Jatijajar Cave

As a child, I grew up in a village located in Central Java, Indonesia. the name is Jatijajar village, which was linked after the name of a cave in the village. 

 

to get into the cave and the village, people can go from Jogja by taking bus in the route to Purwokerto or Cilacap. Stop by in Gombong or Ijo, and just take the small bus to Jatijajar. the bus will take you to just in the front gate of the cave. in the neighborhood, you can also visit other beautiful places like Barat cave, which is just minutes from Jatijajar cave and Petruk cave, few kilometers from Jatijajar cave. Both caves are still undeveloped caves which means you have to hire local people with torchs, since there isn’t any lamps there, to walk with you there. Beautiful things about those two caves is they serve amazing river on the inside of the cave. Then, a little bit further, there are beach and fish market where you can buy fresh seafood.

Considering to stay more than one day, few cheap hotels are available. about half kilo from the cave, there is Puspita hotel which available with AC and non-AC rooms. you can easily transfer from the hotel to all places by bus or becak, an Indonesian transportation.

 

Any comments or questions are always welcomed.

Beste, 

Nur Chasanah

 

 

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Newton vs Leibniz: the developing of calculus

During the growth of calculus, there are some important inventions. The basic problems of calculus that are used nowadays are mostly based on Newton and Leibniz’s inventions in 17th centuries by two of most greatest mathematician, Isaac Newton (1643-1727) and Gottfried Leibniz (1646-1716). The interesting fact is that Leibniz’s invention, which is so called calculus, was claimed by Newton to be the same with his thought about fluxion (Newton’s version of calculus) in earlier period before Leibniz had invented it.

LEIBNIZ and NEWTON

Leibniz invented the calculus in period between 1672 and 1676 and published his theorem at 1684 for the first time. Newton’s discovery about almost the same theorem came before Leibniz, but he never published anything about it until few mentioning on his book Principia Mathematica in 1687. Again, Newton did not publish his full thought about calculus until 1704 after Leibniz’s invention had been presented in L’Hospital’s book. Their theories indeed similar, but they are famous of their own invention in calculus. Leibniz is well known with his differential and integral calculus. Meanwhile, Newton is famous with his fluxion and binomial theorem to find the coefficient of the various powers of x.

 

LEIBNIZ’S CALCULUS

Nowadays calculus about differential and integral were born from the concept given by Leibniz, who was firstly curious about philosophy. Before introducing the term calculus differential in the late of 17th century, Leibniz himself used the term methodus tangentium directa. Meanwhile, Calculus integral itself was called as methodus tengentium inversa. Hence, the most influential actor behind his successful on his mathematics career was Christian Huygens from whom he mostly got encourage to study mathematics during his stay in Paris from 1672 to 1676.

The notations for integral () and differential () were given through a note on 26 October 1675. Not long after that, Leibniz settled in Hanover and became more focus on calculus. Therefore, in 1682 he announced the existence of his calculus and soon started to publish them on 1684. This paper opened the modern period of calculus. Many translations on some languages are followed by the successful of his theorem. It considered as a fast development of Leibniz calculus.

Leibniz publication contains several rules on differentiation. It needs to keep in mind that Leibniz used the term differentials, not derivatives. He gave some rules that absorbed in school calculus nowadays, such as

Moreover, Leibniz also stated the condition for maximum or minimum and for a point of inflection. These conditions have been used nowadays and introduced in school to solve problems like finding extreme or stationary point(s) of a curve, whether it is maximum, minimum or inflection point.

In the relation with Leibniz’s integral, he had published a paper, Supplementum geometriae dimensoriae … similiterque multiplex construction lineae ex data tangentium conditione, where he explained the inverse relation between differentials and integrals by means of a figure. This statement is very important in the field of calculus and is applied mostly in either mathematics or physic.

In summary, Leibniz had contributed many of his thought for the improvement of standard and basic calculus, which we are using nowadays. His contribution is not only on the understanding of concept and invention of some formula and condition, but also name of the terms.

 

NEWTON’S CALCULUS

According to the history, Newton was first interested in the growing of mathematics. However, he then moved his concentration to physic. Newton started to work on mathematics in 1664 under Barrow at Cambridge. One of his earliest sources on calculus was Latin edition by F. van Schooten of the Geometrie of Descartes which derived him to the used of instead of (derived from Leibniz) which we are using now.

In terms of calculus differentials and integrals, Newton had discovered similar things with Leibniz, but wrote them with different terms and ways. In his book, Principia Mathematica, newton introduced moments, which is the same concept as differentials by Leibniz. Still in the same very famous book, Newton firstly explained some rules of fluxions (Newton’s name for calculus), in a section dealing with the motion of bodies that move against resistance. He worked with problems resulted from the integration differential equations with his notations (he used and instead of and ).

However, there is still one thing in calculus that invented by Newton invented and definitely not ignorable, the binomial theorem which now being known as binomial newton. The main concept of binomial theorem is to determine the coefficient of the various powers of and/or (depends on the main function). In a sense, binomial theorem is used to find the coefficients of each variable combination in expanding of a function, such as . How to get the coefficient is by employing formula, so called combination formula. In some cases, expanding function using binomial theorem is useful.

In summary, beside undeniable fact that Newton had contributed nice thoughts on calculus, people do not really use his notations. However, his so called Binomial Newton remains important and useful for mathematicians, physicians and engineers.

 

INTERESTING FACT

  • Besides the fact that Leibniz published his calculus earlier then Newton did, conflict about which terms and notations were more appropriate to be used was solved in the early years of 19th century when leading mathematicians in English-speaking countries began to adopt Leibniz terms and notations, mainly through the impact of the work of Laplace. However, in minor case, Newton’s notations are still used in present day.
  • During Leibniz and Newton’s growing invention on calculus, they did send each other letters in 1676 through the help of Henry Oldenburg, the secretary of the Royal Society, which were published on the correspondence of Isaac Newton. However, Newton never replied to Leibniz after his second letter.
  • Both of them also had similar difficulty on their publication about differentiating the meaning of either differentials and differences or Newton’s moments (in calculus) and real moments. Moreover, they both were never very consistent in their explanations about differentials and moments.

  

Bibliography

Struik, D.J. A Source Book in Mathematics, 1200-1800. Massachusetts Institute of technology. Cambridge: Harvard University Press, 1969.

 minor taking from Katz, Victor J. History of Mathematics. third editionPearson New International Edition. printed in the USA: Pearson Education Limited, 2014.

Is it concrete (for students) ?

Journal article written by Koeno Gravemeijer which was published at Indonesia Mathematics Society-Journal in Mathematics Education (IndoMS-JME) is talking about how concrete are our manipulatives, and whether it does help the students or not. for me, this is one of the best article for us, as educators, to reflect ourselves. The aim of the article is to persuade the readers (educators) that our common way of making things concrete for students does not work, and that we had better try to follow the other way of making things concrete by trying to connect to what the students know.
As for trying to help students to understand, teachers often use manipulatives or models. However, teachers usually their point of view as an observer and design the ‘material concrete’ models or manipulatives to help the students to make connections with what the teachers know. Whether we notice or not, the way we make something concrete in everyday life is different with the way we do this in mathematics. In everyday life, we give an example which is familiar with others, just like designing models, based on what the teachers know. We may see an activity with such media is representing a material, but instead students do not see the connection between what they do (in activity) with the concept or the material. Hence, teachers and students seem to live in two worlds.
In mathematics, we should pay more attention on students’ point of view as a person whose in charge on solving the problems without bringing in all mathematical knowledge that we have. The article points on the way students see the situation. They have their common senses which are their knowledge. In the other hand, we have our knowledge about mathematics, which is in different world with students’ common sense (something real for students). Here we shall pay more attention. The classroom activity should construct a connection with students’ common senses. The manipulatives should become the scaffoldings (pancingan) for students’ common senses to the materials.
More about examples and explanation can be read on the online IndoMS-JME at http://jims-b.org/.

Using percentage bar to help students solving percentage problems

As i read the journal article by Frans van Galen and Dolly van Erde, it’s convinced me more about the advantages of using percentage bar to teach percentage to primary students. the article is talking about the result of a research conducted by them and 10 Indonesian master students who studied in the Netherland.

in the research, students were asked to solve percentage problem about discount on buying bike. the problem was a very standard problem which usually used in Indonesia also. however, the result showed how most of the students were still struggle with the problem. it is also mentioned from the article that students originally have tried to solve the problem ‘in their head’, but they ended up making mistakes.

A research was conducted to prove that percentage bar is effective. the article claims that based on the research, there are three advantages of the use of percentage bar:

1. by making percentage bar, students make a representation for themselves of the relations between what is given and what is asked

2. percentage bar offers scrap paper for the intermediate steps in the calculation process. it helps the students to compute

3. percentage bar offers a natural entry to calculating via 1%. this last advantage is commonly used in Indonesia also, but in some case, students get confuse by the steps: finding the value of 1% and then finding the value of what is asked. the use of percentage bar together with calculating 1% reduce students’ mistake and make it more clear for students

here are examples of percentage bars.

percentage bar11visuals_3-resized-600[1]

beside the percentage bar, students can also use percentage table or double number line to help them solving percentage problems. however, the article states that the author believe in percentage bar as the most effective approach in helping students to solve percentage problems.

Contoh Soal Problem Solving dan Rubrik Penyelesaian

 Rubrik biasanya dibuat sebagai acuan dalam menentukan skor atau guide dalam proses memberi arahan pada siswa. berikut ini diberikan contoh soal problem solving yang cocok untuk anak SD maupun SMP lengkap dengan rubriknya.

 

SOAL:

Riska melihat seekor katak yang terjatuh ke dalam sumur sedalam 11 m pada pukul 13.45 WIB. Karena penasaran, Riska melihat keadaan katak di bawah sumur.  Ternyata Katak itu sedang berusaha untuk keluar dari sumur dengan melompat dari batu ke batu yang menyusun sumur tersebut. Batu-batu itu memiliki tebal 25 cm.

Riska mengamati bahwa katak itu melompat setinggi 8 batu tetapi selalu terjatuh lagi 2 batu setiap kali melompat.

Kemudian Riska mengkira-kira berapa banyak lompatan yang dibutuhkan katak untuk bisa keluar dari sumur itu. Dapatkah kalian membantu Riska menemukan jawabannya?

Riska juga mengamati katak itu membutuhkan waktu 10 detik untuk melakukan lompatan, dan membutuhkan 1 menit untuk bersiap melompat. Berapa lama waktu yang dibutuhkan katak untuk berhasil keluar dari sumur itu? Pada pukul berapakah katak itu berhasil keluar dari sumur itu?”

 

RUBRIK

Diketahui

Perhitungan

 

 

Riska melihat katak berada di dalam

sumur pada pukul 13.45

Kedalaman sumur adalah 11 m

Tebal batu pada sumur 25 cm

Jumlah lapisan batu di antara katak dan ujung atas   sumur adalah 11 m dibagi 25 cm atau (1100/25) cm = 44 buah.

Dalam setiap lompatan,

Katak mampu melompat setinggi 8 batu,

Tetapi selalu tergelincir sedalam 2 batu.

 

Ditanyakan: jumlah lompatan katak

sampai dia bisa keluar dari sumur

Dalam hal ini, ada banyak cara yang dapat digunakan   oleh siswa. Cara yang paling mudah adalah dengan menggunakan ilustrasi gambar   atau dengan bantuan garis bilangan. Contoh teknik dalam menjawab soal ini   adalah sebagai berikut: jarak katak dari bagian atas sumur:

Sebelum melompat = 44 batu

Lompatan pertama = 44 – 8 + 2 = 38 batu

Lopatan kedua = 38 – 8 + 2 = 32 batu

Lompatan ketiga = 32 – 8 + 2 = 26 batu

Lompatan keempat = 26 – 8 + 2 = 20 batu

Lompatan kelima = 20 – 8 + 2 = 14 batu

Lompatan keenam = 14 – 8 + 2 = 8 batu

Lompatan ketujuh = 8 – 8 = 0 batu

Kesimpulan: karena   pada lompatan ketujuh setelah jarak dikurangi 8 batu sudah menghasilkan 0,   logikanya katak sudah sampai diujung sumur dan mampu berpegangan, jadi katak   tersebut tidak tergelincir. Oleh karena itu, jawaban untuk berapa jumlah   lompatan katak adalah 7.

Katak membutuhkan waktu:

10 detik untuk melakukan lompatan

1 menit untuk bersiap melompat

 

Ditanya: Berapa lama waktu yang

dibutuhkan katak untuk berhasil

keluar dari dalam sumur?

Katak melompat sebanyak 7 kali, artinya waktu yang   diperlukan untuk melakukan lompatan adalah 7 x 10 detik = 70 detik atau 1   menit 10 detik

Karena pada saat Siska melihat ke dalam sumur katak   sedang melakukan lompatan pertama, artinya katak melakukan 6 kali persiapan   melompat (yang dilakukan di antara setiap lompatan). Jadi katak menghabiskan   waktu sebanyak 6 x 1 menit = 6 menit

Total waktu yang dihabiskan katak sampai dia dapat   keluar dari sumur adalah 1 menit 10 detik + 6 menit = 7 menit dan 10 detik.

Ditanya: pukul berapakah katak

tersebut berhasil keluar dari dalam sumur?

Katak pertama kali diketahui berada dalam sumur pada   pukul 13.45, kemudian menghabiskan waktu 7 menit dan 10 detik sampai dia   mampu keluar dari dalam sumur. Jadi, katak tersebut keluar dari dalam sumur   pukul 13.45 + 00.07 = 13.52

Jawaban akhir

Katak mampu keluar dari dalam   sumur pada pukul 13.52

 

Proses diskusi kelompok menyelesaikan soal problem solving

<a href="” title=”Proses diskusi kelompok menyelesaikan soal problem solving”>Proses diskusi kelompok menyelesaikan soal problem solving

Video ini memperlihatkan proses diskusi kelompok bagaimana kelompok focus mencoba menyelesaikan persoalan problem solving. siswa-siswa yang berpartisipasi adalah siswa-siswi SDN 98 Palembang. soal yang diberikan adalah sebagai berikut:

Riska melihat seekor katak yang terjatuh ke dalam sumur sedalam 11 m pada pukul 13.45 WIB. Karena penasaran, Riska melihat keadaan katak di bawah sumur.  Ternyata Katak itu sedang berusaha untuk keluar dari sumur dengan melompat dari… batu ke batu yang menyusun sumur tersebut. Batu-batu itu memiliki tebal 25 cm.
Riska mengamati bahwa katak itu melompat setinggi 8 batu tetapi selalu terjatuh lagi 2 batu setiap kali melompat.
Kemudian Riska mengkira-kira berapa banyak lompatan yang dibutuhkan katak untuk bisa keluar dari sumur itu.

Dapatkah kalian membantu Riska menemukan jawabannya?

wawancara untuk memahi cara berpikir siswa dalam menyelesaikan soal

<a href=”” title=”wawancara untuk memahi cara berpikir siswa dalam menyelesaikan soal”>wawancara untuk memahi cara berpikir siswa dalam menyelesaikan soal

video ini memperlihatkan pada kita bahwa siswa yang terbiasa memecahkan soal formal, ketika diberi masalah sehari-hari yang sederhana, terkadang siswa kesulitan dalam menyelesaikannya. hal ini karena siswa tidak dibiasakan mengimplementasikan konsep dan ilmu yg mereka miliki pada permasalahan realistic dalam kehidupan sehari-hari mereka. inilah mengapa sangat penting bagi kita sebagai pendidik untuk tidak hanya menekankan pada operasi formal matematika, tetapi juga implementasinya dalam kehidupan sehingga siswa tidak lagi mempertanyakan fungsi dari belajar matematika.