Answer :
The temperature of the coffee at time [tex]$t$[/tex] can be determined by subtracting the total temperature drop from the initial temperature. Initially, the coffee is at
[tex]$$T(0)=120^\circ F.$$[/tex]
The rate at which the temperature decreases is given by
[tex]$$r(t) = 55 e^{-0.03t^2} \quad \text{(degrees Fahrenheit per minute)}.$$[/tex]
The total drop in temperature over the time interval from [tex]$t=0$[/tex] to [tex]$t=1$[/tex] minute is obtained by integrating this rate:
[tex]$$\Delta T = \int_0^1 55 e^{-0.03t^2} \, dt.$$[/tex]
Evaluating this integral gives approximately:
[tex]$$\int_0^1 55 e^{-0.03t^2} \, dt \approx 54.455 \quad \text{degrees Fahrenheit}.$$[/tex]
Thus, the temperature at [tex]$t=1$[/tex] minute is:
[tex]$$T(1) = 120 - \Delta T = 120 - 54.455 \approx 65.545^\circ F.$$[/tex]
Rounding to one decimal place, the temperature is approximately [tex]$65.5^\circ F$[/tex].
Therefore, the correct answer is:
(C) [tex]$65.5^\circ F$[/tex].
[tex]$$T(0)=120^\circ F.$$[/tex]
The rate at which the temperature decreases is given by
[tex]$$r(t) = 55 e^{-0.03t^2} \quad \text{(degrees Fahrenheit per minute)}.$$[/tex]
The total drop in temperature over the time interval from [tex]$t=0$[/tex] to [tex]$t=1$[/tex] minute is obtained by integrating this rate:
[tex]$$\Delta T = \int_0^1 55 e^{-0.03t^2} \, dt.$$[/tex]
Evaluating this integral gives approximately:
[tex]$$\int_0^1 55 e^{-0.03t^2} \, dt \approx 54.455 \quad \text{degrees Fahrenheit}.$$[/tex]
Thus, the temperature at [tex]$t=1$[/tex] minute is:
[tex]$$T(1) = 120 - \Delta T = 120 - 54.455 \approx 65.545^\circ F.$$[/tex]
Rounding to one decimal place, the temperature is approximately [tex]$65.5^\circ F$[/tex].
Therefore, the correct answer is:
(C) [tex]$65.5^\circ F$[/tex].
