3u Mathematics Marathon V 1.1 (1 Viewer)

[Riviet]

Next question:

If tan(A+B)=1 and tan(A-B)=1/7, find all possible values of tanA and tanB.

 

[Sober]

(i) part is actually

Spoiler

8 book, can be arranged in 8! ways, since there are identical ones, then 8!/(4!x4!)=70

Good point, rather unthinking of me.

Next question:

If tan(A+B)=1 and tan(A-B)=1/7, find all possible values of tanA and tanB.

Spoiler

(1) tan(A+B) = 1
(2) tan(A-B) = 1/7

Let a=tan(A) and b=tan(b)

(1) ( a + b ) / ( 1 - ab ) = 1
(2) ( a - b ) / ( 1 + ab) = 1/7

Simultaneously solving yields two solutions:

tan(A) = 1/2, tan(B) = 1/3
and
tan(A) = -2, tan(B) = -3

 

[pLuvia]

Feel free to post the next question Or anyone

 

[Sober]

Feel free to post the next question Or anyone

Express the area of an anulus in terms of the largest interval that fits inside it.

 

[Riviet]

Express the area of an anulus in terms of the largest interval that fits inside it.

Spoiler

Isn't it just the area of the circle that the annulus is inscribed within?

 

[Sober]

Refer to the image. Interestingly, x does not uniquely define the anulus but it does define its area.

 

[Riviet]

Spoiler

As the annulus approaches the area of the circle, the longest interval approaches zero and as the annulus approaches an area of zero, the longest interval approaches the diameter of the circle.

 

[Sober]

Spoiler

As the annulus approaches the area of the circle, the longest interval approaches zero and as the annulus approaches an area of zero, the longest interval approaches the diameter of the circle.

I mean like area=f(x), solve f(x).

 

[sasquatch]

I think i might have it?

Spoiler

The largest interval that fits within an anulus MUST be at a tangent to the inner circle. This means that the angle between the radius and the tangent is 90 degrees.

Also because the tangent is also a chord of the outer circle, and the line joining this chord is perpendicular to it, the radius of the inner circle bisects the line with x length.

View attachment 12744

Let the inner cricle have a radius of r₁ and the outer circle hve a radius of r₂.

Please see the diagram for more info.

A = πr₂² - πr₁²
= π(r₂² - r₁² )

Now using pythagoras theorem,

r₂² = r₁² + (1/2x)²

Substituting the above equation into the first,

A = π(r₁² + (1/2x)² - r₁²)
= π(1/2x)²
=πx²/4

 

[Sober]

That's right slim, now on to you.

 

[doiyoubi]

no one seems to be posting a question so ive got a question

The speed vm/s of a particle moving on a straight line is given by:
v²=64-16x-8x²
where the displacement from a fixed point O is x
(a)Find an expression for the acceleration and show the motion is SHM
(b)Find the period of the motion
(c)Find the amplitude of the motion

 

[Mountain.Dew]

no one seems to be posting a question so ive got a question

The speed vm/s of a particle moving on a straight line is given by:
v²=64-16x-8x²
where the displacement from a fixed point O is x
(a)Find an expression for the acceleration and show the motion is SHM
(b)Find the period of the motion
(c)Find the amplitude of the motion

heres my 2 cents...

Spoiler

v²=64-16x-8x²

(a) v²=64-16x-8x²

(1/2)v²=32-8x-4x²

d (1/2)v²/dx=d(32-8x-4x²)/dx

knowing that d²x/dt² = d (1/2)v²/dx, then

d²x/dt² = -8-8x

a = -8(x+1) <== this is in the form a = -n^2(x-a), where n = sqrt8, a = -1.

Therefore, it is SHM.

(b) knowing that n = sqrt8, THEN the period, T = 2π/n = 2π/sqrt8 = 2sqrt8π/8 = πsqrt8/4 seconds

(c) the amplitude occurs when v = 0

so, back to original equation:

v²=64-16x-8x², sub v = 0

0 = 64-16x-8x²
0 = 8 - 2x - x²
x² + 2x - 8 = 0

(x - 4)(x + 2) = 0

therefore x = 4, -2 ==> the two extremes are 4 and -2, so the amplitude = 4 - (-2) / 2 = 3

therefore amplitude = 3

Alternatively,

knowing that the centre of motion is when x = -1, we know that 4 and - 2 are both 3 units away from -1. therefore, amplitude is 3.

Alternatively,

simply get v²=64-16x-8x² in the form: v = sqrt(a² - x²). we realise that a² = 9, so a = 3.

 

[Mountain.Dew]

now for my obligation to keep the ball rolling:

we are playing texas holdem. there are 3 player on the table, and a dealer.

the dealer deals out 2 cards to each player. after the inital round of betting, one card is discarded off the top of deck, followed by 3 'community cards' laid face up on the middle of the table.

now: the question:

1) what is the number of possible 2 card hands for:
(i) the 1st player?
(ii) the 2nd player?
(iii) the 3rd player?

2) what is the number of possible 3 'community cards' that could occur?

3) 2 more cards are to be laid down after more rounds of betting. before the 4th card is laid down, another card off the top of deck is discarded - then the 4th card is laid down. after another betting round, the same process applies for the 5th and final card.

(i) what are the possible 5 community cards that could occur?
(ii) how many possible poker hands can each player make with their inital hand and the community cards?

if u dont know the rules of texas holdem, please look here: http://www.texasholdem-poker.com/beginnersintro.php

 

[pLuvia]

Bump, that question is confusing

 

[SoulSearcher]

now for my obligation to keep the ball rolling:

we are playing texas holdem. there are 3 player on the table, and a dealer.

the dealer deals out 2 cards to each player. after the inital round of betting, one card is discarded off the top of deck, followed by 3 'community cards' laid face up on the middle of the table.

now: the question:

1) what is the number of possible 2 card hands for:
(i) the 1st player?
(ii) the 2nd player?
(iii) the 3rd player?

2) what is the number of possible 3 'community cards' that could occur?

3) 2 more cards are to be laid down after more rounds of betting. before the 4th card is laid down, another card off the top of deck is discarded - then the 4th card is laid down. after another betting round, the same process applies for the 5th and final card.

(i) what are the possible 5 community cards that could occur?
(ii) how many possible poker hands can each player make with their inital hand and the community cards?

if u dont know the rules of texas holdem, please look here: http://www.texasholdem-poker.com/beginnersintro.php

I'll attempt, but I wouldn't be surprised if I'm well off

Spoiler

1) i) Number of possible hands for 1st player is ⁵²C₂ = 1326

ii) Number of possible hands for 2nd player is ⁵⁰C₂ = 1225

iii) Number of possible hands for 3rd player is ⁴⁸C₂ = 1128

2) Possible 3 community cards that can appear is ⁴⁷C₃ = 16215

3) i) Possible 5 community cards that can occur is ⁴⁷C₃ * ⁴³C₁ * ⁴¹C₁

= 16215 * 43 * 41
= 28587045 (that doesn't sound right )

ii) Possible number of poker hands for 1st player is equal to ⁵²C₂ * ⁵C₃ = 13260

Possible number of poker hands for 2nd player is equal to ⁵⁰C₂ * ⁵C₃ = 12250

Possible number of poker hands for 3rd player is equal to ⁴⁸C₂ * ⁵C₃ = 11280

I'm not too sure about question 3, the answers sound wrong

 

[SoulSearcher]

I forgot to post up a question.

Next Question:

Use the substitution u = x - 1/x to show that ₁ ∫ ^{(√6+√2)/2} (1+x²)/(1+x⁴) dx = pi/4√2

 

[Riviet]

I remember this one from Cambridge 3 unit, it works out nicely too.

 

[SoulSearcher]

Ahh yeah, taxed that question, can't think of any myself at this time of night

 

[Riviet]

Next Question:

Use the substitution u = x - 1/x to show that ₁ ∫ ^{(√6+√2)/2} (1+x²)/(1+x⁴) dx = pi/4√2

Spoiler

let u=x-1/x
u²=x²-2+x^-2
u²+2=x²+x^-2
du=1+x^-2dx
When x=(√6+√2)/2, u=(2+2√3)/(√6+√2)
When x=1, u=0
.'. ₁ ∫ ^{(√6+√2)/2} (1+x²)/(1+x⁴) dx=₁ ∫ ^{(√6+√2)/2} (1+x²)/x²(x²+x^-2) dx
=₀ ∫ ^{(2+2√3)/(√6+√2)}du/(2+u²)
=(1/√2)[tan^-1(u/√2)]^{(2+2√3)/(√6+√2)}₀
=(1/√2).tan^-1[(2+2√3)/(2√3+2)]
=pi/4√2, as required.

Next Question:
Use the substitution u=2x to find ∫ 4/(1+4x²) dx

 

[SoulSearcher]

Next Question:
Use the substitution u=2x to find ∫ 4/(1+4x²) dx

Another late night Riviet?

Spoiler

∫ 4/(1+4x²) dx
u = 2x
u² = 4x²
du/dx = 2
∫ 4/(1+4x²) dx
= 2 ∫ 2/(1+4x²) dx
= 2 ∫ 1/(1+u²) du
= 2tan^-1u + C
= 2tan^-12x + C

I'll post a question up later today.