Preliminary mathematics marathon (1 Viewer)

Trebla · Apr 23, 2010

hscishard said:
I'll try the second one.
f'(x) = -k/x^2

Therefore tangent is
$y= -\frac{k}{x_{1}^{2}}(x-x_{1})+y_{1}$
y-intercept:
$y= \frac{k}{x_{1}} + y_{1}$
x intercept
$x_{1}(1+x_{1}/k)$

Area of triangle = 1/2AB
B is perp height.
$=x_{1}/2(\frac{k}{x_{1}}+y_{1})(1+\frac{x_{1}}{k})$

Check your working again, the x-intercept isn't quite correct. You're on the right track though. The trick in this is to realise that x₁y₁ = k since P is on the curve.

Also, just because it comes from Conics in Ext2 doesn't mean that Ext1 people aren't capable of answering it. There is nothing exclusive to Ext2 required to solve this problem (i.e. a 2U method can solve this problem)

Trebla · Apr 23, 2010

ohexploitable said:
I was thinking of like a logical approach...

A = 1/2 ab

height = y
length = x

xy=k

therefore A = 1/2 k

since k is constant, area is constant

amirite?

Not quite. If you use xy = k then they must be points on the hyperbola but you've defined them as the x-intercepts and y-intercepts of the tangent.

hscishard · Apr 23, 2010

Aha!

$1/3 S = 2/3^{2}+5/3^{3}+8/3^4...$

$S-\frac{1}{3}S= 2/3 - 2/3^2+5/3^2 - 5/3^3+8/3^3...$

Noticing a pattern, 2/3^2-5/3^2 = 1/3^2(3); -5/3^3 + 8/3^3 = 1/3^3(3)

$Therefore S-S/3= 2/3+\sum \frac{3}{3^{n}}$ Top of sum of is infinity.bottom is 2

The sum of is: 1/3 / (1 - 1/3) [a/1-r]
= 1/2
$\frac{2}{3}S = \frac{2}{3}+\frac{1}{2}$

2/3S = 7/6

S= 21/12 = 7/4, therefore my first assumption is correct

hscishard · Apr 23, 2010

Trebla said:
Check your working again, the x-intercept isn't quite correct. You're on the right track though. The trick in this is to realise that x₁y₁ = k since P is on the curve.

Also, just because it comes from Conics in Ext2 doesn't mean that Ext1 people aren't capable of answering it. There is nothing exclusive to Ext2 required to solve this problem (i.e. a 2U method can solve this problem)

Ooo I see where I got wrong. Would never have thought of the k part. thx

nikkifc · Apr 23, 2010

A piece of wire is 70 cm long. One section is bent into an equilateral triangle and the other into a rectangle that is three times as long as it is wide. Find the lengths of the two pieces of wire if the sum of the areas of the two shapes is a minimum.

Nice question. Definitely something that the 2U and even some 3U candidates would have trouble with.

Contrary to many peoples' responses here (who fell for the obvious red herring), the two lengths of the triangle and the rectangle are: 6.84m and 63.2m (to 3 sf) respectively to yield the minimum sum of areas.

hscishard · Apr 24, 2010

nikkifc said:
Nice question. Definitely something that the 2U and even some 3U candidates would have trouble with.

Contrary to many peoples' responses here (who fell for the obvious red herring), the two lengths of the triangle and the rectangle are: 6.84m and 63.2m (to 3 sf) respectively to yield the minimum sum of areas.

Are you sure?

Your total area is 189.48 m^2.
My one is 116.23 m^2

Tell us how you did it.

ohexploitable · May 18, 2010

ohexploitable said:
This marathon ain't so good cos liek most ppl haven't done as much as us, liek my school has only done basic algebra, and geometry.

I think people know more stuff now...

$\\\text{Solve for }0^{\circ}\leq\theta\leq360^{\circ} \\ \\5\sin^{2}\theta-4\sin\theta\cos\theta+3cos^{2}\theta=2$

hscishard · May 18, 2010

ohexploitable said:
I think people know more stuff now...

$\\\text{Solve for }0^{\circ}\leq\theta\leq360^{\circ} \\ \\5\sin^{2}\theta-4\sin\theta\cos\theta+3cos^{2}\theta=2$

$5sin^2x-4sinxcosx+3cos^2x=2(sin^2x+cos^2x)$
$3sin^2x-4sinxcosx+cos^2x=0$
$(3sinx-cosx)(sinx-cosx=0)$
$sinx=cosx$
x=45 and 225. After dividing both sides by cosx
$3sinx=cosx$
$tanx=1/3$
x = 18.43 and 198.43 (2.dp)

I hate it when I get an irrational no. Feels like you've done something wrong.
I really gotta learn the cross method for factorising

ohexploitable · May 18, 2010

^^

Haha, it took me ages before I thought about sin²θ + cos²θ.

ohexploitable · May 18, 2010

New Question:

Differentiate...

$y=x^{x^{x}}$

hscishard · May 18, 2010

A bushwalker,B,heads for a campsite,C, on a bearing of 230. At the same time he spots a rock formation,R, in a direction 300. From his map, he knows that the distance from the campsite to the rock formationis 5km on a bearing of 22. How far is the walker from the campsite? (2d.p.)

Please say how long your working out was.

Anyone know where to find more of these contruction questions?

ohexploitable · May 18, 2010

^^

I suck at interpreting those questions

Is it 5.27 km?

hscishard · May 18, 2010

ohexploitable said:
New Question:

Differentiate...

$y=x^{x^{x}}$

That isn't prelim. I'm sure for that.

$x^{x^{x}}[logx(logx+1)+x^x]$

What the hell do you do man?

hscishard · May 18, 2010

ohexploitable said:
^^

I suck at interpreting those questions

Is it 5.27 km?

Yea. Did you have to use simulatenous for that question?

ohexploitable · May 18, 2010

I didn't think it was prelim, I found it in an American college textbook.

Anyway...

First you differentiate y=x^x,

$\begin{align*} y&=x^{x} \\\ln y&=x\ln x \end{align*}$

then d/dx both sides and you should end up with,

$\frac{dy}{dx}=x^{x}(\ln x+1)$

so then you differentiate the whole thing,

$\begin{align*} y&=x^{x^{x}} \\\ln y&=x^{x}\ln x \end{align*}$

then d/dx boths sides and substitute the result you got earlier and you should end up with,

$\frac{dy}{dx}=x^{x}(\ln x+1)\ln x+x^{x-1}$

ohexploitable · May 18, 2010

hscishard said:
Yea. Did you have to use simulatenous for that question?

Didn't use simultaneous equations, just drew the triangle and used sine rule.

hscishard · May 18, 2010

ohexploitable said:
Didn't use simultaneous equations, just drew the triangle and used sine rule.

Fuck. My working out was 1 page for that question. Couldn't see the alternate angles.

hscishard · May 18, 2010

ohexploitable said:
I didn't think it was prelim, I found it in an American college textbook.

Anyway...

First you differentiate y=x^x,

$\begin{align*} y&=x^{x} \\\ln y&=x\ln x \end{align*}$

then d/dx both sides and you should end up with,

$\frac{dy}{dx}=x^{x}(\ln x+1)$

so then you differentiate the whole thing,

$\begin{align*} y&=x^{x^{x}} \\\ln y&=x^{x}\ln x \end{align*}$

then d/dx boths sides and substitute the result you got earlier and you should end up with,

$\frac{dy}{dx}=x^{x}(\ln x+1)\ln x+x^{x-1}$

I did the first few steps. But used e. Never knew you can bring out the x^x down. though it was for numbers only.

How many maths textbooks do you have lol. I have 7. LOL

ohexploitable · May 18, 2010

I only have Cambridge but I frequently borrow some textbooks from the library.

hscishard · May 18, 2010

God. Wtf you used 4-unit differentiation. Obviously thats not 3 unit prelim.

I haven't done that..but I think the last lines of working out should be
$1/y\cdot dy/dx$
Hence you have to times everything by x^x^x.
Thats what my book has told me

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Preliminary mathematics marathon (1 Viewer)

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