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) $n(t)=n_0e^{-0.073t}$ b) after 5 days, 417 g will remain (round to the nearest whole number as needed.) c) after 5 days, the rate of change is g/day (round to one decimal place as needed.)

Explanation:

Step1: Recall the general form of a first - order differential equation solution

The differential equation $\frac{dN}{dt}=- 0.073N$ is a first - order separable differential equation of the form $\frac{dN}{N}=-0.073dt$. Integrating both sides, $\int\frac{dN}{N}=-\int0.073dt$. We get $\ln N=-0.073t + C$. When $t = 0$, $N = N_0$, so $\ln N_0=C$. Then $N(t)=N_0e^{-0.073t}$.

Step2: Calculate the amount remaining after 5 days

Given $N_0 = 600$ g and $t = 5$. Substitute into $N(t)=N_0e^{-0.073t}$, so $N(5)=600e^{-0.073\times5}=600e^{-0.365}\approx417$ g.

Step3: Find the rate of change after 5 days

We know that $\frac{dN}{dt}=-0.073N$. First, find $N(5)\approx417$ g. Then $\frac{dN}{dt}\big|_{t = 5}=-0.073\times417\approx - 30.4$ g/day.

Step4: Find the half - life

We want to find $t$ when $N(t)=\frac{N_0}{2}$. Since $N(t)=N_0e^{-0.073t}$, then $\frac{N_0}{2}=N_0e^{-0.073t}$. Canceling out $N_0$ (since $N_0
eq0$), we get $\frac{1}{2}=e^{-0.073t}$. Taking the natural logarithm of both sides, $\ln\frac{1}{2}=-0.073t$. So $t=\frac{\ln\frac{1}{2}}{-0.073}=\frac{-\ln 2}{-0.073}\approx9.5$ days.

Answer:

a) $N(t)=N_0e^{-0.073t}$
b) 417 g
c) - 30.4 g/day
d) 9.5 days