VCE Methods Question Thread!

[IndefatigableLover]

Hey euler and others

can someone please help me

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2009mmCAS2-w.pdf

question 4d ii, 4e i and 4e ii. Been stuck for ages

Q4 dii).

You know what is so if you take the antiderivative of that then you'll end up with t= ___ but this answer is in terms of 'h'... if you transpose to make 'h' the subject so that it then becomes h=_____ then you've got the right answer! You can also leave out the '+c' when you take the anti-derivative since c=0 (due to t=0 and h=0).

Spoiler

From b, so:


At t=0 hours, h = 0m since the vat is 'initially empty', so:

Transposing and you'll get:

Q4 e i).
Looking at the diagram, you can see that the height from 'O' to the base of the statue is 14 metres (8+6 = 14) however at the start of the information we are told that after 9:00am the statue is being lowered at a rate of 1m per hour. As a result the distance from 'O' to the base of the statue will be given by 14-t (where 't' is the time it gets lowered since it's proportional).

Q4 e ii). This is simply making two equations equalling the same thing since they come to the same point (touch). Using your answer from Q4 dii) and Q4 ei), you make these two equal the same thing and then solve it
Watch out for the given units as well since it asks for the time and not for 't'!

EDIT: Go read kinslayer's post!

[keltingmeith]

Hey euler and others

A testament to all the questions I've been consistently answering over the past few days instead of the exam study I should've been doing, hahah.

Anyway, this is one of the infamous triangle questions that have plagued methods for who knows how long. I'm not sure when the triangle began, but whenever it did, people did horribly, and so ever since triangles like this have been in textbooks, exams and SACs everywhere for methods (and textbooks for spec, but not really exams).

So:


From b, so:


At t=0 hours, h = 6m, so:


But, we wanted it in terms of t, so we'll solve for h to get:


Okay, I'm hungry, so I'll let indefatigamabelel do the other two.

EDIT: Scrap my working - checked the assessment report, my working's stuff. Read the question wrong because I jumped in halfway. Indefatigable's right, it should be . However, you should still remember the constant of integration (your plus c), and then remove it by subbing in the point (0, 0).

[MNM101]

For this extended response question ;
The depth of water at the entrance to a harbour 't' hours after high tide is 'D' metres, where D=p+q cos(rt) ^degrees for suitable constants p,q,r. At high tide the depth is 7m, at low tide 6 hours later, the depth is 3m.

Show that r=30and find the values of p and q


The topic is circular functions and I have no idea how to approach this question 😳

[kinslayer]

Hey euler and others

can someone please help me

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2009mmCAS2-w.pdf

question 4d ii, 4e i and 4e ii. Been stuck for ages

4dii



4ei

The base of the statue is intially (at 9am) 6m above the top of the vat which is 8m tall, so 14m above the ground. It descends at a rate of 1m per hour. So the height of the base of the statue above ground level t hours after 9am is is 14 - t metres.

4eii

Equate the height of the acid at time t with the height of the base of the statue at time t:

Editting as I go along..

Hey, that's not fair. 

edit: VVVVV just joking

[IndefatigableLover]

Hey, that's not fair. 

HAHAHA sorry kinslayer... I'll just leave mine as it is now that you've fully solved it though! I was just explaining it first before doing calculations LOL

[LiquidPaperz]

just little bit of a problem im facing is when i decide to do methods hw, i say try and get 8A and Chapter review done, but i end up spending more hours then i wish. Not that im academically less capable or anything, but i like to think about it and write out the equation im using etc.

Anyone else facing this issue haha?

[keltingmeith]

Hey, that's not fair. 

First thing I learned at Monash - students from JMSS don't play fair.

Liquid - that's not really an issue, I'm one of those weirdoes that likes to do everything, too. Hell, my students are probably sick of me telling them to draw graphs of the questions their doing, hahah.

[kinslayer]

First thing I learned at Monash - students from JMSS don't play fair.

They steal my seat on the bus too.

[Rod]

A testament to all the questions I've been consistently answering over the past few days instead of the exam study I should've been doing, hahah.

Anyway, this is one of the infamous triangle questions that have plagued methods for who knows how long. I'm not sure when the triangle began, but whenever it did, people did horribly, and so ever since triangles like this have been in textbooks, exams and SACs everywhere for methods (and textbooks for spec, but not really exams).

So:


From b, so:


At t=0 hours, h = 6m, so:


But, we wanted it in terms of t, so we'll solve for h to get:


Okay, I'm hungry, so I'll let indefatigamabelel do the other two.

EDIT: Scrap my working - checked the assessment report, my working's stuff. Read the question wrong because I jumped in halfway. Indefatigable's right, it should be . However, you should still remember the constant of integration (your plus c), and then remove it by subbing in the point (0, 0).

Wait, do you have to knwo antidirivatives for all these questions? If so, I won't need to know how to do them because we haven't gone through it yet.. So confused..

[keltingmeith]

Yes - you need to know how to anti-derive things for these questions. Don't worry if you don't know about them yet, you'll learn 'em soon enough.

And then you get to probability, my favourite part of the course. <3

[Rod]

Ah fuck it, wasted ages staring at this question and wasted your time too, sorry guys. Didn't know you would have to use anti-derivatives for it.

[keltingmeith]

Don't stress, any time not staring at my physics work is time well spent, IMO.

[Rod]

Thank you euler, kingslay and IL. I respected all your posts .

Wish me luck for tomorrow!!

[kinslayer]

Thank you euler, kingslay and IL. I respected all your posts .

Wish me luck for tomorrow!!

Gluck 

[keltingmeith]

Firstly, this question tricks you by putting the "show that r = 30" first - you actually need to find p and q first.

So, firstly, what ARE p and q? Well, p is the vertical translation, that's nothing new. High tide is the highest point, and low tide is the lowest point, so if we take those two and average them out we should get the vertical translation.



Okay, that's it for what you knew before circular function, sorry. Now, for q. q is what we call the "amplitude" of the function. No, if a circular function represents a "circle", you can think of the amplitude as this circle's "radius". The amplitude is actually the vertical distance from the vertical translation to the functions highest or lowest point. So, if we take the vertical distance, and take away the highest or lowest point, we'll get the distance between the two, yeah? So:



Both gave us 2, so we'll say that q = 2.

So far, our function is . Now, we just need to show that r = 30. Well, we know that t is the time in hours AFTER high tide. So initially, we have high tide. So, at t = 0, D = 7. Now, if we sub this in, it'll work, but this doesn't show us that r = 30, because that'll work for ANY value of r... However, it does say that 6 hours later it's low tide! So, let's work with that:



Which is true - so, you've shown that r = 30, answering the question!