Informative line

### Rate Of Growth Decay

Learn Differential Equation of Exponential Growth and Decay Calculus, practice radioactive decay & exponential growth differential equation.

# Rate of Growth of Population

• In many natural phenomenon quantities grow at a rate which is proportional to their present size.
• If $$y=f(t)$$ is population of bacteria at time t, we can expect the rate of  growth $$f'(t)$$  to be proportional to $$f(t)$$, this means that

$$f'(t) \propto f(t)$$

#### If population of a city is given by $$'y'$$, then rate of  growth of population is given by

A $$\dfrac{dy}{dt} \propto y$$

B $$\dfrac{dy}{dt} \propto \dfrac{1}{y}$$

C $$\dfrac{dy}{dt} \propto y^2$$

D $$\dfrac{dy}{dt} \propto y^3$$

×

If $$y= f(t)$$ is the population of city at time t, then rate of growth is given by

$$f'(t) \propto f(t)$$

$$\dfrac{dy}{dt} \propto y$$

### If population of a city is given by $$'y'$$, then rate of  growth of population is given by

A

$$\dfrac{dy}{dt} \propto y$$

.

B

$$\dfrac{dy}{dt} \propto \dfrac{1}{y}$$

C

$$\dfrac{dy}{dt} \propto y^2$$

D

$$\dfrac{dy}{dt} \propto y^3$$

Option A is Correct

# Exponential Growth

• In many natural phenomenon quantities grow at a rate which is proportional to their present size.
• If $$y= f(t)$$ is population of bacteria at time 't' , we can expect the rate of growth $$f'(t)$$ to be proportional to $$f(t)$$ , this mean that $$f'(t) = k f(t)$$ for some constant  k.
• Let $$f(t) = y$$ then we have $$\dfrac{dy}{dt} = k\,y \to$$  then equation is called the  law of natural growth $$(k>0)$$ .
• It is a  differential equation  whose solution is very easy to guess.
• The solution to differential equitation  $$\dfrac{dy}{dt} = k\,y$$  is  $$y(t) = y(0) e^{k\,t}$$
• In the context of population growth , if $$P(t)$$ is the size of population at time 't' we say $$\dfrac{dP}{dt} = k\,P$$ or  $$\dfrac{1}{P} \dfrac{dP}{dt} =k$$ ,

it is called the relative growth rate or it is growth rate divided  by population .

• We say that the relative growth rate of population is constant.

#### A bacteria culture initially contains 100 cells and grows at a rate proportional to its size . After an hour the population has increased to 420, find the number of bacteria after 3 hours? [ $$ln4.2=1.4351,\;e^{4.3053}=74.091$$ ]

A $$8042$$

B $$7409$$

C $$981$$

D $$5023$$

×

$$P(t) =P(0) e^{k\,t}$$

where $$P(t)$$ is the  population  (of bacteria) at time 't'

$$P(0) =$$ initial population (at t=0)

k = relative growth rate

Given that  $$P(0) = 100,\,\,\,P(1) = 420$$

$$\therefore P(t) = 100\,e^{k\,t}$$ and  $$P(1) = 100\,e^k$$

$$\Rightarrow 420 = 100\,e^k$$

Solving  for k we get $$e^k = \dfrac{420}{100}$$

$$\Rightarrow e^k = 4.2$$

$$\Rightarrow k = \ell n \,\,4.2 = 1.4351$$

$$\therefore P(t) = 100 \,e^{(1.4351\,t)}$$

$$P(3) = 100 \,e^{1.4351×3}= 100× 74.0914 \cong 7409$$

### A bacteria culture initially contains 100 cells and grows at a rate proportional to its size . After an hour the population has increased to 420, find the number of bacteria after 3 hours? [ $$ln4.2=1.4351,\;e^{4.3053}=74.091$$ ]

A

$$8042$$

.

B

$$7409$$

C

$$981$$

D

$$5023$$

Option B is Correct

• In many natural phenomenon quantities decay at a rate proportional to their size.
• In general if $$y(t)$$ is the value of quantity $$y$$ at time $$'t'$$ and if rate of change of $$y$$with respect to $$'t'$$ is proportional  to its  size $$y(t)$$  at any time, then

$$\dfrac{dy}{dt} = k\,y$$     $$(k<0)$$

This is called law of natural decay.

• Radioactive Decay is the process by which some radioactive substances decay by spontaneous emission of radiation.
• If $$m(t)$$ is the mass remaining  form an initial mass $$m_0$$ of substance after time $$'t'$$, then relative decay rate is $$\dfrac{-1}{m} \dfrac{dm}{dt}$$ is constant.

$$\therefore \, \dfrac{dm}{dt} = k\,m$$      $$(k<0)$$

• The solution to this differential equitation is $$m(t) = m_0\,e^{k\,t}$$
• The time required for half of any quantity to decay is called half life.

#### The half life of caesium (a radioactive material) is 30 years. How much of the sample will remain after 100 years if we have 100 mg sample in the beginning? [ $$e^{-2.31}=0.099261,\;ln{\dfrac{1}2}=0.6931$$ ]

A $$8.3426\,mg$$

B $$9.9261\,mg$$

C $$75.1234 \,mg$$

D $$.0567\,mg$$

×

For radioactive decay,  $$m(t) = m_0 \,e^{k\,t}$$  $$(k<0)$$

where $$m(t) =$$ mass  at time $$t$$

$$m(0) =$$ initial mass

In this case , $$m_0 = 100, \,\,m(30) = \dfrac{100}{2} = 50$$

$$\therefore m(t) = 100\,e^{k\,t}$$ and $$m(30)= 100\,e^{30\,k} =50$$

$$\Rightarrow 30 \,k =\ell n\dfrac{1}{2}$$

$$\Rightarrow \,k =\dfrac{1}{30}\ell n\dfrac{1}{2}= -0.0231$$

$$\therefore m(t) = 100\,e^{-0.231\,t}$$

$$\Rightarrow m(100) = 100\,e^{-0.0231× 100} = 100× e^{-2.31}$$

$$\Rightarrow m(100)= 9.9261\,mg$$

### The half life of caesium (a radioactive material) is 30 years. How much of the sample will remain after 100 years if we have 100 mg sample in the beginning? [ $$e^{-2.31}=0.099261,\;ln{\dfrac{1}2}=0.6931$$ ]

A

$$8.3426\,mg$$

.

B

$$9.9261\,mg$$

C

$$75.1234 \,mg$$

D

$$.0567\,mg$$

Option B is Correct

#### A sample of tritium -3 (a radioactive material) decayed 94.5% of its original amount after a year. What is the half- life of tritium - 3?  [ $$ln\dfrac{1}{2}=-0.6931$$ ]

A $$15.129\,{\text{yrs}}$$

B $$12.246\,{\text{yrs}}$$

C $$5.483\,{\text{yrs}}$$

D $$2.794\,{\text{yrs}}$$

×

For radioactive decay,  $$m(t) = m_0 \,e^{k\,t}$$    $$(k<0)$$

and half life is the time at which the material is reduced to half  of original quantity.

In this case, $$m(1) = 94.57 \,{\text{%}}$$ of $$m_0$$

$$\Rightarrow m(1) = 0.945\,m_0$$

$$\therefore m(t) = m_0\,e^{k\,t}$$

$$\Rightarrow m(1) = m_0 \,e^{k×1}$$

$$\Rightarrow 0.945\,m_0 = m_0\,e^k$$

$$\Rightarrow e^k = 0.945$$

$$\Rightarrow k= \ell n (0.945) = -0.0566$$

Let 't' be the half life , then  $$m(t) = \dfrac{m_0}{2}$$

$$\Rightarrow m(t) = m_0 \,e^{-0.0566\,t}$$

$$\Rightarrow \dfrac{m_0}{2} = m_0\,e^{-0.0566\,t}$$

$$\Rightarrow e^{-0.0566\,t }=\dfrac{1}{2}$$

$$\Rightarrow -0.0566\,t = \ell n \dfrac{1}{2}$$

$$\Rightarrow t \cong 12.246 \,\,{\text{yrs}}$$

### A sample of tritium -3 (a radioactive material) decayed 94.5% of its original amount after a year. What is the half- life of tritium - 3?  [ $$ln\dfrac{1}{2}=-0.6931$$ ]

A

$$15.129\,{\text{yrs}}$$

.

B

$$12.246\,{\text{yrs}}$$

C

$$5.483\,{\text{yrs}}$$

D

$$2.794\,{\text{yrs}}$$

Option B is Correct