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We already have discussed linear versus exponential growth/decay in Chapter 8.
Exponential growth and decay are good examples of exponential functions.
A certain quantity (amount of money, amount of bacteria, population of a town...) changes with time.
Time is the independent variable, and the quantity of interest is the dependent variable
We now do just the same in the general framework of functions.
For linear functions, dependent variable changes by the same fixed amount
when the independent variable changes by 1 unit.
This amount is the slope aka rate of change
For exponential functions
the dependent variable changes by the same relative amount (same fraction of what it was)Linear function:
\(y=mx+b\)
\(m\) -- slope, the amount of change,
\(b\) -- \(y\)-intercept, value at \(x=0\)
Exponential function:
\(y=b\times(1+r)^x\)
\(r\) -- relative change
\(b\) -- \(y\)-intercept, value at \(x=0\)
\(y=b\times(1+r)^x\)
When \(x\) changes by 1,
\(y\) becomes \( b\times(1+r)^{x+1} = b \times (1+r)^x \times (1+r) \),
that is \(y\) multiplies by \((1+r)\),
becomes \(y\times(1+r)=y+y\times r\), changes by \(y\times r\).
This is the change by the same relative amount.
Note that the rate \(r\) may well be positive or negative.
The simplest function in this world is the constant function
\(y=b\)
The dependent variable \(y\) takes the same value of \(b\) whatever is the value of the independent variable.
The graph of this function is a horizontal line
This function is a special case of both linear
\(y=mx+b\) with \(m=0\) gives us \(y=b\)
and exponential function:
\(y=b\times (1+r)^x \) with \(r=0\) also gives us \(y=b\).
Zero rate of change makes \(y\) to stay constant (never change).
Instead of our
\(y=b\times (1+r)^x \)
They write
\(Q=Q_0 \times (1+r)^t \)
where
\(Q\) = value of the exponentially growing or decaying quantity
\(Q_0\)= initial value of the quantity (at \(t=0\) )
\(r\)= fractional growth/decay rate
\(t\)=time
\(Q=Q_0 \times (1+r)^t \)
This function models many growth/decay processes which take place in real world
This is growth if \(r>0\), and decay if \(r<0\)
Important:
The units of time in \(r\) and in \(t\) must be the same
Examples: If the quantity increases by \(10\%\) per year, then \(r=0.1\), and \(t\) must be the number of years,
while if the quantity decays by \(50\%\) per minute, then \(r=-0.5\), and and \(t\) must be the number of minutes.
A certain drug breaks down in a human body at a rate of \(15\%\) per hour. The initial amount of the drug in the blood stream is \(8\) mg. Create an exponential function to model the situation described.
Solution.
The variables are \(Q\), the amount of the drug in the bloodstream, and time \(t\), measured in hours.
We thus have
\(Q=Q_0 \times (1+r)^t \)
with \(Q_0=8\) and \(r=-0.15\).
Answer.
\(Q=8\times (0.85)^t \)
A certain drug breaks down in a human body at a rate of \(15\%\) per hour. The initial amount of the drug in the blood stream is \(8\) mg. Some values.
Since we have a function
\(Q=8\times (0.85)^t \)
which models the situation, we can find the projected values, amounts of the drug in the bloodstream,
at various moments of time:
When \(t=1\), one hour after the initial moment,
\(Q=8\times (0.85)^1 =6.8 mg \)
When \(t=24\), one day after the initial moment,
\(Q=8\times (0.85)^{24} =0.16 mg \)
A certain drug breaks down in a human body at a rate of \(15\%\) per hour. The initial amount of the drug in the blood stream is \(8\) mg. Visualization.
Since we have a function
\(Q=8\times (0.85)^t \)
which models the situation, we can also visualize the situation with a graph .
Sometimes it is convenient to have the formula
\(Q=Q_0 \times (1+r)^t \)
in terms of \(T_{\text{double}}\) or \(T_{\text{half}}\):
\(Q=Q_0 \times 2^{t/T_{\text{double}}} \)
\(Q=Q_0 \times \left(\frac{1}{2}\right)^{t/T_{\text{half}}} \)
Think about these formulas as a special choice of units of time:
we measure time in \(T_{\text{double}}\), and \(r=1\) for growth;
we measure time in \(T_{\text{half}}\), and \(r=-1/2\) for decay.
We have:
\(Q=Q_0 \times (1+r)^t \), equivalently \( \frac{Q}{Q_0}=(1+r)^t \)
You may want to find out when \(Q\) reaches a certain value.
\(\log_{10} \left( \frac{Q}{Q_0} \right) = t \times \log_{10} (1+r) \)
Finally, \(t=\frac{ \log_{10} \left( \frac{Q}{Q_0} \right)}{\log_{10} (1+r)} \)
One gets exact formulas for \(T_{\text{double}}\) and \(T_{\text{half}}\) from that:
\(T_{\text{half}} = -\frac{ \log_{10}2}{\log_{10}(1+r)}\) and \(T_{\text{double}} = \frac{\log_{10}2}{\log_{10}(1+r)}\)
A certain drug breaks down in a human body at a rate of \(15\%\) per hour. The initial amount of the drug in the blood stream is \(8\) mg. Find the half-life time of the drug.
Solution.
We apply the half-life formula
\(T_{\text{half}} = -\frac{ \log_{10}2}{\log_{10}(1+r)}\)
with \(r=-0.15\) to calculate
\(T_{\text{half}} = -\frac{ \log_{10}2}{\log_{10}(0.85)} \approx 4.3\) hours
Remark. The approximate formula aka "rule of 70" gives us
\(T_{\text{half}} \approx \frac{70}{15} = 4.7\)
which is not too bad an approximation.
A certain drug breaks down in a human body at a rate of \(15\%\) per hour. The initial amount of the drug in the blood stream is \(8\) mg. How long after the drug was administered \(10\%\) of the drug is left in the bloodstream?
Solution.
Recall our exponential function
\(Q=Q_0\times (0.85)^t \)
We want \(t\) such that \(Q/Q_0 = 0.1\):
\(0.1= (0.85)^t \)
and we have to solve it for \(t\):
\(\log_{10}(0.1) = t \log_{10}(0.85) \)
\(t= \frac{\log_{10}(0.1)}{\log_{10}(0.85)} \approx 14 \) hours
A certain drug breaks down in a human body at a rate of \(15\%\) per hour. The initial amount of the drug in the blood stream is \(8\) mg. How long after the drug was administered the amount of the drug in the bloodstream is \(2\)mg?
Solution.
Recall our exponential function
\(Q=8\times (0.85)^t \)
We want \(t\) such that \(Q= 2\):
\(2= 8 \times (0.85)^t \) or \(0.25= (0.85)^t \)
and we have to solve it for \(t\):
\(\log_{10}(0.25) = t \log_{10}(0.85) \)
\(t= \frac{\log_{10}(0.25)}{\log_{10}(0.85)} \approx 8.53 \) hours