Author Topic: most awesome question thread :) (Read 9255 times) Tweet Share

brightsky · « **Reply #60 on:** January 09, 2010, 06:26:33 pm »

lolz!

superflya · « **Reply #61 on:** January 09, 2010, 06:30:06 pm »

Quote from: TrueTears on January 09, 2010, 06:25:48 pm

Quote from: superflya on January 09, 2010, 06:24:27 pm
Quote from: brightsky on January 09, 2010, 06:21:31 pm
Lol, I have that book too.

all this time i assumed u finished skool brightsky. lol
all this time i assumed u finished skool superflya. lol

:O. beats me TT

superflya · « **Reply #62 on:** January 09, 2010, 06:31:18 pm »

Quote from: brightsky on January 09, 2010, 06:26:33 pm

lolz!

im still confused...are u in yr 12 this year??

TrueTears · « **Reply #63 on:** January 09, 2010, 06:32:44 pm »

Quote from: superflya on January 09, 2010, 06:30:06 pm

Quote from: TrueTears on January 09, 2010, 06:25:48 pm
Quote from: superflya on January 09, 2010, 06:24:27 pm
Quote from: brightsky on January 09, 2010, 06:21:31 pm
Lol, I have that book too.

all this time i assumed u finished skool brightsky. lol
all this time i assumed u finished skool superflya. lol

:O. beats me TT

im still confused...are u in yr 12 this year??

lol this is ridiculous

superflya · « **Reply #64 on:** January 09, 2010, 06:41:48 pm »

Quote from: TrueTears on January 09, 2010, 06:32:44 pm

Quote from: superflya on January 09, 2010, 06:30:06 pm
Quote from: TrueTears on January 09, 2010, 06:25:48 pm
Quote from: superflya on January 09, 2010, 06:24:27 pm
Quote from: brightsky on January 09, 2010, 06:21:31 pm
Lol, I have that book too.

hahaha, unfortunately.

all this time i assumed u finished skool brightsky. lol
all this time i assumed u finished skool superflya. lol

:O. beats me TT
im still confused...are u in yr 12 this year??

lol this is ridiculous

haha unfortunately

superflya · « **Reply #65 on:** January 09, 2010, 07:57:26 pm »

if the books answer is wrong with this one ill be so annoyed.
sketch and evaluate the following integral:

$\int_{-1}^{2}\sin^{-1}\frac{x}{2}\ dx$

i got $\frac{7\pi}{6}-4+\sqrt{3}$

TrueTears · « **Reply #66 on:** January 09, 2010, 08:17:37 pm »

Quote from: superflya on January 09, 2010, 07:57:26 pm

if the books answer is wrong with this one ill be so annoyed.
sketch and evaluate the following integral:

$\int_{-1}^{2}\sin^{-1}\frac{x}{2}\ dx$

i got $\frac{7\pi}{6}-4+\sqrt{3}$

Let $y = \sin^{-1} \frac{x}{2} \implies x = 2\sin(y)$

$\int_{-1}^2 \sin^{-1} \frac{x}{2} dx = \left[2 \cdot \frac{\pi}{2} - \int_0^{\frac{\pi}{2}} 2\sin(y) dy\right] - \left[ \left(1 \cdot \frac{\pi}{6}\right) -\left |\int_{-\frac{\pi}{6}}^0 2\sin(y) dy \right | \right] = \frac{5\pi}{6} - \sqrt{3}$

That required a lot of integrating along the $y$ axis and sketching.

I guess an easier way is to do:

$\frac{d}{dx}\left(x\sin^{-1}\frac{x}{2}\right) = \sin^{-1}\frac{x}{2} + \frac{x}{\sqrt{4-x^2}}$

Then integrate both sides and find the integral of $\sin^{-1}\frac{x}{2}$

superflya · « **Reply #67 on:** January 09, 2010, 08:22:51 pm »

>.< thanks TT .

TrueTears · « **Reply #68 on:** January 09, 2010, 08:26:46 pm »

nps, I see how you got your answer.

heh you did everything right, except at the end you added the area's together, however the question didn't ask for the area

It wanted the signed area haha

Since from -1 to 0 sin^-1 is negative, you gotta subtract it to find the signed area.

superflya · « **Reply #69 on:** January 09, 2010, 08:33:41 pm »

lol im a douce.
thanks again

mangopop · « **Reply #70 on:** January 13, 2010, 11:35:21 pm »

Prove that if the diagonals of a parallelogram are of equal length then the parallelogram is a rectangle.

Argh, vector proofs! I'm a bit lost with how to start this one.

kamil9876 · « **Reply #71 on:** January 14, 2010, 01:50:52 am »

Hint: $|a+b|^2=(a+b).(a+b)$

superflya · « **Reply #72 on:** January 14, 2010, 01:56:37 am »

Lol shoodve named it superflyas most awesome question thread

brightsky · « **Reply #73 on:** January 14, 2010, 10:14:57 am »

Quote from: mangopop on January 13, 2010, 11:35:21 pm

Prove that if the diagonals of a parallelogram are of equal length then the parallelogram is a rectangle.

Argh, vector proofs! I'm a bit lost with how to start this one.

Probs not the right way to do it, but....

Suppose you have a parallelogram $ABCD$ where the diagonals $AC = BD$ .

Consider the triangles $DBC$ and $DBA$ :

Remember that opposite sides of a parallelogram are parallel.

$\because DB = DB$ (common), $\angle BDC = \angle DBA$ (alternate angles are equal), $\angle BDA = \angle DBC$ (alternate angles are equal) .

$\therefore \bigtriangleup DBC \eqiv \bigtriangleup DBA$ (ASA)

So $AD = BC$ (corresponding sides of congruent triangles are equal)

Consider the triangles $DBC$ and $ACD$ .

$\because AD = BC, DC = DC$ (common), $AC = BC$ (given)

$\therefore \bigtriangleup DBC \equiv \bigtriangleup ACD$ (SSS)

So $\angle ADC = \angle BCD$ (corresponding angles of congruent triangles are equal [/tex]

However, $\angle ADC + \angle BCD = 180 \circ$ (co-interior angles are supplementary)

Substitute $\angle ADC = \angle BCD$ into the equation.

$\angle ADC + \angle ADC = 180 \circ$

$\implies \angle ADC = 90 \circ$

Substitute back into $\angle ADC = \angle BCD$ .

$\implies \angle BCD = 90 \circ$

Because co-interior angles are supplementary, that means $\angle BAD$ and $\angle ABD$ are both equal to $90 \circ$ .

As the definition of a rectangle is a parallelogram with four interior right angles,

Hence if the diagonals of a parallelogram are equal, then the parallelogram is a rectangle.

kamil9876 · « **Reply #74 on:** January 14, 2010, 01:32:02 pm »

Just to illustrate how vectors can trivialise geometry:

let $a$ and $b$ be two adjacent sides of the parralogram.

We have that

$|a+b|=|a-b|$

$|a+b|^2=|a-b|^2$

$(a+b).(a+b)=(a-b).(a-b)$

$a.a+2a.b+b.b=a.a-2a.b+b.b$

$4a.b=0$

$a.b=0$

Thus $a$ and $b$ are perpendicular as required.

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