VCE Methods Question Thread!

[TheCommando]

Hi i got f(pie) wrong for question 4c
Where did i go wrong in my working out and is there a easier and quicker way to do it
http://imgur.com/a/FhWsL
http://imgur.com/mWmaplZ

[Shadowxo]

Don't have the question, but looking at your working, you forgot to multiply the π/4 by 2, so the bit inside the cos would be 2π +π/2, not 2π + π/4
Since this is just 2π + π/2 you can replace it with π/2 (2π is just a rotation around a circle so makes no difference)
You know that cos(π/2) =0 so you end up with √3
Note: if it was π/4 you'd use the memorised table of values (should know sin cos and tan of π/6, π/4 and π/3, as well as 0,π/2,π,3π/2 and 2π but you can just visualise these last ones to find them)
Also, with your second last step in your working,
cos(9π/4) = cos(9π/4 - 2π) not cos(2π-9π/4) (in this case it doesn't matter as cos(x)=cos(-x))
You can add or subtract 2π however many times you like from inside a trig function

[johnhalo]

Hello,
Can someone explain to me the reasoning behind 1c?

Thanks

[zhen]

Hello,
Can someone explain to me the reasoning behind 1c?

Thanks

There are 2 ways to do it. The first way is to find the derivative of the function and draw up a table containing the derivative at (1,0) and the derivatives around this point. It the derivative of the function a bit before 1 is positive and the derivative a but after 1 is negative, then it's a local maximum.

Alternatively, you could take the second derivative of the function. Then sub in x=1 and if the second derivative is less than zero, then it's a local max.

[Cyka]

Zhen, Why they hell is your methods SAC on related rates??? It's not even on the methods course anymore.

[zhen]

Zhen, Why they hell is your methods SAC on related rates??? It's not even on the methods course anymore.

I'm assuming it's because my school thinks that a normal methods differentiation SAC isn't difficult enough, so everyone will be getting high marks, since most people at my school are really good at methods, so they won't be able to seperate everyone and properly rank everyone. This is just speculation, but I think it would be a legitimate reason to have a SAC beyond the syllabus. It's either that or maybe my school thinks that learning something more difficult will help improve our ability. I'm honestly not too sure why, but these are my guesses.

[occidit]

I'm assuming it's because my school thinks that a normal methods differentiation SAC isn't difficult enough, so everyone will be getting high marks, since most people at my school are really good at methods, so they won't be able to seperate everyone and properly rank everyone. This is just speculation, but I think it would be a legitimate reason to have a SAC beyond the syllabus. It's either that or maybe my school thinks that learning something more difficult will help improve our ability. I'm honestly not too sure why, but these are my guesses.

To prepare students better for the exam, instead of making our differentiation questions difficult, we'll make them off the course

[captkirk]

Hey guys, this may be a stupid question. I don't know why I can't do this.

How do you make

'y=1/2(2x-7)(x^2-7x+1)^-1/2

Into fraction form (standard)

[Sine]

Zhen, Why they hell is your methods SAC on related rates??? It's not even on the methods course anymore.

It isn't specifically on the methods course anymore. However it is just an application of the chain rule so it may very well still come up although highly unlikely.

Also Zhen goes to MHS so sacs for methods and spec are much harder than the normal yr 12 exam

Hey guys, this may be a stupid question. I don't know why I can't do this.

How do you make

'y=1/2(2x-7)(x^2-7x+1)^-1/2

Into fraction form (standard)

I'm sorry i'm not entirely sure what you mean by fraction form?

[captkirk]

I'm sorry i'm not entirely sure what you mean by fraction form?

I mean how do you get
'y=1/2(2x-7)(x^2-7x+1)^-1/2' into a fraction. Like convert it into a fraction from Its current form.

[Shadowxo]

I mean how do you get
'y=1/2(2x-7)(x^2-7x+1)^-1/2' into a fraction. Like convert it into a fraction from Its current form.

a^-1/2 = 1/a^1/2 so

[Guideme]

I need help with tangent derivatives. They are so difficult.
So here are the two questions i am stuck on :
find the the derivative of:
1. (tan(x))^2
2. (tan(5x))^3


Please provide full working out thank you



According to the worksheet the answers are 1. (2sin(x))/(cos(x))^3  and 2. (15(sin(5x))^2)/ (cos(5x))^4

[zhen]

I need help with tangent derivatives. They are so difficult.
So here are the two questions i am stuck on :
find the the derivative of:
1. (tan(x))^2
2. (tan(5x))^3


Please provide full working out thank you



According to the worksheet the answers are 1. (2sin(x))/(cos(x))^3  and 2. (15(sin(5x))^2)/ (cos(5x))^4

This is basically the chain rule. This is only for 1, but if you know how to do 1, then you can do 2.

[johnhalo]

Hello,

I just need help with the attached question

Thanks

[zhen]

Hello,

I just need help with the attached question

Thanks

I'm not too confident with this, but I'll give it a try.


Note that there are going to be multiple ways to transform this graph to get the desired result, so this may not be the answer in the book. Also, someone should check my working, since I'm not too confident with transformations.