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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: monokekie on July 03, 2009, 11:27:36 am

Title: guys~ SOS
Post by: monokekie on July 03, 2009, 11:27:36 am: heyy

for cambridge chapter 9 review short answer questions number 2 (a),

could somebody kindly confirm whether the answer is correct or no?

:) tkx
Title: Re: guys~ SOS
Post by: Damo17 on July 03, 2009, 11:50:25 am: Quote from: monokekie on July 03, 2009, 11:27:36 am
heyy

for cambridge chapter 9 review short answer questions number 2 (a),

could somebody kindly confirm whether the answer is correct or no?

:) tkx

yes it is correct, do you want me to show working out?
Title: Re: guys~ SOS
Post by: monokekie on July 03, 2009, 11:58:23 am: arhh thanks !

i don't get why sin 5∏ = -1 during the working out of the C value since i got c= -1..

if possible, your workout will be appreciated :)
Title: Re: guys~ SOS
Post by: Damo17 on July 03, 2009, 12:01:24 pm: $\frac{dy}{dx}= \pi cos (2 \pi x)$ , $y=-1$ when $x= \frac{5}{2}$

$y= \int [\pi cos (2 \pi x) ]dx = \frac{\pi sin (2 \pi x)}{2 \pi} + c = \frac{1}{2} sin(2 \pi x)+c$

$-1 = \frac{1}{2}sin(5 \pi) + c$ , $sin(5\pi)= sin (\pi)=0$

$\therefore c=-1$

$\therefore y= \frac{1}{2} sin (2 \pi x) -1$
Title: Re: guys~ SOS
Post by: monokekie on July 03, 2009, 12:04:05 pm: i see :) thanks! Damo!( u type so fast btw)

and also, for question (d), i wonder why wouldn't the Tan inverse rule be used on solving the question, but ratherly, log... ???

anyone? lol
Title: Re: guys~ SOS
Post by: d0minicz on July 03, 2009, 12:12:18 pm: because theres the x on the top =]
$\frac {dy}{dx} = \frac {x}{1+x^2}$
Let $u = 1+x^2$
= $\frac {1}{2} log_e ({1+x^2}) + C$
and just find C.
Title: Re: guys~ SOS
Post by: Damo17 on July 03, 2009, 12:14:06 pm: Quote from: monokekie on July 03, 2009, 12:04:05 pm
i see :) thanks! Damo!

and also, for question (d), i wonder why wouldn't the Tan inverse rule be used on solving the question, but ratherly, log... ???

anyone? lol

nps :)

for d)

$\frac{dy}{dx}= \frac{x}{1+x^2}$ , $y(0)=1$

y has to be a log since there is an x on the top as $\frac{d}{dx} (tan^{-1} \frac{x}{a}) = \frac{a}{a^2+x^2}$ so an x can not be on top for a inverse tan function.
Title: Re: guys~ SOS
Post by: monokekie on July 03, 2009, 12:22:28 pm: i got it now :) thanks sooo much d0minicz and Damo17 !