Given
Average Fahrenheit temperature in a certain city during a typical 365- day year
[tex]y=32\sin[\frac{2\pi}{365}(x-96)]+21[/tex]
Find
On what day the temperature is increasing fastest
Explanation
for maximising y
[tex]\frac{dy}{dx}=0[/tex]
so , now we find the first derivative
[tex]\begin{gathered} y=32\sin[\frac{2\pi}{365}(x-96)]+21 \\ \\ \frac{dy}{dx}=32\cos[\frac{2\pi}{365}(x-96)]\times\frac{2\pi}{365} \end{gathered}[/tex]
put dy/dx = 0
[tex]\begin{gathered} \frac{dy}{dx}=0 \\ \\ 32\cos[\frac{2\pi}{365}(x-96)]\times\frac{2\pi}{365}=0 \\ \\ \cos[\frac{2\pi}{365}(x-96)]=0 \\ \\ since\text{ the function is maximum when }\theta=0 \\ so\text{ , }\frac{dy}{dx}\text{ is maximum , when } \\ [\frac{2\pi}{365}(x-96)]=0 \\ \\ x-96=0 \\ x=96 \end{gathered}[/tex]
so , at 96 day temperature is increasing the fastest
b) rate of change of temperature per day ,
[tex]\frac{dy}{dx}=32\cos[\frac{2\pi}{365}(x-96)]\times\frac{2\pi}{365}[/tex]
put x = 96
[tex]\begin{gathered} \frac{dy}{dx}=32\cos[\frac{2\pi}{365}(x-96)]\times\frac{2\pi}{365} \\ \\ \frac{dy}{dx}=32\cos[\frac{2\pi}{365}(96-96)]\times\frac{2\pi}{365} \\ \\ \frac{dy}{dx}=32\cos0\times\frac{2\pi}{365} \\ \\ \frac{dy}{dx}=32\times\frac{2\pi}{365} \\ \\ \frac{dy}{dx}=0.55085460227\approx0.55 \end{gathered}[/tex]
