Bonding actually comes after reactions, equations and stoichiometry in my program, but I reckon that I can explain equations better if I explain bonding first. Plus, as you've seen in my Spec posts, I don't always 100% care about doing stuff in the right order.
Back to the topic of bonding. I'm not talking about relationships here. I'm talking about chemical bonding. Oh, look, did I just make a really bad joke? Moving on...
There are three main types of bonds: ionic, covalent and metallic. Ionic bonds are generally between metal and non-metal elements, covalent bonds are generally between non-metal elements, and metallic bonds are between metal elements. Bet you couldn't work out the last one for yourself!
All right, lame jokes over. Time to talk about ionic bonding. In my last Chem post, Atomic Structure and the Periodic Table, I mentioned the formation of positive and negative ions. If you've forgotten, here it is, in a nutshell: groups 1, 2 and 13 elements lose electrons to become positive ions and groups 15, 16 and 17 elements gain electrons to become negative ions.
These ions become quite important in ionic bonding, hence the name "ionic bonding." You see, when the elements in group 1, 2 and 13 lose electrons, those electrons have to go somewhere. Similarly, the electrons gained by elements in groups 15, 16 and 17 have to come from somewhere. So it's only logical that the metals in the first groups lose their electrons to the non-metals in the latter groups. When they do this, an ionic bond is formed due to electrostatic attractions between the positive and negative ions formed- remember, opposites attract. In fact, opposites attract so damn well that it takes a lot of energy to separate them, resulting in high melting and boiling points.
Despite this, however, most ionically-bonded molecules are soluble in water because they break up into their ions. This is also partially due to the fact that water is a polar molecule and has one end that is slightly positive and another end which is slightly negative- more on this later. The positive ions are attracted to the slightly more negatively-charged end of the water molecule, and vice versa.
Ionically bonded molecules are also non-conductors in the solid form. This may sound surprising, given that they are made up of charged ions, but bear in mind that to let electricity flow these charged particles need to be able to move freely in order to carry the current. In the molten state, however, these particles are a lot more free to move, and thus many molten ionic compounds are conductors of electricity, as are ionic compounds in aqueous solution.
What about the hardness of ionic compounds? Well, ionic compounds generally exist in a lattice of ionically-bonded compounds. In these lattices, negative ions sit next to positive ions, and vice versa, so that the attractive forces between the oppositely-charged ions can hold the lattice together. The hardness of ionic compounds is due to the strength of the attractive forces throughout the lattice. However, when the lattice is bumped hard enough to knock some ions out of place, ions might end up next to other ions of like charges. The repulsive forces between like ions then cause the lattice to shatter. Therefore, ionically bonded compounds are hard, but brittle.
Sunday, September 30, 2012
Saturday, September 29, 2012
Atomic Structure and the Periodic Table
I'm going to go through this... dot point at a time.
One day, I should go back and add pictures to this blog post. I can't be bothered today though.
Compare the relative charge and relative masses of protons, neutrons and electrons
Atoms are made up of three main particles: protons, neutrons and electrons. (I'm pretty sure that these particles are actually made up of even smaller particles, like protons are made up of a positron and a neutron or something along those lines, but let's not go there.) Positively-charged protons and neutrally-charged neutrons have about the same mass and they stick together in the nucleus in the centre of the atom.
Surrounding the nucleus is an "electron cloud" where all of the electrons hang out, orbiting the nucleus or whatever it is they do there. Electrons are negatively charged and their mass is 1/2000th of that of a proton or a neutron. In an atom, the number of electrons is equal to the number of protons. This keeps the positive and negative charges equal and the overall charge on the atom is neutral.
Identify elements using their atomic number (Z)
The main difference between elements is how many protons are located in their nuclei. The atomic number (or Z) of an element is given by how many protons are in the nucleus of an atom. Therefore, the atomic number and the type of element are related. Any atom with an atomic number of 1 (and therefore only one proton in the nucleus) is a hydrogen atom. Any atom with an atomic number of 2 (and therefore two protons in the nucleus) is a helium atom. And so on.
Explain isotopes using their atomic number (Z) and mass number (A)
Although all atoms of one element have to have the same number of protons, they do not necessarily have to have the same number of neutrons. Atoms with the same number of protons and a different number of neutrons are called isotopes. Examples of isotopes include Carbon-12 and Carbon-14.
Mass number, or A, is the sum of the protons and neutrons of an element. On the periodic table, you might notice that the mass number of most elements is not a whole number. This is because it is the average mass of an atom of that element, taking into consideration all of the isotopes of that atom and how common they are.
Carbon has 6 protons. Carbon-12 has 6 protons and 6 neutrons (6 + 6 = 12). Similarly, Carbon-14 has 6 protons and 8 neutrons (6 + 8 = 14). The mass number of carbon, as listed on the periodic table, is 12.01. This is because Carbon-12 is much more common than Carbon-13 or Carbon-14.
Use the energy level or shell model of electron structure to write the electron configurations for the first twenty elements
As I said before, the electrons are located in an electron cloud which surrounds the nucleus. However, they aren't all located in the same place in this electron cloud. The electron cloud is divided up into different shells, or energy levels, which hold different numbers of electrons. The first energy level, K, can only hold 2 electrons. When it's full, the next lot of electrons start moving into the second energy level, L. L can only hold 8 electrons. When it's full, electrons start moving into the energy level M. M can actually hold up to 18 electrons, but after the 8th electron, subsequent electrons move into shell N instead. Shell M only gets filled up as you progress through the transition metals.
Writing electron configurations is simple. Just write the number of electrons in each shell. Let's take potassium as an example. Potassium has 19 electrons. The first two will go in energy level K, the next 8 in energy level L, the next 8 in energy level M and then the last one will go in energy level N even though M isn't full yet (due to that strange rule that I mentioned before). Therefore, the electron configuration of potassium is as follows:
2, 8, 8, 1
In 2AB Chemistry you only need to go up to Calcium, which is the element right before the transition metals, so you don't need to worry about backfilling shell M too much just yet.
Explain the relationship between position on the Periodic Table and number of valence electrons of elements in groups 1-2 and 13-18
Valence electrons are basically the electrons in the outer shell of an atom. They contribute to the chemical properties of an atom.
Electrons with one valence electron are located in group 1 of the periodic table, and electrons with two valence electrons are located in group 2. Then electrons with 3-8 valence electrons are located in groups 13-18 of the periodic table, respectively. The periodic table is actually quite an ingenious invention.
Explain the relationship between the number of valence electrons and chemical properties of elements in groups 1-2 and 13-18
As I mentioned before, valence electrons contribute to the chemical properties of an atom. You see, all atoms want to attain a stable configuration of a full outer shell (either 2 or 8 valence electrons). This is the same configuration as the elements in group 18.
Since the elements in group 18 already have the electron configuration that they want, they are relatively non-reactive and tend not to form bonds with other elements. Elements that are very close to having a full outer shell (Groups 1 and 17), however, are a different story.
Group 1 elements only have one valence electron. If they lose this valence electron, then they'll be happy. Group 1 elements with more electrons and therefore more shells tend to be more reactive as their one valence electron will be further away from the nucleus and therefore not as strongly affected by the positively-charged protons in the nucleus.
Group 17 elements have 7 valence electrons. If they gain just one electron, they'll be happy. Group 7 elements with fewer electrons and therefore fewer shells tend to be more reactive as the positively-charged protons can more easily attract electrons without too many electron clouds in the way.
Group 2 and 13 elements also lose electrons while group 15 and 16 elements gain electrons. Group 14 elements are a different story, however. It may seem that they can either lose or gain four electrons, which some do occasionally, but more often than not, they tend to share electrons with non-metal elements, forming extremely strong bonds.
Explain the formation of positive and negative ions for elements in groups 1-2 and 13-18
As I said before, group 1, 2 and 13 elements tend to lose 1, 2 or 3 electrons in order to get a full outer shell and become stable. The loss of electrons means that there are now more protons than electrons, and therefore more positive than negative charges. The atoms are now positively charged ions. If one electron is lost, the atom is a +1 ion. If two electrons are lost, the atom is a +2 ion and so on.
Elements in groups 15, 16 and 17, on the other hand, tend to gain 3, 2 or 1 electrons in order to get a full outer shell. Gaining electrons means the formation of negative ions rather than positive ions. Gaining 3 electrons makes a -3 ion, gaining 2 makes a -2 ion, and so on.
Classification of Matter
In Chem we were going through the Chemistry program to check what we remember and what we don't remember. We were meant to quiz each other in pairs, but I had another idea (i.e. writing it up on a blog). Here I go...
This post will be a quickie because the first heading on my program only has two dot points underneath it which kind of overlap anyway.
So... pure substances and mixtures. What's the difference?
Well, a pure substance is made up of entirely one thing- whether it be one element, or one compound. An example would be pure gold, which is made up entirely of the atom Au. Another example would be a salt crystal which is made up entirely of NaCl. A mixture is made up of multiple pure substances. An example would be a salt solution because that has both NaCl and H2O.
Pure substances and mixtures can be broken down further. Pure substances, as alluded to before, can be broken up into elements and compounds. Elements only have one type of atom (e.g. Fe, Au, Ag, C) while compounds have multiple different types of atoms (e.g. CO, NaCl). Meanwhile, mixtures can be broken down into homogeneous and heterogeneous mixtures. (By the way, the only way I was able to spell those two words correctly is because Google Chrome has spell check.)
Both kinds of mixtures are made up of multiple pure substances mixed together. The difference between homogeneous and heterogeneous mixtures, however, is whether or not the proportion between pure substances changes throughout the mixture. In a homogeneous mixture (homo = same), the proportion between pure substances is the same throughout the whole mixture. An example would be salt water- the ratio between salt and water is the same no matter which part of the mixture you're looking at. In a heterogeneous mixture, on the other hand, the proportion between pure substances might change throughout the mixture. An example would be a choc-chip biscuit. The ratio between chocolate chips and biscuit dough differs depending on which part of the biscuit you're looking at.
Generally, if you can see the parts which make up a mixture, it's probably a heterogeneous mixture. If you can't, it's probably a homogeneous mixture.
This post will be a quickie because the first heading on my program only has two dot points underneath it which kind of overlap anyway.
So... pure substances and mixtures. What's the difference?
Well, a pure substance is made up of entirely one thing- whether it be one element, or one compound. An example would be pure gold, which is made up entirely of the atom Au. Another example would be a salt crystal which is made up entirely of NaCl. A mixture is made up of multiple pure substances. An example would be a salt solution because that has both NaCl and H2O.
Pure substances and mixtures can be broken down further. Pure substances, as alluded to before, can be broken up into elements and compounds. Elements only have one type of atom (e.g. Fe, Au, Ag, C) while compounds have multiple different types of atoms (e.g. CO, NaCl). Meanwhile, mixtures can be broken down into homogeneous and heterogeneous mixtures. (By the way, the only way I was able to spell those two words correctly is because Google Chrome has spell check.)
Both kinds of mixtures are made up of multiple pure substances mixed together. The difference between homogeneous and heterogeneous mixtures, however, is whether or not the proportion between pure substances changes throughout the mixture. In a homogeneous mixture (homo = same), the proportion between pure substances is the same throughout the whole mixture. An example would be salt water- the ratio between salt and water is the same no matter which part of the mixture you're looking at. In a heterogeneous mixture, on the other hand, the proportion between pure substances might change throughout the mixture. An example would be a choc-chip biscuit. The ratio between chocolate chips and biscuit dough differs depending on which part of the biscuit you're looking at.
Generally, if you can see the parts which make up a mixture, it's probably a heterogeneous mixture. If you can't, it's probably a homogeneous mixture.
Projection Vectors
In that same worksheet that the question about closest approach came from, there was also a question about projection vectors.
Find the projection of 4i - 3j onto the vector 12i + 8j.
First I had to find out what a projection vector was. So I asked my Spec teacher.
With the vectors 4i - 3j and 12i + 8j, as illustrated below:
Find the projection of 4i - 3j onto the vector 12i + 8j.
First I had to find out what a projection vector was. So I asked my Spec teacher.
With the vectors 4i - 3j and 12i + 8j, as illustrated below:
The projection vector is the purple part on the next diagram:
The projection of a vector onto another is the effect of that vector in the direction of the other vector. (I'm not 100% sure what this means, but I'll get there eventually.)
As can be seen from the diagram, you can draw a line between the 4i - 3j vector and a particular point on the 12i + 8j vector which is perpendicular to the 12i + 8j vector. (It's a bit like those closest approach questions.) The particular point where the lines intersect has a position vector identical to the projection vector of 4i - 3j on 12i + 8j. Therefore, finding the projection vector is all about finding this particular point.
Hopefully that wasn't too confusing.
As you can see above, we have a nice little right-angled triangle. Here it is, rotated because it's easier to understand (in my opinion, anyway).
Let us call 4i - 3j b, and the purple line x.
Using simple trigonometry cos(theta) = |x|/|b|. Therefore |x| = |b| cos(theta).
Now we need to work out cos(theta). Remember, theta in this case is the angle between 12i + 8j (which I'm going to call "a," by the way), and 4i - 3j (which I called "b" not too long ago).
Let's use the scalar product equation to find cos(theta). a(dot)b = |a||b|cos(theta) can be rearranged to cos(theta) = (a(dot)b)/(|a||b|).
Therefore x = (|b|(a(dot)b))/(|a||b|). The |b| then cancels out to give (a(dot)b)/|a|.
This only gives us the magnitude of line x, however. We also need to find the direction, which is in the same direction as line a.
As I'm sure you probably already know if you're also doing 3AB Spec, if you want to find a vector of a certain magnitude in the same direction of another vector, you just need to multiply the desired magnitude by the unit vector (since the unit vector has a magnitude of 1). Therefore, we need to multiply the magnitude of x by the unit vector of a.
The unit vector of a can be represented by the symbol â, or by a/|a| (i.e. vector a divided by the magnitude of a).
Therefore the projection vector is as follows:
((a(dot)b)/|a|)/(a/|a|)
which can be simplified to:
which can be simplified even further to:
(â.b)â
Hopefully my explanations weren't too complex and you now at least have some kind of understanding of projection vectors! Yeah... it was a real Eureka moment yesterday when I understood...
Friday, September 28, 2012
Clarinets
I have 15 minutes left of maths, so I'm going to use it to promote my instrument. The clarinet.
The clarinet is a woodwind instrument. It is made of wood but cheaper models may be made of plastic. As a single reed instrument, you need a single reed (which I believe is made out of cane) to play it. (FYI saxophones also have a single reed, but their reeds are larger. Oboes and bassoons use a double reed.)
Clarinets are extremely versatile due to their huge range, in terms of pitch and dynamics. The clarinet's pitch range is perhaps the largest of all of the woodwind instruments. The standard range extends from the concert D middle line of the bass stave to the high concert F 3 leger lines and a space above the treble staff. The extreme range extends all the way up to the high concert Bb above that. In short: the standard range is 3 octaves and a minor third, and the extreme range is 3 octaves and a minor 6th. Although the clarinet may not be the loudest instrument in the orchestra, it makes up for it for having a large range of dynamics and can play very quietly, which makes the clarinet useful for sections of music that fade away to nothing.
Clarinets are used in a range of ensembles: symphony orchestras, wind orchestras and jazz bands, as well as chamber ensembles such as wind quintets and clarinet quintets. Symphony orchestras only contain two clarinets. Wind orchestras can contain a lot of clarinets grouped into 1st, 2nd, 3rd and sometimes 4th. Chamber ensembles only have one clarinet.
Here is a list of some clarinet music popular among clarinet players (some of which I'll be playing in my performance exam this semester):
The clarinet is a woodwind instrument. It is made of wood but cheaper models may be made of plastic. As a single reed instrument, you need a single reed (which I believe is made out of cane) to play it. (FYI saxophones also have a single reed, but their reeds are larger. Oboes and bassoons use a double reed.)
Clarinets are extremely versatile due to their huge range, in terms of pitch and dynamics. The clarinet's pitch range is perhaps the largest of all of the woodwind instruments. The standard range extends from the concert D middle line of the bass stave to the high concert F 3 leger lines and a space above the treble staff. The extreme range extends all the way up to the high concert Bb above that. In short: the standard range is 3 octaves and a minor third, and the extreme range is 3 octaves and a minor 6th. Although the clarinet may not be the loudest instrument in the orchestra, it makes up for it for having a large range of dynamics and can play very quietly, which makes the clarinet useful for sections of music that fade away to nothing.
Clarinets are used in a range of ensembles: symphony orchestras, wind orchestras and jazz bands, as well as chamber ensembles such as wind quintets and clarinet quintets. Symphony orchestras only contain two clarinets. Wind orchestras can contain a lot of clarinets grouped into 1st, 2nd, 3rd and sometimes 4th. Chamber ensembles only have one clarinet.
Here is a list of some clarinet music popular among clarinet players (some of which I'll be playing in my performance exam this semester):
- Francis Poulenc- Sonata for Clarinet and Piano (I'm playing the second and third movements this semester)
- Gerald Finzi- Five Bagatelles for Clarinet and Piano (I played the last movement last semester)
- Wolfgang Amadeus Mozart- Clarinet Quintet (I played the first movement last semester)
- Johannes Brahms- First Sonata for Clarinet and Piano
- Johannes Brahms- Second Sonata for Clarinet and Piano (I'm playing the second movement this semester)
- Malcolm Arnold- Sonatina for Clarinet and Piano
- Camille Saint-Saens- Sonata for Clarinet and Piano (I played the first movement last semester)
How I learn
Here I'm going to write about how I learn. I hope that I don't sound too forceful or demanding when I attempt to explain how I learn, because I don't want to seem like I'm saying how you should learn, I want to say what works for me and might work for you.
When I learn something, I try and relate it to stuff that I've already learnt. When I was learning about vectors, I was just like, "Hey, this is just like cartesian coordinates, but you're sticking an i on the end of the x-coordinate and a j on the end of the y-coordinate!" When I was learning about imaginary numbers, I was like, "Hey, this is like cartesian coordinates or vectors, but you don't stick anything on the end of the first number, but you do have to stick an i or a j on the second number!" When you relate what you're learning to stuff you already know, that stops it from being new or scary.
When I learn something that appears to require memorisation, I always try and find ways not to simply memorise. I instead try and understand as best I can so that the task of memorisation does not seem so great. An example of this is mathematical formulae. If you understand why the formulae work and can twist them and rearrange them in different ways, you don't have to memorise so much. Also, your understanding of the subject grows when you try and understand rather than try and memorise. I have another example, from chemistry, that may seem rather complicated, but I think actually helps because it builds lots of connections in my brain. Building lots of connections is good because if one connection fails, you still have lots of others that you can fall back on.
To my example: Chemistry students might be familiar with the mnemonic "AN OIL RIG CAT." It means "At the anode occurs oxidation (which is loss), while reduction (which is gain) occurs at the cathode." Although I did have to memorise that the anode is positive and the cathode is negative, I actually found that from there I didn't need to blindly memorise which one oxidation takes place at and which one reduction takes place at. You see, the cathode is negative because it receives electrons from the power source. Reduction is a process involving the gain of electrons to make the atom or ion become more negative (i.e. causing a reduction in the oxidation number). The cathode, that is receiving electrons from the battery, becomes the one-stop-shop for atoms that want to be reduced! Now, the mnemonic is easy enough to remember, but I find that understanding more will probably be more beneficial in the long run.
Another useful tip would be the one that my Year 6 teacher told our class: "Look for an easier method." If something sounds complicated, try and break it down into simple steps that you understand.
This might sound weird coming from me, since I'm a perfectionist, but learning from your mistakes helps too. When you get something wrong, and you don't know why you got it wrong, try and find out why. In the process, you will learn something new.
Finally, here is a little something I wrote titled "The Student's Creed:"
I am here to learn.
I shall not memorise for the sake of tests, but I shall internalise to add to my own rapidly expanding bank of knowledge.
All hatred and fear shall be subordinated to my important task- to learn.
I shall attempt to learn from my failures, for it is in their rectification that I may grow, and what is life without growth?
I shall see tests as the continuation of my learning, rather than the culmination of it, for it is the journey that matters more than the destination.
All this I shall do, and without delay.
When I learn something, I try and relate it to stuff that I've already learnt. When I was learning about vectors, I was just like, "Hey, this is just like cartesian coordinates, but you're sticking an i on the end of the x-coordinate and a j on the end of the y-coordinate!" When I was learning about imaginary numbers, I was like, "Hey, this is like cartesian coordinates or vectors, but you don't stick anything on the end of the first number, but you do have to stick an i or a j on the second number!" When you relate what you're learning to stuff you already know, that stops it from being new or scary.
When I learn something that appears to require memorisation, I always try and find ways not to simply memorise. I instead try and understand as best I can so that the task of memorisation does not seem so great. An example of this is mathematical formulae. If you understand why the formulae work and can twist them and rearrange them in different ways, you don't have to memorise so much. Also, your understanding of the subject grows when you try and understand rather than try and memorise. I have another example, from chemistry, that may seem rather complicated, but I think actually helps because it builds lots of connections in my brain. Building lots of connections is good because if one connection fails, you still have lots of others that you can fall back on.
To my example: Chemistry students might be familiar with the mnemonic "AN OIL RIG CAT." It means "At the anode occurs oxidation (which is loss), while reduction (which is gain) occurs at the cathode." Although I did have to memorise that the anode is positive and the cathode is negative, I actually found that from there I didn't need to blindly memorise which one oxidation takes place at and which one reduction takes place at. You see, the cathode is negative because it receives electrons from the power source. Reduction is a process involving the gain of electrons to make the atom or ion become more negative (i.e. causing a reduction in the oxidation number). The cathode, that is receiving electrons from the battery, becomes the one-stop-shop for atoms that want to be reduced! Now, the mnemonic is easy enough to remember, but I find that understanding more will probably be more beneficial in the long run.
Another useful tip would be the one that my Year 6 teacher told our class: "Look for an easier method." If something sounds complicated, try and break it down into simple steps that you understand.
This might sound weird coming from me, since I'm a perfectionist, but learning from your mistakes helps too. When you get something wrong, and you don't know why you got it wrong, try and find out why. In the process, you will learn something new.
Finally, here is a little something I wrote titled "The Student's Creed:"
I am here to learn.
I shall not memorise for the sake of tests, but I shall internalise to add to my own rapidly expanding bank of knowledge.
All hatred and fear shall be subordinated to my important task- to learn.
I shall attempt to learn from my failures, for it is in their rectification that I may grow, and what is life without growth?
I shall see tests as the continuation of my learning, rather than the culmination of it, for it is the journey that matters more than the destination.
All this I shall do, and without delay.
Central Limit Theorem and Confidence Intervals
Okay, first of all, I hope that the people who monitor the school computers have nothing against Blogspot, because I am kind of typing this up in the middle of maths class.
And I also hope that the people either side of me won't shun me for being nerdy enough to write about maths two days after the 3CD mock exam.
Anyway...
I'm going to write about sampling. Just because I can, or rather I'm not sure I can and want your feedback on this. Actually, I was going to write about Spec but considering that I'm in 3CD class right now I might as well write something about 3CD.
First thing I'm going to tell you about is the Central Limit Theorem. Some smart person, or perhaps a bunch of smart people, worked out that if you take a bunch of samples from a larger population and work out their means, the means will make a graph similar to that of a normal distribution. It doesn't matter what kind of distribution you had before, the sample means will make a graph similar to a normal distribution.
Example number 1: Let's just say that you recorded the heights of all of the adults in China. The distribution of the heights, like the distribution of pretty much everything else in nature, will almost certainly resemble a normal distribution graph.
Then, if you take samples of, say, 300 people, and worked out the means of each sample (which would take a pretty damn long time), and plotted all of these means on a graph, you would find that the distribution of these means will also resemble a normal distribution. Yay.
And this phenomenon isn't just limited to the normal distribution as you shall see in the next example.
Example 2: Having completed the gargantuan task of measuring all of those people, you have decided to live a quieter life dedicated to coin flipping. You are going to flip a coin 10 times and record the number of heads that you get, and repeat this experiment 1000 times. As the event of getting a head is independent and there are only two possible outcomes (success and failure- i.e. getting a head and getting a tail), this can be represented by the binomial distribution.
Recording the number of heads you get from all 1000 experiments will result in a binomial distribution where n = 1000 and p = 0.5. Now, if you take samples of 50 experiments, record the mean number of heads for each sample of 50 experiments, and plot these on a graph, you will also get...
A normal distribution.
It works pretty much the same way for any other type of distribution you can think of. If you want the graph of the means to more closely resemble a normal distribution, try and make the size of the sample larger and the probability of success (in regards to the binomial distribution at least) closer to 0.5. It's recommended that you stick to samples of size 30 or greater and a probability of at least 0.1.
Now, here's the nice part. The mean of the sample means is the same as the population mean, which makes things easy when it comes to calculating the mean.
The standard deviation isn't quite so easy, but it's not very hard either. Just divide the population standard deviation by the square root of the sample size, i.e. (sigma)/sqrt(n).
So, what does it all mean?
Why bother working out the sample mean and standard deviation?
Well, using these figures, you can work out the probability that the mean of a sample will be smaller or greater than a figure. It's simple enough. The sample means are normally distributed, with the mean of the sample means equal to the population mean and the standard deviation equal to (population standard deviation)/sqrt(sample size). You can then work out probabilities the way that you would work out probabilities in any other normal distribution.
You can also use these figures to work out mathematically whether a sample mean deviates too far from the population mean for your liking. You do this by working out the confidence intervals for the sample mean, for example 90% or 95% confidence intervals, and testing to see if the population mean lies within these intervals.
To go about working out a confidence interval, you need to find out where the boundaries of the confidence interval lie. You can work this out in terms of how many standard deviations away from the mean. For example, if you wanted to work out a 95% confidence interval, you are essentially trying to find the value of k in P(-k < X < k) = 0.95 in a standard normal distribution (i.e. mean of 0, standard deviation of 1).
Using a calculator, k = 1.96, i.e. 1.96 standard deviations away from the mean.
To work out the 95% confidence mean, therefore, you take the sample mean and add the sample standard deviation multiplied by 1.96 to get the upper boundary:
(sample mean) + (1.96)(sample standard deviation)
Then you take the sample mean and subtract the sample standard deviation multiplied by 1.96 to get the lower boundary:
(sample mean) - (1.96)(sample standard deviation)
Now that you have your confidence interval, you can do other things with it. For example, you can check to see if the population mean lies within this confidence interval.
By the way, if you're not working out the 95% confidence interval, you would substitute whatever percentage you need (as a decimal) into the P(-k < X <k) equation. You would then plug in whatever answer you got from that into the next two equations.
And then there's one other kind of question. Questions like, "How large do you need a sample to be if you want to be x% confident that the sample mean lies within y units of the population mean?" In this case, you would first work out how many standard deviations away from the mean that you need (i.e. use the P(-k < X < k) equation).
Now you just need to remember that the sample standard deviation is (population standard deviation/sqrt(n)).
(No. of standard deviations away from the mean)(population standard deviation/sqrt(n)) = y
You can then rearrange this to find n.
Someday I might need to clarify this by giving examples using actual numbers because it is getting rather messy using variables. Also I might need to rewrite the equations so that they are easy to read. Watch this space...
And I also hope that the people either side of me won't shun me for being nerdy enough to write about maths two days after the 3CD mock exam.
Anyway...
I'm going to write about sampling. Just because I can, or rather I'm not sure I can and want your feedback on this. Actually, I was going to write about Spec but considering that I'm in 3CD class right now I might as well write something about 3CD.
First thing I'm going to tell you about is the Central Limit Theorem. Some smart person, or perhaps a bunch of smart people, worked out that if you take a bunch of samples from a larger population and work out their means, the means will make a graph similar to that of a normal distribution. It doesn't matter what kind of distribution you had before, the sample means will make a graph similar to a normal distribution.
Example number 1: Let's just say that you recorded the heights of all of the adults in China. The distribution of the heights, like the distribution of pretty much everything else in nature, will almost certainly resemble a normal distribution graph.
Then, if you take samples of, say, 300 people, and worked out the means of each sample (which would take a pretty damn long time), and plotted all of these means on a graph, you would find that the distribution of these means will also resemble a normal distribution. Yay.
And this phenomenon isn't just limited to the normal distribution as you shall see in the next example.
Example 2: Having completed the gargantuan task of measuring all of those people, you have decided to live a quieter life dedicated to coin flipping. You are going to flip a coin 10 times and record the number of heads that you get, and repeat this experiment 1000 times. As the event of getting a head is independent and there are only two possible outcomes (success and failure- i.e. getting a head and getting a tail), this can be represented by the binomial distribution.
Recording the number of heads you get from all 1000 experiments will result in a binomial distribution where n = 1000 and p = 0.5. Now, if you take samples of 50 experiments, record the mean number of heads for each sample of 50 experiments, and plot these on a graph, you will also get...
A normal distribution.
It works pretty much the same way for any other type of distribution you can think of. If you want the graph of the means to more closely resemble a normal distribution, try and make the size of the sample larger and the probability of success (in regards to the binomial distribution at least) closer to 0.5. It's recommended that you stick to samples of size 30 or greater and a probability of at least 0.1.
Now, here's the nice part. The mean of the sample means is the same as the population mean, which makes things easy when it comes to calculating the mean.
The standard deviation isn't quite so easy, but it's not very hard either. Just divide the population standard deviation by the square root of the sample size, i.e. (sigma)/sqrt(n).
So, what does it all mean?
Why bother working out the sample mean and standard deviation?
Well, using these figures, you can work out the probability that the mean of a sample will be smaller or greater than a figure. It's simple enough. The sample means are normally distributed, with the mean of the sample means equal to the population mean and the standard deviation equal to (population standard deviation)/sqrt(sample size). You can then work out probabilities the way that you would work out probabilities in any other normal distribution.
You can also use these figures to work out mathematically whether a sample mean deviates too far from the population mean for your liking. You do this by working out the confidence intervals for the sample mean, for example 90% or 95% confidence intervals, and testing to see if the population mean lies within these intervals.
To go about working out a confidence interval, you need to find out where the boundaries of the confidence interval lie. You can work this out in terms of how many standard deviations away from the mean. For example, if you wanted to work out a 95% confidence interval, you are essentially trying to find the value of k in P(-k < X < k) = 0.95 in a standard normal distribution (i.e. mean of 0, standard deviation of 1).
Using a calculator, k = 1.96, i.e. 1.96 standard deviations away from the mean.
To work out the 95% confidence mean, therefore, you take the sample mean and add the sample standard deviation multiplied by 1.96 to get the upper boundary:
(sample mean) + (1.96)(sample standard deviation)
Then you take the sample mean and subtract the sample standard deviation multiplied by 1.96 to get the lower boundary:
(sample mean) - (1.96)(sample standard deviation)
Now that you have your confidence interval, you can do other things with it. For example, you can check to see if the population mean lies within this confidence interval.
By the way, if you're not working out the 95% confidence interval, you would substitute whatever percentage you need (as a decimal) into the P(-k < X <k) equation. You would then plug in whatever answer you got from that into the next two equations.
And then there's one other kind of question. Questions like, "How large do you need a sample to be if you want to be x% confident that the sample mean lies within y units of the population mean?" In this case, you would first work out how many standard deviations away from the mean that you need (i.e. use the P(-k < X < k) equation).
Now you just need to remember that the sample standard deviation is (population standard deviation/sqrt(n)).
(No. of standard deviations away from the mean)(population standard deviation/sqrt(n)) = y
You can then rearrange this to find n.
Someday I might need to clarify this by giving examples using actual numbers because it is getting rather messy using variables. Also I might need to rewrite the equations so that they are easy to read. Watch this space...
Thursday, September 27, 2012
Scalar Product- Closest Approach
Okay, this seems a bit too complex for a first post, but I wanted to share what I learned today. Or at least what I think I've learned. If I'm totally wrong, please do share because then I could learn from it. One day, I'll go back and write about topics that lead up to this.
I was struggling with this question in my Spec homework, but I think I've got it now.
Here's the question:
Two walkers Harry and Claire are at (30, -17) km and (18, 15) km at 9.30am Saturday, travelling with velocities of (6, 4) km/h and (8, -2) km/h respectively.
a) Find the closest distance that they come to each other.
b) Find the time when they are closest.
c) Find the positions of Harry and Claire when they are closest.
d) Find Harry's position relative to Claire's position as a distance and bearing when they are closest.
The first thing that I did was do some adjusting to make Claire appear to "stand still" at the origin while Harry moved. It's easier to work with an object moving near a still object than with two moving objects.
To do this, I subtracted Claire's position vector from both Harry and Claire's position vectors:
r(C) = (18i + 15j) - (18i + 15j) = 0i + 0j (the origin)
r(H) = (30i - 17j) - (18i + 15j) = 12i - 32j (this is the same as the position vector of Harry relative to Claire)
Next I made Claire "stand still" by subtracting Claire's velocity vector from both Harry and Claire's velocity vectors:
V(C) = (8i - 2j) - (8i - 2j) = 0i + 0j (at rest)
V(H) = (6i + 4j) - (8i - 2j) = -2i + 6j (this is the same as the velocity vector of Harry relative to Claire)
Now we have C at rest at the origin, and H at (12i - 32j) km travelling with the velocity of (-2i + 6j) km/h, as in the diagram below (not to scale):
I was struggling with this question in my Spec homework, but I think I've got it now.
Here's the question:
Two walkers Harry and Claire are at (30, -17) km and (18, 15) km at 9.30am Saturday, travelling with velocities of (6, 4) km/h and (8, -2) km/h respectively.
a) Find the closest distance that they come to each other.
b) Find the time when they are closest.
c) Find the positions of Harry and Claire when they are closest.
d) Find Harry's position relative to Claire's position as a distance and bearing when they are closest.
The first thing that I did was do some adjusting to make Claire appear to "stand still" at the origin while Harry moved. It's easier to work with an object moving near a still object than with two moving objects.
To do this, I subtracted Claire's position vector from both Harry and Claire's position vectors:
r(C) = (18i + 15j) - (18i + 15j) = 0i + 0j (the origin)
r(H) = (30i - 17j) - (18i + 15j) = 12i - 32j (this is the same as the position vector of Harry relative to Claire)
Next I made Claire "stand still" by subtracting Claire's velocity vector from both Harry and Claire's velocity vectors:
V(C) = (8i - 2j) - (8i - 2j) = 0i + 0j (at rest)
V(H) = (6i + 4j) - (8i - 2j) = -2i + 6j (this is the same as the velocity vector of Harry relative to Claire)
Now we have C at rest at the origin, and H at (12i - 32j) km travelling with the velocity of (-2i + 6j) km/h, as in the diagram below (not to scale):
Now, as you hopefully already know, when working out closest approach, you need to find the spot where the line between the point C and the line H is perpendicular to the line H. If that didn't make sense, here is a diagram:
The thick blue line is the shortest distance between C and the line H.
But how should we go about working out how long this line is? First, we let the point where the two lines meet be Point P.
Next, we work out the vector equation for the red line, which is relatively easy:
H = 12i - 32j + t(-2i + 6j) (where t is time in hours)
H = (12 - 2t)i + (6t - 32)j
Now we have to find the position vector of point P. Since P lies along the line H, the position vector of P is simply (12 - 2t)i + (6t - 32)j. We just need to find out what t is, and we'll get to that eventually.
Now, since CP (which is equal to the position vector of P since C is at the origin) is perpendicular to HP, CP(dot)HP is equal to 0. This gives us the following equation:
((12 - 2t)i + (6t - 32)j)(dot)(-2i + 6j) = 0
You might want to know why I don't have to work out what HP is and why I'm just using the velocity vector of line H. Let's just say that two lines, A and B, are perpendicular, that is, they meet at a 90 degree angle. Line A will also be perpendicular to any other line that is parallel to B due to corresponding angles in parallel lines, as can be seen in the diagram below. The vector equation of the line is (point on the line) + (scalar multiple of direction of line). Since only the direction contributes to the gradient of a line, the position vector is irrelevant if you only care about parallel and perpendicular lines, and therefore it can be omitted in this case.
Sorry if that was a terrible explanation. Anyway, back to the original problem at hand:
((12 - 2t)i + (6t - 32)j)(dot)(-2i + 6j) = 0 expands to
-24 + 4t + 36t - 192 = 0 which then simplifies to
40t = 216
t = 5.4
Yay, we now have a t value, so we can work out what the position vector of P is by substituting this value in! Since the position vector of P is equal to CP, finding the magnitude of this will give you the shortest distance!
CP = (12 - 2(5.4))i + (6(5.4) - 32)j
| CP | = 1.26km (2 d.p.)
a) The closest distance between Harry and Claire is 1.26km (2 d.p.).
You can also use the t-value from before to get answer the second part. Just convert 5.4 hours to hours and minutes and add it to 9.30 am, and away you go!
5.4 hours = 5 hours 24 minutes
9.30am + 5 hours 24 minutes = 2.54pm
b) The time at which they are closest is 2.54pm.
It's easy to find the position vectors of Harry and Claire when they are closest. First you have to go way back to the beginning of the question- before you stopped Claire and confined her to the origin. The vector equation for Claire's original movement is (30i - 17j) + t(6i + 4j) and the vector equation for Harry's original movement is (18i + 15j) + t(8i - 2j). Substituting our t-value into these equations will give you the position vectors of Harry and Claire when they are closest.
c) Harry is at (62.4, 4.6) while Claire is at (61.2, 4.2).
The last part isn't too hard either. All you have to do is work out the differences in x-values and the differences in y-values between Harry and Claire. By doing some simple subtraction you can see that Harry is 1.2 km east and 0.4 km north of Claire. I then put this into a simple diagram:
This part should hopefully be really simple. To find the position of Harry relative to Claire as a distance, just use Pythagoras' theorem: sqrt(0.4^2 + 1.2^2). This should hopefully give the same minimum distance as in part a). The bearing is likewise not too difficult. Just use arctan(0.4/1.2) to get the angle. Then, since the angle is in the first quadrant, subtract the angle from 90 degrees to get your bearing.
d) Harry's position relative to Claire's position is 1.26km on a bearing of 71.57 degrees (2 d.p.)
Hopefully my method helped you and I'm not teaching you any wrong methods! If any of you find anything that needs fixing, please tell me so that I don't confuse a bunch of other people...
Year 11 Misadventures
So, what's this blog about?
In a nutshell: I learn best by writing, and they say that one of the best ways to learn something is to teach it to someone. This is what I am going to attempt to do with this blog, as I try and explain what I'm learning in my school subjects to you, my readers.
If you have any questions, feel free to ask. You will most likely raise my understanding and yours by doing so.
Of course, if you're studying a different subject, I might be able to add you as a co-author so that you can blog about that subject. We can all learn from each other!
Let the learning begin! Er, well, let the learning continue, since I assume you were learning before you read this blog...
In a nutshell: I learn best by writing, and they say that one of the best ways to learn something is to teach it to someone. This is what I am going to attempt to do with this blog, as I try and explain what I'm learning in my school subjects to you, my readers.
If you have any questions, feel free to ask. You will most likely raise my understanding and yours by doing so.
Of course, if you're studying a different subject, I might be able to add you as a co-author so that you can blog about that subject. We can all learn from each other!
Let the learning begin! Er, well, let the learning continue, since I assume you were learning before you read this blog...
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