help on application calculus physical world questions (1 Viewer)

[**freakstar87**]

it's from a maths textbook(mif) rather than a trial but i keep getting the answer wrong yet it sounds so easy.

Q1. A chemical reaction causes the amount of chlorine to be reduced at a rate proportional to the amount of chlorine present at any one time. If the amount of chlorine is given by the formula A=Aoe^-kt, and 100mL reduces to 65L after 5 minutes, find
a) the amount of chlorine after 12 minutes
b) how long it will take for the chlorine to reduce to 10L

Q2. The decay of radium is proportional to its mass. If 100kg of radium takes 5years to decay ti 95kg
a) show that the mass of radium is given by M=100e^-0.01t -->(already answered this question correctly)
b) find its mass after 10 years
c) find its half life (the time taken for the radium to halve its mass)

 

[klaw]

Q1. A chemical reaction causes the amount of chlorine to be reduced at a rate proportional to the amount of chlorine present at any one time. If the amount of chlorine is given by the formula A=Aoe^-kt, and 100mL reduces to 65L after 5 minutes, find
a) the amount of chlorine after 12 minutes
A=A₀e^-kt
When t=0, A=100
.: A₀=100
When t=5, A=65
.:65=100[/sub]e^-k5
-5k=ln (13/20)
k=0.086

.:A=100e^-0.0086t
When t=12,
A=100e^-0.0086*12
A=35.56L

 

[yook]

i assume you did a typo when you wrote 100mL [ie. meant to be 100L?!]

dont have the book so not 100% sure this is the answer.. coulda done a careless mistake..

question 1

100 = Ao.e^(-k0)
therefore, Ao = 100

65 = 100e^(-5k)
[ln (65/100)] / - 5 = k
k = 0.086156583

a) A = 100e^(-0.0862 x 12)
A= 35.56248135

b) 10 = 100 e^ (-0.0862 x t)
t = [ln (1/10)]/ -0.0862
t = 26.72558512

(ie. 27 minutes)

 

[Jago]

1. at t = 0
100 = A_oe^0
A_o = 100

at t = 5
65/100 = e^-5k
k = ln(65/100) / -5

at t = 12
A = 100e^-12k <--- (sub in k and find the value)

10 = 100e^-kt
t = ln(0.1)/-k <--- (sub in k and find the value)
t = ?


Edit: hahaha beaten, twice.

 

[klaw]

b) how long it will take for the chlorine to reduce to 10L
When A=10,
10=100e^-0.0086t
ln (1/10)=-0.0086t
t=26.7
.: 27 mins

 

[yook]

question 2

b) mass after 10 years
m = 100 e^ (.01 x 10)
m = 90.4837418

c) half life
50 = 100 e ^(-0.01 x t)

t = [ln (1/2)] / -0.01
t= 69.31471806

 

[klaw]

Q2.
b) find its mass after 10 years
c) find its half life (the time taken for the radium to halve its mass

M=100e^-0.01t
b)so at t=10, M=100 e^-0.1 kg
c)1/2*100 = 50kg.
50=100 e^-0.01t
1\2=e^-0.01 t
ln (1\2)=-0.01 t
-100 ln (1\2)=t
t=100 ln 2 years

 

[yook]

hehe jago.. i felt that too..
but then i felt better that he didnt do part b..

dont know wat the point of this post is..

 

[**freakstar87**]

i assume you did a typo when you wrote 100mL [ie. meant to be 100L?!]

dont have the book so not 100% sure this is the answer.. coulda done a careless mistake..

no worries. the answer i got from the book is right. and yeah it was 100L.
thanks for the help

 

[klaw]

hahahah me too after I posted 1b) and Q2 and saw that I got beaten