Login

Welcome, Guest. Please login or register.

April 07, 2026, 02:29:42 pm

Author Topic: methods vcaa 2014  (Read 2260 times)  Share 

0 Members and 1 Guest are viewing this topic.

minhalgill

  • Trendsetter
  • **
  • Posts: 128
  • Respect: 0
methods vcaa 2014
« on: October 12, 2018, 07:43:33 pm »
0
so the way i did this question was i found the area under the triangle, then divided it by 10 (11-1) and got 3 which i added to 4 (which is where the y value of the graph start)

is this a correct method?

Bri MT

  • VIC MVP - 2018
  • Administrator
  • ATAR Notes Legend
  • *****
  • Posts: 4719
  • invest in wellbeing so it can invest in you
  • Respect: +3677
Re: methods vcaa 2014
« Reply #1 on: October 12, 2018, 09:46:05 pm »
0
it works for this question, but you should also know how to find the average value through integration :)

sailinginwater

  • Trendsetter
  • **
  • Posts: 104
  • Respect: 0
Re: methods vcaa 2014
« Reply #2 on: October 12, 2018, 11:26:06 pm »
0
it works for this question, but you should also know how to find the average value through integration :)
In the above example is there a way to do it by integration even though we don't know the equation of the graph?

Sine

  • Werewolf
  • National Moderator
  • Great Wonder of ATAR Notes
  • *****
  • Posts: 5132
  • Respect: +2103
Re: methods vcaa 2014
« Reply #3 on: October 13, 2018, 02:59:14 am »
+4
In the above example is there a way to do it by integration even though we don't know the equation of the graph?
You should be able to find the equation of a line when you are given two complete coordinates.

sailinginwater

  • Trendsetter
  • **
  • Posts: 104
  • Respect: 0
Re: methods vcaa 2014
« Reply #4 on: October 13, 2018, 08:54:04 am »
0
You should be able to find the equation of a line when you are given two complete coordinates.
So would we first find the equation of both the lines, and then integrate each line with y = 4 separately?

Bri MT

  • VIC MVP - 2018
  • Administrator
  • ATAR Notes Legend
  • *****
  • Posts: 4719
  • invest in wellbeing so it can invest in you
  • Respect: +3677
Re: methods vcaa 2014
« Reply #5 on: October 13, 2018, 10:33:09 am »
+1
So would we first find the equation of both the lines, and then integrate each line with y = 4 separately?

That is a method you should definitely be able to use

If you are able to recognise that the graph is symmetric you could also just find one of the areas and then multiply by two

minhalgill

  • Trendsetter
  • **
  • Posts: 128
  • Respect: 0
Re: methods vcaa 2014
« Reply #6 on: October 14, 2018, 10:31:37 pm »
0
That is a method you should definitely be able to use

If you are able to recognise that the graph is symmetric you could also just find one of the areas and then multiply by two

if we arent able to find the functions of the lines, is there another way we can do it?

Lear

  • MOTM: JUL 18
  • Part of the furniture
  • *****
  • Posts: 1170
  • Respect: +328
Re: methods vcaa 2014
« Reply #7 on: October 14, 2018, 10:36:43 pm »
0
Yes. Find the area under the curve by using geometry. Note: That graph is simply a triangle on top of a rectangle. Add the area of that triangle to the area of the rectangle underneath and multiply it by 1/b-a to get average value.
2018: ATAR: 99.35
Subjects
English: 44
Methods: 43
Further Maths: 50
Chemistry: 46
Legal: 40
2019: Bachelor of Medical Science and Doctor of Medicine @ Monash

Bri MT

  • VIC MVP - 2018
  • Administrator
  • ATAR Notes Legend
  • *****
  • Posts: 4719
  • invest in wellbeing so it can invest in you
  • Respect: +3677
Re: methods vcaa 2014
« Reply #8 on: October 15, 2018, 09:35:21 am »
0
if we arent able to find the functions of the lines, is there another way we can do it?

If you aren't able to find the equation of the a linear graph based on having two points I recommend you learn that skill asap.
You can complete this question without it,  but finding equations of lines is definitely an expected skill

minhalgill

  • Trendsetter
  • **
  • Posts: 128
  • Respect: 0
Re: methods vcaa 2014
« Reply #9 on: October 15, 2018, 05:53:08 pm »
0
If you aren't able to find the equation of the a linear graph based on having two points I recommend you learn that skill asap.
You can complete this question without it,  but finding equations of lines is definitely an expected skill


oh no, i definelty know how to do that, was just asking for future questions for more difficult graphs.

Yes. Find the area under the curve by using geometry. Note: That graph is simply a triangle on top of a rectangle. Add the area of that triangle to the area of the rectangle underneath and multiply it by 1/b-a to get average value.

thanks for clearing it up!