VCE Methods Question Thread!

[darklight]

Thanks for the help. Your method is all correct but I realised you made a calculation error when dividing d/2 by 80. It equals d/160, making the answer 160x/x+80


EDIT: whoops fixed it before me

[kbed]

Hey guys, how do you find the determinant of a 3 x 3 matrix?

[polar]

BORED already? Want to learn a bit of maths? :P

scroll down to the one by calvin climb titled 'determining (n x n) matrix determinants'

[b^3]

Hey guys, how do you find the determinant of a 3 x 3 matrix?

Just remember you won't need to know how to do it by hand for the methods course, if you are asked then its just using the calc to find it, just don't spend too much time on it unless you really want to know how to do it by hand.

[e^1]

Hi guys,

Stumbled a bit on this one.



But integrating otherwise by expanding first:



Though I would have taken the first integrated expression as a constant, why does the former have that number, while the expanded one does not? Is there something in the process of integrating without expanding that it considers or ignores?

Thanks

Note: Latex a bit slow for some reason.

[b^3]

When we integrate the function we get a family of anti-derivatives, thats why we add the plus C on, so effectively for the first one we still don't have a single anti-derivative, since we can still move it up or down by the +C, same goes for the second one. If we had an initial condition we could solve it out to find that they would give the same.

The other way to look at it is Let , and so both have a constant added that will give the vertical translation.

[vashappenin]

Hey I really need help with this question: (ex2G, essential methods 3/4)

2. Consider the simultaneous equations and x+y+z=6
b.Let z= λ. Solve the equations to give the solutions in terms of λ

2a asked to subtract the 2nd equation from the 1st to find y in terms of z if that helps!

I'm really confused and it's freaking me out! Haha I feel a little awks posting a question like this when others have already moved on to other chapters! Ohwells.. I feel like I'm screwed because I'm struggling with orientation hw :S
Please help me D:

EDIT: Latex screwed up my second equation, so I'll just type it out w/o latex

[polar]

So far, we know that the set of solutions in terms of is and , but we still need to find in terms of , to do that, substitute the 'new' y and z into equation 2 (equation 1 is more work)



thus,

[BubbleWrapMan]

It should probably be noted that parameterised solutions to simultaneous equations don't come up in methods exams

[vashappenin]

So far, we know that the set of solutions in terms of is and , but we still need to find in terms of , to do that, substitute the 'new' y and z into equation 2 (equation 1 is more work)



thus,

Thanks! That makes so much more sense now

It should probably be noted that parameterised solutions to simultaneous equations don't come up in methods exams

Ok I sound really thick, but what are parameterised solutions? :S

[Homer]

parametrized solutions are a set of solutions in terms of the parameter which in this case was

[vashappenin]

Ok just a warning that I'll probably be posting quite frequently throughout the year during the timeframe that I'm while I'm doing methods homework/etc.. and I might post the dumbest questions, but please don't judge!

Solve these simultaneous equations, giving your answer in terms of a parameter

and

[FlorianK]

Ok just a warning that I'll probably be posting quite frequently throughout the year during the timeframe that I'm while I'm doing methods homework/etc.. and I might post the dumbest questions, but please don't judge!

Solve these simultaneous equations, giving your answer in terms of a parameter

and

you add up both equations and you get 2z=10
so z=5

[vashappenin]

you add up both equations and you get 2z=10
so z=5

Yeah I understood that part
But don't you need to do anything after that? To give the answer in terms of ?

[BubbleWrapMan]

Put z = 5 back into the equation, then you should get x = y - 1 (I think)

If you let y = t, then your solutions are

x = t - 1
y = t
z = 5