Part 3 of 4 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 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 $t=0$. 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) $N(t)=N_{0} e^{-0.062 t}$ 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 $\square$ g/day. (Round to one decimal place as needed.)

Step 1 ：Given that the rate of decay of the chemical substance is $6.2 \%$ per day, this can be represented as $-0.062$ in the rate of change equation $\frac{dN}{dt} = -0.062N$.

Step 2 ：Let $N_0$ represent the initial amount of the substance at $t=0$. The exponential function that models the decay is $N(t) = N_0 e^{-0.062t}$.

Step 3 ：Suppose that $700g$ of the substance is present at $t=0$. To find out how much will remain after 5 days, we substitute $N_0 = 700$ and $t = 5$ into the decay function to get $N(5) = 700e^{-0.062 \times 5} \approx 513g$.

Step 4 ：To find the rate of change of the amount of the substance after 5 days, we substitute $N(5) = 513$ into the rate of change equation to get $\frac{dN}{dt} = -0.062 \times 513 \approx -31.8g/day$.

Step 5 ：\(\boxed{-31.8}\) g/day is the rate of change of the amount of the substance after 5 days.

Step 6 ：To find out after how many days will half of the original $700g$ of the substance remain, we set $N(t) = \frac{N_0}{2} = 350g$ and solve for $t$ in the decay function. This gives $t = \frac{\ln(\frac{N_0}{2})}{-0.062} \approx 11.1$ days.