VCE Specialist 3/4 Question Thread!

[Gogo14]

Can someone help me explain why x does not equal pi? I found that it did in my workings, but in the answers there is no pi.
http://imgur.com/FipYIGD
http://imgur.com/fG1pqR1

[RuiAce]

Can someone help me explain why x does not equal pi? I found that it did in my workings, but in the answers there is no pi.
http://imgur.com/FipYIGD
http://imgur.com/fG1pqR1

Two possible explainations.
1. Because in your working you had to square, there's a chance you popped out an extra answer that shouldn't be there. That's why in general if we don't have to square something, we choose not to.

Because you squared, you have to go back and check that all of your solutions work. When plugging back, we find that pi does not work as sin(pi)+cos(pi) = -1

2. You chose to assume that \(\cos x = \sqrt{1-\sin^2x}\) for ALL \(x \in [0, 2\pi] \). This is a mistake, as when x=pi we have

\(\cos x = \cos \pi = -1\)
\(\sqrt{1-\sin^2x} = \sqrt{1-\sin^2 \pi}=1\)

The second explanation is stronger than the first here.

[Syndicate]

Can someone help me explain why x does not equal pi? I found that it did in my workings, but in the answers there is no pi.
http://imgur.com/FipYIGD
http://imgur.com/fG1pqR1

I believe a better way of solving this question would be by expressing \( sinx+cosx=1\) in \(Rsin(x+\alpha) \) or \( Rcos(x- \alpha) \) form.





With this method you won't get any extraneous solutions, as there aren't any squares involved.

[Gogo14]

I believe a better way of solving this question would be by expressing \( sinx+cosx=1\) in \(Rsin(x+\alpha) \) or \( Rcos(x- \alpha) \) form.





With this method you won't get any extraneous solutions, as there aren't any squares involved.

Isnt that just similar to converting to polar form in complex number. What is the equivalent process called for vectors?

[Syndicate]

Isnt that just similar to converting to polar form in complex number. What is the equivalent process called for vectors?

Not sure about the similarity between the polar form of a complex number and this. The technique used here is to simplify the expression into a single circular function.

I am not sure which process you are talking about (complex numbers are vectors). Vectors are already part of the polar coordinate system. You just need other processes to work out their magnitude and angle. However, there are the rectangular (I, j, k) and polar notations (magnitude, angle).

[Gogo14]

Not sure about the similarity between the polar form of a complex number and this. The technique used here is to simplify the expression into a single circular function.

I am not sure which process you are talking about (complex numbers are vectors). Vectors are already part of the polar coordinate system. You just need other processes to work out their magnitude and angle. However, there are the rectangular (I, j, k) and polar notations (magnitude, angle).

Wait, complex numbers are vectors?!????Also how would you do it for something which has 3 dimensions (i,j,k)?

[RuiAce]

Wait, complex numbers are vectors?!????Also how would you do it for something which has 3 dimensions (i,j,k)?

I wouldn't say they "are" vectors, but complex numbers can definitely be represented as a 2D vector.

There is no analogue for a 3D vector.

[Quantum44]

I was working through the textbook for specialist maths doing the vectors chapter to get a head start on preparing for the first SAC and I was struggling with this question so I'm hoping someone can help me.

Given that a = i -  j + 2k, b = i + 2j + mk and c = 3i + nj + k are linearly dependent, express m in terms of n in simplest fraction form.
 
I used simultaneous equations on my CAS to get m = (2n - 9)/(n + 3) but the answers say it's (7n + 9)/(6 - n) and I cannot see where I went wrong.

[nyggfany]

I was working through the textbook for specialist maths doing the vectors chapter to get a head start on preparing for the first SAC and I was struggling with this question so I'm hoping someone can help me.

Given that a = i -  j + 2k, b = i + 2j + mk and c = 3i + nj + k are linearly dependent, express m in terms of n in simplest fraction form.
 
I used simultaneous equations on my CAS to get m = (2n - 9)/(n + 3) but the answers say it's (7n + 9)/(6 - n) and I cannot see where I went wrong.

Don't worry about the answer from the textbook. The answer from the worked solutions is (2n-9)/(n-3) so you should be fine

[Quantum44]

Don't worry about the answer from the textbook. The answer from the worked solutions is (2n-9)/(n-3) so you should be fine

Thanks, I was quite confused for a while

[keltingmeith]

Wait, complex numbers are vectors?!?

Lots of things are vectors. sin(x) and cos(x) are actually an example of two linearly independent vectors. Hell, even scalars can be considered a vector (which is just absolutely hilarious and downright confusing, so don't think too hard on it). I once had a lecturer tell me that if you want to work in maths, unless you're doing weird combinatorics type stuff, you'll be working in a vector space.

[RuiAce]

Lots of things are vectors. sin(x) and cos(x) are actually an example of two linearly independent vectors. Hell, even scalars can be considered a vector (which is just absolutely hilarious and downright confusing, so don't think too hard on it). I once had a lecturer tell me that if you want to work in maths, unless you're doing weird combinatorics type stuff, you'll be working in a vector space.

Surely for a high school student they only care about n-tuples and things with "magnitude and direction" though

[keltingmeith]

Surely for a high school student they only care about n-tuples and things with "magnitude and direction" though

In terms of direct testing, yes. But understanding that a vector can be represented by lots of different things is certainly helpful in many different scenarios - particularly for when they hit kinematics and start using what look like scalars as vectors.

[Quantum44]

Where can I buy the worked solutions for the textbook?

[deStudent]

Need help with Q1 and Q2b: http://m.imgur.com/1rRQsZg

For Q1) I got the answer correct but had to guess one step which I still don't understand. I made cos(3t) and sin(3t) the subject, since the equation for a circle is sint squared + cost squared = 1, how does squaring cos3t and sin3t allow me to still get the correct answer?

Q2b) The answer has x = 4 but didn't show their method. I subbed the points and solved for the gradient which is undefined. How would I show that the line is x=4 mathematically?

Thanks