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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: hard on March 11, 2009, 10:56:11 pm

Title: help with log/exp q's plz
Post by: hard on March 11, 2009, 10:56:11 pm: This is from essentials

A radioactive substance is decaying such that the amount A g at time t years is given by the formula A=A₀ $e^kt$ . If when t=1, A=60.7 and when t=6, A=5, find the values of the costants A₀ and k.

heinamenn

solve all values of u: $2u^4+38u^2-16u^3-24u=0$
Title: Re: help with log/exp q's plz
Post by: Damo17 on March 11, 2009, 11:44:21 pm: Quote from: hard on March 11, 2009, 10:56:11 pm
solve all values of u: $2u^4+38u^2-16u^3-24u=0$

$u(2u^3-16u^2+38u-24)=0$

Find factors for $2u^3-16u^2+38u-24$ .
$2(1)^3-16(1)^2+38(1)-24=0$
so 1 is a factor.

synthetics division/ long division yields:
$2u^3-16u^2+38u-24=(u-1)(2u-14+24)$
$=2(u-1)(u-7+12)$
$=2(u-1)(u-4)(u-3)$

So $2u^4+38u^2-16u^3-24u= 2u(u-1)(u-4)(u-3)$

$\therefore$ $u=0,1,3,4$
Title: Re: help with log/exp q's plz
Post by: hard on March 11, 2009, 11:53:20 pm: thanks i got up to the factorizing but i totally forgot about the long division.
Title: Re: help with log/exp q's plz
Post by: mano91 on March 13, 2009, 09:54:05 pm: hey there, it is quite hard to get above 40.
u need a+'s, a's arent good enough to get 40+
Title: Re: help with log/exp q's plz
Post by: mano91 on March 13, 2009, 10:00:27 pm: sorry, wrong thread. :(
Title: Re: help with log/exp q's plz
Post by: kurrymuncher on March 13, 2009, 10:01:01 pm: lol
Title: Re: help with log/exp q's plz
Post by: hard on March 13, 2009, 10:13:21 pm: mm

another q guys.

8(1-e^-0.26t)

when t = 0

draw the graph, but i seem to get y = 0 cause 8(1-1) = 0

but the answer says other wise and also the calculator shows that the x asymptote is y=8 but really there's no +8, it's a minus?
Title: Re: help with log/exp q's plz
Post by: hard on March 13, 2009, 10:15:57 pm: Quote from: dekoyl on March 13, 2009, 10:15:17 pm
Quote from: hard on March 13, 2009, 10:13:21 pm
mm

another q guys.

8(1-e^-0.26t)

when t = 0

draw the graph, but i seem to get y = 0 cause 8(1-1) = 0

but the answer says other wise and also the calculator shows that the x asymptote is y=8 but really there's no +8, it's a minus?
$8^{0} = 1$ not 0.
but it's 8*(o)?
Title: Re: help with log/exp q's plz
Post by: dekoyl on March 13, 2009, 10:17:16 pm: ^Yeah, sorry. My mistake :P I didn't read it correctly.
Title: Re: help with log/exp q's plz
Post by: dekoyl on March 13, 2009, 10:21:09 pm: $y=8(1-e^{-0.26t})$

$y= -8e^{0.26t} + 8$
$\implies y=8$ is asymptote.

So are you trying to find the y-value when t = 0?
If it's y-value when t = 0, then I think y = 0 as well.

What's the answer by the way?
Title: Re: help with log/exp q's plz
Post by: hard on March 13, 2009, 10:29:14 pm: Quote from: dekoyl on March 13, 2009, 10:21:09 pm
$y=8(1-e^{-0.26t})$

$y= -8e^{0.26t} + 8$
$\implies y=8$ is asymptote.

So are you trying to find the y-value when t = 0?
If it's y-value when t = 0, then I think y = 0 as well.

What's the answer by the way?

well the q is actually $y= -8(1-e^{0.2t}$

fuck it works out, sweet thanks bro