Login

Welcome, Guest. Please login or register.

March 28, 2024, 11:53:04 pm

Author Topic: VCE Methods Question Thread!  (Read 4802337 times)  Share 

0 Members and 6 Guests are viewing this topic.

keltingmeith

  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 5493
  • he/him - they is also fine
  • Respect: +1292
Re: VCE Methods Question Thread!
« Reply #18825 on: October 02, 2020, 12:03:25 am »
+3
Hello everyone, could someone clear up part C)

Cross-sectional area = the area if you cut into the shape. For example, the cross-sectional area of a cube will be a square, of a sphere will be a circle, of a cylinder will be a circle if you cut it short-ways, or a rectangle if you cut it long-ways. To go in reverse is easy for the cylinder if you cut it short-ways, you can find the volume by multiplying the circular cross-section by the height of the cylinder (in fact, that's what the formula of the volume of a cylinder is - V=pi*r^2*h)

With that in mind, how much papier mache can you make with a 0.5 m * 0.5 m * 30 cm box? And how much of that can you put into the volume of that ramp?

james.358

  • MOTM: OCT 20
  • Forum Regular
  • **
  • Posts: 93
  • Respect: +110
Re: VCE Methods Question Thread!
« Reply #18826 on: October 03, 2020, 04:42:13 pm »
+7
Hey Larry, welcome to the Atar Notes forums!

Wow this question looks wayyyy beyond the scope of the Maths Methods course, but I will attempt to answer it.

For this question you will need to use the results from both parts a) and b), and do quite a bit of algebraic manipulation for it to simplify, as well as being familiar with the trigonometric formulae

The full working is attached in the image below.

Hope this helps!
James
VCE Class of 2022: 99.90 ATAR
Monash Medical School Class of 2026

RAW 50 Methods & Specialist High Yield Resources

Corey King

  • Trendsetter
  • **
  • Posts: 133
  • Respect: +3
Re: VCE Methods Question Thread!
« Reply #18827 on: October 04, 2020, 04:10:53 pm »
0
Hey guys,
I don't understand why the book states (x-a) and (y-0) in the middle of this (https://gyazo.com/dc410472548eba8a407eb79dafac38bf) screenshot.
When I go through the algebraic process I get this (attached file)
Where is my mistake?
I get -b=-b in the end, which is true, but I'm clearly not getting what they're doing.
Many thanks,
Corey :)

james.358

  • MOTM: OCT 20
  • Forum Regular
  • **
  • Posts: 93
  • Respect: +110
Re: VCE Methods Question Thread!
« Reply #18828 on: October 04, 2020, 05:51:01 pm »
+8
Hey Corey!

Great to see that you're not taking formulas in the book for granted, and deriving it on your own instead.

Your first step in finding the gradient is correct using the typical gradient formula. For the next step you have to apply the point gradient form.

Recall that if you know a point (x1, y1) and the gradient, you can write a linear equation in the y - y1 = m (x - x1) form. Now you can sub in the gradient and the point (a, 0)

Your equation becomes:
y - y1 = m (x - x1)
y - 0 = - b/a (x - a)
y = - b/a (x - a)

Your mistake was trying to sub in both of your points, and in the process got rid of the x and y. I'll leave you to explore the rest of this formula. Let us know how you go and come back on this thread if you have any other questions!

Hope this helps,
James
VCE Class of 2022: 99.90 ATAR
Monash Medical School Class of 2026

RAW 50 Methods & Specialist High Yield Resources

Corey King

  • Trendsetter
  • **
  • Posts: 133
  • Respect: +3
Re: VCE Methods Question Thread!
« Reply #18829 on: October 05, 2020, 10:52:35 am »
0
Hey Corey!

Great to see that you're not taking formulas in the book for granted, and deriving it on your own instead.

Your first step in finding the gradient is correct using the typical gradient formula. For the next step you have to apply the point gradient form.

Recall that if you know a point (x1, y1) and the gradient, you can write a linear equation in the y - y1 = m (x - x1) form. Now you can sub in the gradient and the point (a, 0)

Your equation becomes:
y - y1 = m (x - x1)
y - 0 = - b/a (x - a)
y = - b/a (x - a)

Your mistake was trying to sub in both of your points, and in the process got rid of the x and y. I'll leave you to explore the rest of this formula. Let us know how you go and come back on this thread if you have any other questions!

Hope this helps,
James


Ah I see. I kept substituting either both x1 y2 or y1 and x2 instead for some reason.
I think the change of labelling mixed me up as well. Changing the y2 to just y and x2 to just x. I can see why they do that though :)
Thank you for the help James!
Corey

Corey King

  • Trendsetter
  • **
  • Posts: 133
  • Respect: +3
Re: VCE Methods Question Thread!
« Reply #18830 on: October 05, 2020, 11:07:59 am »
0
Hey guys,
I got indeterminate as an answer for the gradient of a straight line graph.
I'm just wondering what I put as the gradient in the point-gradient formula now? Do I just write zero, and treat it as such?
Many thanks,
Corey

Corey King

  • Trendsetter
  • **
  • Posts: 133
  • Respect: +3
Re: VCE Methods Question Thread!
« Reply #18831 on: October 05, 2020, 11:08:58 am »
0
Hey guys,
I got Undefined* as an answer for the gradient of a straight line graph.
I'm just wondering what I put as the gradient in the point-gradient formula now? Do I just write zero, and treat it as such?
Many thanks,
Corey

p0kem0n21

  • Trailblazer
  • *
  • Posts: 41
  • Respect: +28
Re: VCE Methods Question Thread!
« Reply #18832 on: October 05, 2020, 02:16:17 pm »
+3


If the gradient for a straight line is undefined, then that should mean that it is a vertical line of the formula x=a, where a is any real number (or basically the x-coordinate for your two points). An undefined gradient is not the same as when your gradient is equal to zero. If the gradient of the line was zero, then it would be a horizontal line instead of equation y=b (where b is the y-coordinate of your points).

svnflower

  • Trailblazer
  • *
  • Posts: 37
  • Respect: +1
Re: VCE Methods Question Thread!
« Reply #18833 on: October 05, 2020, 02:20:53 pm »
0
 :) Hello,

How do we graph y=tan2x given a y=tanx graph?

I don't quite understand how to graph trig functions with power 2 e.g. cos2x and sin2x

fun_jirachi

  • MOTM: AUG 18
  • HSC Moderator
  • Part of the furniture
  • *****
  • Posts: 1068
  • All doom and Gloom.
  • Respect: +710
Re: VCE Methods Question Thread!
« Reply #18834 on: October 05, 2020, 02:37:52 pm »
+8
Hey there!

General tips for graphing \((f(x))^2\) given \(f(x)\):
- Start by drawing \(|f(x)|\) (reflecting all negative parts of \(f(x)\) around the x-axis, since \(\sqrt{ (f(x))^2} = |f(x)|\)
- \((f(x))^2\) will be below \(|f(x)|\) where \(|f(x)| < 1\), and \((f(x))^2\) will be above \(|f(x)|\) where \(|f(x)| > 1\)
- \((f(x))^2\) will always be greater than 0 for all x in the domain of \(|f(x)|\)
- Turn all cusps formed by \(|f(x)|\) into smooth turning points (like on a parabola)
- Keep all vertical asymptotes
- Square all horizontal asymptotes (move them accordingly)
- Indicate clearly the slope of \((f(x))^2\) - \(\frac{d}{dx}  (f(x))^2 = 2f'(x)f(x)\).

Can't think of anything else atm, but hope this is enough to get you started :)
« Last Edit: October 05, 2020, 02:40:00 pm by fun_jirachi »
Spoiler
HSC 2018: Mod Hist [88] | 2U Maths [98]
HSC 2019: Physics [92] | Chemistry [93] | English Adv [87] | 3U Maths [98] | 4U Maths [97]
ATAR: 99.05

UCAT: 3310 - VR [740] | DM [890] | QR [880] | AR [800]
Guide Links:
Subject Acceleration (2018)
UCAT Question Compilation/FAQ (2020)
Asking good questions

svnflower

  • Trailblazer
  • *
  • Posts: 37
  • Respect: +1
Re: VCE Methods Question Thread!
« Reply #18835 on: October 07, 2020, 09:15:43 am »
0
Hey there!

General tips for graphing \((f(x))^2\) given \(f(x)\):
- Start by drawing \(|f(x)|\) (reflecting all negative parts of \(f(x)\) around the x-axis, since \(\sqrt{ (f(x))^2} = |f(x)|\)
- \((f(x))^2\) will be below \(|f(x)|\) where \(|f(x)| < 1\), and \((f(x))^2\) will be above \(|f(x)|\) where \(|f(x)| > 1\)
- \((f(x))^2\) will always be greater than 0 for all x in the domain of \(|f(x)|\)
- Turn all cusps formed by \(|f(x)|\) into smooth turning points (like on a parabola)
- Keep all vertical asymptotes
- Square all horizontal asymptotes (move them accordingly)
- Indicate clearly the slope of \((f(x))^2\) - \(\frac{d}{dx}  (f(x))^2 = 2f'(x)f(x)\).

Can't think of anything else atm, but hope this is enough to get you started :)

Thankyou so much, helps a lot!!

Also, what does it mean for a function to be defined? Is it the values where the curve is above the x-axis?

p0kem0n21

  • Trailblazer
  • *
  • Posts: 41
  • Respect: +28
Re: VCE Methods Question Thread!
« Reply #18836 on: October 07, 2020, 09:57:04 am »
+3
Thankyou so much, helps a lot!!

Also, what does it mean for a function to be defined? Is it the values where the curve is above the x-axis?

A function is defined for a particular x-value/input when it's output 'makes sense'. For example, if I substituted x=1 in the equation you gave, you would get the square root of -9, which is not possible when dealing with real numbers. Thus, it is undefined for x=1. Similarly, it will undefined for all x-values for which the output of the expression under the square root is negative.

Answer given below in the spoiler  :)

Spoiler
defined for x∈(-∞,-2] u [4,∞)
« Last Edit: October 07, 2020, 10:06:53 am by p0kem0n21 »

Rose34

  • Trendsetter
  • **
  • Posts: 158
  • Respect: +2
Re: VCE Methods Question Thread!
« Reply #18837 on: October 08, 2020, 08:12:06 pm »
0
Hello everyone,

I was wondering if anyone knows where I can find free/affordable practice SACs for Math Methods Unit 1&2?

Thanks in advance!

t101

  • Victorian
  • Adventurer
  • *
  • Posts: 15
  • Respect: 0
Re: VCE Methods Question Thread!
« Reply #18838 on: October 08, 2020, 09:07:06 pm »
0
question -
2008 VCAA exam 2 q3a
'a. Find the time, to the nearest minute, that Tasmania has to fi nd an antidote (that is, a cure for the toxin).'

the maths gives 191.67 minutes and VCAA gives the rounded answer as 192 minutes. Buttt, Tasmania would be dead, then, right?
Why isn't it 191?
Are there any other VCAA examples of a similar nature?

Corey King

  • Trendsetter
  • **
  • Posts: 133
  • Respect: +3
Re: VCE Methods Question Thread!
« Reply #18839 on: October 09, 2020, 02:41:19 pm »
0
Hey fellas :)
I have yet to encounter this method of solution before. ( https://gyazo.com/2d627e5a74becf140504196d220fb8ec )
What's up with this third variable and what is this method of solution called? Is there somewhere in the book where they explain it?
Many thanks :)
Corey