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.