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Calculus, 2ADV EQ-Bank 13

A quantity of radioactive material decays according to the equation

\(\dfrac{d M}{d t}=-k M\),

where \(M\) is the mass of the material in kilograms, \(t\) is the time in years and \(k\) is a constant.

  1. Verify that  \(M=A e^{-k t}\)  is a solution to this equation, where \(A\) is a constant.   (1 mark)

  2. The time for half of the material to decay is 300 years. The initial amount of material is 20 kg .
  3. Find the amount of material remaining after 1000 years.   (3 marks)

Show Answers Only

a.    \(M=A e^{-k t}\)

\(\dfrac{d M}{d t}=-k A e^{-k t}=-k M \ \left( \text{since} \ M=A e^{-k t}\right)\)
 

b.    \(1.98\text{ kg}\)

Show Worked Solution

a.    \(M=A e^{-k t}\)

\(\dfrac{d M}{d t}=-k A e^{-k t}=-k M \ \left( \text{since} \ M=A e^{-k t}\right)\)
 

b.    \(\text{When} \ t=0, M=20:\)

\(20=A e^{-k(0)}\ \ \Rightarrow\ \ A=20\)

\(M=20 e^{-k t}\)
 

\(\text{When }\ t=300, M=10:\)

\(10\) \(=20 e^{-300 k}\)
\(\dfrac{1}{2}\) \(=e^{-300 k}\)
\(-300 k\) \(=\ln \frac{1}{2}\)
\(-300k\) \(=- \ln2\)
\(k\) \(=\dfrac{\ln 2}{300}\)

 
\(\text{When } \ t=1000:\)

\(M=20 e^{-1000 k}=20 e^{-\tfrac{10}{3} \ln 2}=1.984 \ldots\) 

\(\therefore \ \text{The amount remaining after 1000 years} \approx 1.98\text{ kg}\)