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

SpiralFlex · Oct 19, 2012

Leffife said:
Wow I really love that question, very unique in someway. Is there any chance that I could save her for you? No terms and conditions applied

No, too late she has been saved. Ate her.

GoldyOrNugget · Oct 19, 2012

Another one!

Tangent Points

Point A lies at the coordinate (3; 8).
Find all the points P on the curve y=x^2 such that line PA is tangent to y=x^2.

Solution:

Screen Shot 2012-10-19 at 9.27.27 PM.jpg

GoldyOrNugget · Oct 19, 2012

One more...

Screen Shot 2012-10-19 at 9.22.26 PM.jpg

twinklegal19 · Oct 19, 2012

just a question, how do we check if the stationary point at x=sqrt(x)/2r is a maximum point? What do we have to write for that?

iBibah · Oct 19, 2012

Sy123 said:
SOLUTION

Very well written question, while people to Spiral's one Ill think of my own

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

Part 2:

SpiralFlex · Oct 19, 2012

twinklegal19 said:
just a question, how do we check if the stationary point at x=sqrt(x)/2r is a maximum point? What do we have to write for that?

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

habitres · Oct 19, 2012

differentiate. If = 0 then it is a turning point at x.
test values either side to determine if max or minimum. Just think logically and you should be fine

twinklegal19 · Oct 19, 2012

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

dammit

can't believe that escaped my mind lol

twinklegal19 · Oct 19, 2012

I must redeem myself with iBibah's second question

(inb4 I fail that as well)

Sy123 · Oct 19, 2012

$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

Carrotsticks · Oct 19, 2012

That is a good question.

SpiralFlex · Oct 19, 2012

Impressive, this would be an excellent complex no. question.

SpiralFlex · Oct 19, 2012

Sy, get rid of pi r^2 or it'll give it away.

GoldyOrNugget · Oct 19, 2012

SpiralFlex said:
Impressive, this would be an excellent complex no. question.

It's quite similar to MX2 HSC 2007 8c).

LoveHateSchool · Oct 19, 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

Mate, is that actually a 2U Q? Haha :/

SpiralFlex · Oct 19, 2012

LoveHateSchool said:
Mate, is that actually a 2U Q? Haha :/

It can be done for 2U level but requires something a bit tricky for 2U. Clue: You need to rearrange something.

deswa1 · Oct 19, 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

Beautiful. Lol I just did a VERY similar question in 4U with complex numbers haha.

Very well done- I like

Sy123 · Oct 19, 2012

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

Part 2:

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=1-100 \times \frac{Total}{A_{circle}} \approx = 0.10899....=10.9 $percent$$

SpiralFlex · Oct 19, 2012

Another clue, what does the k actually represent.

kenkap · Oct 19, 2012

whats ITT?

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

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