Author Topic: Mathematics Question Thread (Read 1626427 times) Tweet Share

sweetiepi · « **Reply #2655 on:** October 01, 2017, 09:39:46 pm »

Quote from: jaskirat on October 01, 2017, 09:27:10 pm

where did you move it to?

Hey there! I believe Rui moved it to here: https://atarnotes.com/forum/index.php?topic=173885.msg985661#new

bdobrin · « **Reply #2656 on:** October 02, 2017, 05:16:54 pm »

Hey jake,

I was just wondering if you could help me with this HSC question from the 2004 exam.

1) The limiting sum of the geometric series: '1-Tan² theta + Tan⁴ theta' is Cos² theta. For what values of theta in the interval -π/2 < theta < π/2 does the limiting sum of the series exist?

Thanks,
Ben

RuiAce · « **Reply #2657 on:** October 02, 2017, 05:22:26 pm »

Quote from: bdobrin on October 02, 2017, 05:16:54 pm

Hey jake,

I was just wondering if you could help me with this HSC question from the 2004 exam.

1) The limiting sum of the geometric series: '1-Tan² theta + Tan⁴ theta' is Cos² theta. For what values of theta in the interval -π/2 < theta < π/2 does the limiting sum of the series exist?

Thanks,
Ben

$\text{The common ratio of this geometric series is }-\tan^2x$
$\text{Recall that the common ratio exists when the common ratio satisfies }|r| < 1\\ \text{so here, we require }|-\tan^2 x|<1$
________________________
$\text{Note that }|-\tan^2 x| = |\tan^2 x| = |\tan x|^2\text{ which we substitute in.}\\ \text{Square rooting both sides of }|\tan x|^2 < 1\text{ gives }|\tan x| < 1$
The square rooting was justified as all quantities are positive.
$\text{Solving the absolute value equation gives }-1 < \tan x < 1\\ \text{and therefore }-\frac\pi 4 < x < \frac\pi4\\ \text{(from exact values)}$

Piza · « **Reply #2658 on:** October 02, 2017, 06:13:20 pm »

Hi guys, I am having quite some trouble with the following questions:

1. A cubic curve has a minimum turning point at the point (-2, -15), a maximum turning point at (1,12) and it cuts the y axis at 5. What is the equation of the cubic?

I know that the constant will be 5 and that the curve will be negative, given the location of the minimum and maximum turning points. I tried subbing in the points (differentiated curve equation) and doing simultaneous equations but it didn't work out for me.

Spoiler

Answer: y = -2x³ - 3x² + 12x + 5

2. A sheet of paper for a poster has an area of 6m². The margins at the top and bottom are 0.75m and at the sides 0.5m. Show that the dimensions for maximum printed area are 2m in length and 3m in width.
(I saw an answer to a similar question on this forum but I had trouble applying it)

3. Find the smallest distance between the graphs of y = x² - 4x + 12 and y = 2x +1

Spoiler

2/5 √5

RuiAce · « **Reply #2659 on:** October 02, 2017, 06:27:14 pm »

Quote from: Piza on October 02, 2017, 06:13:20 pm

Hi guys, I am having quite some trouble with the following questions:

1. A cubic curve has a minimum turning point at the point (-2, -15), a maximum turning point at (1,12) and it cuts the y axis at 5. What is the equation of the cubic?

I know that the constant will be 5 and that the curve will be negative, given the location of the minimum and maximum turning points. I tried subbing in the points (differentiated curve equation) and doing simultaneous equations but it didn't work out for me.

Spoiler
Answer: y = -2x³ - 3x² + 12x + 5

$\text{The equation of the general cubic is }y=ax^3+bx^2+cx+d\\ \text{and we wish to find our values for }a,b,c,d$
$\text{When }x=0, y=5.\text{ This immediately gives }d=5\\ \therefore y=ax^3+bx^2+cx+5$
$\text{The graph passes through }(-2,-15)\text{ and }(1,12).\text{ Substituting in these points,}\\ \begin{align*}-15&= -8a+4b-2c+5\tag{1}\\ 12&=a+b+c+5\tag{2}\end{align*}$
$(1)+2\times(2)\text{ gives}\\ \begin{align*}9&=-6a+6b+15\\-1&=-a+b \tag{3}\end{align*}$
__________________________________________________
$\text{We have }\frac{dy}{dx} = 3ax^2+2bx+c\\ \text{As stationary points occur at }x=-2, 1\text{ we have }\frac{dy}{dx} = 0\text{ there.}$
$\begin{align*}0&=12a-4b+c \tag{4} \\ 0&=3a+2b+c\tag{5}\end{align*}$
$(4)-(5)\text{ gives }9a-6b=0 \tag{6}$
__________________________________________________
$\text{Solving }(3)\text{ with }(6)\text{ gives }a=-2,b=-3\\ \text{Plugging back into any of (1),(2),(4),(5), will give }c=12$
$\text{So }\boxed{y=-2x^3-3x^2+12x+5}$
Note that this question could've been done a whole variety of ways due to the 4 simultaneous equations involving a,b,c

RuiAce · « **Reply #2660 on:** October 02, 2017, 06:45:40 pm »

Quote from: Piza on October 02, 2017, 06:13:20 pm

3. Find the smallest distance between the graphs of y = x² - 4x + 12 and y = 2x +1

Spoiler
2/5 √5

$\textbf{This question is Extension 1 material due to the need for parametric representation.}$
$\text{Parametrise an arbitrary point on }y=x^2-4x+12\\ \text{by the point }P(t,t^2-4t+12)$
$\text{We wish to minimise the perpendicular distance between }P\text{ and }2x-y+1=0\\ \text{Using the appropriate formula}\\ \begin{align*}D&= \frac{|2t-(t^2-4t+12)+1|}{\sqrt{2^2+1^2}}\\ &= \frac{|-t^2+6t-11|}{\sqrt5}\\ &= \frac{t^2-6t+11}{\sqrt5}\end{align*}$
$\text{From here, follow the usual process.}\\ \frac{dD}{dt} = \frac{2t-6}{\sqrt5}\text{ and }\frac{d^2D}{dt^2} = \frac2{\sqrt5} > 0\\ \text{so we know it will be a min}$
$\text{Solving }\frac{dD}{dt}=0\text{ gives }t=3\text{ and when }t=3, \, D=\frac2{\sqrt5} = \frac{2\sqrt5}{5}$
The other question is VERY ambiguous. Please link to the question that looked "similar".

pokemonlv10 · « **Reply #2661 on:** October 02, 2017, 06:53:03 pm »

MY friend sent me this question and i'm stumped. Any ideas for this?

RuiAce · « **Reply #2662 on:** October 02, 2017, 06:57:03 pm »

Quote from: pokemonlv10 on October 02, 2017, 06:53:03 pm

MY friend sent me this question and i'm stumped. Any ideas for this?

This question has gone well into Extension 1 territory and I will not address it in this thread. Please post it in the more relevant thread if you wish for an answer.

pokemonlv10 · « **Reply #2663 on:** October 02, 2017, 06:59:38 pm »

Quote from: RuiAce on October 02, 2017, 06:57:03 pm

This question has gone well into Extension 1 territory and I will not address it in this thread. Please post it in the more relevant thread if you wish for an answer.

Ah alright, i'll move it over to there

Piza · « **Reply #2664 on:** October 02, 2017, 07:53:51 pm »

Quote from: RuiAce on October 02, 2017, 06:45:40 pm

The other question is VERY ambiguous. Please link to the question that looked "similar".

Hi RuiAce, thank you so much for answering my questions

Here is the question which looked similar: https://atarnotes.com/forum/index.php?topic=164548.msg867873#msg867873

RuiAce · « **Reply #2665 on:** October 02, 2017, 09:14:24 pm »

Quote from: Piza on October 02, 2017, 07:53:51 pm

Hi RuiAce, thank you so much for answering my questions

Here is the question which looked similar: https://atarnotes.com/forum/index.php?topic=164548.msg867873#msg867873

Hm, I apologise about overlooking something earlier. I had done something which had actually gotten the right answer, but was beyond the scope of the HSC

However, I remain my stance that this question is ambiguous. They relate the dimensions of the "maximum-printed-area" to that of the entire poster, which was quite annoyingly confusing. The way I read the question, it had appeared as though it was just the dimensions of the maximum printed area that required, not the poster.

$\text{The question is a 'flipped' version of the old one}\\ \text{in that we're now interested in the inner area, not the outer area.}$
$\text{Let }x\text{ and }y\text{ be the dimensions of the printing area}\\ \text{Then, }\boxed{x+1}\text{ and }\boxed{y+1.5}\text{ are the dimensions of the paper itself.}$
$\text{The area of the paper itself is 6, so we have}\\ (x+1)(y+1.5) = 6 \implies \boxed{y=\frac{6}{x+1}-1.5}$
$\text{We wish to maximise the area of the printing area, which is }A=xy\\ \text{So we want to obtain the maximum for }\boxed{A=x\left(\frac{6}{x+1}-1.5\right)}$
$\text{This can be differentiated as is, but I will rearrange it a bit just to avoid the product/quotient rule}\\ \begin{align*}A&= 6\frac{x}{x+1}-1.5x\\ &= 6\left(\frac{x+1}{x+1}-\frac1{x+1}\right)-1.5x\\ &= 6 - \frac{6}{x+1}-1.5x\\ \therefore \frac{dA}{dx}&= \frac{6}{(x+1)^2}-1.5\end{align*}$
$\text{Setting }\frac{dA}{dx}=0\text{ gives }\frac{6}{1.5}=(x+1)^2\implies x +1 = \pm 2\\ \text{Discarding the negative solution, }\boxed{x=1}\\ \text{and substituting back into the earlier relationship, }\boxed{y=1.5}$
$\text{So our required dimensions are }x+1=2\text{ and }y+1.5=3\\ \text{i.e. }2\times 3$

bdobrin · « **Reply #2666 on:** October 03, 2017, 04:19:41 pm »

Hi Jake,

just wondering if you could help me on this little integration question.

1.) Find the integral of 'sec²(x) + sec(x)•tan(x)

Cheers

Eric11267 · « **Reply #2667 on:** October 03, 2017, 04:28:03 pm »

Quote from: bdobrin on October 03, 2017, 04:19:41 pm

Hi Jake,

just wondering if you could help me on this little integration question.

1.) Find the integral of 'sec²(x) + sec(x)•tan(x)

Cheers

I do VCE so I'm not sure if this applies to you but I would split it up into two different integrals and integrate sec² as normal. For the other half I would put it in terms of sin and cos and use the substition u=cos(x)

RuiAce · « **Reply #2668 on:** October 03, 2017, 04:31:26 pm »

Quote from: bdobrin on October 03, 2017, 04:19:41 pm

Hi Jake,

just wondering if you could help me on this little integration question.

1.) Find the integral of 'sec²(x) + sec(x)•tan(x)

Cheers

$\int \sec^2x\,dx = \tan x+C\text{ is a known result.}$
$\text{The following standard integral was provided in the past:}\\ \int \sec ax \tan ax\, dx = \frac1a \sec ax + C\\ \text{It has since been removed, and therefore you do not need to worry about this question.}$
$\text{However for completedness sake}\\ \text{Continuing, here we'd just have }\int \sec x \tan x \,dx = \sec x + C$
$\text{So }\int (\sec^2 x + \sec x \tan x) \,dx = \tan x + \sec x + C$

Quote from: Eric11267 on October 03, 2017, 04:28:03 pm

I do VCE so I'm not sure if this applies to you but I would split it up into two different integrals and integrate sec² as normal. For the other half I would put it in terms of sin and cos and use the substition u=cos(x)

If you look at the syllabus (when you are free), integration by substitution is only expected at the Extension 1 level. (Usually the substitution is also provided.)

Mary_a · « **Reply #2669 on:** October 03, 2017, 04:34:28 pm »

Hey,

I know that the answer to studying for maths is practice, practice, practice!

It's just I've done about 3 hsc papers in the last week and I've gotten around 60% every time, and I don't know how to improve that?
Should i just keep going with hsc papers?

I guess I'm just a little bit upset because maths is the subject I put the most work into and it's my worst subject.

Thanks guys,

Mary

Search the forums now!

We have moved!

Author Topic: Mathematics Question Thread (Read 1626427 times) Tweet Share

sweetiepi

Re: Mathematics Question Thread

bdobrin

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

Piza

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

pokemonlv10

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

pokemonlv10

Re: Mathematics Question Thread

Piza

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

bdobrin

Re: Mathematics Question Thread

Eric11267

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

Mary_a

Re: Mathematics Question Thread

Recent Posts

User:
Password: