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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: superflya on January 02, 2010, 10:12:53 pm

Title: most awesome question thread :)
Post by: superflya on January 02, 2010, 10:12:53 pm: $\int_{-2}^{-1}\frac {e^x}{1-e^x}\, dx$

Mod Edit: Fixed up incorrect LaTeX
Title: Re: most awesome question thread :)
Post by: GerrySly on January 02, 2010, 10:21:53 pm: Recognition...

$\begin{align*} \int_{-2}^{-1} \frac{e^x}{1-e^x}.dx&=-\int_{-2}^{-1} \frac{-e^x}{1-e^x}.dx\\ &=\left [-\log_e{(1-e^x)}\right ]_{-2}^{-1}\\ &=\log_e{(1-e^{-2})}-\log_e{(1-e^{-1})}\\ &=\log_e{\left ( \frac{1-e^{-2}}{1-e^{-1}}\right )}\\ &=\log_e{\left ( \frac{1+e}{e}\right )}\\ &=\log_e{(1+e)}-\log_e{e}\\ &=\log_e{(1+e)}-1 \end{align*}$
Title: Re: most awesome question thread :)
Post by: TrueTears on January 02, 2010, 10:23:04 pm: If inspection fails.

Let $u = e^x$

Then how do you integrate $\frac{u}{1-u}$ (long divide)
Title: Re: most awesome question thread :)
Post by: superflya on January 02, 2010, 10:41:24 pm: thanks gerry

getting stumped at the after the long division part TT :S
Title: Re: most awesome question thread :)
Post by: TrueTears on January 02, 2010, 10:45:42 pm: $\frac{u}{1-u} = \frac{-1}{u-1}-1$

The antiderivative of $\frac{-1}{u-1}$ is $-\log_e(u-1)$
Title: Re: most awesome question thread :)
Post by: superflya on January 02, 2010, 10:52:16 pm: legend :)
Title: Re: most awesome question thread :)
Post by: superflya on January 02, 2010, 11:11:28 pm: $U_{n}=\int_{0}^{\frac{\pi}{4}}\tan ^{n} x dx$

express $U_{n}+U_{n-2}$ in terms of n

hence, show $U_{6}=\frac{13}{15}-\frac{\pi}{4}$

Mod Edit: Merged triple post into one.
Title: Re: most awesome question thread :)
Post by: TrueTears on January 02, 2010, 11:40:40 pm: Quote from: superflya on January 02, 2010, 11:11:28 pm
$U_{n}=\int_{0}^{\frac{\pi}{4}}\tan ^{n} x dx$

express $U_{n}+U_{n-2}$ in terms of n

hence, show $U_{6}=\frac{13}{15}-\frac{\pi}{4}$

Mod Edit: Merged triple post into one.

$U_n+U_{n-2} = \int_0^{\frac{\pi}{4}} \tan^n(x)+\tan^{n-2}(x)\,dx$

$=\int_0^\frac{\pi}{4} \tan^{n-2}(x)(1+\tan^2(x))\,dx$

$=\int_0^\frac{\pi}{4} \tan^{n-2}(x)\sec^2(x)\,dx$

Let $u=\tan(x)\Rightarrow \frac{du}{dx} = \sec^2(x)$

$dx = \frac{du}{\sec^2(x)}$

$\Rightarrow U_n+U_{n-2} = \int_0^{1} tan^{n-2}(x) du=\int_0^1 u^{n-2}du$

$\therefore U_n+U_{n-2} = \left[ \frac{u^{n-1}}{n-1}\right]_0^1 = \frac{1}{n-1}$

Hopefully I didn't make any mistakes, stupid computer keeps on freezing for like 10-15 seconds each min or so.
Title: Re: most awesome question thread :)
Post by: superflya on January 02, 2010, 11:45:43 pm: thats right ;)
ur too quick for it TT :P
Title: Re: most awesome question thread :)
Post by: TrueTears on January 02, 2010, 11:54:13 pm: cool, do you still need help on the U_6, i think it should be fairly straightforward if you understood how to do the other part :)
Title: Re: most awesome question thread :)
Post by: superflya on January 03, 2010, 12:00:31 am: i got the answer but cood u show me ur workings, i probs used triple the amount of necessary steps :P

the essentials explanations are really getting to me, they show u s***t all
$\int \sin ^{4}x dx$
u use the double angle formulae for that right? i can do it using the books jibbed methods but how wood u do it?
:)
Title: Re: most awesome question thread :)
Post by: TrueTears on January 03, 2010, 12:12:16 am: $\int \sin^4(x) dx = \int \left(-\frac{1}{2}\left(\cos(2x)-1\right)\right)^2 dx$

Is that how you did it? I also used double angle formula, it is the only spesh method.

~~Or you can use complex numbers... but that's outside the spesh course.~~

won't work for this one
Title: Re: most awesome question thread :)
Post by: superflya on January 03, 2010, 12:20:12 am: yea i ended up with that integration but took a while.
wanted to finish the 2 chapters on antidifferentiation and its apps in 2 days but doesnt look lyk happening :P
Title: Re: most awesome question thread :)
Post by: TrueTears on January 03, 2010, 12:22:41 am: Quote from: superflya on January 03, 2010, 12:20:12 am
yea i ended up with that integration but took a while.
wanted to finish the 2 chapters on antidifferentiation and its apps in 2 days but doesnt look lyk happening :P
Yeah it sure does, but it is right :) Good work!
Title: Re: most awesome question thread :)
Post by: superflya on January 03, 2010, 12:29:14 am: partial fractions look so tempting in the next exercise but ceebs finishing this lol.
Title: Re: most awesome question thread :)
Post by: TrueTears on January 03, 2010, 12:35:01 am: Quote from: superflya on January 03, 2010, 12:29:14 am
partial fractions look so tempting in the next exercise but ceebs finishing this lol.
Gogo the night is still young my friend!!!!!!
Title: Re: most awesome question thread :)
Post by: superflya on January 03, 2010, 12:42:48 am: haha my sleeping patterns are effed so here we go
Title: Re: most awesome question thread :)
Post by: kyzoo on January 03, 2010, 03:34:41 am: Quote from: superflya on January 03, 2010, 12:20:12 am
yea i ended up with that integration but took a while.
wanted to finish the 2 chapters on antidifferentiation and its apps in 2 days but doesnt look lyk happening :P

IMO that is the funnest part of Spesh since I had never seen it before I touched the Spesh book.
Title: Re: most awesome question thread :)
Post by: superflya on January 03, 2010, 04:47:33 am: true it is pretty fun but some questions require a hell of a lot of work
Title: Re: most awesome question thread :)
Post by: superflya on January 03, 2010, 09:28:20 pm: how wood i integrate cosec^2(x-pi/2) dx
didnt know the code for cosec on latex :P
Title: Re: most awesome question thread :)
Post by: /0 on January 03, 2010, 09:30:56 pm: Make a substitution then use $\frac{d}{dx} \cot x = -\csc^2x$
Title: Re: most awesome question thread :)
Post by: TrueTears on January 03, 2010, 09:34:00 pm: Quote from: superflya on January 03, 2010, 09:28:20 pm
how wood i integrate cosec^2(x-pi/2) dx
didnt know the code for cosec on latex :P
$\int \text{cosec}^2\left(x-\frac{\pi}{2}\right) dx = \int \sec^2\left(x) dx = \tan(x)$

Quote from: /0 on January 03, 2010, 09:30:56 pm
Make a substitution then use $\frac{d}{dx} \cot x = -\csc^2x$
/0's way is also good, it is good to remember that 'rule' however try to manipulate your expressions into ones you already know.
Title: Re: most awesome question thread :)
Post by: superflya on January 03, 2010, 09:35:57 pm: i feel lyk an idiot :P
thanks guys
Title: Re: most awesome question thread :)
Post by: /0 on January 03, 2010, 10:06:47 pm: Ah yeah TT's way is better it's more simplified
Title: Re: most awesome question thread :)
Post by: superflya on January 03, 2010, 10:35:24 pm: $\int 7\cos ^{7}t\ dt$
Title: Re: most awesome question thread :)
Post by: /0 on January 03, 2010, 10:45:04 pm: $\int 7\cos^7tdt = 7\int \cos t \cos^6t dt = 7\int\cos t \ (1-\sin^2t)^3dt$ Then substitute
Title: Re: most awesome question thread :)
Post by: superflya on January 03, 2010, 10:47:09 pm: ahh got it :) thanks
Title: Re: most awesome question thread :)
Post by: superflya on January 04, 2010, 10:05:39 pm: $\int_{0}^{1}\frac{e^{2x}}{1+e^{x}}\ dx$

why dont i ever get a substitution that works?
Title: Re: most awesome question thread :)
Post by: TrueTears on January 04, 2010, 10:07:01 pm: $\frac{e^{2x}}{1+e^x} = \frac{1}{e^x+1} +e^x-1$

Try now?
Title: Re: most awesome question thread :)
Post by: superflya on January 04, 2010, 10:11:16 pm: lifesaver :)

btw howd u break it down to that, ive tried n failed :P
Title: Re: most awesome question thread :)
Post by: TrueTears on January 04, 2010, 10:12:58 pm: Quote from: superflya on January 04, 2010, 10:11:16 pm
lifesaver :)

btw howd u break it down to that, ive tried n failed :P
Long division.

Let $u = e^{x}$
Title: Re: most awesome question thread :)
Post by: superflya on January 04, 2010, 10:21:42 pm: got it. thanks ;)
Title: Re: most awesome question thread :)
Post by: superflya on January 09, 2010, 03:24:34 pm: why cant i do this >.<

$\int \frac{x}{\sqrt{4-x^{2}}}\ dx$

sin inverse?
Title: Re: most awesome question thread :)
Post by: /0 on January 09, 2010, 03:35:23 pm: In general if you see anything of the form $\sqrt{a^2-x^2}$ , try the substitution $x=a\sin{\theta}$ , $\theta \in [-\frac{\pi}{2},\frac{\pi}{2}]$

Let $x = 2\sin{\theta}$ with $\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ , then $dx=2\cos{\theta}d\theta$

$\int \frac{x}{\sqrt{4-x^2}}dx = \int \frac{2\sin{\theta}}{\sqrt{4-4\sin^2{\theta}}}(2\cos\theta d\theta)$

$=\int \frac{4\sin{\theta}\cos\theta}{2\sqrt{1-\sin^2\theta}}d\theta$

$=\int \frac{4\sin\theta\cos\theta}{2\cos\theta} d\theta$

$=2\int \sin\theta d\theta$

$=-2\cos\theta +C$

$=-2\cos \left(\sin^{-1}\frac{x}{2}\right)+C$

$=-2\left(\frac{\sqrt{4-x^2}}{2}\right)+C$

$=-\sqrt{4-x^2}+C$

With $\sqrt{a^2+x^2}$ try $x=a\tan{\theta}$ , $\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$

With $\sqrt{x^2-a^2}$ try $x=a\sec{\theta}$ , $\theta \in [0,\frac{\pi}{2}) \cup [\pi, \frac{3\pi}{2})$
Title: Re: most awesome question thread :)
Post by: Ahmad on January 09, 2010, 03:38:22 pm: That's a perfectly reasonable approach, another approach would be noticing that the x in the numerator is the derivative of x^2 upto a constant, which suggests the substitution u = x^2.
Title: Re: most awesome question thread :)
Post by: Damo17 on January 09, 2010, 03:38:32 pm: Quote from: superflya on January 09, 2010, 03:24:34 pm
why cant i do this >.<

$\int \frac{x}{\sqrt{4-x^{2}}}\ dx$

sin inverse?

x is on top of the fraction so it can not be sin inverse.

Let $u=4-x^2$

$\frac{du}{dx}=-2x$ so $x=-\frac{du}{2dx}$

so $\int \frac{\frac{-du}{2dx}}{\sqrt{u}}dx= -\frac{1}{2} \int u^{-\frac{1}{2}}du=-u^{\frac{1}{2}}+C=-\sqrt{4-x^2}+C$
Title: Re: most awesome question thread :)
Post by: /0 on January 09, 2010, 03:42:07 pm: Quote from: Ahmad on January 09, 2010, 03:38:22 pm
That's a perfectly reasonable approach, another approach would be noticing that the x in the numerator is the derivative of x^2 upto a constant, which suggests the substitution u = x^2.

ah, of course! thanks ahmad

*facepalm*
Title: Re: most awesome question thread :)
Post by: Damo17 on January 09, 2010, 03:43:55 pm: Quote from: /0 on January 09, 2010, 03:42:07 pm
Quote from: Ahmad on January 09, 2010, 03:38:22 pm
That's a perfectly reasonable approach, another approach would be noticing that the x in the numerator is the derivative of x^2 upto a constant, which suggests the substitution u = x^2.

ah, of course!

*facepalm*

Too much 'Stewart Calculus' would be to blame. ;D
Title: Re: most awesome question thread :)
Post by: /0 on January 09, 2010, 03:49:04 pm: Quote from: Damo17 on January 09, 2010, 03:43:55 pm
Quote from: /0 on January 09, 2010, 03:42:07 pm
Quote from: Ahmad on January 09, 2010, 03:38:22 pm
That's a perfectly reasonable approach, another approach would be noticing that the x in the numerator is the derivative of x^2 upto a constant, which suggests the substitution u = x^2.

ah, of course!

*facepalm*

Too much 'Stewart Calculus' would be to blame. ;D

Lol yes, now I'm a mindless integrating drone XD
Title: Re: most awesome question thread :)
Post by: Ahmad on January 09, 2010, 03:52:52 pm: Quote from: /0 on January 09, 2010, 03:35:23 pm
In general if you see anything of the form $\sqrt{a^2-x^2}$ , try the substitution $x=a\sin{\theta}$

This is a quality suggestion, certainly something to look for. It's good to do integrals in many different ways, gives you an idea of what works and why. :)
Title: Re: most awesome question thread :)
Post by: superflya on January 09, 2010, 03:55:35 pm: thanks guys :)
Title: Re: most awesome question thread :)
Post by: superflya on January 09, 2010, 04:02:35 pm: differentiate $fx=\sin (x)\cos ^{n-1}(x)$

hence verify that $n\int \cos ^{n}x\ dx=\sin (x)\cos ^{n-1}(x)+(n-1)\int\cos ^{n-2}(x)\ dx$
Title: Re: most awesome question thread :)
Post by: kamil9876 on January 09, 2010, 05:28:39 pm: $f'(x)=sin(x ) *\frac{d}{dx} (cos^{n-1}(x))+cos^{n-1}(x ) * \frac{d}{dx}(sin (x))$

let u=cosx

$f'(x)=sin(x ) *\frac{d}{dx} (u^{n-1})+cos^{n-1}(x ) * \frac{d}{dx}(sin (x))$

$=sin(x ) *\frac{du}{dx}*\frac{d}{du} (u^{n-1})+cos^{n-1}(x ) * \frac{d}{dx}(sin (x))$

$=-sin^2(x) * (n-1) (cos(x))^{n-2} + cos^n x$

$=(cos^2(x)-1)*(n-1) (cos(x))^{n-2} + cos^n x$

$=(n-1)cos^n(x) + (1-n)(cos^{n-2} (x))$

$f'(x)=ncos^n(x) + (1-n)(cos^{n-2} (x))$

$\implies ncos^n(x)=f'(x ) + (n-1)cos^{n-2} (x)$

and now antidifferentiate both sides :)
Title: Re: most awesome question thread :)
Post by: brightsky on January 09, 2010, 05:45:29 pm: I suggest you reading: http://vcenotes.com/forum/index.php/topic,7092.0.html. May help a bit.
Title: Re: most awesome question thread :)
Post by: superflya on January 09, 2010, 05:51:50 pm: sweet ;D
Title: Re: most awesome question thread :)
Post by: brightsky on January 09, 2010, 05:52:33 pm: Hmmm...I've got some questions:

1. Why does 'Trinon's Guide to Anti-Derivatives through Derivatives' have:

Quote
We first multiply this equation by $x$ so that we get $x \log_e(x)$ .

2. kamil, can you show the workings from $f(x) = \sin(x)\cos^{n-1}(x)$ to your first equation?

Thanks. :)
Title: Re: most awesome question thread :)
Post by: TrueTears on January 09, 2010, 06:01:55 pm: $\int \log_e(x) dx$

$\frac{d}{dx}\left(x\log_e(x)\right) = \log_e(x) + 1$

See any connections?
Title: Re: most awesome question thread :)
Post by: brightsky on January 09, 2010, 06:09:42 pm: Anymore clues? lol :p
Title: Re: most awesome question thread :)
Post by: TrueTears on January 09, 2010, 06:10:21 pm: $\int \frac{d}{dx}\left(x\log_e(x)\right) dx = \underbrace{\int \log_e(x)dx} + \int 1 dx$
Title: Re: most awesome question thread :)
Post by: brightsky on January 09, 2010, 06:13:42 pm: Ooh! I see..:) Thanks TrueTears.
Title: Re: most awesome question thread :)
Post by: superflya on January 09, 2010, 06:14:09 pm: last 6 or so questions of the final exercise of the integration chapter are a bi-tch :P
Title: Re: most awesome question thread :)
Post by: kamil9876 on January 09, 2010, 06:14:27 pm: Quote from: brightsky on January 09, 2010, 05:52:33 pm

2. kamil, can you show the workings from $f(x) = \sin(x)\cos^{n-1}(x)$ to your first equation?

Thanks. :)

product rule.
Title: Re: most awesome question thread :)
Post by: superflya on January 09, 2010, 06:16:08 pm: Quote from: kamil9876 on January 09, 2010, 06:14:27 pm
Quote from: brightsky on January 09, 2010, 05:52:33 pm

2. kamil, can you show the workings from $f(x) = \sin(x)\cos^{n-1}(x)$ to your first equation?

Thanks. :)

product rule.

pretty obvious :P
Title: Re: most awesome question thread :)
Post by: brightsky on January 09, 2010, 06:17:08 pm: Oh right. Imma bit slow today.. ::)
Title: Re: most awesome question thread :)
Post by: brightsky on January 09, 2010, 06:17:32 pm: superflya, what textbook are you using?
Title: Re: most awesome question thread :)
Post by: TrueTears on January 09, 2010, 06:17:48 pm: Quote from: brightsky on January 09, 2010, 06:17:32 pm
superflya, what textbook are you using?
Essentials.

Yeah I answered for him :)
Title: Re: most awesome question thread :)
Post by: superflya on January 09, 2010, 06:20:12 pm: Quote from: TrueTears on January 09, 2010, 06:17:48 pm
Quote from: brightsky on January 09, 2010, 06:17:32 pm
superflya, what textbook are you using?
Essentials.

Yeah I answered for him :)

yep ;)
Title: Re: most awesome question thread :)
Post by: brightsky on January 09, 2010, 06:21:31 pm: Lol, I have that book too. :)
Title: Re: most awesome question thread :)
Post by: 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
Title: Re: most awesome question thread :)
Post by: 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
Title: Re: most awesome question thread :)
Post by: brightsky on January 09, 2010, 06:26:33 pm: lolz! :D
Title: Re: most awesome question thread :)
Post by: 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 :P
Title: Re: most awesome question thread :)
Post by: superflya on January 09, 2010, 06:31:18 pm: Quote from: brightsky on January 09, 2010, 06:26:33 pm
lolz! :D

im still confused...are u in yr 12 this year??
Title: Re: most awesome question thread :)
Post by: 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. :)

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

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

lol this is ridiculous :P
Title: Re: most awesome question thread :)
Post by: superflya 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 :P
im still confused...are u in yr 12 this year??

lol this is ridiculous :P

haha unfortunately :P
Title: Re: most awesome question thread :)
Post by: 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 $\frac{7\pi}{6}-4+\sqrt{3}$
Title: Re: most awesome question thread :)
Post by: TrueTears 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 $\sin^{-1}\frac{x}{2}$
Title: Re: most awesome question thread :)
Post by: superflya on January 09, 2010, 08:22:51 pm: >.< thanks TT .
Title: Re: most awesome question thread :)
Post by: TrueTears 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 :P

It wanted the signed area haha

Since from -1 to 0 sin^-1 is negative, you gotta subtract it to find the signed area.
Title: Re: most awesome question thread :)
Post by: superflya on January 09, 2010, 08:33:41 pm: lol im a douce.
thanks again :)
Title: Re: most awesome question thread :)
Post by: 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.
Title: Re: most awesome question thread :)
Post by: kamil9876 on January 14, 2010, 01:50:52 am: Hint: $|a+b|^2=(a+b).(a+b)$
Title: Re: most awesome question thread :)
Post by: superflya on January 14, 2010, 01:56:37 am: Lol shoodve named it superflyas most awesome question thread :P
Title: Re: most awesome question thread :)
Post by: brightsky 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.
Title: Re: most awesome question thread :)
Post by: kamil9876 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.
Title: Re: most awesome question thread :)
Post by: TrueTears on January 14, 2010, 04:02:30 pm: Quote from: kamil9876 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.
lol, nice, I remember you showing me this longggg time ago :P
Title: Re: most awesome question thread :)
Post by: mangopop on January 14, 2010, 09:39:05 pm: Thank you brightsky and kamil! :)
Title: Re: most awesome question thread :)
Post by: kamil9876 on January 14, 2010, 09:59:36 pm: Quote
lol, nice, I remember you showing me this longggg time ago :P

lol serious, i can't remember ever doing this. I had to think ~~a bit~~ this time around.
Title: Re: most awesome question thread :)
Post by: superflya on January 15, 2010, 12:21:10 am: ^^ being a tad modest :P
Title: Re: most awesome question thread :)
Post by: kamil9876 on January 15, 2010, 12:31:00 am: Lol I didn't mean a bit literally :P
Title: Re: most awesome question thread :)
Post by: superflya on January 15, 2010, 12:36:57 am: LOL ;D
Title: Re: most awesome question thread :)
Post by: superflya on January 17, 2010, 06:20:23 pm: find area enclosed by $y= \sqrt{x}$ & $y=6-x$ & $y=1$

im not sure as to the order of equations that i have to subtract, its usually top part of area minus bottom but now theres right left and bottom?

Edit: got it -.-
Title: Re: most awesome question thread :)
Post by: superflya on February 17, 2010, 10:38:46 pm: find the point P on the line $x-6y=11$ such that $\underset{OP}{\rightarrow}$ is parallel to the vector $3i+j$
Title: Re: most awesome question thread :)
Post by: QuantumJG on February 17, 2010, 11:29:03 pm: What I did was:

Let P = (x,y) where x,y are a point on the line.

$\vec{OP}$ = (x,y) - (0,0) = (x,y)

$\vec{OP}$ . (3,1) = 1

$\therefore (x,y) . (3,1) = 1 \rightarrow 3x + y = 1$

Now $x = 11 + 6y \rightarrow 3(11 + 6y) + y = 1$

$\therefore 19y + 33 = 1 \rightarrow y = - \dfrac{32}{19}$

$\therefore x = \dfrac{17}{19}$

$\therefore P = ( \dfrac{17}{19}, \dfrac{-32}{19} )$
Title: Re: most awesome question thread :)
Post by: superflya on February 17, 2010, 11:35:58 pm: book says $(-11, \frac{-11}{3})$ :S
Title: Re: most awesome question thread :)
Post by: GerrySly on February 18, 2010, 12:08:20 am: Assuming $a$ is the x-ordinate of our point...

$P(a, \frac{a-11}{6})$ , now we get the gradient of that point and make it equal $\frac{1}{3}$ because that is the gradient of the vector (worded wrong I assume, but that's what I was thinking heh)

$\begin{align*} \frac{\frac{a-11}{6}}{a}&=\frac{a-11}{6a}\\ \frac{1}{3}&=\frac{a-11}{6a}\\ 2a&=a-11\\ a&=-11 \end{align*}$

Sub that back in

$P(-11, -\frac{11}{3})$
Title: Re: most awesome question thread :)
Post by: QuantumJG on February 18, 2010, 12:36:15 am: Quote from: superflya on February 17, 2010, 11:35:58 pm
book says $(-11, \frac{-11}{3})$ :S

I found a much simpler way to solve it and don't listen to my method above.

The solution is easy.

First let's draw a line through the point (3,1) which is $y = \dfrac{x}{3}$ and let's find where $y = \dfrac{x}{6} - \dfrac{11}{6}$ intersects $y = \dfrac{x}{3}$ . Why? Well because the point could always make the vector anti-parallel to (3,1).

So

$\dfrac{x}{6} - \dfrac{11}{6} = \dfrac{x}{3} \rightarrow -\dfrac{x}{6} = \dfrac{11}{6} \rightarrow x = -11$ & $y = \dfrac{11}{3}$

So P = (-11, $\dfrac{11}{3}$ )
Title: Re: most awesome question thread :)
Post by: superflya on February 18, 2010, 07:48:12 pm: -.- all the working and the solution was that straightforward \facepalms

legend quantum :)
Title: Re: most awesome question thread :)
Post by: superflya on February 25, 2010, 06:04:38 pm: the number of real solutions for $x^{4}-x^{3}=\csc ^{2}(x)-\cot ^{2}(x)$ is?
Title: Re: most awesome question thread :)
Post by: moekamo on February 25, 2010, 07:28:09 pm: Quote from: superflya on February 25, 2010, 06:04:38 pm
the number of real solutions for $x^{4}-x^{3}=\csc ^{2}(x)-\cot ^{2}(x)$ is?

$x^{4}-x^{3}=\csc ^{2}(x)-\cot ^{2}(x)$
$\implies x^{4}-x^{3}=1$ (since $1+\cot^2x=\csc^2x \implies 1=\csc^2x - \cot^2x$ )

graphs the curves $y=x^{4}-x^{3}, y=1$ on calc and see how many intersections there are
Title: Re: most awesome question thread :)
Post by: superflya on February 25, 2010, 07:36:34 pm: thanks so much :)
Title: Re: most awesome question thread :)
Post by: qshyrn on February 25, 2010, 07:47:25 pm: Quote from: moekamo on February 25, 2010, 07:28:09 pm
Quote from: superflya on February 25, 2010, 06:04:38 pm
the number of real solutions for $x^{4}-x^{3}=\csc ^{2}(x)-\cot ^{2}(x)$ is?

$x^{4}-x^{3}=\csc ^{2}(x)-\cot ^{2}(x)$
$\implies x^{4}-x^{3}=1$ (since $1+\cot^2x=\csc^2x \implies 1=\csc^2x - \cot^2x$ )

graphs the curves $y=x^{4}-x^{3}, y=1$ on calc and see how many intersections there are
well if ur gonna use the calc, you might as well use the solve function straight away and see how many ansewrs pop up (easier).
or u can do it in ur head by picturing y=x^4-x^3 = x^3(x-1) (quartic graphs look like a parabola and it passes 0 and 1) so when you draw a line y=1 , you can tell that it must cut 2 of the 'parabola arms'
Title: Re: most awesome question thread :)
Post by: superflya on February 25, 2010, 07:58:57 pm: yea just had a mindblank :P tiredness isnt helping.
Title: Re: most awesome question thread :)
Post by: superflya on July 04, 2010, 10:55:07 pm: $\int \sqrt{\tan (x)}\ dx$

ishghwpao;hg;poahwhgw wtfffff?
Title: Re: most awesome question thread :)
Post by: fady_22 on July 05, 2010, 11:44:55 am: Quote from: superflya on July 04, 2010, 10:55:07 pm
$\int \sqrt{\tan (x)}\ dx$

ishghwpao;hg;poahwhgw wtfffff?

I doubt we'd be asked to evaluate that.
Unless it asks for the definite integral, which you can do on the calculator.
Title: Re: most awesome question thread :)
Post by: brightsky on July 05, 2010, 05:25:13 pm: Let $u = \sqrt{\tan(x)}$ , then $du = \frac{\sec^2(x)}{2 \sqrt{\tan(x)}} dx$

In terms of $u$ , this is:

$\frac{u^4 + 1}{2u}$

Hence,

$dx = \frac{2u}{u^4 + 1} du$

Substitute that back in:

$\int \sqrt{\tan(x)} dx = \int \frac{2u^2}{u^4 + 1} du$

Decompose it using partial fractions, should be a tad easier now.
Title: Re: most awesome question thread :)
Post by: superflya on July 05, 2010, 07:07:54 pm: this isnt a spesh question, found it and it looked interesting :P

gahh ive almost given up. ill try that brightsky ;)