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  #101    
Old March 8th, 2011 (11:35 AM).
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AND I GOT MY SAMPLE EXAM PAPER...
http://examinations.ie/schools/Project_Maths_Phase_2_P2_Higher_Level.pdf
Anyone wanna have a go at it?
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  #102    
Old March 9th, 2011 (07:26 AM).
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I'll join, I tutor maths as a part-time job, so feel free to ask for help lol.

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  #103    
Old March 14th, 2011 (10:18 AM).
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I'm surprised no one's mentioned it- HAPPY PI DAY!
Ah, Pi.... How awesome it is.
Only, it seems some people disagree.
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  #104    
Old March 18th, 2011 (03:11 AM).
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Happy seriously late Pi day and St. Patrick's day.....


... Well, a good topic would be to ask of all of you what you think of the concept of Tau, now that's it's nicely brought up. I think there can definitely be decent ways for Tau and Pi to coexist.
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  #105    
Old March 18th, 2011 (06:11 AM).
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Tau and Pi have to co-exist, Or else nothing would work. The universe is one of these things where if they are each constant seperatally, they must be constant together.
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  #106    
Old March 18th, 2011 (11:28 AM).
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Pi is well-known all over the world. Ditching it is just going to confuse people. It would be a huge inconvenience. I have read the article and understood what the guy was thinking, but my opinion is he is just a bit crazy.

I do not see the point in celebrating "Pi day". Sure, the number is useful, but I don't understand why it could have a fanbase. After all, Mathematics is definitely not about just constants, so celebrating it here is actually just an insult to our intelligence.
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  #107    
Old March 18th, 2011 (09:58 PM). Edited March 18th, 2011 by Dude22376.
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Had a look through that test paper. Did some of the questions, the non-Statistics ones because they can be done without stats software.

Q3
Spoiler:

a) Substitute the coordinates of the point onto the equation of the line.
3(4k-2) - 4(3k+1) + 10
= 12k - 6 -12k - 4 + 10
= 0
b) The equation of l1 in the form y = mx + c is y = 3/4 x + 5/2
Gradient of l1 = 3/4
Gradient of l2 = -4/3
Equation of l2 : (y - 3k - 1)/(x - 4k + 2) = -4/3
Simplify: y = -4/3 x + 25/3 k -5/3
c) Substitute the values of x and y: 11 = -4/3 * 3 + 25/3 k -5/3
Simplify: k = 2
(Equation of line: y = 15 - 4/3 x, which (3,11) does lie on)
d) Foot of perpendicular = point of intersection
Equate y: 3/4 x + 5/2 = 15 - 4/3 x
Simplify: x = 6
Substitute into either line: y = 7


Q4
Spoiler:

Equation of line in the form y = mx + c is y = -x/2 + 3
In order for the x- and y-axes to be tangents to the circle, the magnitudes of the x- and y-coordinates of the centre of the circle must be equal.

Proof:
Draw a circle which satisfies the conditions.
Draw a line perpendicular to each point of contact between the circle and the axes towards the centre of the circle.
The x- and y-axes, plus the two lines you just drew, form a square.
Being a circle, the two lines you drew are of equal length. As they are located on lines perpendicular to the x- and y-axes, this means that the magnitudes of the x- and y-coordinates of the centre of the circle must be equal.

This means that the centre of the circle must lie on the line y = x or y = -x.
Consider each case simultaneously:
Equate y: x = -x/2 + 3 or -x = -x/2 + 3 (Note that the first equation uses y = x, the second uses y = -x)
Solve for x: x = 2 or x = -6
Solve for y: y = 2 oy y = 6 (Note that the first equation uses y = x, the second uses y = -x)
This gives us the coordinates of the centres of the cricles, as well as the radius (which is equal to the magnitude of either coordinate of the centre).
Thus the equations of the circles are:
(x-2)^2 + (y-2)^2 = 4
(x+6)^2 + (y-6)^2 = 36


Q6B
Spoiler:

Firstly, notice that the quadrilateral OECD is a cyclic quadrilateral.
Draw a circle in which OECD is inscribed.
We now claim that OC is equivalent to the diameter of the circle.

Proof:
From the centre of the circle (call this point Z), draw three lines connecting it to points O, D and C.
Since length of ZO=ZD=ZC, we can split ZODC into two isosceles triangles, ZOD and ZCD.
Thus angle ZOD=ZDO and angle ZDC=ZCD
Let angle ZOD=x and angle ZCD=y.
Then we can conclude the following about angles:
ZDO=x ZCD=y
OZD=180-2x CZD=180-2y
OZC=360-2x-2y=360-2(x+y)
Note that angle ODC=90=x+y
Thus OZC=180, and hence OC is a straight line.
The length of OC is the combined length of ZO and ZC, i.e. 2 times the radius, or equivalent to the diameter.

Consider the triangles ODC and EDC.
We now claim that angles DOC and DEC are equal.

Proof:
From the centre of the circle (call this point Z), draw three lines connecting it to points O, D and C.
Since length of ZO=ZD=ZC, we can split ZODC into two isosceles triangles, ZOD and ZCD.
Thus angle ZOD=ZDO and angle ZDC=ZCD
Let angle ZDC=a.
Then we can conclude that DOC=ZOD=ZDO=90-a (since ODC=90)
Now draw two lines connecting Z to points E, D and C.
Triangle EDC can be split into 3 equilateral triangles: ZED, ZDC, ZCE.
Maintaining that angle ZDC=a, we can conclude:
ZCD=a and DZC=180-2a
Let angle ZED=b and angle ZEC=c. We can conclude:
DEC=b+c
ZDE=b ZCE=c
DZE=180-2b CZE=180-2c
Consider the sum of angles ZDC,DZE and ZCE:
540-2a-2b-2c=360
Simplifying, b+c=90-a.
Since DOC=90-a and DEC=b+c, angle ZOC=DEC.
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  #108    
Old March 19th, 2011 (07:20 AM).
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Quote originally posted by AdvancedK47:
... Well, a good topic would be to ask of all of you what you think of the concept of Tau, now that's it's nicely brought up. I think there can definitely be decent ways for Tau and Pi to coexist.
I've not yet been taught Radians, or how they're used in Calculus, but I know what they are, and I think that Tau should be used in at least that instance.
I think with basic equations of the circle/sphere (circumference, area, surface area, volume) Pi is fine. It wouldn't really be any easier to remember with Pi or Tau.
Because I don't really use Pi that much (at this early stage) I'm not too bothered. If when I learn calculus or more complex trig, I might use Tau in my head, to understand it better.
Quote originally posted by ShinyMeowth:
I do not see the point in celebrating "Pi day". Sure, the number is useful, but I don't understand why it could have a fanbase. After all, Mathematics is definitely not about just constants, so celebrating it here is actually just an insult to our intelligence.
I "celebrate" it as more of a joke. Because Pi is so big, and so common in nature especially, it's almost like celebrating the beauty of nature and mathematics itself. That, and it gives me an excuse to eat pie.
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  #109    
Old March 20th, 2011 (11:19 AM).
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I will join this club.

While I am here . . .

Anyone able to solve the Laplace Transform of;

1/(s^2 + 1)^0.5
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  #110    
Old March 25th, 2011 (07:57 AM). Edited March 25th, 2011 by Alakazam17.
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Wow, this thread has been around for nearly two months and I'm just noticing it now. I'lll join as a full-pledge member, as I've always been a mathophile. XD

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And as for a math joke, I'll see if I can remember the just of this one made by my first year university professor:

lim sin(x)/n = 6
x->0

Anyone see what he did there? =D
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  #111    
Old March 26th, 2011 (12:14 AM).
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Another one:
ln e = 1
ln e = one
Cancel n and e from both sides:
1 = 0
This works better written out.
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  #112    
Old April 19th, 2011 (02:25 AM). Edited April 19th, 2011 by smile!.
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This place is dying
Anyway, I just had my Calculus finals today. And one of the questions that I was soooo intrigued to solve (but so far haven't managed to do so yet) is this one:

Verify that


I got pi/2, lol.

*whispers* ILPy, it's a pi!
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  #113    
Old April 20th, 2011 (02:14 AM).
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Hmm alright I took a look into it using a computer package.

y = 1/(1 - x^2)^(1/2) + 2*(1 - x^2)^(1/2)

int(y)

ans = 2*asin(x) + x*(1 - x^2)^(1/2) [where asin(x) = arcsin(x)]

Substituting the limits gives;

2*pi/2 + 0 - 0 - 0 = pi
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  #114    
Old April 20th, 2011 (02:21 AM).
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Quote originally posted by Drakow:
Hmm alright I took a look into it using a computer package.

y = 1/(1 - x^2)^(1/2) + 2*(1 - x^2)^(1/2)

int(y)

ans = 2*asin(x) + x*(1 - x^2)^(1/2) [where asin(x) = arcsin(x)]

Substituting the limits gives;

2*pi/2 + 0 - 0 - 0 = pi
Yes, WolframAlpha says the same thing. But how do you prove it? And we didn't learn about arcsin yet. Is there a way to solve it yourself without computer help, do you know?
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  #115    
Old April 20th, 2011 (02:28 AM).
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There's probably some trigonometric property you needed to know lol. I'll check my formula book now . . .

1/(1-x^2)^1/2 = sin^-1(x)

That's the property you needed.
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  #116    
Old April 20th, 2011 (04:17 AM). Edited April 20th, 2011 by smile!.
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Quote originally posted by Drakow:
There's probably some trigonometric property you needed to know lol. I'll check my formula book now . . .

1/(1-x^2)^1/2 = sin^-1(x)

That's the property you needed.
>.< Yeah, I got that. But I got stuck after this


5. 3 sin^-1 (x)], x=0 to 1 + .....?

The 2x^2 should be manipulated somehow. I tried using the identity (integral [du/(a^2-u^2)^1/2] = sin ^-1 u/a), but it doesn't work. I think that's because a is a constant, and not a variable. I used 1/x^2 for a, and 1/x for u.
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  #117    
Old April 20th, 2011 (08:46 AM).
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You give me no choice. Here is an outline to the solution.

Spoiler:
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  #118    
Old April 20th, 2011 (11:12 AM).
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Quote originally posted by Drakow:
You give me no choice. Here is an outline to the solution.

Spoiler:
Umm. But we didn't learn about arcsin yet. Ergo, the solution should be something that doesn't use arcsin at all. Thanks for giving the solution anyway. I appreciate that
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  #119    
Old April 20th, 2011 (11:16 AM).
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I have no idea how to solve it without using those trig properties then V_V
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  #120    
Old April 23rd, 2011 (05:55 PM). Edited April 23rd, 2011 by Dude22376.
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Arcsin is the inverse function of sin, you might have seen it written as sin^-1.
e.g. sin pi/2 = 1, so arcsin 1 = pi/2.
Note that the range of arcsin x is -pi/2 to pi/2, so arcsin 1 = pi/2 and not, say, 5 pi/2 (even though sin 5 pi/2 = 1)

For me, instead of turning t back into x, I changed the limits and evaluated based on t:

Let x = sin t.
dx/dt = cos t
dx = cos t dt

First we change the upper and lower limits from x to t. arcsin 0 = 0 and arcsin 1 = pi/2.







= 0 - 0 + pi - 0
= pi

PROTIP
If you see sqrt(a^2-x^2), try to sub in x = a sin t.
For sqrt(x^2-a^2), sub in x = a sec t.
For sqrt(a^2+x^2), sub in x = a tan t.

@Drakow: Nitpicking here. It's incorrect to use 1 and 0 as the upper/lower limits after subbing x for t. You must either change the limits (as I did), or start off with the indefinite integral for x so that you don't need to write the limits for t.
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  #121    
Old April 24th, 2011 (12:26 AM). Edited April 24th, 2011 by Overlord Drakow.
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Quote originally posted by Dude22376:
Arcsin is the inverse function of sin, you might have seen it written as sin^-1.
e.g. sin pi/2 = 1, so arcsin 1 = pi/2.
Note that the range of arcsin x is -pi/2 to pi/2, so arcsin 1 = pi/2 and not, say, 5 pi/2 (even though sin 5 pi/2 = 1)

For me, instead of turning t back into x, I changed the limits and evaluated based on t:

Let x = sin t.
dx/dt = cos t
dx = cos t dt

First we change the upper and lower limits from x to t. arcsin 0 = 0 and arcsin 1 = pi/2.







= 0 - 0 + pi - 0
= pi

PROTIP
If you see sqrt(a^2-x^2), try to sub in x = a sin t.
For sqrt(x^2-a^2), sub in x = a sec t.
For sqrt(a^2+x^2), sub in x = a tan t.

@Drakow: Nitpicking here. It's incorrect to use 1 and 0 as the upper/lower limits after subbing x for t. You must either change the limits (as I did), or start off with the indefinite integral for x so that you don't need to write the limits for t.
What you say is true, however, because the substitution reverts back to x anyway, I did not bother changing the limits just to change them back to the original limits. Though you are right, I am mathematically wrong. Anyway, I think your solution is probably the one he was expected to give for his exam.
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  #122    
Old April 25th, 2011 (12:58 AM).
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Quote originally posted by Dude22376:
Arcsin is the inverse function of sin, you might have seen it written as sin^-1.
...
...
...

*fails at life* How come I never knew that before? >.<
Okay, so everything makes sense now. Thanks, Drakow & Dude22376 =)
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  #123    
Old April 25th, 2011 (01:38 AM).
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^Dude! I implied that to you in my earlier message when I gave you the property you needed.

Ah, we all make these mistakes. I doubt you'll ever forget that again lol.
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  #124    
Old April 25th, 2011 (10:35 PM).
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^Yeah, I just didn't realize it. lol. XD Thanks again
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  #125    
Old May 4th, 2011 (10:16 AM).
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This club needs a logo:

(http://www.vestdijkfraneker.nl/Wiskunde/math.gif)

If you like math jokes:

(http://www.aquabacon.com/wp-content/main/2009_09/math-jokes.jpg)

Or just google for stuff.
Go solve (infinite) continued fractions or something.

*Teleports out*
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