Answer
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Hint:Here in this question we have to find the exponential decay rate. That is, the rate at which a population of something decays is directly proportional to the negative of the current population at time t. So we can introduce a proportionality constant. Then further applying an integration on both sides and on simplification we get the required result.
Complete step by step answer:
Exponential decay describes the process of reducing an amount by a consistent percentage rate over a period of time. It can be expressed by the formula \[y = a{\left( {1 - b} \right)^x}\] where y is the final amount, a is the original amount, b is the decay factor, and x is the amount of time that has passed. Exponential decays typically start with a differential equation of the form:
\[ \Rightarrow \,\,\dfrac{{dN}}{{dt}}\alpha \, - N(t)\]
That is, the rate at which a population of something decays is directly proportional to the negative of the current population at time t. So we can introduce a proportionality constant \[\alpha \]:
\[ \Rightarrow \,\,\dfrac{{dN}}{{dt}} = - \alpha \,N(t)\]
We will now solve the equation to find a function of \[N(t)\]:
\[ \Rightarrow \,\,\dfrac{{dN}}{{N(t)}} = - \alpha \,dt\]
\[ \Rightarrow \,\,\dfrac{{dN}}{N} = - \alpha \,dt\]
Apply integration both sides, then
\[ \Rightarrow \,\,\int {\dfrac{{dN}}{N}} = \int { - \alpha } \,dt\]
\[ \Rightarrow \,\,\int {\dfrac{{dN}}{N}} = - \alpha \int {dt} \,\]
Using the integration formula \[\int {\dfrac{1}{x}dx = \ln x + c} \] and \[\int {dx = x + c} \], where c is an integrating constant.
\[ \Rightarrow \,\,\ln N = - \alpha t + c\,\]
As we know the logarithm function is the inverse form of exponential function, then
\[ \Rightarrow \,\,N = {e^{ - \alpha \,\,t + c}}\,\]
Or it can be written as:
\[ \therefore \,\,N = A{e^{ - \alpha \,t}}\]
Where A is a constant.
Hence, the general form of exponential decay rate is \[N = A{e^{ - \alpha \,t}}\].
Note:Exponential decay describes the process of reducing an amount by a constant percentage rate over a period of time. The integration is inverse of the differentiation so to cancel differentiation the integration is applied. Likewise the logarithm is inverse of exponential. Hence by using these concepts we obtain answers.
Complete step by step answer:
Exponential decay describes the process of reducing an amount by a consistent percentage rate over a period of time. It can be expressed by the formula \[y = a{\left( {1 - b} \right)^x}\] where y is the final amount, a is the original amount, b is the decay factor, and x is the amount of time that has passed. Exponential decays typically start with a differential equation of the form:
\[ \Rightarrow \,\,\dfrac{{dN}}{{dt}}\alpha \, - N(t)\]
That is, the rate at which a population of something decays is directly proportional to the negative of the current population at time t. So we can introduce a proportionality constant \[\alpha \]:
\[ \Rightarrow \,\,\dfrac{{dN}}{{dt}} = - \alpha \,N(t)\]
We will now solve the equation to find a function of \[N(t)\]:
\[ \Rightarrow \,\,\dfrac{{dN}}{{N(t)}} = - \alpha \,dt\]
\[ \Rightarrow \,\,\dfrac{{dN}}{N} = - \alpha \,dt\]
Apply integration both sides, then
\[ \Rightarrow \,\,\int {\dfrac{{dN}}{N}} = \int { - \alpha } \,dt\]
\[ \Rightarrow \,\,\int {\dfrac{{dN}}{N}} = - \alpha \int {dt} \,\]
Using the integration formula \[\int {\dfrac{1}{x}dx = \ln x + c} \] and \[\int {dx = x + c} \], where c is an integrating constant.
\[ \Rightarrow \,\,\ln N = - \alpha t + c\,\]
As we know the logarithm function is the inverse form of exponential function, then
\[ \Rightarrow \,\,N = {e^{ - \alpha \,\,t + c}}\,\]
Or it can be written as:
\[ \therefore \,\,N = A{e^{ - \alpha \,t}}\]
Where A is a constant.
Hence, the general form of exponential decay rate is \[N = A{e^{ - \alpha \,t}}\].
Note:Exponential decay describes the process of reducing an amount by a constant percentage rate over a period of time. The integration is inverse of the differentiation so to cancel differentiation the integration is applied. Likewise the logarithm is inverse of exponential. Hence by using these concepts we obtain answers.