Problem
A chemical substance has a decay rate of $6.2 \%$ per day. The rate of change of an amount $\mathrm{N}$ of the chemical after $\mathrm{t}$ days is given by $\frac{\mathrm{dN}}{\mathrm{dt}}=-0.062 \mathrm{~N}$. a) Let $\mathrm{N}_{0}$ represent the amount of the substance present at $\mathrm{t}=0$. Find the exponential function that models the decay. b) Suppose that $700 \mathrm{~g}$ of the substance is present at $t=0$. How much will remain after 5 days? c) What is the rate of change of the amount of the substance after 5 days? d) After how many days will half of the original $700 \mathrm{~g}$ of the substance remain? a) $\mathrm{N}(\mathrm{t})=\mathrm{N}_{0} e^{-0.062 t}$ b) After 5 days, $513 \mathrm{~g}$ will remain. (Round to the nearest whole number as needed.) c) After 5 days, the rate of change is $-31.8 \mathrm{~g} /$ day. (Round to one decimal place as needed.) d) Half of the substance will remain after days. (Round to one decimal place as needed.)
Solution
Step 1 ：The rate of change of the amount of the substance is given by the differential equation \(\frac{dN}{dt}=-0.062N\). This is a first order linear differential equation. The general solution to this type of equation is given by \(N(t) = N_0 e^{kt}\), where \(N_0\) is the initial amount of the substance, \(k\) is the rate of decay, and \(t\) is the time. In this case, \(k = -0.062\).
Step 2 ：So, the exponential function that models the decay is \(N(t) = N_0 e^{-0.062t}\).
Step 3 ：Final Answer: The exponential function that models the decay is \(\boxed{N(t) = N_0 e^{-0.062t}}\).
