How do we convert annual growth rate to daily growth rate?

Question

Let's say we have an annual growth rate of $2 \%$, then $p=p_0(1.02)^Y$ where $Y$ is time in years. Would it be safe to say $p=p_0(1.02)^{D/365}$ where $D$ is the time in days (assuming a year is always $365$ days)? If so, could the daily growth rate be found like this (where $r_D$ is the daily growth rate): $$p=p$$ $$p_{0}\left( 1+\frac{r_{D}}{100} \right)^{D}=p_{0}\left( 1.02 \right)^{D/365}$$ $$1+\frac{r_{D}}{100}=1.02^{1/365}$$ $$r_D=0.005426 \%$$

Answer 1

Yes, this is what you want.

However, you can be a bit more clear on this. You correctly found out that $P=P_0 (1.02)^t$, where $t$ is measured in years. If you wanted to find how much growth occurs in one day, well one day is $\frac{1}{365}$ of a year, so plug in $t=\frac{1}{365}$. This gives $$ P_0(1.02)^{1/365}=P_0(1.00005425)=P_0(1+0.00005425) $$ so as a ratio (proportion), we increase by $0.00005425$ each day. If we want this as a percent, this is $0.005425 \%$ each day.

You can think about this in a slightly different way. Say we want to turn out yearly formula into a daily formula. Well, as before we know that $365$ days are $1$ year so that $1Y=365D$. That is, $D=\frac{1}{365}Y$. Then use the properties of exponents: $$ P=P_0(1.02)^Y=P_0(1.02)^{D/365}=P_0\big((1.02)^{1/365}\big)^D=P_0(1.00005425)^D $$ just as before.