VCE Specialist 3/4 Question Thread!

[KATA]

Hi, I don't understand how to do question 16 from multiple choice from last year's exam at all. Is there any way I can find worked solutions?

[atar.notes.user]

how do u do q3 mcq of nht exam? i got     c<kpi/2
but ans is B

[keltingmeith]

Hi, I don't understand how to do question 16 from multiple choice from last year's exam at all. Is there any way I can find worked solutions?

Unfortunately, I don't know of any resources that go through that. But, the way to approach it is like any curvilinear motion question - start by drawing the situation, putting in some vectors, and figure out the graph that describes its motion to see if you can figure out the horizontal part.

how do u do q3 mcq of nht exam? i got     c<kpi/2
but ans is B

You might find this post from a previous user helpful:

There is a youtube video made by a specialist teacher that goes through the entire NHT exam 2.

https://www.youtube.com/watch?v=RmYH3PSZVrE

Hope that helps!

[lzxnl]

Hi, I don't understand how to do question 16 from multiple choice from last year's exam at all. Is there any way I can find worked solutions?

You need to recognise a few things. Firstly, you have to recognise that the question is projectile motion, and in particular constant acceleration. The horizontal velocity is constant because gravity is vertical and air resistance is negligible. The vertical acceleration is -9.8 m/s^2 if you define the upwards direction to be positive (note: you HAVE to pick one of up or down to be positive).

Now, ask yourself, why does the ball actually stop moving? Imagine you threw a ball yourself. It stops moving when it hits the ground (pretend it doesn't bounce). As there is no horizontal acceleration, you'll find how far it travels from the time (horizontal velocity constant). This tells you that we need to impose the condition that the vertical displacement is zero. As this question deals with constant acceleration, and you now know the vertical acceleration, initial vertical velocity and final vertical displacement, you can solve for the time. With the time and horizontal velocity, you can find the horizontal displacement.

[atar.notes.user]

can i pls get help for 2016 exam 2 q 3b section b?

[uhoh]

If arg(x+yi)=pi, does that mean tan(pi)=y/x

[keltingmeith]

can i pls get help for 2016 exam 2 q 3b section b?

You need to play with rates, and think logically about what's happening. The amount of salt in the tank at any one point is y kg - to find concentration, we just need to divide by how much volume of liquid there is (note the question - they said units of kg/L). So, how much liquid is in there?

Well, we started with 100 L. On top of that, we have 20L more going in every minute - but we also have 10L coming out every minute. So, our total volume is going to be 100L + 20L/min - 10L/min. To be able to add these together, they need to have the same units - since t has units of /min, multiplying L/min by will give:



Alternatively, think of it as every minute, 20L goes in, so 20t is the amount in after t minutes.

Putting all of this together should give you the right answer.

If arg(x+yi)=pi, does that mean tan(pi)=y/x

Tbh, if arg(x+yi)=pi, then y=0, and you have a real, negative, number But I see what you're getting at, and the answer is yes, but make sure you watch your quadrants carefully.

[uhoh]

You need to play with rates, and think logically about what's happening. The amount of salt in the tank at any one point is y kg - to find concentration, we just need to divide by how much volume of liquid there is (note the question - they said units of kg/L). So, how much liquid is in there?

Well, we started with 100 L. On top of that, we have 20L more going in every minute - but we also have 10L coming out every minute. So, our total volume is going to be 100L + 20L/min - 10L/min. To be able to add these together, they need to have the same units - since t has units of /min, multiplying L/min by will give:



Alternatively, think of it as every minute, 20L goes in, so 20t is the amount in after t minutes.

Putting all of this together should give you the right answer.

Tbh, if arg(x+yi)=pi, then y=0, and you have a real, negative, number But I see what you're getting at, and the answer is yes, but make sure you watch your quadrants carefully.

Hahah thanks

What do you mean by watching the quadrants carefully? Can you give an eg

[DailyInsanity]

Theorem - Relating to complex numbers in specialist, e.g) NHT 2017 exam 2

A circle can always be drawn with 3 points in a plane on its circumference given the 3 points do not all have the same x or y co-ordinates.

[keltingmeith]

Hahah thanks

What do you mean by watching the quadrants carefully? Can you give an eg

Well, arctan(x) only has a range between -pi/2 and pi/2. So, let's say your complex number is -1+i (in the third quadrant). If all you know about the number is it has an argument of 3pi/4, then you draw the conclusion of tan(y/x)=-1. No biggie yet, but let's say after that they ask you to find the complex number rotated 90 degrees clockwise (which is -1-i). If you're not careful, you might accidentally just apply arctan to -1, which would give you 1-i which you'd then change to 1+i.

If that made sense at all, basically, just anytime you're working with tan, be aware of what quadrant you're in, so you can do dummy checks and don't make silly mistakes.

Theorem - Relating to complex numbers in specialist, e.g) NHT 2017 exam 2

A circle can always be drawn with 3 points in a plane on its circumference given the 3 points do not all have the same x or y co-ordinates.

Did you want confirmation...? In which case, yes, you only need 3 distinct points to make a circle.

[..........]

Can someone please explain the second step of working in this?
Thankyou.

[Shadowxo]

Can someone please explain the second step of working in this?
Thankyou.

They skipped a few steps

[ezferns]

Can someone pls explain why the graph of Arg(z-2)=-pi/4 has a domain of Re(z)>2

[RuiAce]

Can someone pls explain why the graph of Arg(z-2)=-pi/4 has a domain of Re(z)>2

This literally falls out from drawing the ray and reading off it.




In addition, recall that for our purposes \( \text{Re}(z) > 2\) is equivalent to \( x > 2 \)

[TheAspiringDoc]

On the Casio Classpad,
Main -> interactive -> complex
It has three options down the bottom:
CompToPol
CompToTrig (what does this do??)
CompToRect

So if I want to convert polar to Cartesian youI just use the 3rd one, and for Cartesian to polar just use the first one? And never use the 2nd one?
Thanks