My second take on the Volumes of Revolution formula

Another way of looking at the Volumes of Revolution formula is to divide the area up into a series of... cylinders. Yes. You know how when you were learning about how integration "worked" you had those investigations where you had to divide the area up into small rectangles and calculate the combined area and work out an estimate? Well, that's exactly what I'm doing here, except with rectangles rotated 360 degrees about one of the axes because that just totally makes them less obnoxious. (Not really.)

Here's a triangle that wants to become a cone.

Now, one way that you could work out the volume of the cone-to-be is by using the area-of-base-times-height-divided-by-3 formula. But let's make stuff more complicated, because that's so much fun! (Okay, I should probably stop with the sarcasm now.)

One place to start off would be to divide the area between x = 0 and x = 2 into four rectangles and calculate the volume of each corresponding cylinder and then add the volumes together. (The fourth rectangle in the diagram below is that purple line at the bottom that has 0 area. I really should have picked (y = x^2 +3) or something else that would have given me a better area- an actual area, in fact- for the first rectangle. Ah well.)

As can be seen from the above diagram, these rectangles don't give a very good approximation of the area. Look at all those triangle bits being left out! To counter this partially, you have to divide the area into smaller rectangles like so:

And so on, until the rectangles are all so small that all you see is this vague purple blur.

It's easier to work out the area of each cylinder (formed from each rectangle) if you make yourself a nice simple formula and then work from there. The volume of each cylinder is (pi)(r^2)(h), but that's pretty generic so let's look at how it applies to the above diagram.

For the above diagram with the 8 rectangles (or rather, 7 rectangles plus one pathetic excuse of a rectangle), each cylinder is 2/8 = 1/4 units wide. Therefore, h in this case is 1/4 units for every rectangle. r is the tricky one, because it is different for each rectangle. r is equal to the y value at each x-coordinate. For the first rectangle, r = 0, then for the second one r = 1/4, then for the third rectangle r = 1/2, and so on (it sure helps having y = x as your formula! If it was x^2, though, then the rectangles' r values would be 0, then 1/16 and then 1/4 and so on.) Finally, pi is a constant, which is nice.

Now, as I said before, you eventually want to get a whole load of infinitesimally small rectangles, and therefore infinitesimally small cylinders too. Let's say that you want n small rectangles in that space. Therefore, the width of each rectangle is 1/n units. That gives us (1/n)(pi)(r^2). Now we need to work out a general formula for r^2. For the first rectangle, r = f(0), then, in the second rectangle, r = f(1/n), then r = f (2/n) and so on. A general formula for r would be f((p-1)(1/n)), or f((p-1)/n), where p is what number rectangle you're looking at.

To sum up:

Volume of first cylinder = (1/n)(pi)(f(0/n)^2)
Volume of second cylinder: (1/n)(pi)(f(1/n)^2)
Volume of final (nth) cylinder: (1/n)(pi)(f((n-1)/n)^2)

Combined volume of cylinders: (1/n)(pi)(f(0/n)^2 + f(1/n)^2 + f(2/n)^2 ... f((n-1)/n)^2)

Which can further be summed up in the handy dandy summation formula that I just learned! (Too late for the previous investigation, unfortunately, but hey, better late than never!)

Now, to find the exact area under the curve, you want n to be as large as possible- infinitely large, in fact. That's where limits come in! Now, I'm not sure whether I've used limits correctly in the following equation, so as soon as I've published this I'll shoot off an email to my teachers to ask them. Of course, if you're a teacher yourself, or if you're just plain good at maths, then you can tell me whether I'm right or wrong in my notation.

Let's analyse this formula a bit more. f((p-1)/n) gives the y-value for various x-coordinates, since f at something gives the y-value of whatever x-coordinate "something" is. (f((p-1)/n))^2 is therefore where the y^2 part from the volumes of revolution formula comes from. Pi, our good friend, is there. Finally, the 1/n is the width of the tiny little rectangles/cylinders, made even tinier as n approaches infinity. (1/(a large number) = a very small number.) I believe this is where the "dx" part comes from in the volumes of revolution formula. The volumes of revolution formula also has the start and end point marked in, but I'm not sure how to include that in the summation formula above.

And that is a second explanation for the Volumes of Revolution formula, just because one explanation is simply not enough!

My take on the Volumes of Revolution formula

Okay, well, being the nerd I am I feel like explaining how I remember the Volumes of Revolution formula.

Volumes of revolution, if you didn't know, is basically the volume of the solid you get when you rotate some area of a graph 360 degrees around either the x-axis or y-axis. For example, in the diagram below, when you rotate the shaded bit between the lines x = 0, y = 0, x = 1 and y = 2 about the y-axis (or x-axis for that matter), you get a cylinder. If you don't see how this works, then tape some paper onto a pencil and twirl it around a bit.

Here are the two formulae for working out the volume of the solid formed. The top one is for rotating around the x-axis, the bottom one for the y-axis. In the first equation, a and b are the x-coordinates of the "start" and "end" points of the volume to be calculated, while in the second equation, a and b are the y-coordinates of the "start" and "end" points.

You might be thinking, but wait! How do I integrate with respect to x when I only have ys there, and vice versa? Well, you just have to rearrange the equation. For example, if your equation is y = x^2, and you want to find the volume when you rotate an area under that graph around the x-axis, then you have to rearrange it to get y^2 = (x^2)^2 = x^4, and then substitute x^4 into the first equation. Then all you have to do is solve it like a definite integral.

Now, here's the simple(r) and probably very mathematically incorrect explanation as to why this formula works.

I'm going to put the above graph here again, just so you don't have to scroll back up just to see what I'm blabbing on about.

As you can see, this to-be cylinder either has a height of 2 units and a radius of 1, or vice versa, depending on whether you're going to be rotating the shaded bit about the y-axis or x-axis. For now, we're going to rotate it around the y-axis, because I tend to think of cylinders as being long thin things, because my thinking totally conforms to stereotypes like that. To find the area of this cylinder, then, you could just use our good ol' formula back from year 8 or maybe from primary school if you were lucky enough to have teachers willing to teach that stuff there: the good ol' (pi)(r^2)(h) formula for finding out the volume of cylinders.

Now wait! In the above diagram, since we're rotating about the y-axis, the radius is equal to x. That's how we get the (pi)x^2 part of the formula. Now, as for y, and the whole dy and integral and everything else that's in the volume of integration formula, we have to think outside the orange rectangular box above and start thinking about all kinds of weird irregular shapes.

The (pi)(x^2) part gives the area of any given little slither of area rotated around the y-axis. To actually turn that into volume, we have to multiply that by height as well. We could just multiply by the height of the entire area in the above diagram, but remember: most curves do all kinds of weird things. The radius, or x^2, generally changes as you go along the curve, so you have to multiply by the height of infinitesimally small sections at a time to get the volume. Here's a weird curvy thing that I drew to explain this better:

Note that at y = 2, x, and therefore x^2, is much bigger than it is at x = 1. Every little section along this weird curvy thing has a different x-value too. Therefore, to calculate the total volume of this curve between y = 1 and y = 2, we have to calculate the volume of every tiny little slither of volume between these two points and add them all together.

Thankfully, someone smart invented integration, so we don't have to do all that. I think that the "dy" part of the integration formula is a reference to all the small slithers, a bit like how (delta)y is used to symbolise a small change in y.

Therefore, to recap: the (pi)(x^2) in the integration formula is the circular area of one of the little bits of area under the curve rotated around the y-axis, while the dy is the height of each slither. Then the b and a tell you where to start and stop adding up slithers. And then that fancy integration sign is there because it's fun to draw and makes you feel like a maths genius every time you do so.

For the formula for rotating around the x-axis, it works much the same way: the (pi)(y^2) in the integration formula is the circular area of one of the little bits of area under the curve rotated around the x-axis, while the dx is the width of each slither.

Gosh, I could almost be talking about snakes with the many times that I've said the word "slithers!"

Exponentials and Logs

I'm going to very very briefly go over stuff that was covered last year, and then explain the new stuff.

Differentiating exponential equations: Multiply by the derivative of the power of e, keep everything else the same. (e.g. e^(2x) becomes 2e^(2x))
Integrating exponential equations: Divide by the derivative of the power of e, keep everything else the same (e.g. e^(2x) becomes (1/2)e^(2x))

ln (natural log) is a log to base e
A log with no base specified is a log to base 10

Converting logs to exponentials:

Remember, the bottom number (the a in this case) is the base.
According to Wikipedia, the b is called the argument in the logarithmic function (top) and the answer in the exponential function (bottom).
Also according to Wikipedia, the c is called the answer in the logarithmic function and the power in the exponential function.
Though Wikipedia also suggests that the terms "argument" and "power" are interchangeable.

Adding and subtracting logs: You know how when you multiply numbers with powers, you add the powers, and when you divide numbers with powers, you subtract the powers? Well, logs are the opposite. You multiply when you need to add and divide when you need to subtract. I know, I'm terrible at explaining. And there's the power law too, which involves bringing the power around to the front- also a pretty terrible explanation. So here's some formulae, courtesy of Microsoft Word:

Differentiating logarithmic equations: This method ONLY works for natural logs- logs with base e, or ln.

Derivative of the function ln (f(x)) is given by (f '(x))/ (f(x)) - that is, the derivative of the argument divided by the argument (see "converting logs to exponentials," two sections above).

If you don't have a natural log, you need to convert whatever you have into a natural log. To convert an exponential into a natural log, take a natural log of both sides. For example, 2^x = y can be rearranged to ln (2^x) = ln y and then x ln 2 = ln y and finally x = (ln y) / (ln 2). From there you can find dx/dy and then dy/dx.

To convert a log of another base into a natural log, first convert to exponential form and then use the above method.

Integrating to give logarithmic equations: The new stuff! Finally!

Remember how back in 3AB we were told that you can't find the integral of x^(-1) using the old "raise the power by one and divide by the new power" rule because then you'd end up dividing by 0? Well, fear no more, because now we have a new secret weapon! Natural logs!

You see, x^(-1) = 1/x. And the derivative of ln x = 1/x! Therefore, the integral of x^(-1) must be ln x! (EDIT: Actually, it's ln x + c.)

Well, actually, it's slightly more complicated than that. You see, if you want to find the area under the curve of y = x^(-1) you'd want to use integration, which would lead you to using ln x. The thing is, though, the curve of y = x^(-1) exists for negative values as well as positive values, while ln x is undefined for negative values. To get around this, we actually have to write the integral of x^(-1) as ln |x|. No, ln |x| + c. Oops.

Now for some more fancy stuff! Clearly, there are other logarithmic functions out there than ln x, and therefore lots more equations with logs as their integrals. Let's see how we can work this one out.

First of all, put the bit at the bottom into a natural log (probably not really correct terminology, but whatever). That is to say, if you have 2x + 3 at the bottom, make it ln (2x + 3) and move it to the top. Now, hopefully you also have a multiple of the derivative- in this case 2- sitting at the top with it. (If not, you can't integrate.) Then divide this multiple by the derivative.

Wow, today must be a day for terrible explanations. So here's an example that might shed some light on the matter.

(integral sign) (2 + cos 2x)/(4x + sin 2x) dx

Step 1: Put the bit at the bottom into a natural log and move it on top.

(2 + cos 2x) ln (4x + sin 2x) dx

Step 2: Divide by the derivative of the stuff that was originally at the bottom.

((2 + cos 2x) ln (4x + sin 2x))/(4 + 2 cos 2x)
= ((2 + cos 2x) ln (4x + sin 2x))/(2 (2 + cos 2x))
= (1/2) ln(4x + sin 2x)

Step 3: DON'T FORGET THE PLUS C!!!!

= (1/2) ln(4x + sin 2x) + c

(Please note: I don't know for sure if the above method is mathematically correct or whatever... but hey, it works. Just make sure you've got the derivative of the stuff at the bottom at the top, and you should be fine. Otherwise, use the substitution method as I outlined here: http://year11misadventures.blogspot.com.au/2013/05/integrating-by-substitution.html.)

Differential Equations

So what is a differential equation you might ask? Well, it's an equation that involves derivatives. For example,  dy/dx = 2 is a differential equation. To solve an equation like this, we just use integration, and integrate both sides with respect to x, giving us y = 2x + c. (Don't forget the +c!!!)

Solutions to differential equations involving a + c are called general solutions to the equation. If you're given more information, though, like what y is at a certain point, then you can find a particular solution.

My textbook's also reminding me that, technically, when you integrate both sides, you get a +c on both sides. The integral of (dy/dx) is actually y + c and the integral of 2 is 2x + c. You only need to leave in one of the cs, and this c is technically equal to the difference between the two cs.

Now, if you have something like y(dy/dx) = 2x + 3, you still integrate both sides with respect to x... (and yes, I'm sorry that I still don't know what to type in as a good substitute for the integral sign)

(integral sign) y(dy/dx) dx = (integral sign) 2x + 3 dx
(integral sign) y dy = (integral sign) 2x + 3 dx
(y^2)/2 + c = x^2 + 3x + c
(y^2)/2 = x^2 + 3x + c

Now, look closely at the second line. See how the y is with the dy and the x is with the dx? This is what you should aim to do when you rearrange the equation. Although dy/dx isn't really a fraction- it's the limit of a fraction- thinking of it as one might help you to rearrange the equation so that the ys are with the dy and the xs are with the dx.

Wow, that was a relatively short post (for me, anyway). I can't really be bothered doing area under a curve now because you've probably already done it before and I have other stuff to write about like integrating logarithmic functions, 3D vectors, the shapes of molecules, and Shakespeare's Hamlet. I might write about area under a curve once I've revised everything else that I need to (or maybe while I'm procrastinating over revising those things).

Oh, and by the way...

DON'T FORGET THE + C!!!!!!!!!!!!!!!!!!

Integrating by Substitution

This (hopefully) shouldn't take too long to explain. Basically, you generally use integrating by substitution when you've got something in brackets raised to a power and something else sitting outside those brackets that isn't a multiple of the derivative of the stuff in the brackets.

When you integrate by substitution, you set the stuff in the brackets equal to u. For example, if you want to find the integral of 20x(2x + 3)^3, you might then make 2x + 3 = u.

You then need to play with this equation to find two things: an equation for x and an equation for u.

Rearranging the equation, x = (u - 3)/2.
Also, du/dx = 2. Therefore, dx/du = (1/2).

I don't know how to type in integral signs on Blogger, and I don't know what the shorthand is either. So bear with me here.

(integral sign) 20x (2x + 3)^3 dx
= (integral sign) 20((u - 3)/2)(u)^3(dx/du) du
= (integral sign) 10 (u - 3) (u)^3(1/2) du
= (integral sign) (5u - 15)(u)^3 du
= (integral sign) 5u^4 - 15u^3 du
= u^5 - (15/4)u^4 + c
= u^4(u - (15/4)) + c
= (1/4)(u^4)(4u - 15) + c
= (1/4)(2x + 3)^4(8x + 12 - 15) + c
= (1/4)((2x + 3)^4)(8x - 3) + c

You have to make sure that you pop in your substitution for x (in this case x = (u - 3)/2) otherwise you'll end up trying to integrate two variables at once, which I'm told isn't allowed. This is also why you need to have a multiple of the derivative "sitting outside the brackets" in order to integrate. You don't really have to use the substitution method when you do have a multiple of the derivative present, but you still can.

(Well, I'm currently looking at another example which is to find the integral of 8 cos (2x) (sin 2x)^5... but I'm not really sure how to find x in terms of u using the substitution u = sin 2x. At least in this question a multiple of the derivative is present...)

Integrating Trig Functions

Okay. I HATE INTEGRATING TRIG FUNCTIONS. Absolutely hate doing it. But I have to for Spec. So I'm going to explain how to do it.

The Basics

Probably one of the main reasons why I hate integrating trig functions is because I keep mixing up the derivatives of the trig functions with their integrals. Plus you might get some that seem tricky to work out and then you realise that the trig function in question is simply a multiple of one of the three derivatives of tan x.

Here are the integrals that you need to know:

Integral of sin x = -cos x
Integral of cos x = sin x
Integral of (1/(cos x)^2), (sec x)^2 or 1 + (tan x)^2 = tan x

All the other rules work as before. If the trig function's raised to a power, raise the power by one and divide by the new power and the derivative of the trig function- but make sure that a multiple of said derivative is also present. So you can't differentiate (sin x)^3 directly because sin x's derivative, -cos x, isn't "sitting outside the brackets." (There are, however, other ways to differentiate this, which I will talk about later.) This is where I always screw up because I always divide by the integral by accident, which means that I get the positive/negative signs all wrong.

Oh, and don't forget the...

+ c

If you don't have the derivative sitting outside the brackets... (for trig functions only)

If you have sin x or cos x raised to an odd power, you can do a dirty trick as follows...

If you have, say, (sin x)^3, you can split that up into (sin x)(sin x)^2. Now, (sin x)^2 is also equal to 1 - (cos x)^2. This gives you (sin x)(1 - (cos x)^2), which then gives you sin x - sin x(cos x)^2 which can be integrated to -cos x + (1/3)(cos x)^3 + c!

(And yes, while I was working that one out I made my old mistakes of dividing by the integral instead of the derivative and forgetting the plus c... hopefully I don't make those mistakes in tests...)

If you have sin x or cos x raised to an even power, well, there's a dirty trick for that too, but it's not as convenient.

Here are two more trig identities to memorise (I hate memorisation, so that's just another reason why I hate integrating these things). (cos x)^2 = ((1 + cos 2x)/2) and (sin x)^2 = ((1 - cos 2x)/2). You can go ahead and prove them if you want to feel confident that they work. These functions can be integrated.

If you have (cos x)^6 or whatever, (cos x)^6 = ((cos x)^2)^3 = ((1 + cos 2x)/2)^3. Then you have to do the fun, fun work of expanding that and integrating it. And so on. Argh. I hate these things. I've pretty much said everything about them that I have to, though. Now for some more posts on integration by substitution, differential equations and maybe a quickie on area under a curve as well...

Implicit Differentiation and Differentiating Parametric Equations

More differentiation!

I'm going to do parametric equations first, because they're easier.

Differentiating Parametric Equations

Parametric equations are where you have separate equations for x and y in terms of a third variable, like t. For example, you could have x = 2t and y = t^2 + 1. Sometimes you can combine the two functions to make one single function in terms of either x or y. In this example, you could rearrange x = 2t to become t = (x/2), and then substitute that into the y equation to find an equation for y in terms of x. You could then differentiate that way. Sometimes, though, it's rather tiresome and annoying to do all this substitution and whatnot, which is why there's a method for directly differentiating parametric equations. It's a pretty simple method too, which is nice.

First, you find dx/dt (or, if x isn't given in terms of t, just replace the t with whatever variable x is given in terms of). Then you find dy/dt. Now, by the chain rule, dy/dx = (dy/dt)(dt/dx). But wait! you might say. We have dx/dt, but not dt/dx! Well, the handy thing is, dt/dx is simply equal to 1/(dx/dt).

For the above equations, x = 2t and y = t^2 + 1, dx/dt = 2 and dy/dt = 2t. dy/dx = (dy/dt)(dt/dx) = (2t)(1/2). Therefore, dy/dx = t. If you want to find this in terms of x, just use one of your original equations to find what t is. x = 2t can be arranged to t = x/2. Therefore, dy/dx = t = x/2. You can get the same answer using the substitution method that I described above.

Differentiating Implicitly Defined Functions

Most of the time, when you see equations, they're in the form of y in terms of x or x in terms of y. But sometimes, you might get an equation where x and y are on the same side, like in x^2 + y^2 = 16 (which, by the way, is the graph of a circle, centre (0, 0) and radius 4). One way of differentiating such equations is to rearrange them and then differentiate, but sometimes this is hard or even impossible to do. That's where implicit differentiation comes in!

Implicit differentiation, however, is pretty long-winded, so if you're doing a calculator assumed part of a test and you see an implicit differentiation question only worth 1 or 2 marks, use your calculator instead! It's not worth the effort!

Basically, in implicit differentiation, you have to differentiate each term separately with respect to x (or one of the two variables in the equation). Then you have to factor out dy/dx (or dy/dt or whatever the variables are) to get your derivative.

Here's how to differentiate the different terms. Let x = whatever variable you're differentiating with respect to and y = the other variable.

Number on its own: Derivative is 0 (like always)
x, or powers of x: Differentiate as normal
y: Derivative of y is dy/dx. (dy/dx does mean "derivative of y with respect to x"...)
Powers of y: Use the chain rule as well as the above fact. y^2 becomes (2y)(dy/dx)
Mix of x and y: Separate into two groups, one with the x and one with the y. For example, 6xy could become (6x)(y). Then use the product rule. Derivative of 6xy with respect to x would end up being 6y + 6x(dy/dx).

Here's a pretty simple example for implicit differentiation: finding the derivative of x^2 + y^2 = 16.

(d/dx)(x^2) + (d/dx)(y^2) = (d/dx)(16)
2x + 2y(dy/dx) = 0
(dy/dx)(2y) = -2x
(dy/dx) = (-2x)/(2y)
(dy/dx) = -(x/y)

I'm too lazy to find and type up a harder example... but hopefully you get the idea. You have to differentiate each term separately using the rules that I stated above (well, I kind of made them up as I went along so I hope they work all the time), and then rearrange the equation to get an equation for dy/dx in terms of x and y.

If you have any more questions, just ask!

Differentiating Trig Functions

I'm going to skip ahead in the book a bit and go straight to the differentiation and integration bits, because I don't really have to draw diagrams to explain them (and drawing diagrams on Paint is annoying).

Differentiating Trig Functions- First Principles

First principles- the random long-winded way of differentiation which you only do when you're trying to learn the ideas and concepts behind it, but then drop when you learn quicker, easier methods.

I'm not sure whether or not I've explained first principles before on my blog, so I'll try and briefly explain the concepts behind it here. Differentiation is all about finding the gradient, and gradient can be calculated by rise over run (i.e. difference in y-values divided by difference in x-values). Therefore, one way of finding the gradient at any given point along a curve is to find the gradient between that point and another point very very close to it. You can keep adjusting the distances between the two points until there's only a very infinitesimally small difference between the points, and then you'll get the gradient.

Let's say that you're looking for the gradient at the point where x = x on the curve y = f(x). When x = x, the y-coordinate of the curve is given by f(x), so our first point is (x, f(x)). Our next point is h more units along the x-axis, so its x-coordinate is x+h, and its y-coordinate is f(x+h), giving us the point (x + h, f(x+h)). Now, gradient is rise over run, so the gradient formula is (f(x+h)-f(x))/((x+h)-x), which simplifies down to (f(x+h)-f(x))/h. Now, to get the gradient at point x, you want the distance between the two points (h) to equal zero. But wait! If h = 0, then you're dividing by 0, and you can't do that!

That's why we use a limiting equation, as represented by that "lim h->0" thingy.

(I love Word's "insert equation" feature...)

Now, let's put this into action. Chances are, you know that the derivative of y = x^2 is simply 2x. How do we go about proving this with first principles?

Well, on the top line we first substitute f(x + h) for (x + h)^2 (since the function, f(x), is x^2) and f(x) for x^2. Then we expand and simplify this so we have (2hx + h^2)/h, h approaching zero. Now we factor out h to get (h(2x + h))/h and then cancel out the hs so we get 2x + h. Now we don't have to bother about h approaching zero. Instead, we can allow h to equal zero, since we're no longer dividing by h and therefore making h zero won't make the universe explode. That leaves us with 2x as the derivative of x^2.

We can use a similar approach to find out the derivatives of sin x and cos x, but there's a couple of things that you have to know first.

As x approaches zero, (sin x)/x = 1 and (1-cos x)/x = 0. These can be determined by using tables of values (i.e. working out the values of these two functions by calculating for x = 0.01, x =0.001 etc.) or by looking at the graphs of the functions.

Here's how the derivative of sin x can be calculated using first principles:

As for the derivative of cos x... I'll leave that one for you to do as practise! (Actually I'm only saying that because I can't be stuffed typing that one up.)

The derivative of sin x is cos x, and the derivative of cos x is -sin x. REMEMBER THAT. It's helpful knowledge to have on hand.

tan x has a retarded derivative, or rather derivatives (yes, it has more than one acceptable derivative. One way to differentiate tan x is by using the quotient rule, since tan x = (sin x) / (cos x).

When y = (sin x) / (cos x), dy/dx = (cos x cos x - sin x (-sin x)) / (cos x)^2
= ((cos x)^2 + (sin x) ^2) / (cos x)^2
But (cos x)^2 + (sin x)^2 = 1 (Pythagorean identity, which I realise I haven't mentioned yet on this blog)
So dy/dx = 1 / (cos x)^2.

So one possible derivative of tan x is 1 / (cos x)^2.

Now, apart from sine, cosine and tangent, there are three more trig functions (which you don't really need to know very well for now) which are the reciprocals of the first three functions. They are secant (sec), cosecant (cosec) and cotangent (cot). Secant is 1 / cos x, cosecant is 1 / sin x and cotangent is 1 / tan x.

Since sec = 1 / cos x, (sec x)^2 = 1 / (cos x)^2, which is the derivative of tan x. Therefore, another possible derivative of tan x is (sec x)^2.

Also, there's an alternative way of simplifying my second line of working for the derivative of tan x using the quotient rule. This alternative simplification leads to a third acceptable derivative of tan x.

((cos x)^2 + (sin x) ^2) / (cos x)^2 = ((cos x)^2) / (cos x)^2 + ((sin x)^2) / (cos x)^2))
= 1 + (tan x)^2

Now I'm going to summarise the derivatives for the three main functions, because it's handy having everything in one place.

Derivative of sin x = cos x
Derivative of cos x = -sin x
Derivative of tan x = (1 / (cos x)^2), (sec x)^2 or 1 - (tan x)^2, depending on your mood and the question that you're trying to solve

Rules of Differentiating Trig Functions

Differentiating trig functions is annoying (though nowhere near as annoying as integrating them) because you have to make sure not to get mixed up between the derivatives and stuff (my main problem is getting the positive/negative signs the right way around). At least they go by most of the same rules as pretty much every other function with regards to differentiation: you can still use the good ol' product, chain and quotient rules.

I'll just do a quick example with the chain rule because I believe that that's where it's easiest to get mixed up with regards to what order to do the steps in.

Let's use y = (cos 5x)^3 as an example.

Let cos 5x = u, and 5x = s. Therefore ds / dx = 5. u = cos 5x = cos s, so du/ds = - sin s. Finally y = u^3, so dy/du = 3u^2.

Now, due to the chain rule, dy/dx = (dy/du)(du/ds)(ds/dx).
Therefore, dy/dx = (3u^2)(-sin s)(5).
Substituting to get an answer in terms of x, dy/dx = 3((cos 5x)^2)(- sin 5x)(5) = (-15 sin 5x)(cos 5x)^2.

Wow, explaining that in full was more complicated than I was expecting since cos 5x warranted an extra chain as well (you need to multiply by the derivative of 5x). Just remember that the power takes precedence, then the stuff in the brackets. And take care in working out the derivative of the stuff in the brackets too.

That's all from me on differentiating trig functions for now, but if you want more clarification, feel free to ask!

Some more bits and pieces on Polar Coordinates and Complex Numbers

It feels like it's been a while since I last posted about maths, apart from copy-pasting something from my other blog. This post is actually a continuation of that other post, called Complex Numbers and Polar Coordinates. I've covered pretty much all of the basics there (apart from the fact that that post is in sore need of pictures), but there's still some stuff I have yet to cover.

First off, I need to tell you how to find the distance between two points when they are given in polar form, where polar form is given by (distance from origin, angle from positive x-axis). Here's a diagram:

In the above diagram, point A has the polar coordinates (r1, theta1) (yes, I know that those should be subscripts but I can't be bothered opening Word right now just to type in some subscripts and then copy-paste to Blogspot as you can't type them directly into Blogspot) and point B has the polar coordinates (r2, theta2).

The line from the origin to point A is of length r1. The line from the origin to point B is of length r2. Finally, the angle between the two lines is given by theta1 - theta2: the difference between the two angles. Now you have enough information to work out the length of AB using the cosine rule!

AB = sqrt( (r1)^2 + (r2)^2 + 2(r1)(r2)cos(theta1 - theta2))

(One day Blogspot will be able to type in not only subscripts, but square roots signs and the Greek alphabet as well... one day... or maybe I should just install a Greek keyboard on my computer?! Now that's an idea! Why did I not think of that before? Meh, can't be bothered looking for a good one right now... and I can't be bothered trying every one in the pre-installed list to see if any of them actually work...)

By the way, it doesn't matter whether you have theta1 - theta2 or theta2- theta1. The result is the same. This is because theta1 - theta2 is equal to -(theta 2 - theta1), and cos(x) = cos (-x).

Now, as for the graphs of polar equations... well, I've already written about them before in my very brief review of chapters 1-6, but I might as well just copy-paste the info here into a dedicated Polar Coordinates post.


If you have theta = (angle), then the graph is basically a line in the direction of the angle. (i.e. if the equation is theta = pi/4, then you have a line continuing at pi/4 radians, or 45 degrees, as measured anticlockwise from the positive x-axis). If it says anywhere that r doesn't have to be greater than theta, then the line extends in both directions.

If you have r = (constant), then the graph is a circle with a radius equal to the constant.

You can also get inequalities for these as well. Remember, if it's a greater than/ less than sign without the equals bit underneath, you need to draw a dotted line, not a solid line.

Oh yeah, and there are the spirals too, in the form r = k(theta), where k is a constant, r is the magnitude and theta is the angle. Normally you see these in those questions where it asks you to write the equation for the graph. Normally it helps to use the values of r at (pi/2) and/or pi to help you determine k, and, therefore, the equation.

As for complex numbers in polar form, and how to multiply and divide them, I've already written about that in my aforementioned post about Complex Numbers and Polar Coordinates.

That leaves graphing regions in the complex plane. This is pretty simple and is kind of related to the graphs of polar equations.

If you have |z| = k, where k is a constant, the graph is basically a circle with the centre as the origin and radius k. z is any number with a modulus equal to k.

If you have |z - w| = k as your equation, where w is a complex number (normally given in the form a + bi) and k is a constant, then the graph is a circle with centre w and radius k. Be careful: sometimes they'll give an equation in the form |z + w| = k. If you get this, make sure to rearrange the equation to |z - (-w)| = k, otherwise you'll end up with the centre in the wrong place.

Finally you have the annoying ones which are in the form |z - w| = |z - p|, where w and p are both complex numbers. To do these, mark points w and p on the Argand diagram. Then draw a dotted line between w and p. Finally, draw a continuous solid line that bisects the dotted line (bisecting line = a line that crosses another line in its centre and runs perpendicular to that line). Every point on the solid line should then be equidistant (i.e. of equal distance) from w and p.

And that's pretty much it for those two chapters. It kind of helps that I covered loads of stuff before...

Complex Numbers and Polar Coordinates

(This post was cut and pasted from one of my other blogs.)


So... I have this Maths Specialist investigation on complex numbers in polar form, and they say that the best way to learn something is to teach other people, which is what I'm going to do!

These posts really need diagrams... but I'm pretty lazy, so I'll just add diagrams when I feel like it/ when you pressure me to :)

ANYWAY, I'll get to the point.

Polar Coordinates

Most of the time, when you are given coordinates for a graph, you are given the coordinates in rectangular form. Rectangular form is basically (x, y) where x is the number of units to the right and y is the number of units up. Polar coordinates differ quite drastically: they give coordinates in the form (r,θ) where r is the distance from the origin, and θ is the angle anticlockwise from the positive x-axis.

So, how do you go about converting rectangular coordinates to polar form? It's simple...

To find r, the direct distance from the origin, all you have to do is use Pythagoras' theorem. According to Pythagoras' theorem, r^2 = x^2 + y^2, so r is the square root of x^2 + y^2. (This would all make a lot more sense if I put a graph here, but as I said, I'm pretty lazy.)

Finding the angle between r and the x-axis takes a bit more work. Using SOHCAHTOA (sin = opp/hyp, cos = adj/hyp, tan = opp/adj), tan θ is equal to y/x. (Once again, it would be a lot easier to see if I could be bothered drawing a graph.) Therefore, θ is simply the reverse tangent of y/x.

The above was for if the point (x,y) is in the first quadrant of the graph (i.e. x and y are positive). But what happens if the point (x,y) is in a different section of the graph? No worries- you may simply have to add or subtract 180 or 360 to give the anticlockwise angle from the point (x,y) and the positive x-axis. If the graph lies in the 2nd quadrant (i.e. x is negative, y is positive), you need to subtract the angle from 180 degrees. If the graph lies in the 3rd quadrant (i.e. both x and y are negative) then you need to add 180 degrees. If the graph lies in the 4th quadrant (i.e. x is positive, y is negative), then you need to subtract the angle from 360 degrees. If you ever get confused, draw a graph and draw an angle anticlockwise starting from the positive x-axis. Diagrams are a very useful tool for helping your understanding.

By the way, there are two acceptable formats for giving an angle: the angle anticlockwise from the positive x-axis is given as a positive angle, while the angle clockwise from the positive x-axis is given as a negative angle. Therefore, the angle 270 degrees can also be written as -90 degrees.

To convert from polar form to standard Cartesian coordinates, it helps to draw a quick sketch of the triangle with its hypotenuse and angle marked. You can then easily use SOHCAHTOA to find x and y.

Complex Numbers

Complex numbers are made up of two parts: a real part and an imaginary part. They take on the form z = x + yj (i is sometimes substituted for j, but they both mean the same thing). x is the real part, and yj is the imaginary part. j (or i) is equal to the square root of negative one (which is a real number), but x and y themselves are real numbers- one example of a complex number is 5 + 3j.

Let's take any real number and call it a. If you have the square root of a, and you raise it to the power of 2, you end up with a. Similarly, if you take the square root of -1, and you raise it to the power of 2, you should theoretically also end up with -1. As j = -1, j^2 must also equal -1. This is important for complex number arithmetic.

Adding, subtracting and multiplying complex numbers is easy. When adding or subtracting, just add or subtract the real parts together and then add or subtract the imaginary parts. When multiplying, just multiply as you would normally do in algebra. However, if you end up with j^2, make sure to substitute -1 in its place. (Additional substitution may be required for higher powers of j.)

Dividing requires a new concept: the conjugate of a complex number.

Basically, to get the conjugate of a complex number, just change the addition sign to a subtraction sign and vice versa.

When dividing complex numbers, multiply by the conjugate of the bottom number over itself. For example, if you want to do (2 + 3j)/(4 - 6j), multiply by (4 + 6j)/(4 + 6j). This works for two reasons:

Reason 1: Difference of Two Squares.
When you multiply the bottom number by its conjugate, you get a difference of two squares expression. (x + yj)(x - yj) = x^2 + xyj - xyj - (yj)^2. xyj and -xyj then cancel each other out and the resulting expression can be expanded to x^2-(y^2)(j^2). As I have said earlier, j^2 = -1, so (x + yj)(x - yj) = x^2 + y^2 which is a real number! Therefore, after multiplying a fraction by the conjugate of the bottom number over itself, you really end up simply dividing by a real number at the end.

Reason 2: The Conjugate Over Itself is Equal to 1.
When you multiply the fraction by the conjugate of the bottom number over itself, you simplify it so that you are dividing by a real number, but you are not changing the value of the expression. This is because any number over itself is equal to 1. Therefore, when you are multiplying by the conjugate over itself, the value of the expression remains unchanged.

Complex numbers are able to be graphed. The name for the graph is called the Argand diagram but it's actually very similar to any other graph. The main difference is that the x-axis and y-axis are given different names: the x-axis is the real axis and the y-axis is the imaginary axis. To find the location complex number x + yj on the Argand diagram, simply move x units to the right and y units up.

Now on to the next part...

Complex Numbers in Polar Form

Unfortunately, converting complex numbers to polar form isn't as easy as converting normal numbers to polar form. I think that this is because a complex number has two components (real and imaginary), while a normal number only has one real component. Nevertheless, it's not that hard.

First, we have to use a little trigonometry (again, I need a diagram here). Let r be the distance from the origin to the point x + yj on the Argand diagram and θ be the angle counterclockwise between the positive real axis and r.

cos θ = adj/hyp = x / r. Therefore x = r cos θ.
sin θ = opp/hyp = y / r. Therefore y = r sin θ.

Substituting these values into the general complex number form z = x + yj you get:

z = r cos θ + (r sin θ)j = r (cos θ + j sin θ)

Now, using Pythagoras, r = sqrt(x^2 + y^2). Now |z| = sqrt(z multiplied by the conjugate of z). As I stated in the above section about complex numbers, z multiplied by its conjugate is x^2 + y^2. Therefore, |z| = r = sqrt(x^2 + y^2).

Therefore, z = |z| (cos θ + j sin θ).

Some terminology for you: |z| is the modulus of the complex number. θ is the argument (or arg).

Now let's move on to... the powers of complex numbers in polar form.

No, I didn't mean that they had powers, I meant raising them to a power. Don't get too excited.

When you raise z to a power p, the modulus also gets raised to a power p. The argument, on the other hand, simply gets multiplied by the power. It's like the log laws!

log (z^p) = p log z
arg (z^p) = p arg z

Multiplying and Dividing Complex Numbers

Multiplying and dividing complex numbers in polar form are like raising complex numbers in polar form to powers. The modulus gets treated like a normal number while you have to treat the argument like a logarithm.

When multiplying two complex numbers, say z and w, the modulus is simply |z|*|w| = |zw|. The argument is simply arg (z) + arg (w) = arg (zw).

When dividing two complex numbers, say z and w, the modulus is simply |z|/|w| = |z/w|. The argument is simply arg (z) - arg (w) = arg (z/w).

I think that the laws about multiplying, dividing and raising the modulus to a power are the same as those around normal numbers because the modulus is an ordinary real number as well. But why are the above laws surrounding arguments the same as the laws surrounding logs? Let's take a brief glance at two trigonometry laws and see how we can use them to explain multiplying (and possibly dividing too). These laws are:

cos (θ + φ) = cos θ cos φ - sin θ sin φ
sin (θ + φ) = sin θ cos φ + cos θ sin φ

Let z = |z|(cos θ + j sin θ)
Let w = |w|(cos φ + j sin φ)

zw = |z|(cos θ + j sin θ)|w|(cos φ + j sin φ)
= |zw|(cos θcos φ + i sin θcos φ + cos θ i sin φ + i sin θ i sin φ)
= |zw|((cos θcos φ - sin θsin φ)+ i(sin θcos φ + cos θsin φ))

Using the laws stated previously,

zw = |zw|(cos (θ + φ) + sin (θ + φ))

The modulus of z and w have been multiplied together and the arguments have been added together.

There's a website on how to explain why arg(z^p) = p arg (z), but it looks too confusing for me. If you're interested, though, you can find it at http://math.tutorvista.com/trigonometry/de-moivre-s-theorem.html .

Very brief review of chapters 1-6 (Sadler 3CD Spec) - Part 2

Well, since I didn't get anywhere near to finishing part 1 last time, here's part 2.

Before I get started, I was really curious to find out how people know about this blog apart from the people I've told, because I would have thought you'd have to be pretty specific in your searches to get this particular blog in amongst all of the other websites displaying the same sort of information. One thing that came up was this website here: http://library.onlinepatashala.com/SubPlay.aspx?Video=admin@2509. This blog is listed under the list of blogs in the bottom right corner. Seems a bit odd that a blog like this, written in such an obnoxious style, can be listed on an educational website like that. Maybe I should make a poll or something to find out how everyone's finding out about this blog because I am one of those annoying overly curious people.

Anyway, time to get started on speed reviewing chapters 3 to 6!

Basics of 3D Vectors

3D vectors are pretty similar to 2D vectors, apart from the fact that they're obviously in 3D and as such have an extra component for the extra dimension. Instead of good ol' ai + bj, now you're dealing with ai + bj + ck. Pretty much everything else is the same, though- adding, subtracting and finding the magnitude are all the same. Yes, even in 3D you're finding sqrt(a^2 + b^2 + c^2). The procedure for working out the unit vector is also the same, as is the procedure for finding the angle between two lines. (Getting bored of this now, are you?) Yes, I need to write a blog post on how to do all this stuff, but as I said before, this is a brief review so I'm assuming that you know stuff like that from last year's Spec work.

(By the way, check out Sadler's example at the beginning of the section, particularly the guy called Captain Over. "Hi Victor, it's Over, over?" Almost as bad as Rosencrantz and Guildenstern- "I think we're off course." "Of course!" I'm not sure if that's the exact wording but I can't be bothered checking.)

Equations of Lines and Planes

The vector equation of a line in 3D works exactly the same as the equation of a line in 2D. Basically, you have the position vector of one point in the line plus a scalar multiple of the direction of the line, in the form r = a + (lambda)b.

The normal equation of a line in 3D unfortunately doesn't work the same way, as every line has multiple possible lines that are perpendicular to it. The way I helped myself visualise this one was by sticking my arm out and then holding my ruler perpendicular to my arm, and then seeing for myself that there are multiple ways in which this could be achieved. Maybe I'm just weird, or maybe it might help you visualise it too.

However, you can have a normal equation of a plane. This is because a plane (a.k.a. a big flat thing that extends in all directions, NOT the aeroplanes that fly, unfortunately) only has one direction that's perpendicular to it. Therefore, you can use the equation r (dot) n = c where c = a (dot) n, where a is a point on the plane and n is a vector perpendicular to the plane. To find out a vector perpendicular to a plane, you can do several things.

1) Sometimes you might get lucky and they'll say that the plane is perpendicular to a certain vector/plane/whatever. Then you can just use the vector normal that they give you!
2) If you get a vector parallel to the plane in question, then you need to find a vector perpendicular to that vector (i.e. if it's perpendicular to a plane parallel to your plane, then it's also perpendicular to your plane. You can see this in my post about Scalar Product and Closest Approach). To do this, you can use dot product and guess and check. I normally let the vector normal be equal to xi + yj + zk and then set the dot product of that and the parallel vector equal to 0. Then I let one of the variables be equal to 0 and then work out what y is in relation to z (for example, y = z or 2x = z). Then I pick numbers that suit that criteria (e.g. for y = z I might choose to make x = 0, y = 1 and z = 1, and for 2x = z I might make it x = 1, y = 0 and z = 2) and double check to make sure that I've wound up with a vector normal to the plane.
3) If you get given two lines, you need to find a vector normal to both lines. Find the dot product between each line and xi + yj + zk, and you'll wind up with two equations for x, y and z. This isn't enough to give you exact numbers, but it is enough to give you the relationship between the variables if you tweak your equations around a bit and/or set one of the variables equal to 0.
4) If you get given three points, then you need to find two lines by finding the position of one point relative to the other two. Then, just use the methods mentioned above.

By the way, there are also vector and Cartesian equations of a plane. The vector equation is in the form r = a + (lambda)b + (mu)c, where b and c are non-parallel vectors that are parallel to the plane. The Cartesian form is simply in the form ax + by + cz = (constant). One great thing about this form is that ai + bj + ck is a vector perpendicular to whatever plane you're looking at.

Interception between lines and planes are worked out in the same way as 2D vectors.

Quick Notes on Proofs by Deduction

"Proof by Deduction" is basically the fancy term for most of the proofs that you've probably been doing so far- you know, the ones where you say that "this equals this, because of that axiom" or whatever and you keep putting down logical facts and explanations until you prove whatever it is that had to be proved? Okay, terrible explanation. Maybe I should provide an example: trig proofs. Proofs where you say "a equals e, because they are corresponding angles, and e equals f, because they are vertically opposite, so therefore a equals f." Stuff like that. Here are some of the most basic "known truths" which you might use to prove stuff. (Okay, I think they're called "axioms" or something like that, but whatever.)

1. A straight line has an angle of 180 degrees.
2. Angles around a point add up to 360 degrees.
3. Alternate angles are equal.
4. Corresponding angles are equal.
5. Co-interior angles are supplementary (add to 180 degrees).
6. Angles of a triangle add to 180 degrees.
7. Angles of a quadrilateral add to 360 degrees.
8. Angles of an n-sided polygon add to (n - 2)*180 degrees.
9. The exterior angle of a triangle equals the sum of the two interior opposite angles.
10. Vertically opposite angles are equal.

There's also the counter-example method, which is where you prove or disprove something by providing a counter-example to the given statement. This is a rather simple example but if you wanted to prove that not all odd numbers are divisible by 3 you could just say that 25 isn't divisible by 3.

Proofs using vectors are also rather straightforward, except that you have to take care with the direction of the vectors as well as their magnitudes.

Proof by Exhaustion

Basically, a Proof by Exhaustion literally exhausts all of the possibilities. In Proofs by Exhaustion, you first divide all real numbers up into categories and test each category separately according to whatever you're trying to prove. For example, you might need to test all of the odd numbers and all of the even numbers, or all of the multiples of 3, all of the multiples of 3 plus one, and all of the multiples of 3 plus two. How you divide the numbers up also depends on what you're trying to prove.

Proof by Contradiction

These are NASTY. In Proofs by Contradiction, you have to prove that the inverse of whatever statement you're given isn't true, and therefore the original statement must be true. For example, you might be given the lengths of three sides of a triangle and asked to prove that it's not a right-angled triangle. You would go about this by first acting as though it is a right-angled triangle, before showing that that assertion falls flat on its face, and then concluding that the original statement- that the triangle isn't right-angled- must be true. For example, if the side lengths are 8cm, 9cm and 10cm, you could show that 8^2 + 9^2 doesn't equal 10^2 and therefore doesn't conform to Pythagoras' Theorem for right-angled triangles. Therefore, this triangle can't be right-angled. That was a relatively simple example- there are much worse examples to be encountered.

Basics of Differentiating Trig Functions

As x approaches 0, (sin x/x) = 1, and ((1-cos x)/x) = 0.

Then, using first principles (can't be bothered putting in all the working out here), the derivative of sin x is cos x, and the derivative of cos x is (-sin x). On a similar note, the derivative of (-sin x) is (-cos x) and the derivative of (-cos x) is sin x.

There are three possible derivatives for y = tan x. One is (1 / (cos x)^2), another is (sec x)^2, and a third is 1 + (tan x)^2. Just use the one that suits your mood and the question that you're doing.

The product rule, quotient rule and chain rule also work here. With the chain rule, the power takes precedence, i.e. in (cos u)^a, the derivative is a(cos u)^(a-1)(-sin u)(u'). Bear in mind that the derivative of cos u is (-sin u)(u') according to the chain rule (dy/du is -sin u in this case, and then u' is another way of notating du/dx, and then dy/dx = (dy/du)(du/dx) according to the chain rule).

Implicit Differentiation

Implicit differentiation examples are too long and annoying to type up in a brief review, so I'll just tell you the concept here. Implicitly defined functions are functions that don't have one variable on one side and the other on the other side (e.g. x^2 + y^2 = 10). When differentiating these, you can either rearrange the equation (which isn't always possible) and differentiate normally, or you can use implicit differentiation. In implicit differentiation you have to differentiate every component separately in respect to one of the variables (normally x) and then factor out the (dy/dx) parts to get an equation for (dy/dx). Normally the derivative will be in terms of both x and y.

When you have to differentiate y in respect to x, you get (dy/dx). After all, (dy/dx) can be read as "the derivative of y with respect to x." When you've got something like y^3, you have to use the chain rule, so you end up with (3y^2)(dy/dx). Finally, if you have something with x and y multiplied together, you have to use the product rule and separate the x and y. (If there are constants, pair them up with x.) For example, the derivative of 3xy with respect to x is 3y + 3x(dy/dx).

Differentiating Parametric Equations

Thankfully, these aren't too hard to do. Parametric equations are when you have, for example, an equation for x in terms of t and an equation for y in terms of t. You can differentiate both with respect to t so that you end up with an equation for (dy/dt) and one for (dx/dt). Then you can use the Chain Rule to end up with an equation for (dy/dx)! As (dy/dx) = (dy/dt)(dt/dx), all you need to do is get the reciprocal of (dx/dt) (i.e. 1/(dt/dx)) and multiply it by (dy/dt) to get (dy/dx)!

(Finding the second derivative isn't so easy, but I can't remember the process for doing so.)

So that's a review of chapters 1-6 finished! Yay! When I've got some time and a lot more motivation than I do now, I might flesh out some of the topics above.

Very brief review of chapters 1-6 (Sadler 3CD Spec)

Well, I have a test tomorrow, and I've actually slacked off for the past week or so, so now it's time to do some power studying. (Actually I've already done some power studying, now this is just me writing up about it.)

My idea of power studying is going through all of the topics listed in a book or course outline and writing down stuff on each of the dot points or book headings. If there's something that I can't write anything for, then I revise it.

All right. Here we go... a speedy review of chapters 1-6!

Chapter 1: Polar Coordinates

Distance between two points

Just use cosine rule (i.e. c^2 = a^2 + b^2 - 2ab cos (theta)). Polar coordinates give the lengths of two sides of the triangle so a and b are all worked out for you. As for cos (theta), theta is simply the difference between the angles given in the polar coordinates. Then solve to find c. If you get stuck, just draw a diagram and it should all make sense (if it doesn't, then revise the cosine rule and/or polar coordinate basics).

Polar equations and graphs

If you have theta = (angle), then the graph is basically a line in the direction of the angle. (i.e. if the equation is theta = pi/4, then you have a line continuing at pi/4 radians, or 45 degrees, as measured anticlockwise from the positive x-axis). If it says anywhere that r doesn't have to be greater than theta, then the line extends in both directions.

If you have r = (constant), then the graph is a circle with a radius equal to the constant.

You can also get inequalities for these as well. Remember, if it's a greater than/ less than sign without the equals bit underneath, you need to draw a dotted line, not a solid line.

Oh yeah, and there are the spirals too, in the form r = k(theta), where k is a constant, r is the magnitude and theta is the angle. Normally you see these in those questions where it asks you to write the equation for the graph. Normally it helps to use the values of r at (pi/2) and/or pi to help you determine k, and, therefore, the equation.

Chapter 2: Complex Numbers

Polar form of a complex number

Okay, there's an explanation for this, and while I tend to value understanding over memorisation, this brief summary is an exception. Basically, the polar form is in the form r cis (theta) where r is the modulus (i.e. the length between the origin and the location of the number on an Argand diagram), cis is short for "cos theta plus i sin theta" (ties in with the explanation that I'm missing here) and (theta) is the argument, or the angle between the positive x-axis and the line between the origin and the location of the complex number. Normally, theta is stated in radians, and the principal argument (the one normally used) is between negative pi and pi (including positive pi, but not including negative pi).

By the way, that old familiar form of a complex number a + bi is also called the Cartesian or rectangular form of a complex number.

To convert from rectangular form to polar form, first find the modulus by using Pythagoras' theorem (i.e. modulus = sqrt(a^2 + b^2)). Then use SOHCAHTOA to find the angle/argument. Sometimes it helps to draw a diagram to make sure you get the right angle. Remember, the angle you want is the one measured from the positive x-axis.

To convert back, draw a diagram and use SOHCAHTOA or just have a couple of formulas memorised. The ones you'll want are x = r cos (theta) and y = r sin (theta). (Hopefully I've got them right there...)

Multiplying and dividing complex numbers in polar form

Believe it or not, there ARE benefits to polar form- it's much easier to multiply and divide in this form!

To multiply: multiply the moduli, add the arguments.
To divide: divide the moduli, subtract the arguments.

Then just make sure that the argument fits into the definition of a principal argument (between negative pi and positive pi- but not including negative pi). Done!

Complex numbers- regions on a diagram

For any complex number z, if you ever get the equation |z| = (constant), then the diagram is basically a circle with the centre at the origin and a radius equal to the constant. Apparently, there's a fancy word for it- the "locus" of |z| = (constant).

Then, if you have something like |z-w| = (constant), where w is another complex number, then the diagram is basically a circle with its centre at point w and a radius equal to the constant.

And then you can do all kinds of inequality stuff here too.

Okay, stuff what I said about chapters 1 to 6. It's 9.27pm, and I want to get to bed early-ish because I have orchestra tomorrow and I think that sleep will help me more than studying. So good night, everyone :)

(By the way, I found out that I got 30 page views on the 31st of March. I was kind of surprised, to be honest, because I didn't think that anyone but me ever came on here since my blog is so nerdy, and even then I haven't updated my blog in a few months. Hmm. I'm curious now.)