Author Topic: 3U Maths Question Thread (Read 1235928 times) Tweet Share

RuiAce · « **Reply #915 on:** October 24, 2016, 05:54:58 pm »

Quote from: levendibigd on October 24, 2016, 05:43:08 pm

Hey there! Could I get a hand with HSC question Q13a (iii)? I don't understand how they were able to get a value for the amplitude.

Can you please link the paper at least, since we don't even know what year

levendibigd · « **Reply #916 on:** October 24, 2016, 06:43:40 pm »

My apologies! I thought I put the year... HSC 2015
http://www.boardofstudies.nsw.edu.au/hsc_exams/2015/exams/2015-hsc-maths-ext-1.pdf

Mod Edit: Post Merge

WLalex · « **Reply #917 on:** October 24, 2016, 07:25:02 pm »

Quote from: levendibigd on October 24, 2016, 06:43:40 pm

My apologies! I thought I put the year... HSC 2015
http://www.boardofstudies.nsw.edu.au/hsc_exams/2015/exams/2015-hsc-maths-ext-1.pdf

Mod Edit: Post Merge

a is simply two as it is two units from 5 to 7 and 2 units from 5 to 3 (with 5 being the centre)

RuiAce · « **Reply #918 on:** October 24, 2016, 07:29:55 pm »

Quote from: levendibigd on October 24, 2016, 06:43:40 pm

My apologies! I thought I put the year... HSC 2015
http://www.boardofstudies.nsw.edu.au/hsc_exams/2015/exams/2015-hsc-maths-ext-1.pdf

Mod Edit: Post Merge

$\text{The equation given is standard and you should be familiar with it.}\\ \text{At }c=0\text{ it is the equation }v^2=n^2(a^2-x^2)$
$\text{Since our centre of motion is 5, we have }c=5\\ \text{Our amplitude is 2, so we have }a=2$
$\text{Giving us }v^2=n^2(4-(x-5)^2)\\ \text{One way to find }n\text{ now is just to sub }(5,11)\text{ in.}$

levendibigd · « **Reply #919 on:** October 24, 2016, 07:51:01 pm »

Thank you!

yaboiaderler · « **Reply #920 on:** October 24, 2016, 09:18:48 pm »

hey guys!
This is from the 2011 HSC Paper in Question 3. I understand and know how to do part i,ii and iii, but part iv is where I'm stuck at. The solutions that were provided by BOSTES are a tad bit iffy :l

Thanks!

bethjomay · « **Reply #921 on:** October 24, 2016, 09:23:42 pm »

Just wondering if we can use more colloquial terms like 'horizontal line test' in describing and justifying things?

RuiAce · « **Reply #922 on:** October 24, 2016, 09:26:15 pm »

Quote from: bethjomay on October 24, 2016, 09:23:42 pm

Just wondering if we can use more colloquial terms like 'horizontal line test' in describing and justifying things?

To determine if a function has an inverse? Horizontal line test is the most formal you can get.

RuiAce · « **Reply #923 on:** October 24, 2016, 09:33:46 pm »

Quote from: yaboiaderler on October 24, 2016, 09:18:48 pm

hey guys!
This is from the 2011 HSC Paper in Question 3. I understand and know how to do part i,ii and iii, but part iv is where I'm stuck at. The solutions that were provided by BOSTES are a tad bit iffy :l

Thanks!

$\text{The }x\text{-coordinate is fixed. So the locus is clearly }x=\frac12$
$\text{Now, some restrictions are imposed on the }y\text{-coordinate due to the quadratic.}\\ \text{We know that a quadratic either has a minimum value (concave up)}\\ \text{or a maximum value (concave down). Here the latter case holds.}$
$\text{Observe that the maximum value of }y=t-t^2\text{ is }y=\frac14\\ \text{This can be proven in many ways including completing the square}\\ \text{calculus or using the formula }t=-\frac{b}{2a}=\frac12\text{ and sub back in.}$
$\text{So at the very least, there is a condition that }y \le \frac14\\ \text{otherwise }t\text{ does not exist.}$
________________
$\text{But when }y=\frac14\text{ the points }P\text{ and }Q\text{ coincide.}\\ \text{These extreme cases always have to be checked.}$
$\text{Hence }y\neq \frac14\text{ or we'd have a common tangent, and no defined point of intersection.}\\ \text{So }y < \frac14$

bethjomay · « **Reply #924 on:** October 24, 2016, 10:48:37 pm »

Quote from: RuiAce on October 24, 2016, 09:26:15 pm

To determine if a function has an inverse? Horizontal line test is the most formal you can get.

Oh awesome! I thought it was just a shortcut way of defining a function with/without an inverse!

:3 · « **Reply #925 on:** October 24, 2016, 10:58:52 pm »

Sorry if I'm hitting you with so many questions.

Struggling to understand the answers to iii) to v)

RuiAce · « **Reply #926 on:** October 24, 2016, 11:23:56 pm »

Quote from: :3 on October 24, 2016, 10:58:52 pm

Sorry if I'm hitting you with so many questions.

Struggling to understand the answers to iii) to v)

(Image removed from quote.)

(iv) is LITERALLY an application of what you used in part b), with A = 2theta and B = alpha. Make sure to go back to part b) when they tell you to. (Part b) is not part c) (ii), it is part b).)

$\text{The }2\theta\text{ in part (iii) suggests to use double angles EARLIER}\\ \text{rather than later.}\\ \text{Hence, observe that }F(\theta) = \frac{1}{2}\sin 2\theta - \left(\frac{1}{2}+\frac{1}{2}\cos 2\theta\right)\tan \alpha$
$\text{Hence }F^\prime(\theta)=\cos 2\theta + \sin 2\theta \tan \alpha\\ \text{Set }F^\prime(\theta) = 0\text{, rearrange, and done.}$

Quote from: bethjomay on October 24, 2016, 10:48:37 pm

Oh awesome! I thought it was just a shortcut way of defining a function with/without an inverse!

There are more clever ways around it - a function that's monotone will always pass the horizontal line test and has an inverse. But that rule never goes wrong.

Rikahs · « **Reply #927 on:** October 25, 2016, 12:04:41 am »

So my question is:

A missile is fired at a velocity of 720km/hr and lands 2000metres away on the same horizontal plane. Find the angle(s) of elevation of the missile.

I get two angles but which one is correct?

RuiAce · « **Reply #928 on:** October 25, 2016, 12:07:36 am »

Quote from: Rikahs on October 25, 2016, 12:04:41 am

So my question is:

A missile is fired at a velocity of 720km/hr and lands 2000metres away on the same horizontal plane. Find the angle(s) of elevation of the missile.

I get two angles but which one is correct?

There's supposed to be two answers given all the information you said.

Rikahs · « **Reply #929 on:** October 25, 2016, 12:22:57 am »

oh lol ok and also for questions asking "find the angle when the projectile hits the ground" If i find the y velocity and x velocity then just do tan theta, do i have to minus the answer from 180 degrees?

Search the forums now!

We have moved!

Author Topic: 3U Maths Question Thread (Read 1235928 times) Tweet Share

RuiAce

Re: 3U Maths Question Thread

levendibigd

Re: 3U Maths Question Thread

WLalex

Re: 3U Maths Question Thread

RuiAce

Re: 3U Maths Question Thread

levendibigd

Re: 3U Maths Question Thread

yaboiaderler

Re: 3U Maths Question Thread

bethjomay

3U Maths Question Thread

RuiAce

Re: 3U Maths Question Thread

RuiAce

Re: 3U Maths Question Thread

bethjomay

Re: 3U Maths Question Thread

:3

Re: 3U Maths Question Thread

RuiAce

Re: 3U Maths Question Thread

Rikahs

Re: 3U Maths Question Thread

RuiAce

Re: 3U Maths Question Thread

Rikahs

Re: 3U Maths Question Thread

Recent Posts

User:
Password: