Study Strategy

One question a math tutor – or any tutor – might be asked: “How do I prepare for an exam?”   We’ll examine a basic strategy.

Exam preparation is an important process.  Few people like doing it, so a lot less of it gets done than is really needed.  Naturally, people wonder if there’s a “right” way to do it.  Well, there is!

The secret to exam preparation – and the harder the material, the more advantage it affords – is to start early.  For instance, starting the day the course starts is not absurd.  The other point is that you should do it in small daily portions – maybe a half hour a day or less in most cases.  Usually, if you start long before the exam, you can afford to study only 15 – 30 minutes a day for it.

How should you structure these daily study rituals?

The answer is, be casual.  You can afford to be, since you’re starting so early.  Look over your notes.  Spend a few extra readings on the more difficult parts.  Perhaps you want to highlight some points you still don’t understand, so you can bring them up quickly at the next lecture.  (Instructors usually like being asked about earlier lectures:  it shows you’re paying attention.)  You might not bring all the questions up at once; rather, maybe ask one question today, then one next time.

You might check out the course schedule to discover the future content and leaf through the textbook to get a little familiar with it before it’s taught.  Maybe you only spend five minutes doing this each time, but the pre-emptive strikes on course content usually pay off big.  People learn much easier when they’ve already heard of something than when it’s totally new.

Enjoy these half-hour sessions.  Have a cup of tea while you leaf through your notes.  You don’t have to go through your notes in order; using a random “here, then there” selection often works just as well.  Don’t feel you need to cover all your notes each time; spend an honest 30 min, then stop.

If you’re studying for math, you’ll have to attempt questions from the sections you’re looking back on.  Spread it out:  do a couple from this section, then a couple from another one.  Do a few from long ago, then a couple of recent ones.  After an honest half an hour, stop.  You can afford to – as long as you start this habit long before exam time.  Of course, you can adjust your durations to your personal taste; this is only a rough guide.

I don’t recommend listening to music or watching TV while studying.  Real studying can be a lonely business – no doubt about it.  However, if you start early, and do it daily, it should be a lot easier.

I should offer some reasoning for why this method works (or, at least, why it’s better than cramming).  Think of any biological process – muscle development, for example.  Working out for eight hours today won’t help you develop stronger muscles; we all know that.  However, working out for tweny minutes a day, three days a week, for eight weeks (which totals eight hours), will definitely produce substantial results in most cases.

Biology takes time; you can’t rush it.  However, when you give it time, you often get better results than you would have imagined.  You know that your lawn won’t grow measurably today.  However, you know that if you let it grow for two weeks, you might barely be able to pass the mower through it.

Your daily attention to your studies – even if it is only a few minutes (as long as real effort is achieved during that time) reminds your brain that it needs to continue learning that material.  Your brain – similar to your body – only believes something is worth adapting to if it keeps coming up over and over again.  You have to convince it that what you’re studying is important and here to stay.  That’s what the daily study sessions do – even if they’re only 15 minutes to half an hour.

For those who are reading this a couple of days before the big exam:  well, I guess you’ll have to cram for now.  But you’ll know for next time.

Good luck with your midterms.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Horizontal Asymptotes and Holes: some comments about rational function graphs

When tutoring math 12 or calculus, you encounter graphs of rational functions.  Let’s look at a couple of features:

Rational function graphs are defined by (and you get marks for)  the locations of the asymptotes (if any), as well as the x and y intercepts (once again, provided they exist) and holes (if any).  Today, we’ll look at two of these features:  horizontal asymptotes and holes.

First, to holes:  consider the following rational function:

f(x)=((x-1)(x+2))/((x+2)(x-3)) eqn 1

You can see that, since (x+2) is both in the top and the bottom, the function simplifies to

f(x)=(x-1)/(x-3) eqn 2

However, when you cancel, you are really dividing.  Since you can’t divide by zero, you can’t cancel x+2 when x=-2.  At x=-2, the equation remains undefined.  Therefore, you will get a hole there.  The graph of eqn 1, above, will follow the graph of eqn 2 identically, except for a hole at x=-2.

Now, to horizontal asymptotes:  you get them when the degree on top matches the degree on the bottom or is less than the degree on the bottom.

Case 1:  the degrees on top and bottom are the same.  Consider

f(x)=(2x^2 – 2x -3)/(x^2 +17x+11)

To get the horizontal asymptote, divide the coefficient of the highest exponent term on top by the coefficient of the highest exponent term on the bottom.  The horizontal asymptote will be y=2/1, or just y=2.

Case 2:  the degree on the bottom is greater than that on top.

Simple:  the horizontal asymptote is y=0.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC

Adding Vectors: The “chart thing”

Especially when tutoring physics 12, vector addition comes up.  Let’s use the “chart thing”:

I took Physics 12 in 1988.  In that class, we added vectors component-wise.  However, at university they used sine law and cosine law.  I haven’t seen someone use component-wise addition in over twenty years – until this week.  Now, they call it “the chart thing:)”

Let’s imagine we’re adding the following two vectors:vector 1vector 2

Well, first we use trig to resolve each into horizontal (x) and vertical (y) components:

vector 1 resolvedvector 2 resolved

Now, looking above, we see that for the first vector, the x component is pointing left:  it’s negative.  A downward component would be negative as well, but we don’t have any downward components in this problem.

Now, to the chart:

vector x component y component
vector 1 -12.3 8.60
vector 2 12.0 18.5
total: result vector -0.3 27.1

You just add downwards to get the total x and y components of the resultant vector.  We see that the result has -0.3 for its x component and 27.1 for its y component.  We can build it as follows:

resultant vector
Now, use a² + b² = c² to get the hypotenuse. Of course, you’ll still get 27 (two sig digits), since 0.3 is too small to have any significant effect on the vertical component of 27.1. Use 2ndtan(27.1/0.3) – you may also know it as arctan(27.1/0.3) – to get the angle of 89°.
resultant vector solved

Since there were only two significant digits in the original problem, we should give the answer with two.  (This isn’t always true, technically, but it’s a good rule of thumb.)  Hence, we give the answer in two sig digs: 27 m/s 89° N of W.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

 

Inequalities: Phone Plans

Which phone plan to choose?  A little math tutoring can help you decide.

Let’s assume we have two choices:

Plan A $12 per month $0.10 per text
Plan B $18 per month $0.07 per text

How do you decide which plan to use?

Let’s say you’re buying the plan for a month.  Let x be the number of texts you send.  Then your cost is

$12 + $0.10x for Plan A

$18 + $0.07x for Plan B

You pay more up front with Plan B, but its texts are cheaper.  There will be a specific number of texts at which the plans cost the same:

12 + 0.10x =18 + 0.07x

Using algebra, we solve for x:

  1. Subtract 0.07x from both sides to get

    12 + 0.03x = 18

  2. Subtract 12 from from both sides:

    0.03x=6

  3. Divide both sides by 0.03:

    x=200

At 200 texts, the two plans are equal in price for one month.  Any more texts, and Plan B must be cheaper:  it’s less per text.

Now, let’s explore the situation from the point of view of inequalities.  We ask, “For what number of texts will Plan B be cheaper?”

18 + 0.07x ≤ 12 + 0.10x

  1. Subtract 18 from both sides

    0.07x ≤ -6 + 0.10x

  2. Subtract 0.10x from both sides

    -0.03x ≤ -6

  3. Divide both sides by -0.03: you must flip the sign when doing so:

    x ≥ 200

Note that algebraic operations are the same for inequalities as for equations, with one important exception:  you must flip the sign if you divide or multiply by a negative number.  (Not by positive, just by negative).

The two methods make the same conclusion, but from different points of view.  The equation tells us that, at 200 texts, the two plans cost the same.  The inequality, on the other hand, tells us that if we make more than 200 texts, we’ll pay less with Plan B.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Dependent Variable, Independent Variable, Test and Control

When tutoring Biology 12, terms involved in the scientific method need definition.  Let’s look at some.

In an experiment, there is one variable whose value you must wait to see, and (at least) one that you purposely set.  The variable you set is called the independent variable.  The variable you’re curious about is the dependent variable.

A control group – or simply “control” – is one that you leave alone.  You impose changes on the test group by changing the value of the independent variable.  Then, you give the changes time to affect the test group.  Finally, you measure the test group’s dependent variable in order to discover the effect of the changes you imposed.

The typical example to illustrate the roles of these variables is a plant-growing experiment.  If you want to know if using fertilizer accelerates plant growth, your control group will be plants that don’t get fertilizer.  The test group will be plants that do get fertilizer.  The independent variable will be how much fertilizer you give to each plant.  Finally, the dependent variable will be how much each plant grows over the course of the experiment.

There are other terms that relate to the scientific method, but those are generally better documented.  However, we might look at them here in a future post.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC

Neutrons and Isotopes

Tutoring high school chemistry, isotopes are an early topic.  Let’s have a look….

From basic atomic structure, we know that an atom’s mass comes from its number of protons plus its number of neutrons.  Take fluorine, for example:  its mass is 19.  It’s got 9 protons (because its atomic number is 9), so it must have ten neutrons.

But what about chlorine, whose mass is listed at 35.5?  (Possibly, it says 35.453 or something like that, depending on which periodic table you’re using.  By the way:  you can always tell the mass because it’s the one that can have a decimal, whereas the atomic number is always a whole number.)  If chlorine’s mass comes from its protons plus its neutrons, how can its mass be 35.5?  You can only have whole numbers of protons and neutrons.

The answer comes from the concept of an isotope.  Isotopes are like different versions of the same type of atom.  The number of protons is what defines the type of atom you have.  We see chlorine’s atomic number is 17, so it must always have 17 protons.  However, its number of neutrons can vary:  about 75% of chlorine atoms have 18 neutrons, while the other 25% have twenty.  Therefore, 3 out of 4 chlorine atoms (which is 75%) have a mass of 35, while the other 1 out of 4 has a mass of 37.  Let’s take the average:

massave=(35+35+35+37)/4

Leading to:

massave=35.5

So that’s how you get a mass of 35.5.  No chlorine atom actually has a mass of 35.5; some have a mass of 35, while others have mass 37.  More have mass 35, bringing the average to 35.5.

When you’re calculating the number of neutrons, you should round the mass to the nearest whole number, then subtract the atomic number.  For example, to calculate the neutrons in boron (atomic number 5, mass 10.82), we would round the mass to 11 and subtract 5, giving 6 neutrons.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC

Metric System: decimal places and -re

When you tutor sciences, you’re immersed in the metric system.  Let’s talk a bit about it.

The metric system uses multiples of 10.  Therefore, to convert from any metric unit to any other one, you just need to move the decimal point;  you needn’t do any math.   However, you do need to know the metric prefixes, so you know how many places to move the decimal:

kilo
k
hecto
h
deca
da
metre (m)
gram (g)
litre (L)
deci
d
centi
c
milli
m
0.001 0.01 0.1 1 10 100 1000

From the table above, you can see that 1 metre = 0.001 kilometres.  Or, using abbreviations, 1 m = 0.001 km.

Note that the prefixes have their own abbreviations, and the base units (metre, gram and litre) have their own.  mm can’t be confusing because the m in front is the abbreviation for the prefix milli, whereas m in the back is for metre.  Similarly, dL would mean decilitre.

People often ask about the capital L for litre.  It’s true that except for L (and a few other rare instances we won’t get into), you should never use uppercase letters for the metric system.  The reason for the exception with L is that a lowercase l can look like a 1.  To avoid the possible misunderstanding, “l” for litre is either written capital or else in handwriting.

Now, to convert from one unit to the other:  the table tells all.  They key is knowing where the decimal is now, and how many places you’ll have to move it.  For instance, imagine you have 32 cm and you want hm (hectometres). Looking at the chart, you see that from centi to hecto is four jumps left.  (Don’t count centi; count the jumps to hecto).  Therefore, we must move the decimal four jumps left as well.  (In 32, the decimal is of course at the end, since it’s not written.)

32 cm=0.0032 hm

Another example:  how many cm is 2.3m?  Well, looking at the chart, we see centi is two jumps right from metre.  So, move the decimal two jumps right:

2.3m = 230cm

You knew that anyway, since of course there are 100 cm in a metre :)

One more thing:  in metric, it’s -re, not -er.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC

 

Chlamydia and Gonorrhea: infertility risks

Most people have probably heard of chlamydia and gonorrhea.  They are both bacterial STDs.  Since neither is life-threatening (as far as I’ve ever heard), they don’t get as much attention as AIDS or syphilis.

A risk with both chlamydia and gonorrhea is that they can be asymptomatic – i.e., the victim doesn’t have any symptoms, so doesn’t know they’ve caught the disease.  However, both chlamydia and gonorrhea, operating undetected, can cause scarring of the reproductive tract.  In males, the vas deferens (aka ductus deferens) can become scarred over; in females, the same can happen to the oviducts.  The individual can thus be rendered infertile.

Both chlamydia and gonorrhea, being bacterial, are treatable with antibiotics.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Homeostasis, negative feedback, and thyroxin

Tutoring biology, the concept of negative feedback is important to explain.  Negative feedback is used by all organisms to maintain their living state.  You also use it while driving your car.

More or less, the body needs all its processes to proceed at a constant, continuous rate.  Change is the enemy for biological entities.  However, as you interact with your surroundings – eating, exercising, and so on – change is imposed on your internal environment.  Negative feedback means that when a change happens, your body responds to negate that change – i.e., to bring itself back to normal.  Once back to normal, your body stops its negative response.  It won’t respond again until another change brings it out of normal range.

Consider driving a car – specifically, steering.  Imagine the road is straight, but your car starts drifting to the right.  You correct by steering left, to bring the car back on course.  In so doing, you are applying negative feedback.  The car is going too far right, so you oppose that change of course by steering left.  Once the car is back on course, you stop correcting:  you let the wheel slide back to center.  You won’t steer again until the car begins to drift off course once more.  That process of opposite, corrective response is negative feedback.  You can also call it a feedback loop, since you base your reaction on what is already happening.

Biology has many examples of negative feedback, but we’ll look at the one involving the hormone thyroxin:

The hypothalamus – in the brain – monitors the body’s metabolism.  If the metabolism gets too low, the hypothalamus stimulates the anterior pituitary.  In response, the anterior pituitary releases a hormone that causes the thyroid to release thyroxin.  Thyroxin increases the body’s metabolism.

Once the metabolism reaches an acceptable level, the hypothalamus stops stimulating the anterior pituitary.  In turn, the anterior pituitary ceases to release the thyroid-stimulating-hormone.  Therefore, the thyroid stops releasing thyroxin.  The body’s metabolism remains constant – until another external change depresses it again.

Therefore, the connection among the hypothalamus, anterior pituitary, and thyroid gland constitutes a negative feedback system that maintains the body’s metabolism at a constant rate.  This negative feedback system promotes homeostasis.  Homeostasis means the maintenance of a constant internal environment.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Cell efficiency

When you tutor Biology 12, one topic that comes up is cell efficiency.  It’s a bit tricky for some people, because it involves some math.

Putting it simply, imagine a cell is a sphere.  Its volume is what it needs to maintain, whereas its surface area is where it gets its supplies.  You can quickly realize that it’s best to have a big surface area compared to volume (or surface area to volume ratio), so the cell can easily get enough supplies to feed its volume.  Efficiency, in this context, refers to the cell’s surface area to volume ratio:

Eff.=SA/V

As the radius of a cell grows, its surface area grows, but its volume grows more quickly.  Therefore, its surface area to volume ratio decreases:  its efficiency decreases.

Therefore, cells are better off being small – which is why most cannot be seen with the naked eye.  Ultimately, this same principle (of efficiency) is why a rat can run up the side of a building with ease, but a human cannot.  The human’s muscles, since they have such greater volume than the rat’s, are much less efficient.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC