Extension One Revising Game (2 Viewers)

[Trebla]

n³ - (n-1)³ + (n-1)³ - ... -1³ + 1³ - 0³ = n³

is that fine?

Yep, not to mention its a lot quicker than induction...

 

[vds700]

hey guys, this doesnt really have to do with any of the questions, but its a doubt i have so i was wondering if someone can clarify (sorry if im breaking the rules a bit).

Lets say that in a simple harmonic motion we are given a set of initial conditions and we can say something like; in general x= Acos/sin(nt[+/-]alpha) +B. Now lets say the questions says find apha, and we do so by plugging in the intial conditions. Now in most 3u cases we take alpha as the positive value and forget about the various quadrants and where what is positive/negative (A,S,T,C), but sometimes in 4u (and 3u) we have to determine whether alpha is positive or negative....now heres the question (finally!)...the way i do it is by using the intial input values/conditions e.g. at time t=o x(dot)>0 etc. but from what i understand there is a easier way using a unit circle. Firstly, how does this method work and secondly what are the principles behind it...is it kinda like evaluating shm to circular motion???? PLEASE HELP

I may be wrong on this, but I'm prtty sure there should only be one solution of alpha. The reason there is a constant (alpha) is to shift the sine or cosine curve along the t axis, ie because at t = 0, it is not at the origin or one of the endpoints.

If at t = 0, the particle is
-At the origin: x= asin(nt)
-At the the positive extreme: x = acos(nt)
-At the negative extreme: x =-acos(nt)
-Somewhere in between, x = asin(nt + alpha) of x = acos(nt + alpha)

Sometimes in SHM questions, you need to get multiple solutions to trig equations (like in the question i posted earlier. This is where youneed to use ASTC and the unit circle, to find the correct solutions for t.

 

[Pratham]

thanks!

 

[cutemouse]

Yeah you're correct, but the question is if you went through the whole process of solving it and then substituting back into the equation to find that it doesn't work. Most people cannot say straight away that it's no solution.

no solution?

 

[independantz]

Tommy, I don't think your answer for Q9 is correct. The total number of perm is 6!/2!2! = 180

And the total number of Ps together is basically = 5!/2!2! = 30

So the total number of Ps separaed is = 150.

Umm for "the total number of Ps together is basically =5!/2!2!" you don't need to divide by another 2! as you haven't arranged the Ps to begin with. So really it should be 5!/2!

Correct me if i'm wrong.

 

[lyounamu]

Umm for "the total number of Ps together is basically =5!/2!2!" you don't need to divide by another 2! as you haven't arranged the Ps to begin with. So really it should be 5!/2!

Correct me if i'm wrong.

yep, you are right. I misinterpreted my own working.

 

[lolokay]

some calculus questions

1. a) differentiate x^x

b) 'Air is being pumped into a spherical balloon at a rate of 500cm³/s. Find the rate of increase of surface area S when the radius is 20cm' [S = 4pi r², V = 4/3 pi r³]

c) show by differentiating from first principals that the derivative of ln[x] is 1/x using the definition e=lim[n->infinite](1+1/n)ⁿ


2. by considering the graph of each function, evaluate:
a) ∫[0->a*] sqrt(9 - x²) *(0 =< a <=3)

b) ∫[0->a] arctan[x].dx


3. a) ∫1/(1+2e^3x) .dx

b) ∫(1+x)²/(2+x)⁶ .dx

c) ∫1/(1+2tan²x) .dx


hopefully those are all written how I wanted

 

[lyounamu]

some calculus questions

1. a) differentiate x^x

b) 'Air is being pumped into a spherical balloon at a rate of 500cm³/s. Find the rate of increase of surface area S when the radius is 20cm' [S = 4pi r², V = 4/3 pi r³]

c) show by differentiating from first principals that the derivative of ln[x] is 1/x using the definition e=lim[n->infinite](1+1/n)ⁿ


2. by considering the graph of each function, evaluate:
a) ∫[0->a*] sqrt(9 - x²) *(0 =< a <=3)

b) ∫[0->a] arctan[x].dx


3. a) ∫1/(1+2e^3x) .dx

b) ∫(1+x)²/(2+x)⁶ .dx

c) ∫1/(1+2tan²x) .dx


hopefully those are all written how I wanted

1b). dv/dt = 500

V = 4/3pi r^3
dV/dr = 4pi. r^2

A = 4pi . r^2
dA/dr = 8pi . r

dA/dt = dv/dt . dr/dv . dA/dr
= 500 . 1/(4pi r^2) . 8pi . r
= 1000/r
When r = 20, dA/dt = 50

1c)

e=lim[n->infinite](1+1/n)^n

1 = ln(1+1/n)^n
1 = n ln(1+1/n)
1/n = ln(1+1/n)

f(x) = ln(x)

First principle: lim h->0 (f(x+h) - f(x))/h

= (ln(x+h)-lnx)/h
= ln((x+h)/x))/h
= ln(1+h/x)/h
= ln(1+1/(x/h))/h
= (h/x)/h (using the knowledge given)
= 1/x
= 1/x ( after h->0)

 

[kaz1]

some calculus questions

1. a) differentiate x^x

(x²)^x-1

x^2x-2

Is this right?

 

[tacogym27101990]

(x²)^x-1

x^2x-2

Is this right?

no its
x^x(lnx+1)

 

[kaz1]

no its
x^x(lnx+1)

Thanks. Please excuse my noobishness in Maths.

 

[lolokay]

lyounamu: 1b) correct, 1c) I think that's correct as long as you state that (x/h)->infinty, making it equivalent to n

1a) what tacogym said

 

[tacogym27101990]

Thanks.Please excuse my noobishness in Maths.

no worries
ill give you a solution, but mine uses implicit differentiation, so you might not understand:
let y=x^x
then x=log_xy
x=lny/lnx
lny=xlnx
diff implicitly:
y'/y=lnx + x/x
y'=y(lnx+1)
y'=x^x(lnx+1)

theres probably a quicker way to do it, thats just the first way i thought of

 

[lolokay]

if you don't know implicit dif. you can just use

Spoiler

x^x
= e^xlnx
the derivative of which is (1+lnx)e^xlnx
= (1+lnx)x^x

 

[lyounamu]

if you don't know implicit dif. you can just use

Spoiler

x^x
= e^xlnx
the derivative of which is (1+lnx)e^xlnx
= (1+lnx)x^x

Damn, you post up answer too quickly~

 

[independantz]

Q: In how many ways can 5 women and 5 men be arranged in a circle so that the men are separated and two particular women must not be next to a particular man?

The answer is:

Spoiler

Answer: 864

 

[lyounamu]

Q: In how many ways can 5 women and 5 men be arranged in a circle so that the men are separated and two particular women must not be next to a particular man?

The answer is:

Spoiler

Answer: 864

(5! x 4 x 3 x 3!)/10 = 864

 

[M@ster P]

Expand cos (2A + B) and hence prove that 1/4cos3(A) = cos^3(A) - 3/4cos(A).

Putting x = Kcos(A) and giving K a suitable value, use the preceding formula to find the three roots of the equation 27x^3 - 9x = 1.

Hence write down the value of the product cos20.cos60.cos100.cos140

 

[lolokay]

Expand cos (2A + B) and hence prove that 1/4cos3(A) = cos^3(A) - 3/4cos(A).

Putting x = Kcos(A) and giving K a suitable value, use the preceding formula to find the three roots of the equation 27x^3 - 9x = 1.

Hence write down the value of the product cos20.cos60.cos100.cos140

Spoiler

cbf writing out first bit

letting K = 2/3, we get 8cos³A - 6cosA = 1
cos3A = 1/2
3A = 60', 300', 420' ..
A = 20', 100', 140'
x = 2/3[cos20, cos100, cos140]

the product of the roots is 8/27cos20cos100cos140 = 1/27, and cos60 = 1/2
so cos20.cos60.cos100.cos140 = 1/16

 

[M@ster P]

can u explain what you did for that question if you don't mind.

The first part of the question i understand, just the later parts