Author Topic: Mathematics Question Thread (Read 1626779 times) Tweet Share

Raymond_Cen · « **Reply #2160 on:** July 12, 2017, 10:59:14 pm »

Hello! Can you please help me with this question, thanks.
Find the dimensions of he largest rectangle that can be inscribed in the semicircle y=root(4-x^2)

RuiAce · « **Reply #2161 on:** July 12, 2017, 11:06:32 pm »

Quote from: Raymond_Cen on July 12, 2017, 10:59:14 pm

Hello! Can you please help me with this question, thanks.
Find the dimensions of he largest rectangle that can be inscribed in the semicircle y=root(4-x^2)

$\text{One of the sides of the largest rectangle must lie on the }x\text{-axis}\\ \text{Else the area is not maximised (reason being we can basically 'extend' the rectangle)}$
$\text{Let }x\text{ be some relevant positive }x\text{-coordinate}\\ \text{Note that the point }(x,\sqrt{4-x^2})\text{ lies on the semi-circle}$
$\text{Sketch a diagram of the information required}\\ \text{The vertices of the rectangle will be}\\ (\pm x, 0), (\pm x, \sqrt{4-x^2})\\ \text{so that the rectangle has length }2x\text{ and breadth }\sqrt{4-x^2}$
$\text{Hence, the area of the rectangle is }A=2x\sqrt{4-x^2}\\ \text{This is now your standard max-min problem.}\\ \text{Post further working out if you are stuck.}$
$\text{Note that after you've found the value of }x\text{ that gives the maximum}\\ \text{The dimensions are }\ell = 2x, b = \sqrt{4-x^2}$

katnisschung · « **Reply #2162 on:** July 13, 2017, 06:21:57 pm »

Quote from: RuiSmash on July 12, 2017, 10:51:23 pm

You are not required to know how to deal with these in 2U.

cries why was it in a 2 unit trial test!!! is there any way of solving this with 2u knowledge thanks rui

josephmalicdem · « **Reply #2163 on:** July 13, 2017, 07:21:10 pm »

Can anyone please help me with this question? (Trigonometric Functions Topic)
Thanks.

RuiAce · « **Reply #2164 on:** July 13, 2017, 08:24:48 pm »

Quote from: katnisschung on July 13, 2017, 06:21:57 pm

cries why was it in a 2 unit trial test!!! is there any way of solving this with 2u knowledge thanks rui

Nothing appeared obvious to me and when I plugged it into WoiframAlpha it gave me really bizarre results for that one I doubt it.

Whilst some cubic equations are solvable, most aren't. That one in particular requires stuff beyond even 4U unless guided extensively along the way. What trial did it come from?

RuiAce · « **Reply #2165 on:** July 13, 2017, 08:28:19 pm »

Quote from: josephmalicdem on July 13, 2017, 07:21:10 pm

Can anyone please help me with this question? (Trigonometric Functions Topic)
Thanks.

$\text{The derivative involved is a straight application of the product rule}\\ \text{Please post any working if you have trouble with it.}$
$\text{We need to identify that the derivative and the integral cancel each other out}\\ \text{When you integrate something you differentiated, you get back t to that original function with a }+C\\ \begin{align*}\frac{d}{dx}(\sin x - x\cos x)&=x \sin x\\ \therefore \int_0^{\frac{\pi}2}x\sin x\, dx&=\int_0^{\frac\pi2}\frac{d}{dx}(\sin x - x\cos x)dx\\ &=\left[\sin x - x\cos x\right]_0^{\frac\pi2}\\ &= 1- 0\\ &= 1 \end{align*}$

katnisschung · « **Reply #2166 on:** July 13, 2017, 10:53:35 pm »

Quote from: RuiSmash on July 13, 2017, 08:24:48 pm

Nothing appeared obvious to me and when I plugged it into WoiframAlpha it gave me really bizarre results for that one I doubt it.

Whilst some cubic equations are solvable, most aren't. That one in particular requires stuff beyond even 4U unless guided extensively along the way. What trial did it come from?

sydney grammar boys 2015
q15 a i) show that the curve has no stationary points
ii) show that there is a point of inflexion at (3,6)
iii) how many x intercepts does the curve have? justify your answer

RuiAce · « **Reply #2167 on:** July 13, 2017, 11:02:15 pm »

Quote from: katnisschung on July 13, 2017, 10:53:35 pm

sydney grammar boys 2015
q15 a i) show that the curve has no stationary points
ii) show that there is a point of inflexion at (3,6)
iii) how many x intercepts does the curve have? justify your answer

The question never asked you to find the intercepts themselves. It just asked how many there were.

And the answer is 1. Because the curve has no stationary points, it's monotonic increasing. If it's monotonic increasing, carefully visualising it graphically it will only have one x-intercept.

josephmalicdem · « **Reply #2168 on:** July 14, 2017, 03:07:58 pm »

Quote from: RuiSmash on July 13, 2017, 08:28:19 pm

$\text{The derivative involved is a straight application of the product rule}\\ \text{Please post any working if you have trouble with it.}$
$\text{We need to identify that the derivative and the integral cancel each other out}\\ \text{When you integrate something you differentiated, you get back t to that original function with a }+C\\ \begin{align*}\frac{d}{dx}(\sin x - x\cos x)&=x \sin x\\ \therefore \int_0^{\frac{\pi}2}x\sin x\, dx&=\int_0^{\frac\pi2}\frac{d}{dx}(\sin x - x\cos x)dx\\ &=\left[\sin x - x\cos x\right]_0^{\frac\pi2}\\ &= 1- 0\\ &= 1 \end{align*}$

Thanks a lot

josephmalicdem · « **Reply #2169 on:** July 15, 2017, 01:09:27 am »

This is another question I struggle with a bit.

RuiAce · « **Reply #2170 on:** July 15, 2017, 09:07:55 am »

Quote from: josephmalicdem on July 15, 2017, 01:09:27 am

This is another question I struggle with a bit.

Maths in focus and it's dumb wording... not mentioning the fact that the interest is compounded monthly.
$\text{You should follow the usual process for plan B.}\\ \text{Plan A works like this.}\\ \text{Let }A_n\text{ be the amount owing after }n\text{ months}$
$\text{For NSW bank, suppose the monthly payment is }M\\ \text{We then have}\\ \begin{align*}A_0&=5000\\ A_1&=5000-M\\ A_2&=5000-2M\\ A_3&=5000-3M\end{align*}$
$\text{As the interest period only starts in month 4}\\ \text{The recursive approach becomes necessary here.}\\ \begin{align*}A_4&=A_3(1.0075)-M\\ &= (5000-3M)(1.0075)-M\end{align*}$
$\text{Continuing, you will find that}\\ A_n = (5000-3M)(1.0075)^{n-3}-M\left(1+1.0075+1.0075^2+\dots+1.0075^{n-4}\right)\\ \text{Setting }A_{36}=0\text{ will give you the monthly payment}$
$\text{You can then compare the monthly payment in each scenario}\\ \text{to determine which is the better one.}$

_____ · « **Reply #2171 on:** July 15, 2017, 05:16:57 pm »

1.

For shading the area, once I have x <= 3 what do I do? Test points for the 4 regions?

2.

Am I supposed to recognise this as involving the area of a cone and have the formula memorised or is there a better way? The answer uses the volume of a cone (with h = SQRT(2)) and then adds the top part using a regular integration.

Thanks!

RuiAce · « **Reply #2172 on:** July 15, 2017, 05:25:41 pm »

Quote from: _____ on July 15, 2017, 05:16:57 pm

1. (Image removed from quote.)

For shading the area, once I have x <= 3 what do I do? Test points for the 4 regions?

2. (Image removed from quote.)

Am I supposed to recognise this as involving the area of a cone and have the formula memorised or is there a better way? The answer uses the volume of a cone (with h = SQRT(2)) and then adds the top part using a regular integration.

Thanks!

$\text{The region sketched will be the region enclosed}\\ \text{whenever }y=\frac{1}{x-2}\text{ is above the line }y=1\\ \text{Also note that the solution is not just }x\le 3\text{, but rather }\boxed{0 < x \le 3}\\ \text{The algebra of solving this is only seen in Ext 1.}$
WolframAlpha input
______________________________________

You could've used integration for both if you wanted to. Their method is also acceptable.

jaskirat · « **Reply #2173 on:** July 15, 2017, 06:34:50 pm »

Hey guys, i'm currently prepping for trials and just wanted some tips on what the best way to study for the maths exams is

RuiAce · « **Reply #2174 on:** July 15, 2017, 06:48:29 pm »

Quote from: jaskirat on July 15, 2017, 06:34:50 pm

Hey guys, i'm currently prepping for trials and just wanted some tips on what the best way to study for the maths exams is

It is a known fact that the best way to study is to continuously do past papers and to do as many as you can. You can only get better and more well versed against the exam by doing questions that came from exams in the first place.

After that, everything becomes entirely relative. What will work for someone else may not necessarily work for you.

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