VCE Specialist 3/4 Question Thread!

[shiba woof woof]

Okay I might be a bit sloppy in my explanation just now, but for the SA question, I'm pretty sure the "independent" in this case refers to R and C, as in R is independent to C.
But for the total weight of R alone, we have to find Var(4R) instead of Var(R) alone, the terminologies might be confusing but you can sorta get the difference when you do more questions.
Edit: if they somehow add the variables together (the SA question does this, as the total weight is measured instead of measuring the weight of one tyre and then another), instead of calculating it independently like the MCQ (as they measure the mass of chocolate respectively, instead of combining their masses) the variables would be independent to one another, and you'd have to use Var(X1+X2)

I kinda get it more now haha thanks!

I'll try and keep doing more questions similar to this and hopefully start to get the gist of it 

[imactuallyalegend]

Hey guys!
I'm not too sure on how to complete the square with a 4th degree polynomial. Can someone tell me step by step on how to do so?
Example?
z^4-2z^2+4

[LifeisaConstantStruggle]

you can substitute z² with another value, let's say we let w=z²
So you'll get w²-2w+4
From here you can factorise this with the quadratic formula/any way you want.
You'll get 2 complex roots for w, and then, sub z² in, and do the same thing again (factorise either with your calculator, or the quadratic formula works just fine).

[imactuallyalegend]

you can substitute z² with another value, let's say we let w=z²
So you'll get w²-2w+4
From here you can factorise this with the quadratic formula/any way you want.
You'll get 2 complex roots for w, and then, sub z² in, and do the same thing again (factorise either with your calculator, or the quadratic formula works just fine).

I tried doing that but the algebra got messy and seemed too complicated, compared to the answers.
I'm not too sure how they went from the first step to the second step

[LifeisaConstantStruggle]

I tried doing that but the algebra got messy and seemed too complicated, compared to the answers.
I'm not too sure how they went from the first step to the second step

that is just completing the square, done in a fairly unconventional manner (we usually complete the square with a new added value in but for this one it's actually just splitting up z so you can complete the square. You can move the 6z² to the RHS and root both sides to get z alone, and then start solving from there (you'd get a positive and a negative solution for z² then for z).
I'm actually out and I'm typing this on my phone so sorry. will try to send a pic at home.

[j.wang]

The examiners report (2011 vcaa exam 1 q11) said that a mistake when finding the volume was students finding the associated area insteas

what does associated area mean

i can take a pic of what they say but this is moreso a general defi ition question

[Sine]

The examiners report (2011 vcaa exam 1 q11) said that a mistake when finding the volume was students finding the associated area insteas

what does associated area mean

i can take a pic of what they say but this is moreso a general defi ition question

Basically doing the methods equivalent of this question which would be finding the area bounded by the curves y = 0 , y = sin(x) and x= pi/6 (not the volume of the solid of revolution)

[j.wang]

thanks sine! so associated area is finding the area instead of the volume?

also, can i use the same method for both parts of a question (attached)? So in this case, just using the usual method for differential eqns

and in 5b), since we find t in terms of T (antdiff dt/dT), can i sub in T=10 (given in the question directly) to the eqn of t in terms of T or do I have to rearrange for T in terms of t before subbung T=10

[Rieko Ioane]

Hi,

Could I have some help for this? https://imgur.com/a/fkROK

Part e) the answer finds the integral of dx/dV, but isn't this technically the 'concentration'. Wouldn't it make sense to use dx/dt, as I thought the integral of this gives the amount of chemical that has left the tank?

[keltingmeith]

Hi,

Could I have some help for this? https://imgur.com/a/fkROK

Part e) the answer finds the integral of dx/dV, but isn't this technically the 'concentration'. Wouldn't it make sense to use dx/dt, as I thought the integral of this gives the amount of chemical that has left the tank?

Is your answer wrong? Because if I'm reading this right, both of them should be correct - all you want to do is find x, doesn't matter whether you do this by dx/dt or dx/dV, as long as you use the right bounds and setup the integral properly.

[Willba99]

thanks sine! so associated area is finding the area instead of the volume?

Exactly

also, can i use the same method for both parts of a question (attached)? So in this case, just using the usual method for differential eqns

and in 5b), since we find t in terms of T (antdiff dt/dT), can i sub in T=10 (given in the question directly) to the eqn of t in terms of T or do I have to rearrange for T in terms of t before subbung T=10

When subbing in initial conditions to find c, it doesn't matter when the initial condition is subbed in. Generally i sub in the initial conditions as soon as i anti derive so i don't forget but it doesn't matter

[Rieko Ioane]

Is your answer wrong? Because if I'm reading this right, both of them should be correct - all you want to do is find x, doesn't matter whether you do this by dx/dt or dx/dV, as long as you use the right bounds and setup the integral properly.

My answer was wrong, VCAA has 51.6g.

[VanillaRice]

Hi,

Could I have some help for this? https://imgur.com/a/fkROK

Part e) the answer finds the integral of dx/dV, but isn't this technically the 'concentration'. Wouldn't it make sense to use dx/dt, as I thought the integral of this gives the amount of chemical that has left the tank?

dx/dt describes the change in mass of chemical in the tank. So, the integral that you have set up describes the net change in the chemical concentration in the tank from t = 0 to t = 10. In other words, it takes into account both what has come in, and what has left. dx/dt can be thought to be made up of two components i.e. rate in - rate out. Your answer of  34.9g means that overall, 34.9g of chemical has gone into the tank. 
To find the amount of concentration that has left the tank using your method, you must first isolate the 'rate out' term and integrate that from t = 0 to t = 10 instead, which should give you your answer. (Hint: you should be able to find the 'rate out' expression in part b).

Hope this helps

[Rieko Ioane]

Ah right, thanks.

[Rieko Ioane]

dx/dt describes the change in mass of chemical in the tank. So, the integral that you have set up describes the net change in the chemical concentration in the tank from t = 0 to t = 10. In other words, it takes into account both what has come in, and what has left. dx/dt can be thought to be made up of two components i.e. rate in - rate out. Your answer of  34.9g means that overall, 34.9g of chemical has gone into the tank. 
To find the amount of concentration that has left the tank using your method, you must first isolate the 'rate out' term and integrate that from t = 0 to t = 10 instead, which should give you your answer. (Hint: you should be able to find the 'rate out' expression in part b).

Hope this helps

One more thing, I got the correct answer but I tried another way as well but I didnt get the same answer, I'm not sure why https://imgur.com/a/T4wc0

In part ci) they tell us x(t) and when you take its derivative, the first fraction represents dx/dt out, so couldn't you also take the integral of this from 0 to 10 to get the answer for part e? This gives me 580, rather than 51.6.