## University Calculus: Early Transcendentals (3rd Edition)

(a) $$p=6.6\Big(\frac{1}{2}\Big)^{\frac{t}{14}}$$ $6.6$(g) is the initial amount of phosphorus-32 $\frac{1}{2}$ is the rate of change $t$: the amount of time $p$: the amount of phosphorus-32 left after the amount of time $t$ has passed (b) After about 38 days, the amount of phosphorus-32 is 1 gram.
From the exercise, we have: - The half-life of phosphorus-32 is 14 days. - The amount of phosphorus-32 initially. (a) The half-life of phosphorus-32 being 14 days means that after 14 days, the amount of phosphorus-32 present would be reduced by half. In other words, here we have an intial amount of phosphorus-32, which is $6.6g$. After 14 days, what we have left is only $3.3g$ of phosphorus-32. After the next 14 days, what we have left is, again, half of that, which is $1.65g$. Therefore, the rate of change here is $\frac{1}{2}$ for a period of every 14 days. That means, we can estimate the amount of phosphorus-32 left using this function: $$p=6.6\Big(\frac{1}{2}\Big)^{\frac{t}{14}}$$ (We use $\frac{t}{14}$ instead of $t$ because only after 14 days will the amount be left by half. During the meantime, it reduces gradually.) $6.6$(g) is the initial amount of phosphorus-32 $\frac{1}{2}$ is the rate of change $t$: the amount of time $p$: the amount of phosphorus-32 left after the amount of time $t$ has passed (b) We need to find when the remaining amount of phosphorus-32 is $1g$, which in essence is to find $t$ so that $p=1$ To find $t$, we substitute the known values: $$6.6\Big(\frac{1}{2}\Big)^{\frac{t}{14}}=1$$ $$\Big(\frac{1}{2}\Big)^{\frac{t}{14}}=\frac{1}{6.6}=\frac{5}{33}$$ Here, we use graphing calculator and find out that $$\frac{t}{14}\approx2.722$$ $$t\approx38.108\approx38(days)$$ Therefore, we can conclude that after about 38 days, the amount of phosphorus-32 is 1 gram.