ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: WhoTookMyUsername on June 19, 2011, 10:14:20 am

Title: My 1/2 question thread
Post by: WhoTookMyUsername on June 19, 2011, 10:14:20 am: can anyone help me out with interpreting this question? I just can't get my head around it :(

A man has to travel 50km in 4 hours. he does it by walking the first 7 km at $x$ km/h, cycling the next 7 km at 4x km/h and motoring the remainder at $6x+3$ km/h.

Show that $32x^2 - 102x - 35 = 0$

Thanks!

EDIT: Solved :|

It was very easy =.=
Title: Re: My 1/2 question thread
Post by: WhoTookMyUsername on June 19, 2011, 12:08:17 pm: Find the value of $9^\frac {1} {3} + 9^ \frac {1} {9} + 9^ \frac{1} {27} ...<br />$

why...? (question over post)

its a sum to infinity sequence , of the rule $\frac {1} {1-r}$

my first thought was to times the whole thing by 9, in order to get a nice $a<br />$
and then divide by 9 later on

this brought the answer $\frac {3} {2}$

But why does treating $a$ as $9^{1/3}$

and dividing by $1- \frac {1} {3}$ not get the same answer?

which way is right?

EDIT - how the *%ck do i use latex?
Title: Re: My 1/2 question thread
Post by: b^3 on June 19, 2011, 12:32:55 pm: Quote from: Bazza16 on June 19, 2011, 12:08:17 pm
EDIT - how the *%ck do i use latex?
LaTeX help thread -> http://vce.atarnotes.com/forum/index.php?topic=3137.0
LaTeX generator - >http://www.codecogs.com/latex/eqneditor.php the link was on the thread but anyway, credit to Stroodle
Title: Re: My 1/2 question thread
Post by: WhoTookMyUsername on June 19, 2011, 12:48:31 pm: Ah the beauty!
Title: Re: My 1/2 question thread
Post by: WhoTookMyUsername on June 19, 2011, 01:22:08 pm: edited second question, is now completely different

tearing hair out, please prevent me going bald by responding ASAP :D
Title: Re: My 1/2 question thread
Post by: Truck on June 19, 2011, 01:25:47 pm: Quote from: Bazza16 on June 19, 2011, 01:22:08 pm
edited second question, is now completely different

tearing hair out, please prevent me going bald by responding ASAP :D
9^1/3 x 9^1/9 x 9^1/27
=9^(1/3 + 1/9 + 1/27)
Sum to infinity of powers; a=1/3, r=1/3.
1/3 / 1 - 1/3 = 1/2 (skipped working out here).
=> 9^1/2 = 3 = answer.
Title: Re: My 1/2 question thread
Post by: gossamer on June 19, 2011, 01:29:56 pm: Quote from: Bazza16 on June 19, 2011, 12:08:17 pm
Find the value of $9^\frac {1} {3} + 9^ \frac {1} {9} + 9^ \frac{1} {27} ...<br />$

why...? (question over post)

its a sum to infinity sequence , of the rule $\frac {1} {1-r}$

my first thought was to times the whole thing by 9, in order to get a nice $a<br />$
and then divide by 9 later on

this brought the answer $\frac {3} {2}$

But why does treating $a$ as $9^{1/3}$

and dividing by $1- \frac {1} {3}$ not get the same answer?

which way is right?

EDIT - how the *%ck do i use latex?

and now you use the rule (since it is sum, not product). Did it not work out?
Title: Re: My 1/2 question thread
Post by: WhoTookMyUsername on June 19, 2011, 01:42:38 pm: i mean, if i times the whole thing by 9, then sub it in and divide the answer by 9, why is this incorrect? (gives 1.5 which is too small)

EDIT: i have a very small brain capacity

scrap everything
Title: Re: My 1/2 question thread
Post by: Mao on June 19, 2011, 11:16:53 pm: It is not a geometric series. If it was, it would look like $9^{1/3} + 9^{2/3} + 9^{3/3} + 9^{4/3} + \cdots$

This series can be represented by $\sum_{k=1}^{\infty} 9^{1/3^k} = 9^{1/3} + 9^{1/3^2} + 9^{1/3^3} + \cdots$

If this series is infinite (as you have written it), then it is divergent (to positive infinity). We know this because $9^{1/n}>9^0=1$ for any positive n. Thus $\sum_{k=1}^{\infty} 9^{1/3^k}>\sum_{k=1}^{\infty}1$ , the latter sum is divergent (to positive infinity), therefore the first one must be divergent.
Title: Re: My 1/2 question thread
Post by: WhoTookMyUsername on July 31, 2011, 01:56:56 pm: can someone prove that
i just can't do it :'(

secx - sinx = $(tan^2 x + cos^2 x)/(secx + sinx)$

also
how to i copy paste an image or upload a word document?
Title: Re: My 1/2 question thread
Post by: pi on July 31, 2011, 02:23:02 pm: I'll do it a really cheap way, but I like it ;)

So we have the identities:
$\sin^2{x} + \cos^2{x} = 1$
$\sec^2{x} - \tan^2{x} = 1$

$\therefore \sin^2{x} + \cos^2{x} = \sec^2{x} - \tan^2{x}$
$\Rightarrow \cos^2{x} + \tan^2{x} = \sec^2{x} - \sin^2{x}$

No, we look at the LHS of your equation:
$\sec{x} - \sin{x}$
$= \sec{x} - \sin{x} \times \frac{\sec{x} + \sin{x}}{\sec{x} + \sin{x}}$
$= \frac{\sec^2{x} - \sin^2{x}}{\sec{x} + \sin{x}}$ (using the above mix of identities)
$= \frac{\cos^2{x} + \tan^2{x}}{\sec{x} + \sin{x}}$
$= RHS$

QED.
Title: Re: My 1/2 question thread
Post by: pi on July 31, 2011, 02:33:27 pm: I know uploads haven't come though yet, but this sheet I typed up last year is gold for this stuff (attached)
Title: Re: My 1/2 question thread
Post by: WhoTookMyUsername on July 31, 2011, 02:38:25 pm: Quote from: Rohitpi on July 31, 2011, 02:23:02 pm
I'll do it a really cheap way, but I like it ;)

So we have the identities:
$\sin^2{x} + \cos^2{x} = 1$
$\sec^2{x} - \tan^2{x} = 1$

$\therefore \sin^2{x} + \cos^2{x} = \sec^2{x} - \tan^2{x}$
$\Rightarrow \cos^2{x} + \tan^2{x} = \sec^2{x} - \sin^2{x}$

No, we look at the LHS of your equation:
$\sec{x} - \sin{x}$
$= \sec{x} - \sin{x} \times \frac{\sec{x} + \sin{x}}{\sec{x} + \sin{x}}$
$= \frac{\sec^2{x} - \sin^2{x}}{\sec{x} + \sin{x}}$ (using the above mix of identities)
$= \frac{\cos^2{x} + \tan^2{x}}{\sec{x} + \sin{x}}$
$= RHS$

QED.

thanks a lot rohitpi :D

any chance someone could do it the longisher way so i can improve my basic techniuqe (i'm a noob)?:D + how did you even see that :O
I figure i should have been able to do it just with the things i've been taught, more about developing technique, and i'm lost as f^!$
Title: Re: My 1/2 question thread
Post by: pi on July 31, 2011, 04:07:13 pm: Quote from: Bazza16 on July 31, 2011, 02:38:25 pm
thanks a lot rohitpi :D

...how did you even see that :O

As for this question, I'm used to instantly going for pythagorean trig identities whenever I see a $\tan^2{x}$ in a "show" or "prove" question. So I saw where the top part was coming from. Secondly, another good technique to try is if there is a reciprocal of the original function present in it's simplified form, multiply both sides by that reciprocal (in the form of 1) straight away, it means that I don't have to worry about the denominator.

As for a "long" way, I'm not really sure :(
Title: Re: My 1/2 question thread
Post by: xZero on July 31, 2011, 04:15:35 pm: firstly $sec^2x=1+tan^2x$

$\frac{tan^2x+cos^2x}{secx+sinx}$

$=\frac{sec^2x-1+cos^2x}{secx+sinx}$

$=\frac{sec^2x-(1-cos^2x)}{secx+sinx}$

$=\frac{sec^2x-sin^2x}{secx+sinx}$

$=\frac{(secx-sinx)(secx+sinx)}{secx+sinx}$

$=secx-sinx$

LHS=RHS

as required

Edit:wtf i can tweet, share, google+ and like my AN post? :D
Title: Re: My 1/2 question thread
Post by: pi on July 31, 2011, 04:20:23 pm: ^^Nice :)
Title: Re: My 1/2 question thread
Post by: tony3272 on July 31, 2011, 04:22:30 pm: Quote from: xZero on July 31, 2011, 04:15:35 pm

Edit:wtf i can tweet, share, google+ and like my AN post? :D
Lol what's up with that :P
Title: Re: My 1/2 question thread
Post by: Truck on August 01, 2011, 10:54:07 pm: kinda hijacking his thread, but the question is in the same topic soooo...

anyone able to solve;

cos^2(2x) + 2cos(4x)=-1 ,x element of [-pi, 3pi/2]
Title: Re: My 1/2 question thread
Post by: WhoTookMyUsername on September 15, 2011, 08:44:56 pm: URGENT HELP !!!! NEED ANSWER ASP lol

r^2 (sin2x + cos2x) = 3
(x = theta)

where
r = root (x^2 + y^2)
How do you put that in cartesian?

thanks a lot
Title: Re: My 1/2 question thread
Post by: WhoTookMyUsername on September 17, 2011, 11:27:15 am: Another question :)
Determine the locus of a point P which moves so that the difference of the squares of its distances from two fixed points P1(4,0) and P2(-4,0) is constant
Title: Re: My 1/2 question thread
Post by: WhoTookMyUsername on October 14, 2011, 04:27:42 pm: The resultant force when two forces of magnitude 20 kg wt and 20 kg wt act at an angle 60 to each other.
The resultant force is?

(tried everything but can't get right answer!)
Title: Re: My 1/2 question thread
Post by: WhoTookMyUsername on October 24, 2011, 03:07:27 pm: urgent help! i know your all busy, but if anyone doing 1/2, or any person doing 3/4 is kind enough to help would be much appreciated with questions 8, 11, 12, 13

thanks a lot! (pdf attached)
Title: Re: My 1/2 question thread
Post by: xZero on October 24, 2011, 04:26:32 pm: I'll do q8, if i have time tonight I'll have a look at the rest

8 ) there's 2 initial approach, do which ever that suits you the most

approach 1)
$z^2=5+12i$

$z=\pm \sqrt{5+12i}$ , well how the hell do you square root that? simple, all you have to do is write $5+12i$ in terms of $(x+yi)^2$

$5+12i=(x+yi)^2$

$5+12i=x^2-y^2+2xyi$

$\implies 2xy=12$ $x^2-y^2=5$

$x=\frac{6}{y}$

$(\frac{6}{y})^2-y^2=5$

$36-y^4=5y^2$

$y^4+5y^2-36=0$

$(y^2+\frac{5}{2})^2-169/4=0$

$(y^2+\frac{5}{2})^2=169/4$

$y^2+\frac{5}{2}=\pm \frac{13}{2}$ , ignore the negative because from $(x+yi)^2$ , we implied that x and y are integers, even if you take the negative root, your final solution will be the same

$y^2=4$

$y=\pm 2$ , again take negative or positive, still gets you the same solution, in this case i chose positive to keep things neat

$\implies y=2$

$xy=6$

$\implies x=3$

$\implies z=\pm \sqrt{(3+2i)^2$

$\implies z=\pm (3+2i)$

approach 2, let $(z-a-bi)(z-c-di)=z^2-5-12i$ , expand LHS, equate all the coefficients and you'll get to the same quadratic equations $x^2-y^2=5$ but in terms of c and d
Title: Re: My 1/2 question thread
Post by: WhoTookMyUsername on October 25, 2011, 04:17:58 pm: thanks a ton xZero!
great explanation for question 8 :)
(if you could manage to do the other 3 some time soonish that would be awesome)
much appreciated :D
Title: Re: My 1/2 question thread
Post by: xZero on October 25, 2011, 08:51:35 pm: ceebs with 12 and 13... cant take any more physics

11) a)
OK=OR+RK
=OS+SK
=OS-KS, if PK=nPS then KS=(1-n)PS
=OS-(1-n)PS
=OS-(1-n)(-OP+OS)
=OS+(1-n)OP-(1-N)OS
=(1-n)OP+nOS
=(1-n)p+2nr

for the second part of part a), just do the same thing but with OQ and QK, see if you can do it

b)
equate the 2 equations you found in part a), the solution should be n=1/4 and m=1/2
Title: Re: My 1/2 question thread
Post by: WhoTookMyUsername on October 27, 2011, 07:53:20 pm: thanks xZero,
i did a similar thing with OQ and QK, but i'm not sure what to do when i equate the 2 equations? i just get a bunch of random letters in terms of n, m, p etc.

any chance you could show me :)

thanks!

EDIT: worked out thanks