VCE Specialist 3/4 Question Thread!

[Eric11267]

For u to exist (the vector resolute of a in the direction b), does the mod of a have to be smaller than the mod of b?

Nope not required, because its in the direction of b, so the length of b is irrelevant.

[Willba99]

Nope not required, because its in the direction of b, so the length of b is irrelevant.

thank you!

[vasuk]

Could someone help with this question please

a teacher decides to trial a new method for teaching her classes, to see if their test results improve using this method.
she knows from previously they have a test average of 74 and a SD of 8.
after trialling the new method she selects a sample of 25 students and finds the test average is now 77.

a) write down the null and alternative hypotheses, find p value
my answer
H0 : μ = 74
H1 : μ > 74
p value on cas: 0.0303963

b) can she conclude the new method has improved results? testing at 5% level of sig
how do i explain this and what do i say

thanks

[Eric11267]

Could someone help with this question please

a teacher decides to trial a new method for teaching her classes, to see if their test results improve using this method.
she knows from previously they have a test average of 74 and a SD of 8.
after trialling the new method she selects a sample of 25 students and finds the test average is now 77.

a) write down the null and alternative hypotheses, find p value
my answer
H0 : μ = 74
H1 : μ > 74
p value on cas: 0.0303963

b) can she conclude the new method has improved results? testing at 5% level of sig
how do i explain this and what do i say

thanks

You would probably say that since the p value obtained is less than the 5% significance level, then the results are statistically significant and the null hypothesis can be rejected, whilst the alternative one is accepted.

[Syedali_]

With vectors is it alright to take common factor out
ie. 2i + 4j + 8k
as 2( i + 2j + 4k)

or should it remain expanded

Thanks

[Sine]

With vectors is it alright to take common factor out
ie. 2i + 4j + 8k
as 2( i + 2j + 4k)

or should it remain expanded

Thanks

i don't think it matters too much but I personally think the expanded version looks nicer on paper.
However depends on the question, for something like proving two lines are parallel I would definitely factor out to make it look clearer.

[carol1234567890]

Could anyone please post an explanation of how to do implicit differentiation when it involves the product rule etc - just a really long example
I don't understand it
Thanks
So
Much

[Eric11267]

Could anyone please post an explanation of how to do implicit differentiation when it involves the product rule etc - just a really long example
I don't understand it
Thanks
So
Much

One way I like to look at it is if you have a term that needs to be differentiated that has both x and y in it, first differentiate with respect to x, pretending y is a constant, and then differentiate with respect to y, pretending x is a constant and add dy/dx. For example:

In this case you first differentiate with respect to x, with y as a constant, and then you differentiate with respect to y, with x as a constant and add dy/dx
Another example:

Then you would just rearrange to get dy/dx
I hope this makes sense, if its still not clear let me know

[Willba99]

If a parallelogram's four corners are cis functions such that they are defined by z1=acis(b), z2=ccis(d), and so on for z3 and z4, will z1*z2*z3*z4 always =0?

Cheers

[gnaf]

How do I Q4, 5 and 11 in the VCAA NHT spesh exam?

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2017/nht/sm1nht_examrep17.pdf

For Q4, the ans are a=-5, b=1 and a=-2, b=-2. Expanding (z-1-1)(z-1+i)(z+a) then matching coefficient with the given eqn, I only get a=-2, b=-2... how did I NOT get a=-5, b=1?

For Q5, I get u=3pi/4 for both the upper and lower terminals since u=1-x^2 where x=-1/2 and x=1/2...
The ans is pi squared/ 3 btw

For Q11, I get 4api/ 3 since but the answer is 6a

[keltingmeith]

If a parallelogram's four corners are cis functions such that they are defined by z1=acis(b), z2=ccis(d), and so on for z3 and z4, will z1*z2*z3*z4 always =0?

Cheers

Good question - let's do a little exploring!

So, we start with defining our four complex numbers:


And now, we multiply them together, remembering our rules for multiplying complex numbers:


Now, we want to know if this is going to be zero if our four points make a parallelogram. Instead of assuming we have a parallelogram and making them equal 0, I'm going to instead go the opposite way, make them 0, and see if that leads to something that should be a parallelogram. Since we know that each of our modulus can't be zero, this gives us the final equation:



This is where it gets interesting. See, for a complex number to be zero, both the imaginary component and the real component need to be 0. So, if this is true, then all of these angles must add to a number that is 0 for BOTH sin and cos. But, there is no number for which this is true - which means that not only will the product of four complex numbers never be 0 unless they have a 0 modulus, but they will definitely not be 0 if they define a parallelogram unless one of the points is the origin.

[Willba99]

Hey guys! Can someone help me with question 5)c) in Section B of Exam 2, 2016 (VCAA)? I understand that I stuffed up by assuming constant acceleration, but I've no clue how to do the question correctly. I can't understand the worked solutions either.

My computer is being an absolute dog and its not letting me attach an image of my working out, but the link is:
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2016/2016specmath2-w.pdf

Can someone walk me through this question?

Thanks in advance

[Willba99]

Good question - let's do a little exploring!

So, we start with defining our four complex numbers:


And now, we multiply them together, remembering our rules for multiplying complex numbers:


Now, we want to know if this is going to be zero if our four points make a parallelogram. Instead of assuming we have a parallelogram and making them equal 0, I'm going to instead go the opposite way, make them 0, and see if that leads to something that should be a parallelogram. Since we know that each of our modulus can't be zero, this gives us the final equation:



This is where it gets interesting. See, for a complex number to be zero, both the imaginary component and the real component need to be 0. So, if this is true, then all of these angles must add to a number that is 0 for BOTH sin and cos. But, there is no number for which this is true - which means that not only will the product of four complex numbers never be 0 unless they have a 0 modulus, but they will definitely not be 0 if they define a parallelogram unless one of the points is the origin.

thanks so much!!

"Sine" MOD EDIT: removed multiple quotes

[Willba99]

re why I quoted keltingmeith three times:

my computer is being a dog today.

[Sine]

re why I quoted keltingmeith three times:

my computer is being a dog today.

dw removed 2 of the qutoes