Answer
The rate of increase in the number of hours of daylight on March 21 is higher than that on May 21.
Work Step by Step
$$L(t)=12+2.8\sin[\frac{2\pi}{365}(t-80)]$$
1) Here the question asks about the increase of the number of hours of daylight in Philadelphia, so it relates to the rate of change of the number of hours of daylight.
Therefore, again we would find the derivative of formula $L(t)$, which is the rate of change we need to find now. $$L'(t)=[12+2.8\sin[\frac{2\pi}{365}(t-80)]]'$$ $$L'(t)=0+2.8\frac{d\sin[\frac{2\pi}{365}(t-80)]}{dt}$$ $$L'(t)=2.8\frac{d\sin[\frac{2\pi}{365}(t-80)]}{d[\frac{2\pi}{365}(t-80)]}\frac{\frac{2\pi}{365}d(t-80)}{dt}$$ $$L'(t)=2.8\cos[\frac{2\pi}{365}(t-80)]\frac{2\pi}{365}(\frac{dt}{dt}-\frac{d(80)}{dt})$$ $$L'(t)=\frac{5.6\pi}{365}\cos[\frac{2\pi}{365}(t-80)](1-0)$$ $$L'(t)=\frac{5.6\pi}{365}\cos[\frac{2\pi}{365}(t-80)](hours/day)$$
2) Find the rate of increase on March 21.
March 21 is the 80th day of the year, which means $t=80$.
The rate of increase in the number of hours of daylight would be $$L'(80)=\frac{5.6\pi}{365}\cos[\frac{2\pi}{365}(80-80)]$$ $$L'(80)=\frac{5.6\pi}{365}\cos[\frac{2\pi}{365}\times0]$$ $$L'(80)=\frac{5.6\pi}{365}\cos0$$ $$L'(80)=\frac{5.6\pi}{365}\times1$$ $$L'(80)=\frac{5.6\pi}{365}\approx0.048(hours/day)$$
3) Find the rate of increase on May 21.
May 21 is the 141st day of the year, which means $t=141$.
The rate of increase in the number of hours of daylight would be $$L'(141)=\frac{5.6\pi}{365}\cos[\frac{2\pi}{365}(141-80)]$$ $$L'(141)=\frac{5.6\pi}{365}\cos[\frac{2\pi}{365}\times61]$$ $$L'(141)=\frac{5.6\pi}{365}\cos[\frac{122\pi}{365}]$$ $$L'(141)=\frac{5.6\pi}{365}\cos(1.05)$$ $$L'(141)=\frac{5.6\pi}{365}\times0.498$$ $$L'(141)=0.048\times0.498\approx0.024(hours/day)$$
Therefore, the rate of increase in the number of hours of daylight on March 21 is higher than that on May 21.
