Answer
a. day 101 of the year.
b. $0.64^{\circ}F$/day
Work Step by Step
a. Given the equation $y=37sin[\frac{2\pi}{365}(x-101)]+25$, we have $\frac{dy}{dx}=37cos[\frac{2\pi}{365}(x-101)](\frac{2\pi}{365})=\frac{74\pi}{365}cos[\frac{2\pi}{365}(x-101)]$.
The day that the temperature increases the fastest corresponds to the maximum of $\frac{dy}{dx}$ which happens when $\frac{2\pi}{365}(x-101)=0$. This gives $x=101$ -- that is, the 101st day of the year.
b. At $x=101$, we have $\frac{dy}{dx}=\frac{74\pi}{365}cos[\frac{2\pi}{365}(101-101)]=\frac{74\pi}{365}\approx0.64^{\circ}F$/day
