Worked solutions for Maths in Focus: Excercise 6.3, Q13 - 15 (1 Viewer)

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Question 13

a)

M = M₀e^kt
Let M = 1 and 1 is 100% of radium.
Find M₀ which is the initial quantity.

1 = M₀e^{k x 0}
M₀ = 1

Now we must find the value of k.
We know that the radium decays to 50% after 1600 years i.e. k = ?, M = 0.5 & t = 1600

0.5 = e^1600k
ln 0.5 = ln e^1600k
ln 0.5 = 1600k ln e
(ln e = 1)
k = ln 0.5 / 1600 (Remember this value)

Now we must find out much has decayed after 500 years. i.e. M = ?, t = 500

M = e^{(ln 0.5 / 1600) x 500}
M = 0.805245166 ...

But the question asks how much has DECAYED not what's left.

So Amount = 1 - M = 0.194754834 ...

.: The radium has decayed by 19.5% after 500 years.

b)

M = M₀e^kt

We must find out how long it takes for radium to decay by 75% i.e t = ?, M = 0.25

0.25 = e^{(ln 0.5 / 1600) t}
ln 0.25 = ln e^{(ln 0.5 / 1600) t}
ln 0.25 = (ln 0.5 / 1600) t ln e
ln 0.25 = (ln 0.5 / 1600) t
t = ln 0.25 / (ln 0.5 / 1600)
t = 3200

.: It takes 3200 years for radium to decay by 75%



Question 14

a)

P(t) = (t₀)e^-kt
dP(t)/dt = -k(t₀)e^-kt
(k is a constant whereas t is a variable)
dP(t)/dt = -kP(t)

b)

You are given 10% decline (90% remaining population) in 4 years,
i.e k = ?, P = 0.9 & t = 4


P = P₀e^kt
1 = P₀e^{k x 0}
P₀ = 1

P = P₀e^kt
0.9 = e^4k
ln 0.9 = ln e^4k
ln 0.9 = 4k ln e
k = ln 0.9 / 4

Now we must find out the percentage of the population decline after 10 years.
i.e P = ?, k = ln 0.9 / 4 & t = 10

P = e^{(ln 0.9 / 4) x 10}
P = 0.7684334714 ...

But the question asks how much the population has DECLINED not what's remaining.

So Amount = 1 - M = 0.2315665286 ...

.: The population has declined by 23.2% after 10 years.

c)

To find the percentage rate of decline in population after 10 years (NOTE the keyword RATE),
use the differential solution.

i.e dP(t)/dt = -kP(t) = ?, t = 10, k = (ln 0.9 / 4)

dP(t)/dt = -k(t₀)e^-kt
dP(10)/dt = - (ln 0.9 / 4) x 1 x e^{- (ln 0.9 / 4) x 10}
dP(10)/dt = 0.0202406367 ...

P is population and t is the time in year so,

.: The population declines at a rate of 2.02% per year.

d)

Now we must find when the population will fall by 20% (80% remaining),
i.e. t = ?, P = 0.8, k = ln 0.9 / 4

0.8 = e^{(ln 0.9 / 4) t}
ln 0.8 = ln e^{(ln 0.9 / 4) t}
ln 0.8 = (ln 0.9 / 4) t ln e
t = ln 0.8 / (ln 0.9 / 4)
t = 8.47161956 ...

.: The population will fall by 20% after 8.5 years.



Question 15

The container will be 60% empty (40% full) after 5 minutes but we must find k first.
i.e. k = ?, A = 0.4 & t = 5

A = A₀e^kt
0.4 = e^5k
ln 0.4 = ln e^5k
ln 0.4 = 5k ln e
k = ln 0.4 / 5

We must find long it will take until the container is 90% empty (10% full)
i.e. t = ?, k = ln 0.4 / 5 & A = 0.1

A = A₀e^kt
0.1 = e^{(ln 0.4 / 5) x t}
ln 0.1 = ln e^{(ln 0.4 / 5) x t}
ln 0.1 = (ln 0.4 / 5) x t ln e
t = ln 0.1 / (ln 0.4 / 5)
t = 12.56470797 ...

.: It will take 12.6 minutes until the container is 90% empty.