from the imperial treasury. He died in Regensburg in 1630. Besides the works mentioned here, Kepler published numerous smaller works on a variety of subjects.
Taken from http://galileo.rice.edu/sci/kepler.html
In Indonesian:
JOHANNES KEPLER (1571-1630)
Johannes Kepler lahir di Weil der Stadt di Swebia, di bagian barat daya Jerman. Kakek dari pihak ayahnya, Sebald Kepler, seorang tukang yang dihormati yang menjabat sebagai walikota di kotanya; kakek dari pihak ibunya, Melchior guldenmann, seorang pengurus rumah penginapan dan walikota di dekat desa Eltingen. Ayahnya, Heinrich Kepler, “adalah seorang yang tuna susila, kasar, prajurit yang suka bertengkar,” berdasarkan Kepler, dan dia mendiskripsikan ibunya juga serupa tak menyenangkannya. Dari tahun 1574-1576 Johannes tinggal bersama eyangnya; tahun 1576 orang tuanya pindah ke dekat Leonberg, dimana johannes masuk ke Sekolah Latin. Tahun 1584 dia masuk Sekolah Menengah Protestan di Adelberg, dan tahun 1589 dia memulai pendidikannya di Universitas Protestan Tubingen. Disana dia belajar teologi. Dia lulus ujian M.A. tahun 1591.
Michael Maestlin (1550-1635) adalah guru matematika Kepler. Maestlin adalah salah satu dari pendahulu ahli astronomi yang menggunakan teori Heliosentrik Copernicus, meskipun di kuliahnya dia hanya mengajar sistem Ptolemaic. Hanya murid lulusan sekolah menengah saja yang dia perkenalkan, yaitu diantaranya Kepler, dengan rincian teknis dari Sistem Copernican.
Tahun 1594 Kepler diangkat sebagai profesor matematika di Sekolah Menengah Protestan di Graz (di Austria, Syria). Dia juga menunjuk daerah matematika dan membuat kalender. Kepler tinggal di Graz sampai tahun 1600, ketika seluruh Protestan dipaksa untuk berpindah agama ke Katolik atau meninggalkan daerah itu, bagian dari Counter Reformation. Selama 6 tahun Kepler mengajari aritmatik, geometri(ketika disana banyak murid yang tertarik) dan berpidato. Di waktu luangnya dia mengajarkan astronomi dan astrologi pada murid-murid privatnya. Tahun 1597 Kepler menikah dengan Barbara Muller. Di tahun yang sama dia juga menerbitkan karya besarnya yang pertama yaitu “The Cosmographic Mystery”, dimana dia membantah bahwa jarak planet dari matahari di sistem Copernican yang dideterminkan oleh 5 zat padat biasa, jika satu diandaikan orbit sebuah planet yang dibatasi oleh satu zat padat dan menggoreskan pada yang lain.
Kecuali air raksa, hasil gagasan Kepler sungguh luar biasa akurat. Karena bakatnya sebagai ahli matematika yang menunjukkan prestasi maka Kepler diundang oleh Tycho Brahe ke Prague untuk menjadi asistennya dan menghitung orbit baru untuk planet-planet dari pengamatan Tycho. Kepler pindah ke Prague tahun 1600.
Kepler membantu sebagai asisten Tycho Brahe sampai kemudian Brahe meninggal pada tahun 1601 dan kemudian menunjuk pengganti Tycho sebagai kaisar ahli matematika, pangkat paling tinggi di matematika Eropa. Dia sibuk dengan kedudukannya sampai tahun 1612, Kaisar Rudolph II dipecat. Di Prague Kepler menerbitkan buku-buku penting. Tahun 1604 Astronomia pars Optica (“The Optical Part of Astronomy”) muncul, yang menyuguhkan pembiasan atmosfer tetapi juga menyuguhkan lensa-lensa dan memberikan penjelasan modern dari cara kerja mata; tahun 1606 dia menerbitkan De Stella Nova (“Concerning The New Star”); dan pada tahun 1609 Astronomia Nova (“New Astronomy”) muncul, yang berisi dua hukum pertama(planet-planet bergerak dengan lintasan elips dengan matahari berada di salah satu fokusnya dan Luas daerah yang disapu pada selang waktu yang sama akan selalu sama). Mengingat ahli astronomi lain masih mengikuti aturan kuno yang mempelajari planet-planet bermasalah hanya pada ilmu geraknya, Kepler mengambil pendekatan dinamik.
Tahun 1610 Kepler mendengar dan membaca penemuan Galileo. Dia menggubah sebuah surat panjang dukungan yang dia terbitkan dengan cepat sebagai Dissertatio cum Nuncio Sidereo (“Conversation with the Sidereal Messenger”) dan tahun kemudian, dia menemukan penggunaan dari teleskop, dia menerbitkan hasil pengamatannya yaitu satelit Jupiter dengan judul Narratio de Observatis Quatuor Jovis Satellitibus (“Narration about Four Satellites of Jupiter observed”). Sistem itu sangat mendukung Galileo, orang yang penemuannya diragukan atau ditolak oleh beberapa orang. Keduanya yang dicetak ulang dengan cepat di Florence. Kepler pergi untuk menyediakan permulaan dari teori teleskop pada Dioptrice-nya, yang diterbitkan pada tahun 1611.
Selama periode itu Kepler mempunyai tiga orang anak (dua lahir di Graz tetapi meninggal), Susanna(1602), yang menikah dengan asisten Kepler Jacob Bartsch tahun 1630, Friedrich(1604-1611), dan Ludwig(1607-1663). Istri Kepler, Barbara, meninggal pada tahun 1612. Pada tahun itu Kepler menerima jabatan daerah ahli matematika di kota Linz, sampai tahun 1626. Di Linz Kepler menikah dengan Susanna Reuttinger. Pasangan itu menghasilkan 6 anak, 3 meninggal.
Di Linz, Kepler menerbitkan karya yang pertama “Kronologi” dan “Tahun Kelahiran Jesus”, di Jerman pada tahun 1613 dan lebih banyak di Latin tahun 1614: De Vero Anno quo Aeternus Dei Filius Humanam Naturam in Utero Benedictae Virginis Mariae Assumpsit (“Concerning The True Year in which The Sound of God assumed a Human Nature in the Uterus of The Blessed Virgin Mary”). Kepler menunjukkan bahwa kalender orang Kristen terdapat kesalahan selama lima tahun, Jesus lahir di 4 BC, sebuah kesimpulan hingga sekarang dapat diterima secara universal. Diantara tahun 1617-1621 Kepler menerbitkan Epitomo Astronomiae Copernicanae (“Epitome of Copernican Astronomi”) yang menjadi pengaruh perkenalan Heliosentrik Astronomi; tahun 1619 dia menerbitkan Harmonic Mundi (“Harmony of The World”), dimana dia mendapat jarak Heliosentrik dari planet-planet dan periodenya dari mempertimbangkan pertunjukan musik harmoni. Di pekerjaan itu dia menemukan hukum ketiganya, hubungan periode planet untuk rata-rata orbitnya.
Tahun 1615 seorang tukang sihir wanita memburu daerah Kepler, dan ibunya dituduh jadi tersangkanya. Tahun 1620 terjadi perlawanan. Pada akhirnya, anaknya Johannes Kepler, yang menyelamatkan dirinya.
Tahun 1618 permulaan dari 30 tahun perang yang merusak Jerman dan daerah Austria. Jabatan Kepler di Linz menjadi lebih buruk, sebagai Counter Reformation mempengaruhi kaum Protestan di bagian Austria ibukota Linz karena dia adalah pegawai negeri pengadilan Kepler dibebaskan dari dekrit tetapi walaupun demikian dia tetap dianiaya. Saat itu Kepler mencetak Tabulae Rudolphinae (“Rudolphine Tables”), tabel baru berdasarkan pengamatan akurat Tycho Brahe, menghitung keserasian astronomi Kepler. Ketika terjadi pemberontakan, api memusnahkan tempat percetakan, toko, dan banyak edisi cetakan. Prajurit yang ada di dalam kota mengamankan rumah Kepler. Dia dan keluarganya meninggalkan Linz tahun 1626. The Tabulae Rudolphinae diterbitkan di Ulm tahun 1627.
Hingga akhirnya Kepler kehilangan jabatannya dan tidak mendapatkan gajinya. Dia mencoba untuk mendapatkan jabatannya kembali dari berbagai pengadilan dan dikembalikan di Prague untuk membongkar gaji yang dihutang selama bertahun-tahun sebagai kaisar ahli matematika dari perbendaharaan kekaisaran. Dia meninggal di Regensburg tahun 1630. Kepler juga menerbitkan karya-karya yang lebih kecil pada berbagai macam subjek lainnya.
2. Indonesian into English
ILMU KOMPUTER
Disiplin Ilmu Komputer sudah muncul sejak era tahun 1940, seiring dengan berpadunya teori algoritma dan logika matematika, serta ditemukannya komputer elektronik dengan kemampuan penyimpanan program. Adalah Alan Turing dan Kurt Godel, yang pada tahun 1930-an berhasil memadukan algoritma, logika, dan penghitungan matematika serta merealisasikannya dalam sebuah alat atau rule system. Prinsip algoritma yang digunakan adalah dari Ada Lovelace, yang dikembangkan 60 tahun sebelumnya.
Penemu algoritma sendiri yang tercatat dalam sejarah awal adalah dari seorang yang bernama Abu Abdullah Muhammad Ibnu Musa Al-Khwarizmi. Al-Khwarizmi adalah seorang ahli matematika dari
Kemudian John Von Neumann mendemonstrasikan salah satu karya fenomenalnya pada tahun 1945, yaitu sebuah arsitektur komputer yang disebut "von Neumann machine", dimana program disimpan di memori. Arsitektur komputer inilah yang kemudian digunakan oleh computer modern sampai sekarang.
Tahun 1960 adalah babak baru dimulainya formalisasi Ilmu Komputer. Jurusan Ilmu Komputer pada universitas-universitas mulai marak dibangun. Disiplin ilmu baru ini kemudian terkenal dengan sebutan Ilmu Komputer (Computer Science), Teknik Komputer (Computer Engineering), Komputing (Computing), atau Informatika (Informatics).
Dewasa ini banyak sekali peneliti yang mencoba membuat kajian dan melakukan pendefinisian terhadap Ilmu Komputer. Bagaimanapun juga, dasar Ilmu Komputer adalah matematika dan engineering (teknik). Matematika menyumbangkan metode analisa, dan engineering menyumbangkan metode desain pada bidang ini.
CSAB [3] (Computing Sciences Accreditation Board, http://www.csab.org) membuat definisi menarik tentang Ilmu Komputer:
Ilmu Komputer adalah ilmu pengetahuan yang berhubungan dengan computer dan komputasi. Di dalamnya terdapat teoritika, eksperimen, dan pendesainan komponen, serta termasuk didalamnya hal-hal yang berhubungan dengan: teori-teori untuk memahami komputer device, program, dan system Eksperimen untuk pengembangan dan pengetesan konsep Metodologi desain, algoritma, dan tool untuk merealisasikannya,
Metode analisa untuk melakukan pembuktian bahwa realisasi sudah sesuai
dengan requirement yang diminta.
Beberapa definisi lain yang lebih abstrak adalah:
Ilmu Komputer adalah ilmu yang mempelajari tentang representasi pengatahuan (knowledge representation) dan implementasinya. Atau definisi Ilmu Komputer adalah ilmu yang mempelajari tentang abstraksi dan bagaimana mengendalikan kekompleksan.
Kita bisa simpulkan dari persamaan pemakaian terminologi dan hakekat makna dalam definisi yang digunakan para peneliti diatas, bahwa:
Ilmu Komputer adalah ilmu pengetahuan yang berisi tentang teori, metodologi, desain dan implementasi, berhubungan dengan komputasi, komputer, dan algoritmanya dalam perspektif perangkat lunak (software) maupun perangkat keras (hardware).
Ilmu Komputer adalah ilmu yang mempelajari tentang komputer.
Ilmu Komputer adalah ilmu yang mempelajari tentang bagaimana menulis
program komputer.
Taken from:
In English:
COMPUTER SCIENCE
Discipline Computer Science have appeared since 1940, in step with associated of algorithm theory and mathematics logic and also found computer electronik with program storage capacity. He is Alan Turing and Kurt Gudel, which in 1930 successfully combine algorithm, logic and calculation of mathematics and also realize it in an implement or rule system. Algorithm principle that used is from Ada Lovelace, which is developed 60 years ago.
The finder of algorithm which noted in the first history from Abu Abdullah Muhammad Ibn Musa Al Kwarizmi. He is mathematician from Uzbekistan who life in era 770-840 M. In west literature, he more famous with predicate Algorizm. Algorithm came from this. In the act, computer analog was created by Vannevar Bush in 1920, end then computer electronik which developed by Howard Aiken ands Konrad Zuse in 1930.
Then John Von Neumann demonstrate the one of his fenomenal work in 1945, a computer architectur which then used by computer modern until now.
In 1960, new generation is started for fomalitation of Computer Science. The way of Computer Science in the university was established. Discipline this new science then famous with named Computer Science, Computer Engineering, Computing or Informatics.
Nowdays, oodles scientist who try meeting and definite for Computer Science. However, Computer Science principle is mathematics and engineering. Mathematics dedicated analysis method and engineering dedicated desain method for this sector.
Computing Sciences Accreditation Board definite about Computer Science: Computer Science is science that related with computer and computation. There are theoritics, experiment and component desain, and also included the matter that related with: theory-theory to comprehend device computer, program and experiment system for development and examination desain methodology concept, algorithm and tool to realize it, Analysis method for proof that realize is according with requirement that required.
Several other definition for more abstract is: Computer Science is science that learn knowledge representation and this implementation, or definition of Computer Science is science which learn abstraction and how to drive complexation.
We can generalize from terminology usage equation and substance significance definition, that: Computer Science is science that included theory, methodology, desain and implementation, which related with computation, computer and algorithm on the software or hardware.
Computer Science is science to learn how to write computer program.
Computer Science is science to learn using of the applications of computer.
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