Answer :
The problem tells us that the coffee starts at a temperature of
[tex]$$T(0) = 120^\circ F$$[/tex]
and cools down at a rate given by
[tex]$$r(t) = 55 e^{-0.03t^2} \quad \text{(in °F per minute)}.$$[/tex]
To find the temperature at time [tex]$t = 1$[/tex] minute, we first determine the total decrease in temperature over the interval from [tex]$t = 0$[/tex] to [tex]$t = 1$[/tex]. This total decrease is given by the integral:
[tex]$$
\Delta T = \int_0^1 55 e^{-0.03t^2} \, dt.
$$[/tex]
Evaluating this integral, we obtain a temperature drop of approximately
[tex]$$\Delta T \approx 54.4549^\circ F.$$[/tex]
Since the coffee cools down, we subtract this decrease from the initial temperature:
[tex]$$
T(1) = T(0) - \Delta T \approx 120^\circ F - 54.4549^\circ F \approx 65.5451^\circ F.
$$[/tex]
Rounded to one decimal place, the temperature at [tex]$t = 1$[/tex] minute is approximately
[tex]$$65.5^\circ F.$$[/tex]
Thus, the correct answer is option (C) [tex]$65.5^\circ F$[/tex].
[tex]$$T(0) = 120^\circ F$$[/tex]
and cools down at a rate given by
[tex]$$r(t) = 55 e^{-0.03t^2} \quad \text{(in °F per minute)}.$$[/tex]
To find the temperature at time [tex]$t = 1$[/tex] minute, we first determine the total decrease in temperature over the interval from [tex]$t = 0$[/tex] to [tex]$t = 1$[/tex]. This total decrease is given by the integral:
[tex]$$
\Delta T = \int_0^1 55 e^{-0.03t^2} \, dt.
$$[/tex]
Evaluating this integral, we obtain a temperature drop of approximately
[tex]$$\Delta T \approx 54.4549^\circ F.$$[/tex]
Since the coffee cools down, we subtract this decrease from the initial temperature:
[tex]$$
T(1) = T(0) - \Delta T \approx 120^\circ F - 54.4549^\circ F \approx 65.5451^\circ F.
$$[/tex]
Rounded to one decimal place, the temperature at [tex]$t = 1$[/tex] minute is approximately
[tex]$$65.5^\circ F.$$[/tex]
Thus, the correct answer is option (C) [tex]$65.5^\circ F$[/tex].
