ITT: Post Mathematics (2U) Questions/Problems (1 Viewer)

twinklegal19 · Oct 19, 2012

iBibah said:
Well done and thanks. Thanks to carrot for checking it for me.

Part 2:

I got 44%, is that right?

edit: just saw sy123's solution, nvm

Sy123 · Oct 19, 2012

twinklegal19 said:
I got 44%, is that right?

edit: just saw sy123's solution, nvm

Did you add up the sector and the two segments? If so then I may have made a silly mistake

twinklegal19 · Oct 19, 2012

Sy123 said:
a) Call the centre O (for part iii)
Dotted lines are sight lines.

b) 2 lookout towers needed for whole pitch

c)

Refer to diagrams
S refers to sector
Seg refers to segment

$\triangle ABC $ is equilateral$ \\ \therefore \angle AOB=\frac{2\pi}{3} \\ \therefore $(By cosine rule)$ \ \ u=50\sqrt{3}$

$S_{ABC}=0.5u^2 \frac{\pi}{3}=1250\pi$

$Seg_{AB}=0.5(50)^2(2\pi/3-\sin 2\pi /3)=Seg_{AC}=\frac{1250\pi}{3}-625\sqrt{3}$

$Total=1250\pi+\frac{2500\pi}{3}-1250\sqrt{3}=6250/3 \pi-1250\sqrt{3}$

$Percentage=100 \times \frac{Total}{A_{circle}} \approx = 0.55766....$

The question says the percentage of the pitch *not* visible, so does that mean you minus your final answer from 1?

nice solution btw

Sy123 · Oct 19, 2012

twinklegal19 said:
The question says the percentage of the pitch *not* visible, so does that mean you minus your final answer from 1?

nice solution btw

Oh thats true oops. Then yeah obviously yours is correct then (dem silly mistakes)

iBibah · Oct 19, 2012

Sy123 said:
Did you add up the sector and the two segments? If so then I may have made a silly mistake

i got 10.90%.

44% is way too much, remember it's how much is NOT visible by one tower.

Sy123 · Oct 19, 2012

iBibah said:
i got 10.90%.

44% is way too much, remember it's how much is NOT visible by one tower.

You are correct, and my method is correct I made a wrong calculation in the calculator, recalculated and got what you got

iBibah · Oct 19, 2012

Sy123 said:
You are correct, and my method is correct I made a wrong calculation in the calculator, recalculated and got what you got

Nice work

twinklegal19 · Oct 19, 2012

Sy123 said:
You are correct, and my method is correct I made a wrong calculation in the calculator, recalculated and got what you got

yeah this

my page was full of scribbes lol, knew my working was going to be dodgy haha

Sy123 · Oct 19, 2012

SpiralFlex said:
Substitute into f''(x), note that r is positive.

I forgot to write my reasoning for this, but if the situation is like this, are we allowed to assume that the stationary point is a maximum since there is no minimum area (A=0 min)

Am I allowed to say this and not differentiate? Such a pain to do it again

SpiralFlex · Oct 19, 2012

Sy123 said:
I forgot to write my reasoning for this, but if the situation is like this, are we allowed to assume that the stationary point is a maximum since there is no minimum area (A=0 min)

Am I allowed to say this and not differentiate? Such a pain to do it again

No, you MUST always prove maximum/minimum regardless if it says so.

iBibah · Oct 19, 2012

SpiralFlex said:
No, you MUST always prove maximum/minimum regardless if it says so.

This. In 2U they just say "...and justify your answer".

hayhay2012 · Oct 20, 2012

How do you know if something for Simpsons rule should be In rads or not?for eg look at hsc 2000 7b thnks!

medicore · Oct 20, 2012

are these questions like question 9 - 10, cause i can't do majority, oh god.

Carrotsticks · Oct 20, 2012

hayhay2012 said:
How do you know if something for Simpsons rule should be In rads or not?for eg look at hsc 2000 7b thnks!

When dealing with Trigonometry and any sort of measure (including area), ALWAYS use Radians.

qwerty44 · Oct 20, 2012

hayhay2012 said:
How do you know if something for Simpsons rule should be In rads or not?for eg look at hsc 2000 7b thnks!

When doing any calculus (differentiation or integration) with trig function you use radians. Because degrees are not a linear measurement.

Newbit · Oct 20, 2012

How do I integrate secxtanx dx?..... The answer is secx but I don't know to get to it.

iBibah · Oct 20, 2012

Newbit said:
How do I integrate secxtanx dx?..... The answer is secx but I don't know to get to it.

Look at the standard integrals sheet.

Also derive secx and see what you get.

onedaysnotice · Oct 20, 2012

Are you guys gonna bother studying differentiating from first principles? Ceeeeeebs.............

Sy123 · Oct 20, 2012

onedaysnotice said:
Are you guys gonna bother studying differentiating from first principles? Ceeeeeebs.............

Just know basic algebra and the definition of first principles and you should be good to go, on a 2U level they probably wont ask stuff involving substitutions (unless they give it to you), but who knows maybe at the start of Q16 a decent 3 marks to a hard first principles question.

Sy123 · Oct 20, 2012

Sy123 said:
$Take a regular n-sided polygon, each side of which subtends equal angles of $ \theta $ to the centre, with length of r from centre to the vertices. (As shown in the diagram$

$i) \ \ $Explain why $ \theta = \frac{2\pi}{n} $ and hence prove that the area of the polygon A is given by $ A=\frac{n}{2}r^2\sin \frac{2\pi}{n}$

$ii) \ \ $Using the substitution $ k=\frac{2\pi}{n} $ Find $ \lim_{n \to \infty} \frac{nr^2}{2} \sin \frac{2\pi}{n}$

$iii) \ \ $What does this prove?$$

Yes I am very uncreative

SOLUTION

Posting solution since no one would learn if they didnt know

ITT: Post Mathematics (2U) Questions/Problems (1 Viewer)

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