VCE Methods Question Thread!

[Smiley_]

Hahah, serious? Don't remember it at all... Granted, I also don't remember the exam 2 being difficult

[Valyria]

Hey,

The question was left open ended, need some closure pls;

Could someone please help me with 2007 VCAA exam 2, Q2(G)? I would upload a screen shot but I'm not too sure which information from Q2 is relevant.

I spent a few minutes re-reading the stem of the question but I can't seem to find what time would deem Tazzy Jones' recovery of the diamond successful.

So far, I have calculated the total time he takes to recover the diamond (1279/16 minutes) but I don't know why we have to deduct this figure from 80 minutes. Why is 80 minutes the time to beat?

Any help would be appreciated!

[kawfee]

Could someone either explain or direct me (assuming someone has already asked this) to the relevant post for a question in Exam 2 Methods 2013
1)f
The one about Trigg, and images, transformations, dilations.
The answer/VCAA is below...I do not get it.

pg 15 http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2013/2013mmcas2-w.pdf

Thank you

[BLACKCATT]

^Yes please.

[IndefatigableLover]

Hey,

The question was left open ended, need some closure pls;

Could someone please help me with 2007 VCAA exam 2, Q2(G)? I would upload a screen shot but I'm not too sure which information from Q2 is relevant.

I spent a few minutes re-reading the stem of the question but I can't seem to find what time would deem Tazzy Jones' recovery of the diamond successful.

So far, I have calculated the total time he takes to recover the diamond (1279/16 minutes) but I don't know why we have to deduct this figure from 80 minutes. Why is 80 minutes the time to beat?

Any help would be appreciated!

I actually answered question a few pages back Valyria!

Haha just did this paper today and luckily I got this question right!

So essentially from your answer from Question 2e. you would have gotten 8/3 using -> however realistically Tasmania Jones has one shot to get the diamond since he can't drop it halfway and pick it up afterwards and all... so basically your answer from Question 2e is halved to 4/3 (just chose one of the fractions I guess) which is equal to 80 minutes when you convert from hours to minutes giving you the overall total time From there you would just do this:





Therefore he has 1/16 minutes left to spare or 3.75 seconds when you convert from minutes to seconds

Integral can be split into

Second integral is trivial to evaluate, so I'll focus on the first one.

What is our new function relative to f(x)? Well, it's a dilation by 5 from the y-axis, right? This means that (x,y) -> (5x,y).

Thus, we change the terminals from x->5x. So now our terminals are 0 and 5a for the translated graph instead of 0 and a. But since our new graph is dilated by a factor of 5 from the y axis, it is being stretched by that much.


Thus,

Ooh thanks psyxwar! That definitely cleared up my misunderstandings (it was just that dilation part that I had trouble with) :S

[Reus]

I know we need to derive x/sin(x) and the sub in π/2 however what method is used to derive? I get confused with the quotient, product and chain rule. Soz for my noobness haha 

Edit; did it using the quotient rule haha getting somewhere

[IndefatigableLover]

I know we need to derive x/sin(x) and the sub in π/2 however what method is used to derive? I get confused with the quotient, product and chain rule. Soz for my noobness haha 

Well because it's a fraction, you'll be using the quotient rule! However alternatively you could use the product rule if you changed the powers around..

- Quotient Rule

- Product Rule

And from there I'm sure you just apply each rule and sub in as your 'x' value and you'll end up with 1 as your answer

(But yeah personally I'd use the quotient rule since it's easier handling with positive numbers )

[faredcarsking123]

2007 Exam 2 VCAA Q3di

One of their solutions is a a translation of 1 unit left then a reflection in the y-axis.

Shouldnt it just be a translation of 1 unit left?

Also can anyone explain Q5fii to me

Thanks in advance

[Reus]

Well because it's a fraction, you'll be using the quotient rule! However alternatively you could use the product rule if you changed the powers around..

- Quotient Rule

- Product Rule

And from there I'm sure you just apply each rule and sub in as your 'x' value and you'll end up with 1 as your answer

(But yeah personally I'd use the quotient rule since it's easier handling with positive numbers )

Yep! Thanks so much! Another thing how the hell are we meant to know sin() = 1? Special triangles? Again sorry for my stupid questions!

[shadows]

2007 Exam 2 VCAA Q3di

One of their solutions is a a translation of 1 unit left then a reflection in the y-axis.

Shouldnt it just be a translation of 1 unit left?

Also can anyone explain Q5fii to me

Thanks in advance

It specifically tells you that it is translated before reflected.
Reflection in the y (imagine theres a mirror in the y axis, it reflects the x value co-ordinates), So the initial translation of one unit left, say (-1,0)  is reflected to one unit right (1,0)
hope this makes sense.

[keltingmeith]

Yep! Thanks so much! Another thing how the hell are we meant to know sin() = 1? Special triangles? Again sorry for my stupid questions!

Honestly, I just remember it through the graphs... 0, pi/2, pi and 3pi/2 are exact values you sort of just have to know. Either by the graphs, "hand trick" or rote.

[Jason12]

if part of a graph is under the x-axis do we need to put a -ve sign in front of the integral? In one vcaa exam they didn't so it confuses me (VCAA 2007 Q3bii)

[shadows]

Yep! Thanks so much! Another thing how the hell are we meant to know sin() = 1? Special triangles? Again sorry for my stupid questions!

In the unit circle, the x co-ordinate on the circle is the cos value. (cos(0)=1, cos(pi)=-1
the y value on the circle is the sin value (sin(0)=0, sin(3pi/2)=-1)
It can be derived from simple trig. Since the hypotenuse of triangle created in unit circle is always 1. cos(angle)= A/H, where H=1,
so essentially cos(angle)=Adjacent length. (which in the unit circle is the horizontal length)
sin(angle)=O/H, where h=1
so sin(angle)=Opposite length, which is the vertical length of the triangle created.

Tan is just sin/cos. (always gives you the gradient of the line) since gradient is rise/run. (y value/x value)

I strongly suggest you commit to memory the other special values for sin cos and tan. (angles of 0, pi/4,pi/3,pi/2,3pi/4 etc etc) .  I strongly suggest understand how these are derived, as it makes remembering them and applying trig questions so much easier.
Very likely they will ask you something that requires this on Exam 1.

hope i made sense.

[faredcarsking123]

It specifically tells you that it is translated before reflected.
Reflection in the y (imagine theres a mirror in the y axis, it reflects the x value co-ordinates), So the initial translation of one unit left, say (-1,0)  is reflected to one unit right (1,0)
hope this makes sense.

I see what you mean but I sketched both graphs and a translation of 1 unit left without a reflection already brings in to the same shape as the other graph

[Reus]

In the unit circle, the x co-ordinate on the circle is the cos value. (cos(0)=1, cos(pi)=-1
the y value on the circle is the sin value (sin(0)=0, sin(3pi/2)=-1)
It can be derived from simple trig. Since the hypotenuse of triangle created in unit circle is always 1. cos(angle)= A/H, where H=1,
so essentially cos(angle)=Adjacent length. (which in the unit circle is the horizontal length)
sin(angle)=O/H, where h=1
so sin(angle)=Opposite length, which is the vertical length of the triangle created.

Tan is just sin/cos. (always gives you the gradient of the line) since gradient is rise/run. (y value/x value)

I strongly suggest you commit to memory the other special values for sin cos and tan. (angles of 0, pi/4,pi/3,pi/2,3pi/4 etc etc)
Very likely they will ask you something that requires this on Exam 1.

hope i made sense.

Honestly, I just remember it through the graphs... 0, pi/2, pi and 3pi/2 are exact values you sort of just have to know. Either by the graphs, "hand trick" or rote.

Thanks!! You guys are really helping But I guess I'll just have to memorise the simple triangles and basics such as sin(pi/2)=1 etc...
Any other essentials that I should rote for exam 1?