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#### isaacdelatorre

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« Reply #465 on: October 10, 2016, 01:18:42 pm »
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Hey guys,

I'm not sure where to start and have been blankly staring at it for a while now.

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#### RuiAce

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« Reply #466 on: October 10, 2016, 01:31:56 pm »
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Hey guys,

I'm not sure where to start and have been blankly staring at it for a while now.

$\textit{Observe that this question says to EXPLAIN.}$
$\text{1}{4}\text{ is representative of 15 minutes according to the question.}\\ \text{Note that if Xuan or Yvette have waited for more than 15 minutes, they will leave.}\\ \text{Hence, the times that they arrive must not be further apart by more than 15 minutes.}$
$\text{The time taken for Xuan is represented by }x\\ \text{If Yvette arrives 15 minutes later, Xuan will go.}\\ \text{So }x-y \le \frac{1}{4}\\ \text{Similarly, if Yvette was earlier we'd have }y-x\le \frac{1}{4}$
____________________________
$\text{This is simply too hard to visualise.}\\ \text{On your graph, sketch }x-y\le \frac14 \iff y\ge x-\frac14\\ \text{and also sketch }y-x\le \frac{1}{4}\iff y \le x+\frac{1}{4}$
$\text{Observe that the brown area is reflective of the probability we need.}\\ \text{It is the region bound by the square and the two regions }x-y\le\frac14, y-x\le \frac14$
$\text{To determine the area, we can split the hexagon into two triangles and a rectangle.}\\ \text{The two triangles each have area }A=\frac{1}{2}\times \frac14\times \frac14\\ \text{The rectangle's area is harder.}$
$\text{For the rectangle, by either Pythagoras' theorem or trigonometry}\\ \ell = 0.75\sqrt{2}\\ b=0.25\sqrt{2}\\ \text{So the area is }2\times 0.75\times 0.25\\ \text{Add everything up, and you have your answer.}$
I want to let you have a go at (iii) first. Use a graph similar to the graph in (ii). Come back with whatever progress you have.
« Last Edit: July 15, 2017, 05:29:37 pm by RuiAce »

#### isaacdelatorre

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« Reply #467 on: October 10, 2016, 02:42:55 pm »
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$\textit{Observe that this question says to EXPLAIN.}$
$\text{1}{4}\text{ is representative of 15 minutes according to the question.}\\ \text{Note that if Xuan or Yvette have waited for more than 15 minutes, they will leave.}\\ \text{Hence, the times that they arrive must not be further apart by more than 15 minutes.}$
$\text{The time taken for Xuan is represented by }x\\ \text{If Yvette arrives 15 minutes later, Xuan will go.}\\ \text{So }x-y \le \frac{1}{4}\\ \text{Similarly, if Yvette was earlier we'd have }y-x\le \frac{1}{4}$
____________________________
$\text{This is simply too hard to visualise.}\\ \text{On your graph, sketch }x-y\le \frac14 \iff y\ge x-\frac14\\ \text{and also sketch }y-x\le \frac{1}{4}\iff y \le x+\frac{1}{4}$(Image removed from quote.)
$\text{Observe that the brown area is reflective of the probability we need.}\\ \text{It is the region bound by the square and the two regions }x-y\le\frac14, y-x\le \frac14$
$\text{To determine the area, we can split the hexagon into two triangles and a rectangle.}\\ \text{The two triangles each have area }A=\frac{1}{2}\times \frac14\times \frac14\\ \text{The rectangle's area is harder.}$
$\text{For the rectangle, by either Pythagoras' theorem or trigonometry}\\ \ell = 0.75\sqrt{2}\\ b=0.25\sqrt{2}\\ \text{So the area is }2\times 0.75\times 0.25\\ \text{Add everything up, and you have your answer.}$
I want to let you have a go at (iii) first. Use a graph similar to the graph in (ii). Come back with whatever progress you have.

Awesome!! Thanks for the help Rui.

It took a few tries... and apparently a whole hour; but I got it in the end.

Thanks
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#### asd987

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« Reply #468 on: October 10, 2016, 10:10:56 pm »
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Hi, I'm can some please complete this question for me.
find the stationary points on the curve y=(x-4)(x+2)^2 and hence sketch the curve

#### jakesilove

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« Reply #469 on: October 10, 2016, 10:18:21 pm »
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Hi, I'm can some please complete this question for me.
find the stationary points on the curve y=(x-4)(x+2)^2 and hence sketch the curve

Hey! So, we can expand the graph out to

$y=x^3-12x-16$

First, we find the derivative.

$y'=3x^2-12$

We set it equal to zero, to find stationary point (where the gradient is zero)

$3x^2-12=0$
$x=2, -2$

We find the second derivative to determine the nature of the stationary points.

$y''=6x$
$x=2, y''=12>0$

Therefore, x=2 is a min

$y''=6x$
$x=-2, y''=-12<0$

Therefore, x=-2 is a max.

We know the x-intercepts (from the original graph, at x=4, and x=2). We can sub in our turning points to find y values. Finally, we know that it is a 'positive' cubic, is we get +x^3. Therefore, the graph will look like

Let me know if I can clarify anything!
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#### asd987

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« Reply #470 on: October 10, 2016, 11:03:06 pm »
+1
TYVM i understand. i just didn't know you had to expand the equation of the graph

#### RuiAce

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« Reply #471 on: October 10, 2016, 11:07:27 pm »
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TYVM i understand. i just didn't know you had to expand the equation of the graph
$\text{It can be commenced without the expansion}\\ \text{The product rule and chain rule are both needed to differentiate.}\\ \frac{dy}{dx}=2(x-4)(x+2)+(x+2)^2$
$\text{Then to solve }\frac{dy}{dx}=0\\ \text{Factorise out }(x-2):\\ (x+2)[2(x-4)+(x+2)]=0 \iff (x-2)(3x-6)=0\\ \text{Which still solves to give }x=\pm 2$

#### anotherworld2b

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« Reply #472 on: October 10, 2016, 11:50:32 pm »
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I was wondering how do we simply imaginary numbers? We just started doing them but im already confused.
I was wondering if i could get help understanding how to graph cubics and tan graphs please

#### BPunjabi

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« Reply #473 on: October 11, 2016, 12:07:18 am »
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I was wondering how do we simply imaginary numbers? We just started doing them but im already confused.
I was wondering if i could get help understanding how to graph cubics and tan graphs please

Wait is this any maths course or just 2 Unit cause I swear I have seen nothing like that before.
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#### RuiAce

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« Reply #474 on: October 11, 2016, 12:08:58 am »
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Wait is this any maths course or just 2 Unit cause I swear I have seen nothing like that before.
It's not in 2U.

#### BPunjabi

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« Reply #475 on: October 11, 2016, 12:09:44 am »
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It's not in 2U.

Thank the heavens...
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#### BPunjabi

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##### Re: 96 in 2U Maths: Ask me Anything!
« Reply #476 on: October 11, 2016, 12:18:29 am »
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Hey Liiz:

HEY Im jake (obviously not haha). Im a year 12 student who completed my 2u HSC mathematics course in year 11 and lm just happy to help out here. Now, this type of question is amongst one of the most difficult ones in geometrical applications of calculus, and unfortunately in HSC exams there WILL be harder ones. But don't worry, once you have practised enough, you will begin to seize some tricks to approach these questions.

BEFORE YOU DO ANYTHING, DRAW A DIAGRAM WITH LABELS
1. Highlight all USEFUL INFORMATIONS (in this case, highlight rectangular box, square base, no top, 500cm^3 and least area)
2. Appoint two variables to the unknown sides (in this case, I named the side length of the square base as x, and the height of the rectangular box as L)
3. There will be at least one number quantity in every one of these questions in 2u mathematics, so the first equation you should construct, using your name variables to construct an equation that uses the numbers provided by the question
4. Draw out the relationship between the two variables through this equation that you have constructed
5. Construct another equation using your variables and the subject that is asked for in the question (In this case, for example, we constructed an Area equation which directly relates to what we are asked to find)
6. Substitute in the equivalent expression of a variable (In this case, for example, L = 500/x^2, so we substitute any L we see with 500/x^2) to reduce the total number of values down to one, so that we can construct an equation entire out of only one variable, which then allows us to perform differentiation
7. Clean up the equation after the substitution to make life easier
8. Differentiate the equation
9. Let this derivative = 0 to find any stationary points (In an Exam, YOU MUST STATE "LET dy/dx = 0 TO FIND ANY STATIONARY POINTS, OTHERWISE MARKS MAYBE DEDUCTED!!!)
10. Solve the derivative equation and find a value for your variable which will be your stationary point
11. Test both sides to show that a local minimum/maximum occurs at your stationary points
12. Substitute your minimum value back into the area equation (or maximum value if the question asks for maximum area) to find the minimum area (or the maximum area if you substitute in the maximum value)

So here is my solution:
(Image removed from quote.)

Hope you find my solutions clear and useful! If you are confused with anything, dont hesitate to ask!!!

Best Regards
Jacky He

LOL WTF how were you guys so good at math? Did you constantly study!
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#### Alize

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« Reply #477 on: October 11, 2016, 11:47:21 am »
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Hey,
Could someone please explain to me how solve the following?
7sin3x=2x-1

#### RuiAce

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« Reply #478 on: October 11, 2016, 11:55:41 am »
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Hey,
Could someone please explain to me how solve the following?
7sin3x=2x-1

$\textbf{You can't.}\\ \text{There is no known algebraic method in the entire world of maths that can be used}\\ \text{to solve }\sin x = ax+b$
$\text{You can only }\textbf{approximate}\text{ the solution to }7\sin 3x=2x-1\\ \text{To do so, you need to sketch}\\ y=7\sin 3x\\ y=2x-1\\ \text{to a }\textbf{perfect scale}.$
$\text{Once you have your graph, you read the }x\text{-coordinates}\\ \text{of your point of intersection.}\\ \text{This will allow you to approximate a solution.}$

#### Aussie1Italia2

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« Reply #479 on: October 11, 2016, 12:05:25 pm »
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Hey! I feel like this would be really obvious but I'm struggling with a few questions like find the gradient of the tangent to the curve: y= square root of x at (4,2) and find the gradient of the normal to the curve y=x2 + 5 at (-2,9)
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