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Practical applications
Example 1
According to the law of cooling proven by experiments, the time needed for an object with temperature \latex{ T0 } in an environment with temperature \latex{ K } to cool down to temperature \latex{ T } is
\latex{t=k\times\ln{\left(\frac{T_0-K}{T-K}\right)},}
where\latex{ k } is a characteristic constant of the material cooling down and time is obtained in minutes.
How much time is needed for the roasted piglet with temperature \latex{85^{\circ}\text{C}} to cool down to the temperature
How much time is needed for the roasted piglet with temperature \latex{85^{\circ}\text{C}} to cool down to the temperature
- \latex{70^{\circ}\text{C}};
- \latex{36^{\circ}\text{C}};
- \latex{25^{\circ}\text{C}}
in a room with temperature \latex{21^{\circ}\text{C}} and \latex{k = 12.5}?
Solution
By substituting the data into the relation the values can be calculated with the help of a calculator:
- \latex{t=12.5\times\ln\frac{85-21}{70-21}=12.5\times\ln\frac{64}{49}\approx3.34} minutes;
- \latex{t=12.5\times\ln\frac{85-21}{38-21}=12.5\times\ln\frac{64}{17}\approx16.57} minutes;
- \latex{t=12.5\times\ln\frac{85-21}{25-21}=12.5\times\ln\frac{64}{4}\approx34.66} minutes;
Example 2
A photocopier costs \latex{ 220 } euros. During use the value of the machine decreases by \latex{ 20 }% every year. When the value of the machine decreases below \latex{ 50 } euros, in the case of a bigger malfunction it is no longer worth getting it repaired.
In how many years can we say that if the machine breaks down, then it is practical to replace it with a new one?
Solution
According to the condition the value of the machine will be
in \latex{ 1 } year \latex{220 × 0.8} euros,
in \latex{ 2 } years \latex{(220 × 0.8)× 0.8} euros \latex{= 220 × 0.8^2} euros,
\latex{\vdots}
in \latex{ x } years \latex{220 × 0.8^x} euros.
in \latex{ 2 } years \latex{(220 × 0.8)× 0.8} euros \latex{= 220 × 0.8^2} euros,
\latex{\vdots}
in \latex{ x } years \latex{220 × 0.8^x} euros.
We are looking for \latex{ x } for which
\latex{\begin{align*}220\times0.8^x=50,\\ 0.8^x=0.2273.\end{align*}}
Let us take the logarithm of both sides with base \latex{ 10 }
\latex{\begin{align*}\log{0.8}^x=\log{0.2273},\\ x\times\log{0.8}=\log{0.2273},\\ x=\frac{\log{0.2273}}{\log{0.8}}\end{align*}}.
When calculating using a calculator
\latex{x\approx6.64}.
The value of the copy machine decreases to \latex{ 50 } euros in about \latex{ 6.64 } years, after this it is not worth getting it repaired if it breaks down, it is practical to replace it with a new one.
⯁ ⯁ ⯁
If at time zero there had been a mass of \latex{m_0} radioactive material then at time t there will be a mass of
\latex{m(t)=m_0\times e^{-k\times t}}
radioactive material left, where \latex{ k } is the constant specific to the material.
Carbon-\latex{ 14 } is a radioactive isotope of carbon with mass number \latex{ 14 }, and it can also be found in living organisms. In view of the mass of the carbon-\latex{ 14 } in an archaeological find the age of the find can be given.
The decay half-life of the carbon-\latex{ 14 } isotope is \latex{ 5715 } years (it takes the radioactive material to decrease in mass to half its value this many years).
The decay half-life of the carbon-\latex{ 14 } isotope is \latex{ 5715 } years (it takes the radioactive material to decrease in mass to half its value this many years).
Example 3
While examining an archaeological find, \latex{ 4\, g } out of \latex{ 12\, g } of carbon was radioactive. About how old is the find in years?
Solution
In view of the decay half-life let us calculate the value \latex{ k } specific to \latex{^{14}\text{C}}:
\latex{\frac{m_0}{2}=m_0\times e^{-5,715\times k}, \\ \frac{1}{2}=e^{-5,715\times k}.}
Let us take the logarithm of both sides with base \latex{ e }:
\latex{\ln{\frac{1}{2}}=\ln e^{-5,715\times k},}
\latex{\ln\frac{1}{2}=-5,715\times k}.
\latex{\ln\frac{1}{2}=-5,715\times k}.
When calculating using a calculator:
\latex{k=\frac{\ln{\frac{1}{2}}}{-5715}=0.000121 \left(\frac{1}{\text{year}}\right)}.
The law of decay in the case of carbon-\latex{ 14 }:
\latex{m(t)=m_0\times e^{-0.000121\times t}}.
According to the measurement results
\latex{4=12\times e^{-0.000121\times t},}
\latex{\frac{1}{3}=e^{-0.000121\times t}.}
\latex{\frac{1}{3}=e^{-0.000121\times t}.}
Let us take the logarithm of both sides with base \latex{ e }:
\latex{\ln{\frac{1}{3}}=\ln{e^{-0.000121\times t}},}
\latex{\ln{\frac{1}{3}}=-0.000121\times t,}
\latex{\frac{\ln{\frac{1}{3}}}{-0.000121}}.
\latex{\ln{\frac{1}{3}}=-0.000121\times t,}
\latex{\frac{\ln{\frac{1}{3}}}{-0.000121}}.
When calculating using a calculator:
\latex{t\approx 9079.4}.
So the archaeological find is about \latex{ 9,079 } years old.
The age of the Egyptian mummies was also estimated with the help of the carbon-\latex{ 14 } isotope.
Note: One of the sensational results of the carbon-\latex{ 14 } age determination method was when in \latex{ 1988 } it could be established that the Shroud of Turin, which is said to be the burial shroud of Jesus of Nazareth, could have been created in about the \latex{ 14 }th century. Scientists are still debating about the precision of the method.
Example 4
The table contains the data of \latex{ 3 } years regarding the number of deaths in relation to drug abuse in a country.
\latex{ 2005 }
\latex{ 2006 }
\latex{ 2007 }
\latex{ 204 }
\latex{ 288 }
\latex{ 339 }
Let us give a forecast for the year \latex{ 2020 }.
Solution
Since it is a quickly increasing process, let us perform the approximation with the function \latex{h(t) = h_0 × e^{k × t}} . Let \latex{h_0} be the initial data from \latex{2005: h_0 = 204}. Time is obviously measured in years. We are looking for k that is specific to the process.
Based on the data of \latex{ 2005 } and \latex{ 2006 }:
\latex{\begin{align*}288=204\times e^{k\times 1},\\ \frac{288}{204}=e^k, \end{align*}}
when calculating using a calculator:
\latex{k=\ln\frac{288}{204}\approx0.3448\left(\frac{1}{\text{year}}\right)}.
By comparing the data of \latex{ 2005 } and \latex{ 2007 }:
\latex{\begin{align*}339=204\times e^{k\times 2},\\ \frac{339}{204}=e^{2k},\\ 2k=\ln\frac{339}{204} \end{align*}}
When calculating with a calculator:
\latex{k=\frac{1}{2}\times\ln\frac{339}{204}\approx0.2539\left(\frac{1}{\text{year}}\right)}.
The difference between the two values obtained for \latex{ k } shows that we would need more data to refine our model.
We can still draw the attention to the seriousness of the issue if we give the forecast for the year \latex{ 2020 } with the smaller value of \latex{ k }.
\latex{h(15)=204\times e^{0.2539\times 15}\approx9,202}
is the expected number of deaths.
By increasing the amount of data and by changing the approximation function one can achieve that the mathematical model describes the phenomena of society more precisely.
