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Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2825036 times)  Share 

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keltingmeith

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Re: Specialist 3/4 Question Thread!
« Reply #8400 on: November 22, 2016, 12:53:48 pm »
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Asymptotes are allowed to cut the graph and you are allowed to draw it in cutting through in VCE exams.

This is a true fact - not exactly indicative of an asymptote, though. It's like saying the sky is blue - it doesn't tell you what the sky is, just a fact about it.

Asymptotes : The curve approaches,not touching not curving away.

This is what an asymptote is - only need to add that it's a limiting behaviour, i.e. as x goes to infinity (except for vertical asymptotes, which instead sends y to infinity)

To find vertical asymptotes solve the denominator to  = 0* or find logs and equate the terms inside to 0. Horizontal asymptotes

Note*: Be careful as sometimes percieved asymptotes are really just a point of discontinuity if factors in the numerator and denominator can cancel out.

That note is SUPER important. f(x)=(x^2-2x+1)/(x-1) has no asymptotes.

Horizontal/oblique asymptotes let x approach positive or neg infinity.

Note that this and the last approach aren't really facts, either. Their algorithms - a set of steps you follow to arrive at the answer. However, the important part of algorithms is that they only work in specific contexts - in this case, when you have a continuous function that is in its most simplified form. If the function is not continuous, or not in simplified form, they will break and not work.

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Re: Specialist 3/4 Question Thread!
« Reply #8401 on: November 22, 2016, 06:42:00 pm »
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hmmmm is this an VCE explanation or the (real) explanation?

When your only question doesn't get answered ahaha  :)

Gogo14

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Re: Specialist 3/4 Question Thread!
« Reply #8402 on: November 27, 2016, 05:42:31 pm »
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Is rectangular form for complex numbers the same as cartesian form?
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Syndicate

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Re: Specialist 3/4 Question Thread!
« Reply #8403 on: November 27, 2016, 05:50:04 pm »
+1
Is rectangular form for complex numbers the same as cartesian form?

Rectangular form (same as the cartesian form) of a complex number refers to something like 2 +2i, whereas the polar form of a complex number refers to something like 4cis(60).
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Gogo14

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Re: Specialist 3/4 Question Thread!
« Reply #8404 on: December 01, 2016, 08:49:54 pm »
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In the dot product, why does 2 vectors multiplied together give a scalar?
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Re: Specialist 3/4 Question Thread!
« Reply #8405 on: December 01, 2016, 11:54:52 pm »
+1
Use a double angle formula to show that cos(pi/8) = .5 * root(2 + root(2))

Unless I'm mistaken, wouldn't I have to know what cos(pi/16) is to calculate this?


RuiAce

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Re: Specialist 3/4 Question Thread!
« Reply #8406 on: December 01, 2016, 11:58:11 pm »
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In the dot product, why does 2 vectors multiplied together give a scalar?




Gogo14

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Re: Specialist 3/4 Question Thread!
« Reply #8407 on: December 02, 2016, 09:23:07 pm »
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I mean like, a
scalar times a scalar gives a scalar
Vector times a scalar gives a vector
So why does a vector times a vector give a scalar? Why is the direction gone?
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Syndicate

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Re: Specialist 3/4 Question Thread!
« Reply #8408 on: December 03, 2016, 10:18:05 am »
+2
I mean like, a
scalar times a scalar gives a scalar
Vector times a scalar gives a vector
So why does a vector times a vector give a scalar? Why is the direction gone?

Dot product (also known as the scalar product) is the multiplication of the scalar values of two vectors. In the formula |a|b|cos(x), you are getting the magnitude (which is a scalar value) of vectors a and b, and multiplying it by the angle produced between the two vectors.

You might have confused yourself with the cross product (which is not required to be learnt at a VCE level).
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RuiAce

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Re: Specialist 3/4 Question Thread!
« Reply #8409 on: December 03, 2016, 10:22:04 am »
+3
I mean like, a
scalar times a scalar gives a scalar
Vector times a scalar gives a vector
So why does a vector times a vector give a scalar? Why is the direction gone?
Syndicate mostly covered it I reckon.




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Re: Specialist 3/4 Question Thread!
« Reply #8410 on: December 04, 2016, 10:24:16 am »
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You also have to ask yourself: what do you want to achieve from your 'multiplication'? Let's start from the basics.

What is multiplication? At its most fundamental, multiplication by an integer is repeated addition. Multiplication by a rational is considered as multiplying the numerator and multiplying by the reciprocal of the denominator, which is all good if you're comfortable with operations on rational numbers. Multiplying general real numbers is a little more complicated and the easiest way of thinking about this is to approximate the real number as a sequence of rationals and then define the real number as the limit of that sequence. (so in easier terms, suppose you want to find pi * 2. You could do this by approximating pi with better and better rational approximations, called pi_n, and take the limit of the result pi_n * 2).
But note how everything here is motivated by multiplication as repeated addition. You can do the same with vectors because you can define addition of vectors in a very similar way. However, you can't define a similar multiplication because...what does it mean to add (i + j) a total of (2i + j) times? So instead, mathematicians have tried to define operations on vectors that behave like multiplication.

Let's look at the basic properties of multiplication.
a x b = b x a (commutativity)
a x (b + c) = a x b + a x c (distributive property)
a x 0 = 0 (existence of a zero)
a x (b x c + d) = a x b x c + a x d (linearity)
a x a >= 0, zero if and only if a = 0.

There are other properties too that can't be satisfied by these 'products' of vectors (like existence of a reciprocal) so I won't go into those.
Let's look at the dot product. Note that it satisfies all of the above properties. Therefore, it 'behaves' a lot like multiplication. In fact, any operation on two vectors producing a real number that satisfies the above is called an inner product on vectors.

tl;dr, the dot product is an operation constructed to behave like multiplication, but not actually have the same definition as multiplication.
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Re: Specialist 3/4 Question Thread!
« Reply #8411 on: December 04, 2016, 10:57:08 am »
+1
The thing with these axioms is that they are the conditions required for a set to be considered a "field". These always have two operations - addition and multiplication

The twelve axioms satisfied by a field
1. Closure under addition
2. Closure under multiplication
3. Associativity of addition
4. Associativity of multiplication
5. Commutativity of addition
6. Commutativity of multiplication
7. Existence of a zero element (additive identity) usually denoted by 0
8. Existence of a unity (multiplicative identity) usually denoted by 1
9. Existence of an additive inverse
10. Existence of a multiplicative inverse (not necessary for the additive identity)
11 + 12. Left and right distributivity

Which is the point of the above. Multiplication, at its most simplest form, is defined as repeated addition. Just like how exponentiation is repeated multiplication, and tetration is repeated exponentiation (though these aren't strictly necessary for a field).

If we consider our standard number sets:
R is most certainly a field
C is most certainly a field
Q is also a field - all twelve axioms can be checked
But then Z is not a field, as some axioms start breaking down (e.g. there exists no multiplicative inverse in general)

These are just the standard number sets. Note that a vector space such as R3 isn't necessarily a field. (As an aside, a vector space has to be formed over a field).

And here's where the above explanation comes in. If we wanted to treat the set of ordered triples in R3 as a "field" (i.e. your vectors <a, b, c>) then somehow multiplication must be defined. But how can we define multiplication in a way such that all of the axioms are held together? And IF, this was achieved, how would multiplication be useful?
If you figure out a way to make "vector multiplication" defined and useful then that might be a Nobel prize.

In essence, we must never treat vector spaces the same way we treat fields. The axioms of a vector space specifically do not include "multiplication"; there are axioms related to scaling (scalar multiplication) only because the definition of 'scaling' makes sense and is useful. And lastly, we also need to make sure that we don't confuse word choices either.
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Whether or not the 'dot product' was defined to be similar to multiplication though is probably up to a bit of debate. I reckon, if the properties behind the dot product did not exist then it would be less useful (and possibly discarded, just like how complex numbers would've been without the polar form).

In its algebraic form, the dot product a.b = Σajbj is just defined as the sum of the component-wise products. But staring at that doesn't really tell us anything.

If anything, the link to the geometric interpretation a.b=|a||b|cos(θ) is the first thing that makes it more useful to us.

It is also conveniently linked to a matrix multiplication: aTb = a.b if we interpret vectors to be nx1 matrices. The dot product also gained more significance due to the link with projections (aka resolutes)

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Re: Specialist 3/4 Question Thread!
« Reply #8412 on: December 04, 2016, 04:09:06 pm »
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Re: Specialist 3/4 Question Thread!
« Reply #8413 on: December 04, 2016, 04:11:23 pm »
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Prove:
1. |x - y| <= |x| + |y|
2. |x| - |y| <= |x-y|
3. |x + y + z| <= |x| + |y| + |z|

I've got the answers, but I have no idea behind their reasoning. n general, how would I tackle these sorts of problems?

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Re: Specialist 3/4 Question Thread!
« Reply #8414 on: December 04, 2016, 04:38:57 pm »
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Prove:
1. |x - y| <= |x| + |y|
2. |x| - |y| <= |x-y|
3. |x + y + z| <= |x| + |y| + |z|

I've got the answers, but I have no idea behind their reasoning. n general, how would I tackle these sorts of problems?
I see that these questions are closely related to the "Triangle Inequality", which states:
The sum of any two sides of a triangle is always greater than the third side.

For example, with question 3, if we construct a diagram such as the image attached:
We can see that through the use of the triangle inequality that:
|n|+|m| => |L|          (|L| = |n+m|)
and that:
|j|+|k| => |n|             (|n| = |j+k|)

Using these two inequalities that we have found, we can prove that:
|j| + |k| + |m| => |L|       (|L| = |j+k+m|)

Note: j + k + m = L would only hold when j, k and m are in the same direction, hence, a straight line resultant is formed.
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