VCE Specialist 3/4 Question Thread!

[BubbleWrapMan]

Put restrictions on the interval, e.g. try to solve between T = 0 and T = 100, then 100 to 200, etc., until you get a solution.

[SocialRhubarb]

Anyone have any solutions to this?

Yeah, what BubbleWrapMan said. Restrict the possible range of solutions. I don't know how to use a Classpad, but I think restricting it to just t>0 would be enough in this case.

[ashoni]

Ahh okay I see! thanks guys

[Daenerys Targaryen]

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2007specmaths2.pdf
VCAA2007 MC12 and 4(c)

EditL oops 2007*

Still need help with that MC12

[ahat]

Hey, other than just observing the angles (i.e. all 90^o apart), what would be the algebraic approach to solving this problem? Could we do
z^1/4 = 16cis(π/3) and then use DeMoivre's theorem? Thanks.

Spoiler

I'm asking this question because at the start of the year we did a sac with a similar question and this is the method I used. That method seems to be incorrect for this question though. Advice really appreciated.

Spoiler

Still need help with that MC12

I've attached a solution, more convenient than typing The key word in this question is "rate" which instantly tells us that it's an application of the chain rule question.

[Daenerys Targaryen]

Hey, other than just observing the angles (i.e. all 90^o apart), what would be the algebraic approach to solving this problem? Could we do
z^1/4 = 16cis(π/3) and then use DeMoivre's theorem? Thanks.

You can just go and add it onto the Arg

[ahat]

You can just go and add it onto the Arg

How about the modulus though? That's what is confusing me.

[~T]

Still need help with that MC12

Because the shuttle is ascending vertically upwards, then its displacement is along the vertical side of the right-angle triangle drawn. Let's denote that as x. Now, we are looking for the speed, which is the (magnitude of the) rate of change of displacement with respect to time. We have the rate of change of the angle with respect to time, so we use that to form a related rates question.





We can use the right-angled triangle to find a relationship between x (the vertical) and theta. Using the tangent identity, we get:







We have the rate of change of the angle in the question as 0.5, so put these together and you get:





This comes out as A.

[BubbleWrapMan]

Hey, other than just observing the angles (i.e. all 90^o apart), what would be the algebraic approach to solving this problem? Could we do
z^1/4 = 16cis(π/3) and then use DeMoivre's theorem? Thanks.

You can solve the equation for , but adding multiples of is far more efficient.

How about the modulus though? That's what is confusing me.

What do you mean by the modulus?


EDIT: I should really say solve for , since as defined in the question is equal to and we want to find the other fourth roots .

[availn]

How about the modulus though? That's what is confusing me.

If you're talking about the r in r·cis(θ), it's pretty simple. As you said yourself, z^1/4 = 16cis(π/3), so therefore z = 16⁴cis(4π/3). So when you quad root it again, you know that r must still equal 16.

[BubbleWrapMan]

Yes, the modulus of every solution for to the equation is the same.

To see this, let and be solutions to . In other words, and .

Taking the modulus of both sides of each equation we get and , so .

But for any complex number by De Moivre's theorem, so . Since both moduli are non-negative real numbers, we can conclude that . This shows that the roots of all have the same modulus.

[thecreeker]

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2004specmaths2.pdf

Question 5c
Might be a noob move, but could some explain why we have to minus the angle at the gradient from 90? Itute doesn't do a great job of explaining

[ahat]

16⁴cis(4π/3). So when you quad root it again, you know that r must still equal 16.

Hahaha, my gosh, how did I not realise this! (such a perfect answer btw)
Thanks guys, I appreciate it

[availn]

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2004specmaths2.pdf

Question 5c
Might be a noob move, but could some explain why we have to minus the angle at the gradient from 90? Itute doesn't do a great job of explaining

It's because the disk is rotated, and placed on the ground horizontally. So after finding the angle of the gradient, you have to subtract it from 90, as you are rotating the disk.

[thecreeker]

It's because the disk is rotated, and placed on the ground horizontally. So after finding the angle of the gradient, you have to subtract it from 90, as you are rotating the disk.

ahhh that makes sense
Thank you very much and good luck for tomorrow