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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  

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

Illustration Questions

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.

Illustration Questions

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

Radioactive Decay 

  • 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.

Illustration Questions

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

Illustration Questions

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

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