4u Mathematics Marathon V 1.0 (2 Viewers)

[gman03]

Q: integrate 1/(1+e^x) dx

let y = e^x
then dy/y = dx

int 1/(1+e^x) dx
= int dy/y(1+y)
= int dy/y - int dy/(1+y) by partial fraction
= ln (y/(1+y)) + C
= - ln (1+e^-x) + C


Q: integrate 1/(e^x-e^-x) dx

 

[Stan..]

let u = 1 + e^x
du/u-1 = dx
Thus, 1/u (u-1) du
ln(u-1) -lnu + C
ln e^x / 1 + e^x + C

@ Gman
That is incorrect, try differentiating that.

 

[Stan..]

1/ (e^x - e^ -x) dx
e^-x - e^x dx
-e^x - e^-x + c
-(e^x + e^-x) + c

EDIT : Minus signs went loco.

 

[gman03]

let u = 1 + e^x
du/u-1 = dx
Thus, 1/u (u-1) du
ln(u-1) -lnu + C
ln e^x / 1 + e^x + C

@ Gman
That is incorrect, try differentiating that.

- ln (1+e^-x) + C
= ln (1 / (1+e^-x)) + C
= ln (e^x / (e^x+1)) + C

which is same as yours. where is the problemo?

 

[Stan..]

- ln (1+e^-x) + C
= ln (1 / (1+e^-x)) + C
= ln (e^x / (e^x+1)) + C

which is same as yours. where is the problemo?

Touche. Overlooked that.

Next Question:

Proove that the length between two points on a curve is:

/a
l ( 1 + (dy/dx)^2 ) ^ 1/2 dx
/b

 

[Stan..]

No takers?

Alright, it is
L^2 = dx^2 + dy^2
L^2 = 1 + ((f'(x))^2
L = (1 + ((f'(x))^2)^1/2

There is a sum of these minuscule distances between points a and b on this curve so it becomes:

L = Int (a, b) ( 1 + ((f'(x))^2)^1/2 as required

Alright, another easier this time.

I = Int ( e^ax cosbx ) dx

 

[gman03]

I = Int ( e^ax cosbx ) dx

euler's formula way:

I = Int ( e^ax cosbx ) dx
= Re( Int ( e^ax (cosbx + i sin bx) ) dx)
= Re( Int ( e^(ax + ibx) dx)
= Re( Int ( e^(a+ib)x dx)
= Re( e^(a+ib)x /(a+ib))
= Re( e^ax (cosbx + i sin bx)/(a+ib))
= e^ax / (a^2 + b^2) * (a cosbx + b sin bx)

the by parts way:

I = Int ( e^ax cosbx ) dx
= e^ax cosbx / a - -b/a Int ( e^ax sinbx ) dx
= e^ax cosbx / a + b/a [ e^ax sinbx /a - b/a Int ( e^ax cosbx ) dx ]
= e^ax cosbx / a + b/a [ e^ax sinbx /a - b/a I ]
= e^ax cosbx / a + b/a^2 e^ax sinbx - b^2/a^2 I
I (1 + b^2/a^2) = e^ax cosbx / a + b/a^2 e^ax sinbx
I (a^2 + b^2) = e^ax (a cosbx + b sinbx)
I = e^ax (a cosbx + b sinbx) / (a^2 + b^2)

 

[Riviet]

One of u guys gonna post up a new Q?

 

[pLuvia]

I'll post one

This is easy but hey

Sketch the locus of z defined by |z-2-2i|=1, hence,
a) find the range of |z| and the complex number that has the maximum modulus
b) find the range of arg(z), and the complex number that has the minimum argument

 

[Bookie]

man, this stuff is hard.

 

[bananasmoothy]

I'll post one

This is easy but hey

Sketch the locus of z defined by |z-2-2i|=1, hence,
a) find the range of |z| and the complex number that has the maximum modulus
b) find the range of arg(z), and the complex number that has the minimum argument

I feel so far behind... just learnt sketching of complex number locus, and can't recall learning the max modulus and min argument... >.<

Meh, I'll have a go anyway, and then we can all see what not to do...

ummm...
a) ok, so |z-2-2i|=1 would give us a circle with centre (2, 2i), radius 1, which would be located in the first quadrant, yes? So then, range of |z| = 2 ... and that is all. You have heard my words of wisdom for today.

(Am tired. )

 

[Riviet]

I'll post one

This is easy but hey

Sketch the locus of z defined by |z-2-2i|=1, hence,
a) find the range of |z| and the complex number that has the maximum modulus
b) find the range of arg(z), and the complex number that has the minimum argument

The locus of z is a circle with centre (2,2i) and radius 1.

a) I'm not sure what you mean by range but i will just find the max and min |z|
min |z|=sqrt(4+4)-1
=2sqrt2 - 1
max |z|= 2sqrt2+1

b) Once again i'm not sure what you mean by range but i will find the max and min argument of z
min arg(z): tan@=2
@=63.4^o
max arg(z) is the complementary to 90-min arg(z)
=90- 63.4^o
= 26.6^o

Next question
Simplify the following, leaving your answer in exact values:
a) (cospi/7-isinpi/7)⁷
b) [(1-cospi/7+isinpi/7)/(1-cospi/7-isinpi/7)]⁷

 

[pLuvia]

Simplify the following, leaving your answer in exact values:
a) (cospi/7-isinpi/7)⁷
b) [(1-cospi/7+isinpi/7)/(1-cospi/7-isinpi/7)]⁷


a) (cos pi/7 - isin pi/7)⁷
= (cos (-pi/7) + isin (-pi/7))⁷
= cos -pi + isin -pi
= -1 + 0i

Ai, can't do b

 

[SeDaTeD]

The answer is 1 I believe.

Show the modulus of (1-cospi/7+isinpi/7)/(1-cospi/7-isinpi/7) is 1.
Then rewrite it in the form cis(theta) for some theta (which can be found from a diagram)
Raise it to the 7th power.

 

[Riviet]

Interesting way of doing it, SeDaTeD.

The way that i did it is very different... hint: "the magic number is 14"
The hint is very helpful if you know what i mean.

 

[SeDaTeD]

Hehe, I also did the rationalising denominator, making the numerator bit squared in a way. Then i didn't like the idea of expanding and collecting terms of a binomial raised to the 14th power.

 

[Yip]

using double angle theorem for cos and sin to turn [(1-cospi/7+isinpi/7)/(1-cospi/7-isinpi/7)]^7 ie 1-cospi/7=2sin^2[pi/14], isin[pi/7]=2isin[pi/14]cos[pi/14]
into
cis6pi=1 is faster i think

 

[Riviet]

Yip's got the hint, and yes it is equal to 1, now show the working.

 

[Yip]

[(1-cospi/7+isinpi/7)/(1-cospi/7-isinpi/7)]^7
=[ (2sin^2[pi/4]+2isin[pi/14]cos[pi/14])/(2sin^2[pi/4]-2isin[pi/14]cos[pi/14])]^7
=[(sin[pi/14]+icos[pi/14])/(sin[pi/14]-icos[pi/14])]^7
Realizing, you get
[[(sin[pi/14]+icos[pi/14])^2]/1]^7
=[cis[pi/2-pi/14]]^14
=(cis[3pi/7])^14
=cis[6pi]=1

 

[Riviet]

Woot you got it!

You have the honours of posting up next Q... :uhhuh: