exp. growth and decay (2 Viewers)

[currysauce]

1 qu

The half-life of radium is 1600 years.

(a) find the % of radium that will be decayed after 500 years

(b)find the number of years that it will take for 75% to decay

 

[Pace_T]

.5 = e^-(1600k)

k = 0.0004332169879

.'. % = e^(500*k)
=0.805
=80.5% still there
that is,
19.5% decayed

 

[Pace_T]

.75 = e^k(t)

sub in k value u get

.75 = e^-0.0004332169879t

.'. ln0.75/-0.0004332169879 = t
t = 664.05

= 664 years

 

[currysauce]

thanks, do u use Maths in Focus, if so

could u help me with qu 14 and 15... i don't get them

 

[Pace_T]

sure thing
which exercise??


EDIT: ex 6.3 yeh?

 

[currysauce]

yup, thanks alot man

 

[Pace_T]

q 14 (ex 6.3)

a)
we must differentiate p(t0) = e^-kt to see if it equals what they say

since in the initial eqn, it has with respect to t, we know to do that for the equation
.'.
dP(t)/dt = -k(P9t0)e^-kt
= -k(P(t) (just put the eqn from above back in)

b)
.'. .9 = e^k(4)

k = ln.9/4
= -0.026340128
.'. at 10 years,

% = e^-0.026340128*10
=77%
.'. there is a decline of 23%


c) we know that rate = the differential equation with respect to t right?
.'. dp(t)/dt = -k(Pt)
= -(0.026340128)(.77)
= -0.02
= 2% decrease

d)
since fall by 20% means 80% remaining
so
0.8 = e^-kt
0.8 = e^-(0.026340128t)

t = ln.8/-0.026340128
= 8.47
= 8.5 years

 

[Pace_T]

q 15 (ex 6.3)

.'. .4 = e^-k(5)

k = ln.4/-5
= 0.183258146

.'.
.1 = e^-0.183258146t

t = ln.1/-0.1832581
=12.5
= 12.5 minutes

 

[Pace_T]

umm here's some few tips i think would be useful

when calculating k, use all the values, don't round it off, caues its exponential and it could lead in a completely different answer

when the question has the thing decreasining it's to the power -kt, when increasining, to the power of kt (well that's how i do it anyway, im not sure if that's right, i figured it out myself lol)

see how i have no A in any of my solutions, that's because initially, its at full capacity (that is, 100%, ie 1, so i just leave it out. it's probably best to put 1 in the exam though )

thanks,
Pace T.

EDIT: oh and by the way, this is 2 unit

 

[currysauce]

you sir are a god at the moment, thanks!