how big a change is yr 11 maths to yr 10 maths (2 Viewers)

Shadowdude · Jul 15, 2011

Let's see if this works...

Since I can't be bothered - I wrote up the solution my tutor gave to the first integral:

$\int\frac{1+t^2}{1+t^4}dt\\ \textup{Set}\;u=t-t^{-1},\;t\neq 0\\ \textup{Then:}\\ \frac{du}{dt}=1+\frac{1}{t^2}\\ du=\frac{1+t^2}{t^2}dt\\ \therefore dt = \frac{t^2}{1+t^2}du\\ \textup{So:} \int\frac{1+t^2}{1+t^4} = \int\frac{1+t^2}{1+t^4}.\frac{t^2}{1+t^2}du\\ \int\frac{t^2}{1+t^4}du = \int\frac{1}{t^{-2}+t^{2}}du\\ \textup{Note:} u = t-t^{-1} \Leftrightarrow u^2 = t^{-2} - 2 + t^2\\ \int\frac{1}{t^{-2}+t^{2}}du=\int\frac{1}{2 + u^2}du\\ \int\frac{1}{2 + u^2}du=\frac{1}{\sqrt{2}}tan^{-1}\left ( \frac{u}{\sqrt{2}}\right ) + c\\ \therefore \int\frac{1+t^2}{1+t^4} = \frac{1}{\sqrt{2}}tan^{-1}\left ( \frac{t - t^{-1}}{\sqrt{2}}\right ) + c$

AAEldar · Jul 15, 2011

I was right, I just forgot to make it $\sqrt2$ in the arctan and not just 2 D:

bleakarcher · Jul 15, 2011

thatz good lol

bleakarcher · Jul 15, 2011

next question

Shadowdude · Jul 15, 2011

Try this one, this looks interesting:

$\textup{Find\;a\;simplifying\;expression\;for\;the\;product}\;\left ( 1-\frac{1}{2} \right )\left ( 1-\frac{1}{3} \right )...\left ( 1-\frac{1}{n} \right )\\ \textup{Then,\;verify\;its\;validity\;for\;all\;integers}\;n\geq 2$

bleakarcher · Jul 15, 2011

i havent seen that b4 lol

Timothy.Siu · Jul 15, 2011

Shadowdude said:
Try this one, this looks interesting:

$\textup{Find\;a\;simplifying\;expression\;for\;the\;product}\;\left ( 1-\frac{1}{2} \right )\left ( 1-\frac{1}{3} \right )...\left ( 1-\frac{1}{n} \right )\\ \textup{Then,\;verify\;its\;validity\;for\;all\;integers}\;n\geq 2$

This might be right...

$\left ( 1-\frac{1}{2} \right )\left ( 1-\frac{1}{3} \right )...\left ( 1-\frac{1}{n} \right )\\ = \prod_{n=2}^n \left(\frac{n-1}{n}\right)\\=\frac{(n-1)!}{n!}\\=\frac{1}{n}$

Shadowdude · Jul 15, 2011

Just realised I have no solution for that. I just lifted it from a textbook =P

But Timothy, that looks right.

ZombieApocalypse · Jul 15, 2011

I have a question for you guys

If the gradient of the tangent to y=x^1/2 is 1/6 at point H, find the coordinates of H.
I would use latex, but I don't know how

bleakarcher · Jul 15, 2011

y=x^(1/2)
dy/dx=(1/2)x^(-1/2)
When dy/dx=m(T)=1/6,
(1/2)x^(-1/2)=1/6
x^(-1/2)=1/3
x=(1/3)^(-2)=1/(1/3)^2=1/(1/9)=9
y=x^(1/2)
When x=9,
y=9^(1/2)=3
Thus, H(9,3)

ZombieApocalypse · Jul 15, 2011

CORRECT!
Good work.

Shadowdude · Jul 15, 2011

As Timothy.Siu already answered my question, try:

1. $\textup{Sketch\;the\;curve:}\;y^2 = x(1-x)^2$ (this is MX2 stuff, perhaps even harder than MX2 stuff - but if you want to give it a shot, go ahead)

2. $\textup{Find\;the\;area\;of\;the\;'loop',\;made\;by\;the\;curve:}\;y^2 = x(1-x)^2$

SpiralFlex · Jul 15, 2011

Shadowdude said:
As Timothy.Siu already answered my question, try:

1. $\textup{Sketch\;the\;curve:}\;y^2 = x(1-x)^2$ (this is MX2 stuff, perhaps even harder than MX2 stuff - but if you want to give it a shot, go ahead)

2. $\textup{Find\;the\;area\;of\;the\;'loop',\;made\;by\;the\;curve:}\;y^2 = x(1-x)^2$

My teacher taught me how to sketch this graph.

You can't integrate normally can you?

Omnipotence · Jul 15, 2011

Here's a question kiddies;
int(exp(-a*r^2+a*r*x*cos(theta))*r))

SpiralFlex · Jul 15, 2011

Omnipotence said:
Here's a question kiddies;
int(exp(-a*r^2+a*r*x*cos(theta))*r))

Use Latex please...

Omnipotence · Jul 15, 2011

Too lazy..

SpiralFlex · Jul 15, 2011

Clue please Charmander...

Omnipotence · Jul 15, 2011

Double integral and polar co-ordinates...

SpiralFlex · Jul 15, 2011

Omnipotence said:
Double integral and polar co-ordinates...

Haven't done double integrals yet. The only thing I know about them is they find the surface area or something.

bleakarcher · Jul 15, 2011

double integralz?...use ur head, were in high school

how big a change is yr 11 maths to yr 10 maths (2 Viewers)

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