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In mathematical modeling, we choose a familiar general function with properties that suggest that it will model the real-world phenomenon we wish to analyze.In the case of rapid growth, we may choose the exponential growth function: \[y=A_0e^\] where \(A_0\) is equal to the value at time zero, \(e\) is Euler’s constant, and \(k\) is a positive constant that determines the rate (percentage) of growth.In some applications, however, as we will see when we discuss the logistic equation, the logistic model sometimes fits the data better than the exponential model.On the other hand, if a quantity is falling rapidly toward zero, without ever reaching zero, then we should probably choose the exponential decay model.

To find the half-life of a function describing exponential decay, solve the following equation: \(\dfrac A_0=A_0e^\) We find that the half-life depends only on the constant \(k\) and not on the starting quantity \(A_0\).

Express the amount of carbon-14 remaining as a function of time, \(t\). \[\begin A&= A_0e^ \qquad \text\ 0.5A_0&= A_0e^ \qquad \text 0.5A_0 \text f(t)\ 0.5&= e^ \qquad \text A_0\ \ln(0.5)&= 5730k \qquad \text\ k&= \dfrac \qquad \text\ A&= A_0e^ \qquad \text \end\] The function that describes this continuous decay is \(f(t)=A_0e^\).

We observe that the coefficient of \(t\), \(\dfrac≈−1.2097×10^\) is negative, as expected in the case of exponential decay.

The formula is derived as follows \[\begin \dfrac A_0&= A_0e^\ \dfrac&= e^ \qquad \text A_0\ \ln \left (\dfrac \right )&= ktv \qquad \text\ -\ln(2)&= kt \qquad \text\ -\ln(2)k&= t \qquad \text \end\] Since \(t\), the time, is positive, \(k\) must, as expected, be negative.

This gives us the half-life formula \[t=−\dfrac\] Example \(\Page Index\): Finding the Function that Describes Radioactive Decay The half-life of carbon-14 is \(5,730\) years.

We may use the exponential decay model when we are calculating half-life, or the time it takes for a substance to exponentially decay to half of its original quantity.

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