4U Revising Game (1 Viewer)

gurmies · May 10, 2009

harism said:
its not a typo, drongoski.
check it yourself... copy what looks like an 'n' and paste it into word. you will see its a pi, not an n.

i had mistaken it for an 'n' myself. But it was gurmies who knocked some sense into me.

Using logic, he wouldn't go all the way to character map/remember the alt code for pi to make an error

azureus88 · May 10, 2009

It doesnt matter whether its pi or n. Its assumed that they're positive constants anyway so it should work either way.

Anyway, new question:

Find all solutions to z^4 = (z-1)^4

gurmies · May 10, 2009

jet · May 10, 2009

azureus88 · May 11, 2009

[maths]\delta V=\pi\left \{ (4-x+\delta x)^2-(4-x)^2 \right \}(2y)\\=4\pi y(4-x)\delta x\\=2\sqrt{3}\pi(4-x)\sqrt{4-x^2}\delta x\\\\V=\lim_{\delta x\to 0}\sum_{x=-2}^{2}2\sqrt{3}\pi(4-x)\sqrt{4-x^2}\delta x\\=8\sqrt{3}\pi\int_{-2}^{2}\sqrt{4-x^2}\delta x-2\sqrt{3}\pi\int_{-2}^{2}x\sqrt{4-x^2}\delta x\\=8\sqrt{3}\pi(\frac{\pi(2^2)}{2})-0\,\,\,\,(1/2\,$area of circle, odd function)\\=16\sqrt{3}\pi^2[/maths]

azureus88 · May 11, 2009

New Question:

[maths]$Show that$\,\,1+(1+x)+(1+x)^2+...+(1+x)^n=\frac{(1+x)^{n+1}-1}{x}\\$Hence show that$\,\,\binom{n}{r}+\binom{n-1}{r}+\binom{n-2}{r}+...+\binom{r}{r}=\binom{n+1}{r+1}[/maths]

gurmies · May 11, 2009

$1 + (1+x) + (1+x)^{2} + ... + (1+x)^{n} = \frac{(1+x)^{n+1}-1}{1+x-1}=\frac{(1+x)^{n+1}-1}{x} \,\, by \,\, using \,\, sum \,\, of \,\, a \,\, G.P. \,\, with \,\, n+1 \,\, terms. \\\\ \frac{x^{n+1}+^{n+1}C_{n}x^{n}+...+^{n+1}C_{r+1}x^{r+1}+^{n+1}C_{r}x^{r}+...+^{n+1}C_{2}x^{2}+^{n+1}C_{1}x+1-1}{x} \\\\ = x^{n}+ ^{n+1}C_{n}x^{n-1}+...+^{n+1}C_{r+1}x^{r}+...+^{n+1}C_{2}x + n+1 \\\\ Coefficient \,\, of \,\, x^{r} \,\, is \,\, \binom{n+1}{r+1}. \\\\ On \,\, the \,\ LHS, \,\, x^{r} \,\, will \,\, emerge \,\, starting \,\, with \,\, the \,\, r^{th} \,\, power. \,\, From \,\, then \,\, on, \,\, x^{r} will \,\, continue \,\, to \,\, emerge \,\, until \,\, n^{th} \,\, power. \therefore Coefficient \,\, of \,\, x^{r} \,\, is \,\, \binom{r}{r}+\binom{r+1}{r}+...+\binom{n-2}{r}+\binom{n-1}{r}+\binom{n}{r}$

$\therefore \binom{n}{r}+\binom{n-1}{r}+\binom{n-2}{r}+...+\binom{r}{r}=\binom{n+1}{r+1}$

NEW QUESTION:

$Express \,\, \frac{\sin9\theta}{\sin\theta} \,\, as \,\, polynomial \,\, in \cos\theta. \\\\ Hence \,\, deduce \,\, that \,\, \sec^{2}\frac{\pi}{9}+\sec^{2}\frac{2\pi}{9}+\sec^{2}\frac{4\pi}{9}=36 \,\, and \,\, \sec\frac{\pi}{9}\sec\frac{2\pi}{9}\sec\frac{4\pi}{9}=8$

Trebla · May 11, 2009

harism said:
trebla, put us out of our misery.
is it a pi or an n?

thanks.

It is meant to be pi and x doesn't have to be integer valued. Either way, azureus88's got the right idea, just replace n with pi. Well done.

Drongoski · May 11, 2009

harism said:
its not a typo, drongoski.
check it yourself... copy what looks like an 'n' and paste it into word. you will see its a pi, not an n.

i had mistaken it for an 'n' myself. But it was gurmies who knocked some sense into me.

U are right haris; I peered close without my glasses and realise I couldn't see well.

gurmies · May 11, 2009

Drongoski said:
U are right haris; I peered close without my glasses and realise I couldn't see well.

Never underestimate the squint of a person who wears glasses =)

lolokay · May 14, 2009

gurmies said:
NEW QUESTION:

$Express \,\, \frac{\sin9\theta}{\sin\theta} \,\, as \,\, polynomial \,\, in \cos\theta. \\\\ Hence \,\, deduce \,\, that \,\, \sec^{2}\frac{\pi}{9}+\sec^{2}\frac{2\pi}{9}+\sec^{2}\frac{4\pi}{9}=36 \,\, and \,\, \sec\frac{\pi}{9}\sec\frac{2\pi}{9}\sec\frac{4\pi}{9}=8$

$sin3\theta = sin2\theta cos\theta +cos2\theta sin\theta \\ = 2sin\theta cos^2\theta + 2cos^2\theta sin\theta -sin\theta \\ = sin\theta (4cos^2\theta -1)\\ sin9\theta=sin3\theta(4cos^2 3\theta-1)\\ = sin\theta(4cos^2\theta-1)(4cos^2 3\theta-1)\\ cos3\theta = cos2\theta cos\theta -sin2\theta sin\theta \\ = 2cos^3\theta -cos\theta - 2sin^2\theta cos\theta \\ = 2cos^3\theta -cos\theta - 2cos\theta + 2cos^3\theta\\ = 4cos^3\theta - 3cos\theta\\ 4cos^2 3\theta-1 = 64cos^6\theta - 96cos^4\theta + 36cos^2\theta - 1\\ sin9\theta = sin\theta(4cos^2\theta-1)(64cos^6\theta - 96cos^4\theta + 36cos^2\theta - 1)\\ sin9\theta/sin\theta = (4cos^2\theta-1)(64cos^6\theta - 96cos^4\theta + 36cos^2\theta - 1)\\ sin9\theta/sin\theta\text{ has zeroes of }\theta=n\pi /9, n=\pm 1,\pm2,\pm3,\pm4\\ let\ x=cos^2\theta,\ then\ x=cos^2\frac{\pi}{9},cos^2\frac{2\pi}{9},cos^2\frac{\pi}{3},cos^2\frac{4\pi}{9}\\ satisfy\ (4x-1)(64x^3 - 96x^2 + 36x - 1)=0\\$

$as\ these\ are\ the\ only\ unique\ values\ for\ cos^2\theta\\ as\ x=cos^2\frac{\pi}{3} is\ a\ root\ for\ (4x-1)=0,\\ x=cos^2\frac{\pi}{9}, cos^2\frac{2\pi}{9},cos^2\frac{4\pi}{9} are\ the\ 3\ roots\ of\ \\ (64x^3 - 96x^2 + 36x - 1)=0\\ sec^2\frac{\pi}{9}+sec^2\frac{2\pi}{9}+sec^2\frac{4\pi}{9}= \frac{\sum \alpha \beta}{\sum \alpha \beta \gamma }\\ = \frac{\frac{36}{64}}{\frac{1}{64}}\\ = 36\\ sec\frac{\pi}{9}sec\frac{2\pi}{9}sec\frac{4\pi}{9}=\frac{1}{\sqrt{\sum \alpha \beta \gamma }}\\ = \frac{1}{\sqrt{\frac{1}{64}}}\\ =8$

gurmies · May 14, 2009

Nice solution lolokay.

$Express \,\, \cos3\theta \,\, and \,\, \cos2\theta \,\, in \,\, terms \,\, of \,\, \cos\theta \,\, and \,\, show \,\, that \,\, \cos3\theta = \cos2\theta \,\, can \,\, be \,\, expressed \,\, as \,\, 4x^{3}-2x^{2}-3x+1=0, \,\, where \,\, x=\cos\theta. \,\, By \,\, solving \,\, this \,\, equation \,\, in \,\, x, \,\, find \,\, the \,\, exact \,\, value \,\, of \,\, \cos\frac{2\pi}{5}.$

harism · May 16, 2009

hahaha.
nice question there gurms.
my god you have to SEE it here.

jet · May 23, 2009

Time to revive this thread!

jet · May 23, 2009

Knox Trial 2005:
$\text{The base of a certain solid is the area bounded by the parabola }y^2 = 4ax ($for $ y \geq 0, a\geq 0) \text{ and the chord joining } (0,0) $ and $ A(9a, 6a). \text{Cross-sections of this solid, determined by planes } \\ \text{taken perpendicular to the x-axis, are semicircles with the diameter completely } \\ \text{in the base of the solid. By using the method of slicing, find the total volume of the solid formed.}$
And a screenshot!

GUSSSSSSSSSSSSS · May 23, 2009

this seems familiar ......... =S (shuning, namu and maybe gurmies wud understand xD)

gradient of the chord = 6a/9a = 2/3
Equation of the chord: y = (2/3)x

each cross section is a semicircle, Area = (1/2)(pi.r^2)
r= (1/2)(y1 - y2) ........... y1 = sqrt(4ax) .............. y2 = (2/3)x
= (1/2)(2sqrt(ax) - (2/3)x)
= (1/6)(6sqrt(ax) - 2x)

therefore

Area = (pi/2)((1/6)(6sqrt(ax) - 2x))^2
= (pi/72)(36ax - 24xsqrt(ax) + 4x^2)
= (pi/72)(36ax - (24/a)(ax)^(3/2) + 4x^2)

therefore

volume = pi/72 . S(36ax - (24/a)(ax)^(3/2) + 4x^2)dx limits: 9a --> 0
= pi/72 . [18ax^2 - (48/5a^2)(ax)^(5/2) + (4/3)x^3] limits: 9a --> 0
= pi/72 . (1458a^3 - (48/5a^2)(9a^2)^(5/2) + (4/3)(9a)^3) - 0
= pi/72 . (1458a^3 - (48/5a^2)(243a^(5) + 972a^3)
= pi/72 . (2430a^3 - (11664/5)a^3)
= pi/72 . ((486/5)a^3)
= (27pi.a^3)/20

hmmmmmmmmmmmmmmmmmmmm THAT looks a tad weird ...... always make so many typos on computer =[[[[[[

azureus88 · May 23, 2009

[maths]y_1^2=4ax\,\,and\,\,y_2=\frac{2}{3}x\\\delta V=\frac{\pi}{2}(y_1^2-y_2^2)\delta x\\=\frac{\pi}{2}(4ax-\frac{4}{9}x^2)\delta x\\V=\frac{\pi}{2}\int_{0}^{9a}(4ax-\frac{4}{9}x^2)dx\\=2\pi\int_{0}^{9a}(ax-\frac{x^2}{9})dx\\=27\pi a^3[/maths]

i might have misunderstood the question.

GUSSSSSSSSSSSSS · May 23, 2009

azureus88 said:
i might have misunderstood the question.

wat i interpreted it as saying was that the semicircles have their diameters IN that shaded region, so that shaded region is the BASE of a solid, with semi circular cross sections built up from it xD

......if that even makes sense at all =P

jet · May 23, 2009

So did I. The semicircles stick out from the black area out of the screen, in a metaphorical sense.

GUSSSSSSSSSSSSS · May 23, 2009

LOL well i just ammended my solution...saw one of the mistakes .... but it still doesnt look very pretty =[[[ probably heaps more mistakes in there =[[[

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