Question

a chemical substance has a decay rate of 7.3% per day. the rate of change of an amount n of the chemical after t days is given by $\frac{dn}{dt}=-0.073n$. a) let $n_0$ represent the amount of the substance present at $t = 0$. find the exponential function that models the decay. b) suppose that 600 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 600 g of the substance remain? a) $n(t)=$

Explanation:

Step1: Solve the differential equation

We have the first - order differential equation $\frac{dN}{dt}=- 0.073N$. Separate variables: $\frac{dN}{N}=-0.073dt$. Integrate both sides: $\int\frac{dN}{N}=\int - 0.073dt$. The left - hand side integral is $\ln|N|$ and the right - hand side is $-0.073t + C$. So, $\ln|N|=-0.073t + C$. Exponentiating both sides gives $N = Ce^{-0.073t}$. When $t = 0$, $N = N_0$, so $N_0 = C$. Thus, the exponential function is $N(t)=N_0e^{-0.073t}$.

Step2: Find the amount after 5 days

Given $N_0 = 600$ g and $t = 5$. Substitute into $N(t)=N_0e^{-0.073t}$, we get $N(5)=600e^{-0.073\times5}=600e^{-0.365}\approx600\times0.6947 = 416.82$ g.

Step3: Find the rate of change after 5 days

First, we know $\frac{dN}{dt}=-0.073N$. When $t = 5$, $N(5)\approx416.82$ g. So $\frac{dN}{dt}\big|_{t = 5}=-0.073\times416.82\approx - 30.43$ g/day.

Step4: Find the time for half - life

We want to find $t$ when $N(t)=\frac{N_0}{2}$. Given $N(t)=N_0e^{-0.073t}$, set $N(t)=\frac{N_0}{2}$, so $\frac{N_0}{2}=N_0e^{-0.073t}$. Divide both sides by $N_0$ (since $N_0
eq0$), we get $\frac{1}{2}=e^{-0.073t}$. Take the natural logarithm of both sides: $\ln(\frac{1}{2})=-0.073t$. Then $t=\frac{\ln(\frac{1}{2})}{-0.073}=\frac{-\ln(2)}{-0.073}\approx9.49$ days.

Answer:

a) $N(t)=N_0e^{-0.073t}$
b) $416.82$ g
c) $-30.43$ g/day
d) $9.49$ days